From da33268cedbf5295d9956c92678647e5f92e8895 Mon Sep 17 00:00:00 2001 From: fretchen Date: Sat, 4 Jan 2025 10:54:11 +0100 Subject: [PATCH 01/13] More lectures coming in --- amo/lecture3.md | 385 ++++++++++++++++ .../Lecture 3 - The two-level system(1).tex | 362 +++++++++++++++ ...ecture 4 - Atoms in oscillating fields.tex | 335 ++++++++++++++ .../Lecture 5 - The Hydrogen Atom.tex | 339 ++++++++++++++ ...ole approximation in the hydrogen atom.tex | 426 ++++++++++++++++++ ...re 7 - Beyond the boring hydrogen atom.tex | 345 ++++++++++++++ amo/tex_files/Lecture 8 - The Helium atom.tex | 414 +++++++++++++++++ utils/cleanMd.ts | 7 + 8 files changed, 2613 insertions(+) create mode 100644 amo/lecture3.md create mode 100644 amo/tex_files/Lecture 3 - The two-level system(1).tex create mode 100644 amo/tex_files/Lecture 4 - Atoms in oscillating fields.tex create mode 100644 amo/tex_files/Lecture 5 - The Hydrogen Atom.tex create mode 100644 amo/tex_files/Lecture 6 - The dipole approximation in the hydrogen atom.tex create mode 100644 amo/tex_files/Lecture 7 - Beyond the boring hydrogen atom.tex create mode 100644 amo/tex_files/Lecture 8 - The Helium atom.tex diff --git a/amo/lecture3.md b/amo/lecture3.md new file mode 100644 index 0000000..eabc632 --- /dev/null +++ b/amo/lecture3.md @@ -0,0 +1,385 @@ +--- +author: +- Fred Jendrzejewski +- Selim Jochim +bibliography: +- bibliography/converted_to_latex.bib +date: January 04, 2025 +title: Lecture 3 - The two-level system +--- + +We are going to discuss the two-level system, it's static properties +like level splitting at avoided crossings and dynamical properties like +Rabi oscillations. + +After the previous discussions of some basic cooking recipes to quantum +mechanics in last weeks lectures [@Jendrzejewskia] and [@Jendrzejewski], +we will use them to understand the two-level system. A very detailled +discussion can be found in chapter 4 of Ref. [@1]. The importance of the +two-level system is at least three-fold: + +1. It is the simplest system of quantum mechanics as it spans a Hilbert + space of only two states. + +2. It is quite ubiquitous in nature and very widely used in atomic + physics. + +3. The two-level system is another word for the qubit, which is the + fundamental building block of the exploding field of quantum + computing and quantum information science. + +![Examples for two-state systems. a) Benzene: In the ground state, the +electrons are delocalized. b) Ammonia: The nitrogen atom is either found +above or below the hydrogen triangle. The state changes when the +nitrogen atom tunnels. c) Molecular ion : The electron is either +localized near proton 1 or 2. +](figures/Bildschirmfoto-2018-09-28-um-14-39-281/Bildschirmfoto-2018-09-28-um-14-39-281){#217687 +width="0.70\\columnwidth"} + +Some of the many examples for two-level systems that can be found in +nature: + +- Spin of the electron: Up vs. down state + +- Two-level atom with one electron (simplified): Excited vs. ground + state + +- Structures of molecules, e.g., [NH~3~](#fig:twostate) + +- Occupation of mesoscopic capacitors in nanodevices. + +- Current states in superconducting loops. + +- Nitrogen-vacancy centers in diamond. + +# Hamiltonian, Eigenstates and Matrix Notation + +To start out, we will consider two eigenstates +$\left|0\right\rangle$, $\left|1\right\rangle$ +of the Hamiltonian $\hat{H}_0$ with + +$$ + \hat{H}_0\left|0\right\rangle=E_0\left|0\right\rangle, \qquad \hat{H}_0\left|1\right\rangle=E_1\left|1\right\rangle. +$$ + +Quite typically we might think of it as a two-level atom +with states 1 and 2. The eigenstates can be expressed in matrix +notation: + +$$ + \left|0\right\rangle=\left( \begin{array}{c} 1 \\ 0 \end{array} \right), \qquad \left|1\right\rangle=\left( \begin{array}{c} 0 \\ 1 \end{array} \right), +$$ + +so that $\hat{H}_0$ be written as a diagonal matrix +$$ + \hat{H}_0 = \left(\begin{array}{cc} E_0 & 0 \\ 0 & E_1 \end{array}\right). +$$ + +If we would only prepare eigenstates the system would be +rather boring. However, we typically have the ability to change the +Hamiltonian by switching on and off laser or microwave fields [^1]. We +can then write the Hamiltonian in its most general form as: +$$ +\label{Eq:TwoLevelGeneral} +\hat{H} = \frac{\hbar}{2}\left( \begin{array}{cc} \Delta & \Omega_x - i\Omega_y\\ \Omega_x +i\Omega_y & -\Delta \end{array} \right) +$$ + +Sometimes we will also chose the definition: +$$ +\Omega = |\Omega| e^{i\varphi}=\Omega_x + i\Omega_y +$$ + +It is particularly useful for the case in which the +coupling is created by a laser. Another useful way of thinking about the +two-level system is as a spin in a magnetic field. Let us remind us of +the definitions of the of the spin-1/2 matrices: + +$$ +s_x = \frac{\hbar}{2}\left(\begin{array}{cc} +0 & 1\\ +1 & 0 +\end{array} +\right)~ +s_y = \frac{\hbar}{2}\left(\begin{array}{cc} +0 & -i\\ +i & 0 +\end{array} +\right)~s_z =\frac{\hbar}{2} \left(\begin{array}{cc} +1 & 0\\ +0 & -1 +\end{array} +\right) +$$ + +We then obtain: + +$$ +\label{Eq:HamSpin} +\hat{H} = \mathbf{B}\cdot\hat{\mathbf{s}}\text{ with }\mathbf{B} = (\Omega_x, \Omega_y, \Delta) +$$ + +You will go through this calculation in the excercise of +this week. + +## Case of no perturbation $\Omega = 0$ + +This is exactly the case of no applied laser fields that we discussed +previously. We simply removed the energy offset +$E_m = \frac{E_0+E_1}{2}$ and pulled out the factor $\hbar$, such that +$\Delta$ measures a frequency. So we have: + +$$ +E_0 = E_m+ \frac{\hbar}{2}\Delta\\ +E_1 = E_m- \frac{\hbar}{2}\Delta +$$ + +We typically call $\Delta$ the energy difference between +the levels or the **detuning**. + +## Case of no detuning $\Delta = 0$ + +Let us suppose that the diagonal elements are exactly zero. And for +simplicity we will also keep $\Omega_y =0$ as it simply complicates the +calculations without adding much to the discussion at this stage. The +Hamiltonian reads then: + +$$ +\hat{H} = \frac{\hbar}{2}\left( \begin{array}{cc} 0 & \Omega\\ \Omega &0 \end{array} \right) +$$ + +Quite clearly the states $\varphi_{1,2}$ are not the eigenstates of the +system anymore. How should the system be described now ? We can once +again diagonalize the system and write + +$$ +\hat{H}\left|\varphi_{\pm\right\rangle} = E_{\pm}\left|\varphi_\pm\right\rangle\\ +E_{\pm} = \pm\frac{\hbar}{2}\Omega\\ +\left|\varphi_\pm\right\rangle = \frac{\left|0\right\rangle\pm\left|1\right\rangle}{\sqrt{2}} +$$ + +Two important consequences can be understood from this +result: + +1. The coupling of the two states shifts their energy by $\Omega$. This + is the idea of level repulsion. + +2. The coupled states are a superposition of the initial states. + +This is also a motivation the formulation of the 'bare' system for +$\Omega = 0$ and the 'dressed' states for the coupled system. + +## General case + +Quite importantly we can solve the system completely even in the general +case. By diagonalizing Eq. +[\[Eq:TwoLevelGeneral\]](#Eq:TwoLevelGeneral){reference-type="eqref" +reference="Eq:TwoLevelGeneral"} we obtain: + +$$ +\label{eq:Epm} + E_\pm = \pm \frac{\hbar}{2} \sqrt{\Delta^2+|\Omega|^2} +$$ + +The energies can be nicely summarized as in Fig. +[2](#326199){reference-type="ref" reference="326199"} + +![Anticrossing of energy levels. +](figures/Bildschirmfoto-2018-09-28-um-14-35-34/AvoidedCrossing){#326199 +width="0.70\\columnwidth"} + +  + +The Eigenstates then read: + +$$ +\left|\psi_+\right\rangle=\cos\left(\frac{\theta}{2}\right) \mathrm{e}^{-i{\varphi}/{2}}\left|0\right\rangle+\sin\left(\frac{\theta}{2}\right) \mathrm{e}^{i{\varphi}/{2}}\left|1\right\rangle, \label{eq:staticpsiplus} +$$ + +$$ +\left|\psi_-\right\rangle=-\sin\left(\frac{\theta}{2}\right) \mathrm{e}^{-i{\varphi}/{2}}\left|0\right\rangle+\cos\left(\frac{\theta}{2}\right) \mathrm{e}^{i{\varphi}/{2}}\left|1\right\rangle, \label{eq:staticpsiminus} +$$ + +where + +$$ + \label{eq:parameters} +\tan(\theta) = \frac{|\Omega|}{\Delta} +$$ + +# The Bloch sphere + +While we could just discuss the details of the above state in the +abstract, it is extremely helpful to visualize the problem on the Bloch +sphere. The idea of the Bloch sphere is that the we have a complex wave +function of well defined norm and two free parameters. So it seems quite +natural to look for a good representation of it. And this is the Bloch +sphere as drawn in Fig. [3](#613576){reference-type="ref" +reference="613576"}. + +![The presentation of the eigenstate on the Bloch sphere. +](figures/BlochSphereWithVectorForLecture/BlochSphereWithVectorForLecture){#613576 +width="0.70\\columnwidth"} + +We will see especially its usefulness especially as we discuss the +dynamics of the two-state system. + +# Dynamical Aspects + +## Time Evolution of $\left|\psi(t)\right\rangle$ + +After the static case we now want to investigate the dynamical +properties of the two-state system. We calculate the time evolution of +$\left|\psi(t)\right\rangle = c_0(t)\left|0\right\rangle + c_1(t)\left|1\right\rangle$ +with the Schrödinger equation and the perturbed Hamiltonian +[\[Eq:TwoLevelGeneral\]](#Eq:TwoLevelGeneral){reference-type="eqref" +reference="Eq:TwoLevelGeneral"}: + +$$ +i\hbar \frac{d}{dt}\left|\psi(t)\right\rangle=\hat{H}\left|\psi(t)\right\rangle,\\ +i \frac{d}{dt}\left(\begin{array}{c} c_0(t) \\ c_1(t) \end{array}\right) = \frac{1}{2}\left( \begin{array}{cc} \Delta & \Omega \\ \Omega^* & -\Delta \end{array} \right) \left(\begin{array}{c} c_0(t) \\ c_1(t) \end{array} \right). +$$ + +We have two coupled differential equations and we luckily already know +how to solve them as we have calculated the two eigenenergies in the +previous section. For the state +$\left|\psi(t)\right\rangle$ we get + +$$ + \left|\psi(t)\right\rangle=\lambda \mathrm{e}^{-i{E_+}t/{\hbar}} \left|\psi_+\right\rangle + \mu \mathrm{e}^{-i{E_-}t/{\hbar}} \left|\psi_-\right\rangle \label{eq:psitimeevolution} +$$ + +with the factors $\lambda$ and $\mu$, which are defined +by the initial state. The most common question is then what happens to +the system if we start out in the bare state +$\left|0\right\rangle$ and then let it evolve under +coupling with a laser ? So what is the probability to find it in the +other state $\left|1\right\rangle$: + +$$ +P_1(t)=\left|\left\langle 1|\psi(t)\right\rangle\right|^2. +$$ + +As a first step, we have to apply the initial condition +to +[\[eq:psitimeevolution\]](#eq:psitimeevolution){reference-type="eqref" +reference="eq:psitimeevolution"} and express +$\left|\varphi\right\rangle$ in terms of +[\[eq:staticpsiplus\]](#eq:staticpsiplus){reference-type="eqref" +reference="eq:staticpsiplus"} and +[\[eq:staticpsiminus\]](#eq:staticpsiminus){reference-type="eqref" +reference="eq:staticpsiminus"}: + +$$ +\left|\psi(0)\right\rangle \overset{!}{=}& \left|0\right\rangle\\ + = & \mathrm{e}^{i{\varphi}/{2}} \left[ \cos\left( \frac{\theta}{2}\right) \left|\psi_+\right\rangle-\sin\left(\frac{\theta}{2}\right)\left|\psi_-\right\rangle\right] +$$ + +By equating the coefficients we get for $\lambda$ and +$\mu$: + +$$ +\lambda = \mathrm{e}^{i{\varphi}/{2}}\cos\left(\frac{\theta}{2}\right), \qquad \mu = -\mathrm{e}^{i{\varphi}/{2}}\sin\left(\frac{\theta}{2}\right). +$$ + +One thus gets: + +$$ +\hspace{-2mm} P_1(t)=\left|\left\langle 1|\psi(t)\right\rangle\right|^2 \\ += \left|\mathrm{e}^{i\varphi} \sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)\left[\mathrm{e}^{-i{E_+}t/{\hbar}} - \mathrm{e}^{-i{E_-}t/{\hbar}}\right]\right|^2\\ += \sin^2(\theta)\sin^2\left(\frac{E_+-E_-}{2\hbar}t\right) +$$ + +$P_1(t)$ can be expressed with $\Delta$ and $\Omega$ +alone. The obtained relation is called Rabi's formula: + +$$ + P_1(t)=\frac{1}{1+\left(\frac{\Delta}{|\Omega|}\right)^2}\sin^2\left(\sqrt{|\Omega|^2+\Delta^2}\frac{t}{2}\right) +$$ + +![Rabi oscillations +](figures/Bildschirmfoto-2018-09-28-um-14-43-51/RabiOscillation){#833990 +width="0.70\\columnwidth"} + +## Visualization of the dynamics in the spin picture + +While the previous derivation might be the standard one, which certainly +leads to the right results it might not be the most intuitive way of +thinking about the dynamics. They become actually quite transparent in +the spin language and on the Bloch sphere. So let us go back to the +formulation of the Hamiltonian in terms of spins as in Eq. +[\[Eq:HamSpin\]](#Eq:HamSpin){reference-type="eqref" +reference="Eq:HamSpin"}. + +How would the question of the time evolution from $0$ to $1$ and back go +now ? Basically, we would assume that the spin has been initialize into +one of the eigenstates of the $z$-basis and now starts to rotate in some +magnetic field. How ? This can be nicely studied in the Heisenberg +picture, where operators have a time evolution. In the Heisenberg +picture we have: + +$$ +\frac{d}{dt} \hat{s}_i = \frac{i}{\hbar}\left[\hat{H},\hat{s}_i\right]\\ +\frac{d}{dt} \hat{s}_i = \frac{i}{\hbar}\sum_j B_j \left[\hat{s}_j,\hat{s}_i\right]\\ + +$$ + +So to understand we time evolution, we only need to +employ the commutator relationships between the spins: + +$$ += \hbar is_z~~[ s_y, s_z] = \hbar is_x~~[ s_z, s_x] = \hbar is_y +$$ + +For the specific case of $B_x=\Omega$, $B_y = B_z = 0$, +we have then: + +$$ +\frac{d}{dt} \hat{s}_x = 0\\ +\frac{d}{dt} \hat{s}_y = -\Omega \hat{s}_z\\ +\frac{d}{dt} \hat{s}_z = \Omega \hat{s}_y + +$$ + +So applying a field in x-direction leads to a rotation of the spin +around the $x$ axis with velocity $\Omega$. We can now use this general +picture to understand the dynamics as rotations around an axis, which is +defined by the different components of the magnetic field. + +# A few words on the quantum information notation + +The qubit is THE basic ingredient of quantum computers. A nice way to +play around with them is actually the [IBM Quantum +experience](https://quantum-computing.ibm.com/). However, you will +typically not find Pauli matrices etc within these systems. The typical +notation there is: + +- $R_x(\phi)$ is a rotation around the x-axis for an angle $\phi$. + +- Same holds for $R_y$ and $R_z$. + +- $X$ denotes the rotation around the x axis for an angle $\pi$. So it + transforms $\left|1\right\rangle$ into + $\left|0\right\rangle$ and vise versa. + +- $Z$ denotes the rotation around the x axis for an angle $\pi$. So it + transforms $\left|+\right\rangle$ into + $\left|-\right\rangle$ and vise versa. + +The most commonly used gate is actually one that we did not talk about +at all, it is the *Hadamard* gate, which transforms +$\left|1\right\rangle$ into +$\left|-\right\rangle$ and +$\left|0\right\rangle$ into +$\left|+\right\rangle$: + +$$ +\hat{H}\left|1\right\rangle = \left|-\right\rangle ~ \hat{H}\left|0\right\rangle = \left|+\right\rangle\\ +\hat{H}\left|-\right\rangle = \left|1\right\rangle ~ \hat{H}\left|+\right\rangle = \left|0\right\rangle +$$ + +In the [forth +lecture](https://www.authorea.com/326506/emMDRkXxtm44IKqpCtDi6g) we will +see how it is that a time-dependent field can actually couple two atomic +states, which are normally of very different energies. + +[^1]: See the discussions of the next lecture diff --git a/amo/tex_files/Lecture 3 - The two-level system(1).tex b/amo/tex_files/Lecture 3 - The two-level system(1).tex new file mode 100644 index 0000000..7fbe86c --- /dev/null +++ b/amo/tex_files/Lecture 3 - The two-level system(1).tex @@ -0,0 +1,362 @@ +\documentclass[10pt]{article} + +\usepackage{fullpage} +\usepackage{setspace} +\usepackage{parskip} +\usepackage{titlesec} +\usepackage[section]{placeins} +\usepackage{xcolor} +\usepackage{breakcites} +\usepackage{lineno} +\usepackage{hyphenat} + + + + + +\PassOptionsToPackage{hyphens}{url} +\usepackage[colorlinks = true, + linkcolor = blue, + urlcolor = blue, + citecolor = blue, + anchorcolor = blue]{hyperref} +\usepackage{etoolbox} +\makeatletter +\patchcmd\@combinedblfloats{\box\@outputbox}{\unvbox\@outputbox}{}{% + \errmessage{\noexpand\@combinedblfloats could not be patched}% +}% +\makeatother + + +\usepackage[round]{natbib} +\let\cite\citep + + + + +\renewenvironment{abstract} + {{\bfseries\noindent{\abstractname}\par\nobreak}\footnotesize} + {\bigskip} + +\titlespacing{\section}{0pt}{*3}{*1} +\titlespacing{\subsection}{0pt}{*2}{*0.5} +\titlespacing{\subsubsection}{0pt}{*1.5}{0pt} + + +\usepackage{authblk} + + +\usepackage{graphicx} +\usepackage[space]{grffile} +\usepackage{latexsym} +\usepackage{textcomp} +\usepackage{longtable} +\usepackage{tabulary} +\usepackage{booktabs,array,multirow} +\usepackage{amsfonts,amsmath,amssymb} +\providecommand\citet{\cite} +\providecommand\citep{\cite} +\providecommand\citealt{\cite} +% You can conditionalize code for latexml or normal latex using this. +\newif\iflatexml\latexmlfalse +\AtBeginDocument{\DeclareGraphicsExtensions{.pdf,.PDF,.eps,.EPS,.png,.PNG,.tif,.TIF,.jpg,.JPG,.jpeg,.JPEG}} + +\usepackage[utf8]{inputenc} +\usepackage[ngerman,english]{babel} + + + + + + + + +\usepackage{amsmath} +\newcommand{\bra}[1]{\ensuremath{\left\langle#1\right|}} +\newcommand{\ket}[1]{\ensuremath{\left|#1\right\rangle}} +\newcommand{\braket}[1]{\ensuremath{\left\langle#1\right\rangle}} +\newcommand{\rhohat}{\hat{\rho}} +\newcommand{\tr}[1]{\mathrm{tr}(#1)} +\newcommand{\trarb}[2]{\mathrm{tr}_{#1}(#2)} +\newcommand{\vv}[1]{\mathbf{#1}} +\newcommand*\dif{\mathop{}\!\mathrm{d}} +\newcommand{\eexp}[1]{\mathrm{e}^{#1}} + +\begin{document} + +\title{Lecture 3 - The two-level system} + + + +\author[1]{Fred Jendrzejewski}% +\author[2]{Selim Jochim}% +\affil[1]{Kirchhoff-Institut für Physik}% +\affil[2]{Physikalisches Institut der Universität Heidelberg}% + + +\vspace{-1em} + + + + + \date{January 04, 2025} + + +\begingroup +\let\center\flushleft +\let\endcenter\endflushleft +\maketitle +\endgroup + + + + + +\selectlanguage{english} +\begin{abstract} +We are going to discuss the two-level system, it's static properties like level splitting at avoided crossings and dynamical properties like Rabi oscillations.% +\end{abstract}% + + + +\sloppy + + +After the previous discussions of some basic cooking recipes to quantum mechanics in last weeks lectures \cite{Jendrzejewskia} and \cite{Jendrzejewski}, we will use them to understand the two-level system. A very detailled discussion can be found in chapter 4 of Ref. \cite{1}. The importance of the two-level system is at least three-fold: +\begin{enumerate} +\item It is the simplest system of quantum mechanics as it spans a Hilbert space of only two states. +\item It is quite ubiquitous in nature and very widely used in atomic physics. +\item The two-level system is another word for the qubit, which is the fundamental building block of the exploding field of quantum computing and quantum information science. +\end{enumerate}\selectlanguage{english} +\begin{figure}[h!] +\begin{center} +\includegraphics[width=0.70\columnwidth]{figures/Bildschirmfoto-2018-09-28-um-14-39-281/Bildschirmfoto-2018-09-28-um-14-39-281} +\caption{{Examples for two-state systems. a) Benzene: In the ground state, the +electrons are delocalized. b) Ammonia: The nitrogen atom is either found +above or below the hydrogen triangle. The state changes when the +nitrogen atom tunnels. c) Molecular ion : The electron is either +localized near proton 1 or 2. +{\label{217687}}% +}} +\end{center} +\end{figure} + +Some of the many examples for two-level systems that can be found in nature: +\begin{itemize} +\item Spin of the electron: Up vs. down state +\item Two-level atom with one electron (simplified): Excited vs. ground state +\item Structures of molecules, e.g., \hyperref[fig:twostate]{NH\textsubscript{3}} +\item Occupation of mesoscopic capacitors in nanodevices. +\item Current states in superconducting loops. +\item Nitrogen-vacancy centers in diamond. +\end{itemize} + +\section{Hamiltonian, Eigenstates and Matrix Notation} + +To start out, we will consider two eigenstates $\ket{0}$, $\ket{1}$ of the Hamiltonian $\hat{H}_0$ with +\begin{align} + \hat{H}_0\ket{0}=E_0\ket{0}, \qquad \hat{H}_0\ket{1}=E_1\ket{1}. +\end{align} +Quite typically we might think of it as a two-level atom with states 1 and 2. The eigenstates can be expressed in matrix notation: +\begin{align} + \ket{0}=\left( \begin{array}{c} 1 \\ 0 \end{array} \right), \qquad \ket{1}=\left( \begin{array}{c} 0 \\ 1 \end{array} \right), +\end{align} +so that $\hat{H}_0$ be written as a diagonal matrix +\begin{align} + \hat{H}_0 = \left(\begin{array}{cc} E_0 & 0 \\ 0 & E_1 \end{array}\right). +\end{align} +If we would only prepare eigenstates the system would be rather boring. However, we typically have the ability to change the Hamiltonian by switching on and off laser or microwave fields \footnote{See the discussions of the next lecture}. We can then write the Hamiltonian in its most general form as: +\begin{align}\label{Eq:TwoLevelGeneral} +\hat{H} = \frac{\hbar}{2}\left( \begin{array}{cc} \Delta & \Omega_x - i\Omega_y\\ \Omega_x +i\Omega_y & -\Delta \end{array} \right) +\end{align} +Sometimes we will also chose the definition: +\begin{align} +\Omega = |\Omega| e^{i\varphi}=\Omega_x + i\Omega_y +\end{align} +It is particularly useful for the case in which the coupling is created by a laser. Another useful way of thinking about the two-level system is as a spin in a magnetic field. Let us remind us of the definitions of the of the spin-1/2 matrices: +\begin{align} +s_x = \frac{\hbar}{2}\left(\begin{array}{cc} +0 & 1\\ +1 & 0 +\end{array} +\right)~ +s_y = \frac{\hbar}{2}\left(\begin{array}{cc} +0 & -i\\ +i & 0 +\end{array} +\right)~s_z =\frac{\hbar}{2} \left(\begin{array}{cc} +1 & 0\\ +0 & -1 +\end{array} +\right) +\end{align} +We then obtain: +\begin{align}\label{Eq:HamSpin} +\hat{H} = \mathbf{B}\cdot\hat{\mathbf{s}}\text{ with }\mathbf{B} = (\Omega_x, \Omega_y, \Delta) +\end{align} +You will go through this calculation in the excercise of this week. + +\subsection{Case of no perturbation $\Omega = 0$} + +This is exactly the case of no applied laser fields that we discussed previously. We simply removed the energy offset $E_m = \frac{E_0+E_1}{2}$ and pulled out the factor $\hbar$, such that $\Delta$ measures a frequency. So we have: +\begin{align} +E_0 = E_m+ \frac{\hbar}{2}\Delta\\ +E_1 = E_m- \frac{\hbar}{2}\Delta +\end{align} +We typically call $\Delta$ the energy difference between the levels or the \textbf{detuning}. + +\subsection{Case of no detuning $\Delta = 0$} + +Let us suppose that the diagonal elements are exactly zero. And for simplicity we will also keep $\Omega_y =0$ as it simply complicates the calculations without adding much to the discussion at this stage. The Hamiltonian reads then: +\begin{align} +\hat{H} = \frac{\hbar}{2}\left( \begin{array}{cc} 0 & \Omega\\ \Omega &0 \end{array} \right) +\end{align} + +Quite clearly the states $\varphi_{1,2}$ are not the eigenstates of the system anymore. How should the system be described now ? We can once again diagonalize the system and write +\begin{align} +\hat{H}\ket{\varphi_{\pm}} = E_{\pm}\ket{\varphi_\pm}\\ +E_{\pm} = \pm\frac{\hbar}{2}\Omega\\ +\ket{\varphi_\pm} = \frac{\ket{0}\pm\ket{1}}{\sqrt{2}} +\end{align} +Two important consequences can be understood from this result: +\begin{enumerate} +\item The coupling of the two states shifts their energy by $\Omega$. This is the idea of level repulsion. +\item The coupled states are a superposition of the initial states. +\end{enumerate} +This is also a motivation the formulation of the 'bare' system for $\Omega = 0$ and the 'dressed' states for the coupled system. + +\subsection{General case} + +Quite importantly we can solve the system completely even in the general case. By diagonalizing Eq. \eqref{Eq:TwoLevelGeneral} we obtain: +\begin{align}\label{eq:Epm} + E_\pm = \pm \frac{\hbar}{2} \sqrt{\Delta^2+|\Omega|^2} +\end{align} +The energies can be nicely summarized as in Fig. \ref{326199}\selectlanguage{english} +\begin{figure}[h!] +\begin{center} +\includegraphics[width=0.70\columnwidth]{figures/Bildschirmfoto-2018-09-28-um-14-35-34/AvoidedCrossing} +\caption{{Anticrossing of energy levels. +{\label{326199}}% +}} +\end{center} +\end{figure} + +~ + +The Eigenstates then read: +\begin{align} +\ket{\psi_+}&=\cos\left(\frac{\theta}{2}\right) \eexp{-i{\varphi}/{2}}\ket{0}+\sin\left(\frac{\theta}{2}\right) \eexp{i{\varphi}/{2}}\ket{1}, \label{eq:staticpsiplus} +\end{align} +\begin{align} +\ket{\psi_-}&=-\sin\left(\frac{\theta}{2}\right) \eexp{-i{\varphi}/{2}}\ket{0}+\cos\left(\frac{\theta}{2}\right) \eexp{i{\varphi}/{2}}\ket{1}, \label{eq:staticpsiminus} +\end{align} +where +\begin{align} \label{eq:parameters} +\tan(\theta) = \frac{|\Omega|}{\Delta} +\end{align} + +\section{The Bloch sphere} + +While we could just discuss the details of the above state in the abstract, it is extremely helpful to visualize the problem on the Bloch sphere. The idea of the Bloch sphere is that the we have a complex wave function of well defined norm and two free parameters. So it seems quite natural to look for a good representation of it. And this is the Bloch sphere as drawn in Fig. \ref{613576}.\selectlanguage{english} +\begin{figure}[h!] +\begin{center} +\includegraphics[width=0.70\columnwidth]{figures/BlochSphereWithVectorForLecture/BlochSphereWithVectorForLecture} +\caption{{The presentation of the eigenstate on the Bloch sphere. +{\label{613576}}% +}} +\end{center} +\end{figure} + +We will see especially its usefulness especially as we discuss the dynamics of the two-state system. + +\section{Dynamical Aspects} +\subsection{Time Evolution of $\ket{\psi(t)}$} + After the static case we now want to investigate the dynamical properties of the two-state system. We calculate the time evolution of $\ket{\psi(t)} = c_0(t)\ket{0} + c_1(t)\ket{1}$ with the Schr\selectlanguage{ngerman}ödinger equation and the perturbed Hamiltonian \eqref{Eq:TwoLevelGeneral}: +\begin{align} +i\hbar \frac{d}{dt}\ket{\psi(t)}&=\hat{H}\ket{\psi(t)},\\ +i \frac{d}{dt}\left(\begin{array}{c} c_0(t) \\ c_1(t) \end{array}\right) &= \frac{1}{2}\left( \begin{array}{cc} \Delta & \Omega \\ \Omega^* & -\Delta \end{array} \right) \left(\begin{array}{c} c_0(t) \\ c_1(t) \end{array} \right). +\end{align} + +We have two coupled differential equations and we luckily already know how to solve them as we have calculated the two eigenenergies in the previous section. For the state $\ket{\psi(t)}$ we get +\begin{align} + \ket{\psi(t)}=\lambda \eexp{-i{E_+}t/{\hbar}} \ket{\psi_+} + \mu \eexp{-i{E_-}t/{\hbar}} \ket{\psi_-} \label{eq:psitimeevolution} +\end{align} +with the factors $\lambda$ and $\mu$, which are defined by the initial state. The most common question is then what happens to the system if we start out in the bare state $\ket{0}$ and then let it evolve under coupling with a laser ? So what is the probability to find it in the other state $\ket{1}$: +\begin{align} +P_1(t)=\left|\braket{1|\psi(t)}\right|^2. +\end{align} + As a first step, we have to apply the initial condition to \eqref{eq:psitimeevolution} and express $\ket{\varphi}$ in terms of \eqref{eq:staticpsiplus} and \eqref{eq:staticpsiminus}: +\begin{align} +\ket{\psi(0)} \overset{!}{=}& \ket{0}\\ + = & \eexp{i{\varphi}/{2}} \left[ \cos\left( \frac{\theta}{2}\right) \ket{\psi_+}-\sin\left(\frac{\theta}{2}\right)\ket{\psi_-}\right] +\end{align} +By equating the coefficients we get for $\lambda$ and $\mu$: +\begin{align} +\lambda = \eexp{i{\varphi}/{2}}\cos\left(\frac{\theta}{2}\right), \qquad \mu = -\eexp{i{\varphi}/{2}}\sin\left(\frac{\theta}{2}\right). +\end{align} +One thus gets: +\begin{align} +\hspace{-2mm} P_1(t)=&\left|\braket{1|\psi(t)}\right|^2 \\ +=& \left|\eexp{i\varphi} \sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)\left[\eexp{-i{E_+}t/{\hbar}} - \eexp{-i{E_-}t/{\hbar}}\right]\right|^2\\ +=& \sin^2(\theta)\sin^2\left(\frac{E_+-E_-}{2\hbar}t\right) +\end{align} +$P_1(t)$ can be expressed with $\Delta$ and $\Omega$ alone. The obtained relation is called Rabi's formula: +\begin{align} + P_1(t)=\frac{1}{1+\left(\frac{\Delta}{|\Omega|}\right)^2}\sin^2\left(\sqrt{|\Omega|^2+\Delta^2}\frac{t}{2}\right) +\end{align}\selectlanguage{english} +\begin{figure}[h!] +\begin{center} +\includegraphics[width=0.70\columnwidth]{figures/Bildschirmfoto-2018-09-28-um-14-43-51/RabiOscillation} +\caption{{Rabi oscillations +{\label{833990}}% +}} +\end{center} +\end{figure} + +\subsection{Visualization of the dynamics in the spin picture} + +While the previous derivation might be the standard one, which certainly leads to the right results it might not be the most intuitive way of thinking about the dynamics. They become actually quite transparent in the spin language and on the Bloch sphere. So let us go back to the formulation of the Hamiltonian in terms of spins as in Eq. \eqref{Eq:HamSpin}. + +How would the question of the time evolution from $0$ to $1$ and back go now ? Basically, we would assume that the spin has been initialize into one of the eigenstates of the $z$-basis and now starts to rotate in some magnetic field. How ? This can be nicely studied in the Heisenberg picture, where operators have a time evolution. In the Heisenberg picture we have: +\begin{align} +\frac{d}{dt} \hat{s}_i &= \frac{i}{\hbar}\left[\hat{H},\hat{s}_i\right]\\ +\frac{d}{dt} \hat{s}_i &= \frac{i}{\hbar}\sum_j B_j \left[\hat{s}_j,\hat{s}_i\right]\\ + \end{align} +So to understand we time evolution, we only need to employ the commutator relationships between the spins: +\begin{align} +[ s_x, s_y] = \hbar is_z~~[ s_y, s_z] = \hbar is_x~~[ s_z, s_x] = \hbar is_y +\end{align} +For the specific case of $B_x=\Omega$, $B_y = B_z = 0$, we have then: +\begin{align} +\frac{d}{dt} \hat{s}_x &= 0\\ +\frac{d}{dt} \hat{s}_y &= -\Omega \hat{s}_z\\ +\frac{d}{dt} \hat{s}_z &= \Omega \hat{s}_y + \end{align} + + So applying a field in x-direction leads to a rotation of the spin around the $x$ axis with velocity $\Omega$. We can now use this general picture to understand the dynamics as rotations around an axis, which is defined by the different components of the magnetic field. + +\section{A few words on the quantum information notation} + +The qubit is THE basic ingredient of quantum computers. A nice way to play around with them is actually the \href{https://quantum-computing.ibm.com/}{IBM Quantum experience}. However, you will typically not find Pauli matrices etc within these systems. The typical notation there is: +\begin{itemize} +\item $R_x(\phi)$ is a rotation around the x-axis for an angle $\phi$. +\item Same holds for $R_y$ and $R_z$. +\item $X$ denotes the rotation around the x axis for an angle $\pi$. So it transforms $\ket{1}$ into $\ket{0}$ and vise versa. +\item $Z$ denotes the rotation around the x axis for an angle $\pi$. So it transforms $\ket{+}$ into $\ket{-}$ and vise versa. +\end{itemize} +The most commonly used gate is actually one that we did not talk about at all, it is the \textit{Hadamard} gate, which transforms $\ket{1}$ into $\ket{-}$ and $\ket{0}$ into $\ket{+}$: +\begin{align} +\hat{H}\ket{1} &= \ket{-} ~ \hat{H}\ket{0} &= \ket{+}\\ +\hat{H}\ket{-} &= \ket{1} ~ \hat{H}\ket{+} &= \ket{0} +\end{align} + +In the \href{https://www.authorea.com/326506/emMDRkXxtm44IKqpCtDi6g}{forth lecture} we will see how it is that a time-dependent field can actually couple two atomic states, which are normally of very different energies. + +\selectlanguage{english} +\FloatBarrier +\bibliographystyle{plainnat} +\bibliography{bibliography/converted_to_latex.bib% +} + +\end{document} + diff --git a/amo/tex_files/Lecture 4 - Atoms in oscillating fields.tex b/amo/tex_files/Lecture 4 - Atoms in oscillating fields.tex new file mode 100644 index 0000000..4586226 --- /dev/null +++ b/amo/tex_files/Lecture 4 - Atoms in oscillating fields.tex @@ -0,0 +1,335 @@ +\documentclass[10pt]{article} + +\usepackage{fullpage} +\usepackage{setspace} +\usepackage{parskip} +\usepackage{titlesec} +\usepackage[section]{placeins} +\usepackage{xcolor} +\usepackage{breakcites} +\usepackage{lineno} +\usepackage{hyphenat} + + + + + +\PassOptionsToPackage{hyphens}{url} +\usepackage[colorlinks = true, + linkcolor = blue, + urlcolor = blue, + citecolor = blue, + anchorcolor = blue]{hyperref} +\usepackage{etoolbox} +\makeatletter +\patchcmd\@combinedblfloats{\box\@outputbox}{\unvbox\@outputbox}{}{% + \errmessage{\noexpand\@combinedblfloats could not be patched}% +}% +\makeatother + + +\usepackage[round]{natbib} +\let\cite\citep + + + + +\renewenvironment{abstract} + {{\bfseries\noindent{\abstractname}\par\nobreak}\footnotesize} + {\bigskip} + +\titlespacing{\section}{0pt}{*3}{*1} +\titlespacing{\subsection}{0pt}{*2}{*0.5} +\titlespacing{\subsubsection}{0pt}{*1.5}{0pt} + + +\usepackage{authblk} + + +\usepackage{graphicx} +\usepackage[space]{grffile} +\usepackage{latexsym} +\usepackage{textcomp} +\usepackage{longtable} +\usepackage{tabulary} +\usepackage{booktabs,array,multirow} +\usepackage{amsfonts,amsmath,amssymb} +\providecommand\citet{\cite} +\providecommand\citep{\cite} +\providecommand\citealt{\cite} +% You can conditionalize code for latexml or normal latex using this. +\newif\iflatexml\latexmlfalse +\AtBeginDocument{\DeclareGraphicsExtensions{.pdf,.PDF,.eps,.EPS,.png,.PNG,.tif,.TIF,.jpg,.JPG,.jpeg,.JPEG}} + +\usepackage[utf8]{inputenc} +\usepackage[ngerman,english]{babel} + + + + + + + + +\usepackage{amsmath} +\newcommand{\bra}[1]{\ensuremath{\left\langle#1\right|}} +\newcommand{\ket}[1]{\ensuremath{\left|#1\right\rangle}} +\newcommand{\braket}[1]{\ensuremath{\left\langle#1\right\rangle}} +\newcommand{\rhohat}{\hat{\rho}} +\newcommand{\tr}[1]{\mathrm{tr}(#1)} +\newcommand{\trarb}[2]{\mathrm{tr}_{#1}(#2)} +\newcommand{\vv}[1]{\mathbf{#1}} +\newcommand*\dif{\mathop{}\!\mathrm{d}} +\newcommand{\eexp}[1]{\mathrm{e}^{#1}} + +\begin{document} + +\title{Lecture 4 - Atoms in oscillating fields} + + + +\author[1]{Fred Jendrzejewski}% +\author[2]{Selim Jochim}% +\affil[1]{Kirchhoff-Institut für Physik}% +\affil[2]{Physikalisches Institut der Universität Heidelberg}% + + +\vspace{-1em} + + + + + \date{January 04, 2025} + + +\begingroup +\let\center\flushleft +\let\endcenter\endflushleft +\maketitle +\endgroup + + + + + +\selectlanguage{english} +\begin{abstract} +In the lecture, we will see how a time dependent coupling allows us to engineer a new Hamiltonian. Most importantly, we will discuss the resonant coupling of two levels and the decay of a single level to a continuum.% +\end{abstract}% + + + +\sloppy + + +In the last lecture \cite{Jendrzejewski}, we discussed the properties of two coupled levels. However, we did not elaborate at any stage how such a system might emerge in a true atom. Two fundamental questions come to mind: +\begin{enumerate} +\item How is it that a laser allows to treat two atomic levels of very different energies as if they were degenerate ? +\item An atom has many energy levels $E_n$ and most of them are not degenerate. How can we reduce this complicated structure to a two-level system? +\end{enumerate} + +The solution is to resonantly couple two of the atom's levels by applying an external, oscillatory field, which is very nicely discussed in chapter 12 of Ref. \cite{2002} \cite{Cohen_Tannoudji_1998}. We will discuss important and fundamental properties of systems with a time-dependent Hamiltonian. + +We will discuss a simple model for the atom in the oscillatory field. We can write down the Hamiltonian: + +\begin{align} + \hat{H} = \hat{H}_0 + \hat{V}(t). +\end{align} +Here, $\hat{H}_0$ belongs to the atom and $V(t)$ describes the time-dependent field and its interaction with the atom. We assume that $\ket{n}$ is an eigenstate of $\hat{H}_0$ and write: +\begin{align} +\hat{H}_0\ket{n} = E_n \ket{n}. +\end{align} + +If the system is initially prepared in the state $\ket{i}$, so that +\begin{align} +\ket{\psi(t=0)} = \ket{i}, +\end{align} +what is the probability +\begin{align} +P_m(t) = \left|\braket{m|\psi(t)}\right|^2 +\end{align} +to find the system in the state $\ket{m}$ at the time $t$? + +\section{Evolution Equation} +The system $\ket{\psi(t)}$ can be expressed as follows: +\begin{align} +\ket{\psi(t)} = \sum_n \gamma_n(t) \eexp{-i{E_n}t/{\hbar}} \ket{n}, +\end{align} +where the exponential is the time evolution for $\hat{H}_1 =~0$. We plug this equation in the Schr\selectlanguage{ngerman}ödinger equation and get: +\begin{align} +i\hbar \sum_n\left(\dot{\gamma}_n(t)-i\frac{E_n}{\hbar}\gamma_n(t)\right)\eexp{-i{E_n}t/{\hbar}}\ket{n} = \sum_n \gamma_n(t) \eexp{-i{E_n}t/{\hbar}}\left(\hat{H}_0 + \hat{V}\right) \ket{n}\label{eq:timeev}\\ +\Longleftrightarrow i\hbar\sum_n \dot{\gamma}_n(t) \eexp{-i{E_n}t/{\hbar}} \ket{n} + = \sum_n \gamma_n(t) \eexp{-i{E_n}t/{\hbar}} \hat{V} \ket{n} +\end{align} +If we multiply \eqref{eq:timeev} with $\bra{k}$ we obtain a set of coupled differential equations +\begin{align} +i\hbar \dot{\gamma}_k \eexp{-i{E_k}t/{\hbar}} &= \sum_n \gamma_n \eexp{-{E_n}t/{\hbar}}\bra{k}\hat{V}\ket{n},\\ +i\hbar \dot{\gamma}_k &= \sum_n \gamma_n \eexp{-i {(E_n-E_k)}t/{\hbar}} \bra{k} \hat{V}\ket{n} +\end{align} +with initial conditions $\ket{\psi(t=0)}$. They determine the full time evolution. + +The solution of this set of equations depends on the details of the system. However, there are a few important points: + +\begin{itemize} +\item For short enough times, the dynamics are driving by the coupling strength $\bra{k}\hat{V} \ket{n}$. +\item The right-hand sight will oscillate on time scales of $E_n-E_k$ and typically average to zero for long times. +\item If the coupling element is an oscillating field $\propto e^{i\omega_L t}$, it might put certain times on resonance and allow us to avoid the averaging effect. It is exactly this effect, which allows us to isolate specific transitions to a very high degree \footnote{This is the idea behind atomic and optical clocks, which work nowadays at $10^{-18}$.} +\end{itemize} + +We will now see how the two-state system emerges from these approximations and then set-up the perturbative treatment step-by-step. + +\section{Rotating wave approximation} +We will now assume that the coupling term in indeed an oscillating field with frequency $\omega_L$, so it reads: +\begin{align} +\hat{V} = \hat{V}_0 \cos(\omega_Lt) = \frac{\hat{V}_0}{2} \left(e^{i\omega_lt}+e^{-i\omega_lt}\right) +\end{align} +We will further assume the we would like use it to isolate the transition $i\rightarrow f$, which is of frequency $\hbar \omega_0 = E_f - E_i$. The relevant quantity is then the detuning $\delta = \omega_0 - \omega_L$. If it is much smaller than any other energy difference $E_n-E_i$, we directly reduce the system to the following closed system: + +\begin{align} +i\dot{\gamma}_i &= \gamma_f \eexp{-i \delta t} \Omega\\ +i\dot{\gamma}_f &= \gamma_i \eexp{i \delta t}\Omega^* +\end{align} +Here we defined $\Omega = \bra{i} \frac{\hat{V_0}}{2\hbar}\ket{f}$. And to make it really a time-of the same form as the two-level system from the last lecture, we perform the transformation $\gamma_f = \tilde{\gamma}_f e^{i\delta t}$, which reduces the system too: +\begin{align} +i \dot{\gamma}_i &= \Omega \tilde{\gamma}_f \\ +i\dot{\tilde{\gamma}}_f &= \delta \tilde{\gamma}_f + \Omega^* \gamma_i +\end{align} +This has exactly the form of the two-level system that we studied previously. + + + +\subsection{Adiabatic elimination} + +We can now proceed to the quite important case of far detuning, where $\delta \gg \Omega$. In this case, the final state $\ket{f}$ gets barely populated and the time evolution can be approximated to to be zero \cite{lukin}. +\begin{align} +\dot{\tilde{\gamma}}_f = 0 +\end{align} +We can use this equation to eliminate $\gamma$ from the time evolution of the ground state. This approximation is known as \textit{adiabatic elimination}: +\begin{align} +\tilde{\gamma}_f &= \frac{\Omega^*}{\delta}\gamma_i\\ +\Rightarrow i\hbar \dot{\gamma}_i &= \frac{|\Omega|^2}{\delta} \tilde{\gamma}_i +\end{align} +The last equation described the evolution of the initial state with an energy $E_i = \frac{|\Omega|^2}{\delta}$. If the Rabi coupling is created through an oscillating electric field, i.e. a laser, this is know as the \textbf{light shift} or the \textbf{optical dipole potential}. It is this concept that underlies the optical tweezer for which Arthur Ashkin got the nobel prize in the 2018 \cite{2018}. + + +\subsection{Example: Atomic clocks in optical tweezers} + +A neat example that ties the previous concepts together is the recent paper \cite{readout}. The experimental setup is visualized in Fig. \ref{870855}.\selectlanguage{english} +\begin{figure}[h!] +\begin{center} +\includegraphics[width=0.70\columnwidth]{figures/Bildschirmfoto-2019-10-23-um-11-27-17/Bildschirmfoto-2019-10-23-um-11-27-17} +\caption{{Experimental setup of an atomic array optical clock as taken from +\protect\cite{readout}. +{\label{870855}}% +}} +\end{center} +\end{figure} + +While nice examples these clocks are still far away from the best clocks out there, which are based on optical lattice clocks and ions \cite{Ludlow_2015}. + + + +\section{Perturbative Solution} +The more formal student might wonder at which points all these rather hefty approximation are actually valid, which is obviously a very substantial question. So, we will now try to isolate the most important contributions to the complicated system through perturbation theory. For that we will assume that we can write: +\begin{align} +\hat{V}(t) =\lambda \hat{H}_1(t) +\end{align} +, where $\lambda$ is a small parameter. In other words we assume that the initial system $\hat{H}_0$ is only weakly perturbed. +Having identified the small parameter $\lambda$, we make the \textit{perturbative ansatz} +\begin{align} + \gamma_n(t) = \gamma_n^{(0)} + \lambda \gamma_n^{(1)} + \lambda^2 \gamma_n^{(2)} + \cdots +\end{align} +and plug this ansatz in the evolution equations and sort them by terms of equal power in $\lambda$. + +The $0$th order reads +\begin{align} + i\hbar \dot{\gamma}_k^{(0)} = 0. +\end{align} +The $0$th order does not have a time evolution since we prepared it in an eigenstate of $\hat{H}_0$. Any evolution arises due the coupling, which is at least of order $\lambda$. + +So, for the $1$st order we get +\begin{align} \label{eq:1storderapprox} +i\hbar \dot{\gamma}_k^{(1)} = \sum_n \gamma_n^{(0)} \eexp{-i(E_n-E_k)t/{\hbar}}\bra{k}\hat{H}_1\ket{n}. +\end{align} + +\subsection{First Order Solution (Born Approximation)} +For the initial conditions $\psi(t=0)=\ket{i}$ we get +\begin{align} +\gamma_k^{(0)}(t) = \delta_{ik}. +\end{align} +We plug this in the $1$st order approximation \eqref{eq:1storderapprox} and obtain the rate for the system to go to the final state $\ket{f}$: +% +\begin{align} +i \hbar\dot{\gamma}^{(1)} = \eexp{i(E_f-E_i)t/{\hbar}} \bra{f}\hat{H}_1 \ket{i} +\end{align} +Integration with $\gamma_f^{(1)}(t=0) = 0$ yields +\begin{align}\label{eq:gammaf1} +\gamma_f^{(1)} = \frac{1}{i\hbar}\int\limits_0^t \eexp{i(E_f-E_i)t'/{\hbar}} \bra{f} \hat{H}_1(t')\ket{i} \dif t', +\end{align} +so that we obtain the probability for ending up in the final state: +\begin{align} +P_{i\to f}(t) = \lambda^2\left| \gamma_f^{(1)}(t)\right|^2. +\end{align} +Note that $ P_{i\to f}(t) \ll 1$ is the condition for this approximation to be valid! + +\textbf{Example 1: Constant Perturbation.}\selectlanguage{english} +\begin{figure}[h!] +\begin{center} +\includegraphics[width=0.70\columnwidth]{figures/Bildschirmfoto-2018-09-28-um-15-46-04/Bildschirmfoto-2018-09-28-um-15-46-04} +\caption{{Sketch of a constant perturbation +{\label{723552}}% +}} +\end{center} +\end{figure} + +We apply a constant perturbation in the time interval $\left[0,T\right]$, as shown in +\ref{723552}. If we use \eqref{eq:gammaf1} and set $\hbar \omega_0 = E_f-E_i$, we get +\begin{align} +\gamma_f^{(1)}(t\geq T) = \frac{1}{i \hbar} \bra{f}\hat{H}_1\ket{i} \frac{\eexp{i\omega_0 T}-1}{i\omega_0}, +\end{align} +and therefore +\begin{align} +P_{i\to f} = \frac{1}{\hbar^2}\left|\bra{f}\hat{V}\ket{i}\right|^2 \underbrace{\frac{\sin^2\left(\omega_0\frac{T}{2}\right)}{\left(\frac{\omega_0}{2}\right)^2}}_{\mathrm{y}(\omega_0,T)}. +\end{align} +A sketch of $\mathrm{y}(\omega_0,T)$ is shown in \ref{615128}.\selectlanguage{english} +\begin{figure}[h!] +\begin{center} +\includegraphics[width=0.70\columnwidth]{figures/Bildschirmfoto-2018-09-28-um-15-54-58/Bildschirmfoto-2018-09-28-um-15-54-58} +\caption{{A sketch of y +{\label{615128}}% +}} +\end{center} +\end{figure} + +We can push this calculation to the extreme case of $T\rightarrow \infty$. This results in a delta function, which is peaked round $\omega_0 = 0$ and we can write: +\begin{align} +P_{i\to f} = T\frac{2\pi}{\hbar^2}\left|\bra{f}\hat{V}\ket{i}\right|^2\delta(\omega_0) +\end{align} +This is the celebrated \textbf{Fermi's golden rule}. + +\textbf{Example 2: Sinusoidal Perturbation.} +For the perturbation +\begin{align} +\hat{H}_1(t) = \left\{ \begin{array}{ccl} \hat{H}_1\eexp{-i\omega t} && \text{for}\; 0 < t < T \\ 0 &&\text{otherwise}\end{array} \right. +\end{align} +we obtain the probability +\begin{align} +P_{i\to f} (t \geq T) = \frac{1}{\hbar^2} \left|\bra{f}\hat{V}\ket{i}\right|^2 \mathrm{y}(\omega_0 - \omega, T). +\end{align} + +At $\omega = \left|E_f - E_i\right|/\hbar$ we are on resonance. + +In the +\href{https://www.authorea.com/users/143341/articles/326514-lecture-5-the-hydrogen-atom}{fifth +lecture}, we will start to dive into the hydrogen atom. + +\selectlanguage{english} +\FloatBarrier +\nocite{*} + +\bibliographystyle{plainnat} +\bibliography{bibliography/converted_to_latex.bib% +} + +\end{document} + diff --git a/amo/tex_files/Lecture 5 - The Hydrogen Atom.tex b/amo/tex_files/Lecture 5 - The Hydrogen Atom.tex new file mode 100644 index 0000000..5513c8f --- /dev/null +++ b/amo/tex_files/Lecture 5 - The Hydrogen Atom.tex @@ -0,0 +1,339 @@ +\documentclass[10pt]{article} + +\usepackage{fullpage} +\usepackage{setspace} +\usepackage{parskip} +\usepackage{titlesec} +\usepackage[section]{placeins} +\usepackage{xcolor} +\usepackage{breakcites} +\usepackage{lineno} +\usepackage{hyphenat} + + + + + +\PassOptionsToPackage{hyphens}{url} +\usepackage[colorlinks = true, + linkcolor = blue, + urlcolor = blue, + citecolor = blue, + anchorcolor = blue]{hyperref} +\usepackage{etoolbox} +\makeatletter +\patchcmd\@combinedblfloats{\box\@outputbox}{\unvbox\@outputbox}{}{% + \errmessage{\noexpand\@combinedblfloats could not be patched}% +}% +\makeatother + + +\usepackage[round]{natbib} +\let\cite\citep + + + + +\renewenvironment{abstract} + {{\bfseries\noindent{\abstractname}\par\nobreak}\footnotesize} + {\bigskip} + +\titlespacing{\section}{0pt}{*3}{*1} +\titlespacing{\subsection}{0pt}{*2}{*0.5} +\titlespacing{\subsubsection}{0pt}{*1.5}{0pt} + + +\usepackage{authblk} + + +\usepackage{graphicx} +\usepackage[space]{grffile} +\usepackage{latexsym} +\usepackage{textcomp} +\usepackage{longtable} +\usepackage{tabulary} +\usepackage{booktabs,array,multirow} +\usepackage{amsfonts,amsmath,amssymb} +\providecommand\citet{\cite} +\providecommand\citep{\cite} +\providecommand\citealt{\cite} +% You can conditionalize code for latexml or normal latex using this. +\newif\iflatexml\latexmlfalse +\AtBeginDocument{\DeclareGraphicsExtensions{.pdf,.PDF,.eps,.EPS,.png,.PNG,.tif,.TIF,.jpg,.JPG,.jpeg,.JPEG}} + +\usepackage[utf8]{inputenc} +\usepackage[ngerman,english]{babel} + + + + + + + + +\usepackage{siunitx} +\usepackage{amsmath} +\newcommand{\bra}[1]{\ensuremath{\left\langle#1\right|}} +\newcommand{\ket}[1]{\ensuremath{\left|#1\right\rangle}} +\newcommand{\braket}[1]{\ensuremath{\left\langle#1\right\rangle}} +\newcommand{\rhohat}{\hat{\rho}} +\newcommand{\tr}[1]{\mathrm{tr}(#1)} +\newcommand{\trarb}[2]{\mathrm{tr}_{#1}(#2)} +\newcommand{\vv}[1]{\mathbf{#1}} +\newcommand*\dif{\mathop{}\!\mathrm{d}} +\newcommand{\eexp}[1]{\mathrm{e}^{#1}} + +\begin{document} + +\title{Lecture 5 - The Hydrogen Atom} + + + +\author[1]{Fred Jendrzejewski}% +\author[2]{Selim Jochim}% +\affil[1]{Kirchhoff-Institut für Physik}% +\affil[2]{Physikalisches Institut der Universität Heidelberg}% + + +\vspace{-1em} + + + + + \date{January 04, 2025} + + +\begingroup +\let\center\flushleft +\let\endcenter\endflushleft +\maketitle +\endgroup + + + + + +\selectlanguage{english} +\begin{abstract} +In this lecture we will first discuss the diagonalization of the harmonic oscillator and then discuss the main properties of the hydrogen atom.% +\end{abstract}% + + + +\sloppy + + +In the previous lectures we have seen how to treat eigenstates of the two-level system and then how we can derive its effective emergence from some complex level structure if we \href{https://www.authorea.com/users/143341/articles/326506-lecture-4-atoms-in-oscillating-fields}{apply oscillating fields}. + +Today, we will increase the complexity towards the harmonic oscillator and the hydrogen atom. + + +\section{The harmonic oscillator} + +The harmonic oscillator is another great toy model to understand certain properties of quantum mechanical systems. Most importantly, it is a great introduction into the properties of bound systems and ladder operators. The basic Hamiltonian comes along in a rather innocent fashion, namely: + +\begin{align}\label{Eq:HamHO} +\hat{H} &= \frac{\hat{p}^2}{2m}+ \frac{m\omega^2}{2}\hat{x}^2 +\end{align} +The two variables $\hat{p}$ and $\hat{x}$ are non-commuting $[\hat{x}, \hat{p}] = i\hbar$, so they cannot be measured at the same time. We would now like to put the operator into a diagonal form such that it reads something like: +\begin{align}\label{Eq:HamHO} +\hat{H} &= \sum_n \epsilon_n \ket{n}\bra{n} +\end{align} + +We will follow he quite closely the discussion of Ref. \cite{interactions}. + +\subsection{The ladder operators} +We would like to get the spectrum first. So make the equation look a bit nicer we will define $\hat{p} = \hat{P} \sqrt{m\omega}$ and $\hat{x} = \frac{\hat{X}}{\sqrt{m\omega}}$ such that we have: +\begin{align}\label{Eq:HamHO} +\hat{H} &= \frac{\omega}{2}\left(\hat{P}^2 + \hat{X}^2\right) +\end{align} +\footnote{The commutator between $\hat{X}$ and $\hat{P}$ is still as for $x$ and $p$.} The next step is then to define the ladder operators: +\begin{align} +\hat{a} = \frac{1}{\sqrt{2\hbar}}\left(\hat{X}+i\hat{P}\right)\\ +\hat{a}^\dag = \frac{1}{\sqrt{2\hbar}}\left(\hat{X}-i\hat{P}\right)\\ +\end{align} +At this stage we can just try to rewrite the Hamiltonian in terms of the operators, such that: +\begin{align} +\hat{a}^\dag \hat{a} &= \frac{1}{2\hbar}(\hat{X}-i\hat{P})(\hat{X}+i\hat{P})\\ +&= \frac{1}{2\hbar}(\hat{X}^2 +\hat{P}^2 -\hbar)\\ + \frac{1}{2}(X^2 +\hat{P}^2 ) &= \hbar \left(\hat{a}^\dag \hat{a}-\frac{1}{2}\right) +\end{align} +So the Hamiltonian can now be written as: +\begin{align} +\hat{H} &= \hbar \omega \left(\hat{N} + \frac{1}{2}\right)\text{ with } \hat{N} = a^\dag a +\end{align} +At this stage we have diagonalized the Hamiltonian, what remains to be understood is the the values that $\hat{a}^\dag a$ can take. + +\subsection{Action of the ladder operators in the Fock basis} + +We would like to understand the basis, which is defined by: +\begin{align} +\hat{N} \ket{n} = n \ket{n} +\end{align} +The non-commutation between $\hat{X}$ and $\hat{P}$ is translated to the ladder operators as: +\begin{align} +[\hat{a}, \hat{a}^\dag] &= \frac{1}{2\hbar}[\hat{X}+iP,\hat{X}-i\hat{P}] = 1\\ +~[\hat{N}, a] &= -\hat{a}\\ +~[\hat{N}, a^\dag] &= a^\dag +\end{align} +From these relationship we can show then that: +\begin{align} +\hat{a}\ket{n} = \sqrt{n}\ket{n-1}\\ +\hat{a}^\dag \ket{n} = \sqrt{n+1}\ket{n+1}\\ +\end{align} +These relations are the motivation for the name ladder operators as they connect the different eigenstates. And they are raising/lowering the quantum number by one. Finally we have to find the lower limit. And this is quite naturally 0 as $n = \bra{n}\hat{N}\ket{n} = \bra{\psi_1}\ket{\psi_1}\geq 0$. So we can construct the full basis by just defining the action of the lowering operator on the zero element $a\ket{0} = 0$ and the other operators are then constructed as: +\begin{align} +\ket{n} = \frac{(a^\dag)^n}{\sqrt{n!}}\ket{0} +\end{align} + +\subsection{Spatial representation of the eigenstates} + +While we now have the spectrum it would be really nice to obtain the spatial properties of the different states. For that we have to project them onto the x basis. Let us start out with the ground state for which we have $\hat{a}\ket{0}= 0$: +\begin{align} +\bra{x}\frac{1}{\sqrt{2\hbar}}\left(\sqrt{m\omega}\hat{x} +i \frac{1}{\sqrt{m\omega}}\hat{p}\right)\ket{0}= 0\\ +\left(\sqrt{\frac{m\omega}{\hbar}}x + \sqrt{\frac{\hbar}{m\omega}}\partial_x\right)\psi_0(x)= 0\\ +\Rightarrow \psi_0(x) \propto e^{-\frac{x^2}{2a_{HO}^2}} +\end{align} +This also introduces the typical distance in the quantum harmonic oscillator which is given by $a_{HO} =\sqrt{\hbar/m\omega}$. The other states are solutions to the defining equations: +\begin{align} +\psi_n(x) = \frac{1}{\sqrt{n!}2^n}\left(\sqrt{m\omega}x - \frac{1}{\sqrt{m\omega}}\frac{d}{dx}\right)^n \psi_0(x)\\ +\psi_n(x) = \frac{1}{\sqrt{n!}2^n}H_n(x) \psi_0(x)\\ +\end{align} +where $H_n(x)$ are the Hermite polynoms. + +\section{The hamiltonian of the hydrogen atom} + +The hydrogen atom plays at central role in atomic physics as it is \textit{the} basic ingredient of atomic structures. It describes a single \textit{electron}, which is bound to the nucleus of a single \textit{proton}. As such it is the simplest of all atoms and can be described analytically within high precision. This has motivated an enormous body of literature on the problem, which derives all imaginable properties in nauseating detail. Therefore, we will focus here on the main properties and only sketch the derivations, while we will reference to the more technical details. + +For the hydrogen atom as shown in \ref{261310}, we can write down the Hamiltonian +\begin{align} +\hat{H}=\frac{{{\hat{\vec{p}}}^2_\text{p}}}{2m_\text{p}} + \frac{{\hat{\vec{p}}}^2_\text{e}}{2m_\text{e}} - \frac{Ze^2}{4\pi\epsilon_0 r}, +\end{align} +where $Ze$ is the nuclear charge. To solve the problem, we have to find the right Hilbert space. We can not solve the problem of the electron alone. If we do a separation of coordinates, i.e., we separate the Hamiltonian into the the center of mass and the relative motion, we get +\begin{align} +\hat{H} = \underbrace{\frac{{\hat{\vec{p}}}^2_{\textrm{cm}}}{2M}}_{\hat{H}_{\textrm{cm}}} + \underbrace{\frac{{\hat{\vec{p}}}^2_\text{r}}{2\mu}- \frac{Ze^2}{4\pi\epsilon_0r}}_{\hat{H}_{\text{atom}}} \label{eq:hydrogencmatomsplit} +\end{align} +with the reduced mass $1/\mu=1/m_\text{e}+1/m_\text{p}$. +If the state of the hydrogen atom $\ket{\psi}$ is an eigenstate of $\hat{H}$, we can write +\begin{align} +\hat{H}\ket{\psi}=&\left( \hat{H}_\textrm{cm}+\hat{H}_{\text{atom}} \right)\ket{\psi_\textrm{cm}}\otimes \ket{\psi_\text{atom}} \label{eq:hydrogencmatom}\\ +=& \left( E_{\text{kin}} + E_\text{atom} \right) \ket{\psi}. +\end{align} +Both states in \eqref{eq:hydrogencmatom} are eigenstates of the system. %, e.g. particle being in momentum eigenstate +The state $\ket{\psi}$ can be split up as shown since the two degrees of freedom are generally not entangled.% Hilbert space\selectlanguage{english} +\begin{figure}[h!] +\begin{center} +\includegraphics[width=0.70\columnwidth]{figures/Bildschirmfoto-2018-09-28-um-16-07-07/Bildschirmfoto-2018-09-28-um-16-07-07} +\caption{{Sketch of the hydrogen atom with the relative coordinate and the +coordinates of the proton and the electron. +{\label{261310}}% +}} +\end{center} +\end{figure} + + + +The wave function of the system then reads: +\begin{align} +\psi(\vec{R},\vec{r}) =& \left( \bra{R} \otimes \bra{r}\right)\left( \ket{\psi_\textrm{cm}} \otimes \ket{\psi_{\text{atom}}}\right)\\ +=& \psi(\vec{R}) \cdot \psi (\vec{r}) +\end{align} +% +Our goal is now to find the eigenfunctions and eigenenergies of $\hat{H}_\text{atom}$. In order to further divide the Hilbert space, we can use the symmetries. + +\section{Conservation of orbital angular momentum} + +$\hat{H}_\text{atom}$ possesses spherical symmetry, which implies that \textbf{orbital angular momentum} $\hat{\vec{L}}$ is conserved. It is defined as: +\begin{align} +\hat{\vec{L}}=\hat{\vec{r}} \times \hat{\vec{p}} +\end{align} +In other words, we have: +\begin{align} +[\hat{H}_\text{atom}, \hat{\vec{L}}] = 0 +\end{align} +Let us show first that the kinetic term commutes with the angular momentum operator, +We will employ the commutator relationships for position and momentum $[x_i, p_j]=i\hbar$ and the relationship $[A,BC] = [A,B]C+B[A,C]$ and $[f(x), p_x] = [x,p_x]\frac{\partial f(x)}{\partial x}$. So we obtain: +\begin{align} +[p_x^2+p_y^2+p_z^2, xp_y - yp_x]&= [p_x^2,xp_y]-[p_y^2,yp_x] \\ + &= [p_x^2,x]p_y-[p_y^2,y] p_x\\ + &=i\hbar 2 p_xp_y-2i\hbar p_y p_x\\ + &= 0 +\end{align} +Analog calculations show that $L_y$ and $L_z$ commute. In a similiar fashion we can verify that the potential term commutes with the different components of $\hat{\vec{L}}$ +\begin{align} +[\frac{1}{r}, xp_y -yp_x] &= [\frac{1}{r}, xp_y]-[\frac{1}{r}, yp_x]\\ +&= x[\frac{1}{r}, p_y]-y[\frac{1}{r}, p_x]\\ +&= -x \frac{yi\hbar}{2r^{3/2}}+y\frac{xi\hbar}{2r^{3/2}}\\ +&=0 +\end{align} +We can therefore decompose the eigenfunctions of the hydrogen atom over the eigenbasis of the angular momentum operator. A detailled discussion of the properties of $\vec{L}$ can be found in Appendix B of \cite{Hertel_2015}. To find the eigenbasis, we first need to identify the commutation relationships between the components of $\hat{\vec{L}}$. We can calculate them following commutation relationships: +\begin{align} +[L_x, L_y] &= [yp_z - zp_y, zp_x - xp_z]\\ +&=[yp_z, zp_x]-[yp_z,xp_z]- [zp_y, zp_x] + [zp_y,xp_z]\\ +&=[yp_z, zp_x] + [zp_y,xp_z]\\ +&=[yp_z, z]p_x +x[zp_y,p_z]\\ +&=-i\hbar yp_x +i\hbar xp_y\\ +&= i\hbar L_z +\end{align} +This relationship holds for all the other components too and we have in general: +\begin{align} +[L_i, L_j] = i\hbar \epsilon_{ijk}L_k +\end{align} +The orbital angular momentum is therefore part of the large family of angular momentum operators, which also comprises spin etc. In particular the different components are not independent, and therefore we cannot form a basis out the three components. A suitable choice is actually to use the following combinations: +\begin{align} +\hat{\vec{L}}^2\ket{l,m_l} =& \hbar^2 l (l+1)\ket{l,m_l}\\ +\hat{L}_z\ket{l,m_l} =& \hbar m_l \ket{l,m_l} +\end{align} +\begin{itemize} +\item $l$ is a non-negative integer and it is called the \textbf{orbital angular momentum quantum number}. +\item $m_l$ takes values $-l, -l+1, ..., l-1, l$ and it is sometimes called the \textbf{projection of the angular momentum}. +\end{itemize} + +\subsection{Eigenfunction of the angular momentum operators} + +Having identified the relevant operators it would be nice to obtain a space representation of them. This works especially nicely in spherical coordinates. There, we get +\begin{align} +\hat{L}_z&= - i \hbar \partial_{\phi}\\ +\hat{\vec{L}}^2 &= - \hbar^2 \left[\frac{1}{\sin(\theta)}\partial_{\theta} \left( \sin(\theta) \partial_\theta\right) + \frac{1}{\sin^2(\theta)} \partial_{\phi\phi} \right]. +\end{align} +The corresponding wave functions are +\begin{align} +\braket{\theta, \phi | l,m_l} = Y_{lm}(\theta,\phi). +\end{align} + +Where $Y_{lm}(\theta, \phi)$ are the \textbf{spherical harmonics}. + +\section{The radial wave equation} + +Given that we now know that the angular momentum is conserved for the hydrogen atom, we can actually rewrite the Hamltonian \ref{eq:hydrogencmatomsplit} in terms of the angular momentum as we find: +\begin{align} +\hat{H}_\text{atom} = \hat{H}_r + \frac{\hat{L}}{2\mu r^2}+V(r) \\ +\hat{H}_r = -\frac{\hbar^2}{2\mu}\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) +\end{align} +We can now separate out the angular part and decompose it over the eigenfunctions of $\hat{\vec{L}}$, such that we make the ansatz \footnote{ Only if the system is in a well-defined angular momentum state, we can write it down like this.}: +\begin{align} +\psi (r,\theta,\phi) = R(r) Y_{lm}(\theta,\phi) +\end{align} + +We can plug this separated ansatz in the Schr\selectlanguage{ngerman}ödinger equation. We already solved the angular in the discussion of the angular momentum and for the radial part we obtain: +\begin{align} +-\frac{\hbar^2}{2\mu}\frac{1}{r}\frac{d^2(rR(r))}{dr^2} - \frac{Ze^2}{4\pi\epsilon_0 r} R(r) + \frac{\hbar^2}{2\mu}\frac{l(l+1)}{r^2}R(r) = ER(r) +\end{align} +% +Substituting $R(r)=u(r)/r$ leads to +\begin{align} +-\frac{\hbar^2}{2\mu}\frac{d^2}{dr^2}u(r) +\underbrace{ \left( -\frac{Ze^2}{4\pi\epsilon_0 r} + \frac{\hbar^2}{2\mu} \frac{l(l+1)}{r^2} \right)}_{V_{\text{eff}}} u(r) = E \, u(r),\end{align} +which is known as the ``radial wave equation''. It is a very general result for \emph{any} central potential. It can also be used to describe unbound states ($E>0$) that occur during scattering. + +In the \href{https://www.authorea.com/users/143341/articles/326674-lecture-6-the-dipole-approximation-in-the-hydrogen-atom}{next lecture} we will look into the energy scales of the hydrogen atom and then start coupling different levels. + +\selectlanguage{english} +\FloatBarrier +\nocite{*} + +\bibliographystyle{plainnat} +\bibliography{bibliography/converted_to_latex.bib% +} + +\end{document} + diff --git a/amo/tex_files/Lecture 6 - The dipole approximation in the hydrogen atom.tex b/amo/tex_files/Lecture 6 - The dipole approximation in the hydrogen atom.tex new file mode 100644 index 0000000..4d9ee2f --- /dev/null +++ b/amo/tex_files/Lecture 6 - The dipole approximation in the hydrogen atom.tex @@ -0,0 +1,426 @@ +\documentclass[10pt]{article} + +\usepackage{fullpage} +\usepackage{setspace} +\usepackage{parskip} +\usepackage{titlesec} +\usepackage[section]{placeins} +\usepackage{xcolor} +\usepackage{breakcites} +\usepackage{lineno} +\usepackage{hyphenat} + + + + + +\PassOptionsToPackage{hyphens}{url} +\usepackage[colorlinks = true, + linkcolor = blue, + urlcolor = blue, + citecolor = blue, + anchorcolor = blue]{hyperref} +\usepackage{etoolbox} +\makeatletter +\patchcmd\@combinedblfloats{\box\@outputbox}{\unvbox\@outputbox}{}{% + \errmessage{\noexpand\@combinedblfloats could not be patched}% +}% +\makeatother + + +\usepackage[round]{natbib} +\let\cite\citep + + + + +\renewenvironment{abstract} + {{\bfseries\noindent{\abstractname}\par\nobreak}\footnotesize} + {\bigskip} + +\titlespacing{\section}{0pt}{*3}{*1} +\titlespacing{\subsection}{0pt}{*2}{*0.5} +\titlespacing{\subsubsection}{0pt}{*1.5}{0pt} + + +\usepackage{authblk} + + +\usepackage{graphicx} +\usepackage[space]{grffile} +\usepackage{latexsym} +\usepackage{textcomp} +\usepackage{longtable} +\usepackage{tabulary} +\usepackage{booktabs,array,multirow} +\usepackage{amsfonts,amsmath,amssymb} +\providecommand\citet{\cite} +\providecommand\citep{\cite} +\providecommand\citealt{\cite} +% You can conditionalize code for latexml or normal latex using this. +\newif\iflatexml\latexmlfalse +\AtBeginDocument{\DeclareGraphicsExtensions{.pdf,.PDF,.eps,.EPS,.png,.PNG,.tif,.TIF,.jpg,.JPG,.jpeg,.JPEG}} + +\usepackage[utf8]{inputenc} +\usepackage[ngerman,english]{babel} + + + + + + + + +\usepackage{siunitx} +\usepackage{amsmath} +\newcommand{\bra}[1]{\ensuremath{\left\langle#1\right|}} +\newcommand{\ket}[1]{\ensuremath{\left|#1\right\rangle}} +\newcommand{\braket}[1]{\ensuremath{\left\langle#1\right\rangle}} +\newcommand{\rhohat}{\hat{\rho}} +\newcommand{\tr}[1]{\mathrm{tr}(#1)} +\newcommand{\trarb}[2]{\mathrm{tr}_{#1}(#2)} +\newcommand{\vv}[1]{\mathbf{#1}} +\newcommand*\dif{\mathop{}\!\mathrm{d}} +\newcommand{\eexp}[1]{\mathrm{e}^{#1}} +\newcommand{\Hzero}{\hat{H}_0} +\newcommand{\Wop}{\hat{W}} +\newcommand{\aOs}{\tilde{a}_{0}} + +\begin{document} + +\title{Lecture 6 - The dipole approximation in the hydrogen atom} + + + +\author[1]{Fred Jendrzejewski}% +\author[2]{Selim Jochim}% +\affil[1]{Kirchhoff-Institut für Physik}% +\affil[2]{Physikalisches Institut für Physik der Universität Heidelberg}% + + +\vspace{-1em} + + + + + \date{January 04, 2025} + + +\begingroup +\let\center\flushleft +\let\endcenter\endflushleft +\maketitle +\endgroup + + + + + +\selectlanguage{english} +\begin{abstract} +We will continue with some properties of the hydrogen atom. First compare it to the harmonic oscillator, then look into dipole transitions and end with the coupling to static magnetic fields.% +\end{abstract}% + + + +\sloppy + + +In the last lecture \cite{atom} we discussed the basic properties of the hydrogen atom and found its eigenstates. We will now summarize the most important properties and look into its orbitals. From that we will understand the understand the interaction with electromagnetic waves and introduce the selection rules for dipole transitions. + +\section{The energies of Hydrogen and its wavefunctions} +In the last lecture, we looked into hydrogen and saw that we could write it's Hamiltonian as: +\begin{align} +\hat{H}_\text{atom} = \hat{H}_r + \frac{\hat{L}}{2\mu r^2}+V(r) \\ +\hat{H}_r = -\frac{\hbar^2}{2\mu}\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) +\end{align} +We could then separate out the angular part and decompose it as: +\begin{align} +\psi (r,\theta,\phi) = \frac{u(r)}{r} Y_{lm}(\theta,\phi) +\end{align} +The radial wave equation reads then: +\begin{align}\label{Eq:RadWF} +-\frac{\hbar^2}{2\mu}\frac{d^2}{dr^2}u(r) +\underbrace{ \left( -\frac{Ze^2}{4\pi\epsilon_0 r} + \frac{\hbar^2}{2\mu} \frac{l(l+1)}{r^2} \right)}_{V_{\text{eff}}} u(r) = E \, u(r),\end{align} + +\subsection{Energy scales} +We can now make \eqref{Eq:RadWF} dimensionless, by rewriting: +\begin{align} +r = \rho \aOs +\end{align} +So we rewrite: +\begin{align} +-\frac{\hbar^2}{2\mu \aOs^2}\frac{d^2}{d\rho^2}u(r) + \left( -\frac{Ze^2}{4\pi\epsilon_0\aOs }\frac{1}{\rho} + \frac{\hbar^2}{2\mu \aOs^2} \frac{l(l+1)}{\rho^2} \right) u(r) = E \, u(r), +\end{align} +This allows us to measure energies in units of: +\begin{align} +E &= \epsilon R_{y,\textrm{m}}\\ +R_{y,\textrm{m}} &= -\frac{\hbar^2}{2\mu \aOs^2} +\end{align} +The equation reads then: +\begin{align} +\frac{d^2}{d\rho^2}u(\rho) + \left( \frac{\mu Ze^2 \aOs}{\hbar^2 4\pi\epsilon_0}\frac{2}{\rho} - \frac{l(l+1)}{\rho^2} \right) u(\rho) = \epsilon u(\rho), +\end{align} +If we finally set +\begin{align} +\aOs &=\frac{4\pi\epsilon_0 \hbar^2}{\mu Z e^2} +\end{align} +We obtain the especially elegant formulation: +\begin{align} +\frac{d^2}{d\rho^2}u(\rho) + \left( \frac{2}{\rho} - \frac{l(l+1)}{\rho^2} \right) u(\rho) = \epsilon u(\rho), +\end{align} +We typically call $\aOs$ the \textbf{Bohr radius} for an atom with reduced mass $\mu$ and with a nucleus with charge number $Z$. $R_{y,\textrm{m}}$ is the \textbf{Rydberg energy} of such an atom. + +The universal constant is defined for the infinite mass limit $\mu \approx m_e$ and for $Z=1$. As a length scale we introduce the Bohr radius for infinite nuclear mass +\begin{align} +a_0 &= \frac{4\pi\epsilon_0\hbar^2}{m_e e^2} = \text{\num{0.5} \text{angstrom}} = \text{\SI{0.05}{\nano\meter}}. +\end{align} +The energy scale reads: +\begin{align} +R_{y,\infty} &= \frac{m_e e^4}{32 \pi^2 \epsilon_0^2 \hbar^2}\\ +&\approx \SI{2.179e-18}{J}\\ +& \approx e \times\SI{13.6}{eV}\\ +&\approx h \times\SI{3289}{T\hertz} +\end{align} + So if we excite the hydrogen atom for time scales of a few attoseconds, we will coherently create superposition states of all existing levels. But which ones ? And at which frequency ? + +\subsection{Solution of the radial wave equation} +At this stage we can have a look into the energy landscape:\selectlanguage{english} +\begin{figure}[h!] +\begin{center} +\includegraphics[width=0.70\columnwidth]{figures/Bildschirmfoto-2018-09-28-um-16-21-46/Bildschirmfoto-2018-09-28-um-16-21-46} +\caption{{Energy potential of the hydrogen atom +{\label{951159}}% +}} +\end{center} +\end{figure} + + + + +The energies read then +\begin{align} +E_n = -\frac{R_{y,\textrm{m}}}{n^2} \qquad \text{with} \qquad n=1,2,3,\cdots +\end{align} +for $l=0$ and +\begin{align} +E_n = -\frac{R_{y,\textrm{m}}}{n^2} \qquad \text{with} \qquad n=2,3,4,\cdots +\end{align} +for $l=1$. Despite the different effective potentials (see \ref{951159}), we get the same eigenstates. This looks like an accidental degeneracy. +Actually, there is a hidden symmetry which comes from the so-called ``Runge-Lenz'' vector. It only occurs in an attractive $1/r$-potential \cite{atom}. This vector reads: +\begin{equation} +\mathbf{A} =\mathbf{p}\times\mathbf{L}-\mathbf{r} +\end{equation} + +Finally, we can also visualize the radial wavefunctions for the hydrogen atom as shown in Fig. \ref{785001}.\selectlanguage{english} +\begin{figure}[h!] +\begin{center} +\includegraphics[width=0.70\columnwidth]{figures/Bildschirmfoto-2018-10-29-um-08-04-45/Bildschirmfoto-2018-10-29-um-08-04-45} +\caption{{Radial wavefunctions +{\label{785001}}% +}} +\end{center} +\end{figure} + +Associated with these radial wavefunctions, we also have the angular profiles. Where $Y_{lm}(\theta, \phi)$ are the \textbf{spherical harmonics} as shown in Fig. \ref{175742}.\selectlanguage{english} +\begin{figure}[h!] +\begin{center} +\includegraphics[width=0.70\columnwidth]{figures/Bildschirmfoto-2018-10-18-um-09-04-42/Bildschirmfoto-2018-10-18-um-09-04-42} +\caption{{The spherical harmonics. Fig is taken from Ref.~\protect\cite{Demtr_der_2018} +{\label{175742}}% +}} +\end{center} +\end{figure} + +Their shape is especially important for understanding the possibility of coupling different orbits through electromagnetic waves. + + + + + +\section{The electric dipole approximation}\selectlanguage{english} +\begin{figure}[h!] +\begin{center} +\includegraphics[width=0.70\columnwidth]{figures/Bildschirmfoto-2018-09-29-um-21-46-38/Bildschirmfoto-2018-09-29-um-21-46-38} +\caption{{Interaction between an atom and an electromagnetic wave +\(\vec{E}\) with wave vector \(\vec{k}\). The states +\(\text{|g>}\) and~\(\text{|e>}\) stand for the ground and +excited state and \(\hbar\omega_0\) is the energy of the resonant +transition between the states. +{\label{823292}}% +}} +\end{center} +\end{figure} + + + +We consider an atom which is located in a radiation field. +By resonant coupling with the frequency $\omega_0$, it can go from the ground state $\ket{g}$ to the excited state $\ket{e}$ (see \ref{823292}). + +The potential energy of a charge distribution in a homogeneous electromagnetic field $\vec{E}$ is: +\begin{align} +E_\text{pot} = \sum_i q_i \vec{r}_i\cdot \vec{E}. +\end{align} +%Multipole expansion. Only one part left: +If the upper limit of the sum is 2, we obtain the dipole moment +\begin{align} +\vec{D} = e \vec{r}. +\end{align} +For the hydrogen atom, the distance corresponds to the Bohr radius.\selectlanguage{english} +\begin{figure}[h!] +\begin{center} +\includegraphics[width=0.70\columnwidth]{figures/Bildschirmfoto-2018-09-29-um-21-49-06/Bildschirmfoto-2018-09-29-um-21-49-06} +\caption{{A charge distribution in an electromagnetic field +\$\textbackslash{}vec\{E\}\$ +{\label{241421}}% +}} +\end{center} +\end{figure} + +\textbf{Note.} Apart from the monopole, the dipole potential is the lowest order term of the multipole expansion of the scalar potential $\phi$: +% Distance between particles small compared to range of field distribution we are in. +\begin{align} +\phi \left( \vec{r} \right) =& \frac{1}{4\pi\epsilon_0}\frac{\vec{D}\cdot\vec{r}}{|\vec{r}|^3}\\ +\vec{E}(\vec{r})=& \vec{\nabla}\phi(\vec{r}) = \frac{ 3 \left(\vec{D}\cdot \vec{r}\right) \vec{r}/{|\vec{r}|^2}- \vec{D}}{4\pi\epsilon_0|\vec{r}|^3}. +\end{align} + +For the dipole approximation we consider the size of the atom and compare it to the wavelength $\lambda$ of the electromagnetic field: +\begin{align} +\braket{|r|} \sim 1\text{angstrom}\ll \lambda \sim 10^3\text{angstrom} +\end{align} + +\begin{itemize} +\item Therefore, we assume that the field is homogeneous in space and omit the spatial dependence: +\begin{align} +E(r,t) \approx E(t) +\end{align} +\item The correction term resulting from the semi-classical dipole approximation then is +\begin{align} +\hat{H}_1(t)=-e\hat{\vec{r}} \cdot \vec{E}(t) = -\hat{\vec{D}} \cdot \vec{E}(t) +\end{align} +\item Why can the magnetic field be ignored in this approximation? The velocity of an electron is $\sim \alpha c$. The hydrogen atom only has small relativistic corrections. If we compare the modulus of the magnetic and the electric field, we get: +\begin{align} +\left| \vec{B} \right| = \frac{|\vec{E}|}{c} +\end{align} +The electric field contribution thus dominates. +\end{itemize} +% +Now we choose +\begin{align} +\vec{E} = E_0 \vec{\epsilon} \cos \left(\omega t - \vec{k} \cdot \vec{r}\right) +\end{align} +and do time-\-de\-pen\-dent perturbation theory (see \cite{Jendrzejewski}): + +\begin{align} +\ket{\psi(t)} = \gamma_1(t) \eexp{-iE_1t/\hbar} \ket{1} +& \gamma_2(t) \eexp{-iE_2t/\hbar} \ket{2}\\ ++&\sum_{n=3}^\infty \gamma_n \eexp{-iE_nt/\hbar} \ket{n} +\end{align} +As initial condition we choose +\begin{align} + \gamma_i(0) = \left\{ \begin{array}{ccc} 1 &\text{for}& i=1 \\ 0 &\text{for}& i>1 \end{array} \right. +\end{align} +% +We write $\omega_0 = (E_2-E_1)/\hbar$ and get to first order $\hat{vec{D}}$: + +\begin{align} +\gamma_2(t) = \overbrace{\frac{E_0}{2\hbar} \braket{2|\hat{\vec{D}}\cdot \vec{\epsilon}\,|1}}^{\text{Rabi frequency }\Omega} \underbrace{\left(\frac{\eexp{i(\omega_0 + \omega)t}-1}{\omega_0 + \omega} + \frac{\eexp{i(\omega_0 - \omega)t}-1}{\omega_0 - \omega}\right)}_{\text{time evolution of the system}} +\end{align} +% +The term before the round brackets is called dipole matrix element: +% +\begin{align}\label{Eq:DipOp} +\braket{2|\hat{\vec{D}}\cdot \vec{\epsilon}\,|1} =e \int \psi_2\left(\vec{r}\right) \cdot \vec{r} \cdot \vec{\epsilon} \cdot \psi_1\left(\vec{r}\right) \dif \vec{r}. +\end{align}\selectlanguage{english} +\begin{figure}[h!] +\begin{center} +\includegraphics[width=0.70\columnwidth]{figures/Bildschirmfoto-2018-10-31-um-11-54-46/SelectionRules} +\caption{{Coupling of different orbitals. +{\label{708926}}% +}} +\end{center} +\end{figure} + +\section{Selection rules} +We can now look into the allowed transition in the atom as they are what we will typically observe within experiments. + +\subsection{Change of parity} +The parity operator is defined as: +\begin{align} +\hat{P}\psi(\vec{r}) = \psi(-\vec{r}) +\end{align} +For the eigenfunction we have: +\begin{align} +\hat{P} \psi(\vec{r}) = \lambda \psi(\vec{r})\\ +\lambda = \pm 1 +\end{align} +The eigenvalues are called \textit{odd} and \textit{even}. From the definition of the dipole operator we can see that it is of odd parity. What about the parity of the states that it is coupling ? If they have both the same parity than the whole integral will disappear and no dipole transition can appear. + +We can become more concrete for the given eigenfunctions as we have within spherical coordinates: +\begin{align} +(r, \theta, \phi) \rightarrow (r, \pi -\theta, \phi+\pi) +\end{align} +For the orbitals of the hydrogen atom we then have explicitly: +\begin{align} +\hat{P}\psi_{nlm}(r, \theta, \phi) &= R_{nl}(r)Y_{lm}(\pi -\theta, \phi+\pi)\\ +&= (-1)^l R_{nl}(r)Y_{lm}(, \theta, \phi) +\end{align} +This gives us the first selection rule that the \textbf{orbital angular momentum has to change for dipole transitions} $\Delta l = \pm 1$. +\begin{itemize} +\item $s$ orbitals are only coupled to $p$ orbitals through dipole transitions. +\item $p$ orbitals are only coupled to $s$ and $d$ orbitals through dipole transitions. +\end{itemize} + + + +\subsection{Coupling for linearly polarized light} +Having established the need for parity change, we also need to investigate the influence of the polarization of the light, which enters the dipole operator through the vector $\epsilon$. In the simplest case the light has linear polarization ($\pi$ polarized) and we can write: +\begin{align} +\vec{E}(t) = \vec{e}_zE_0 \cos(\omega t +\varphi) +\end{align} +This means that the dipole transition element \eqref{Eq:DipOp} is now given by: +\begin{align} +\bra{2}\vec{D}\cdot\vec{e}_z\ket{1} = e \int \psi_2(\vec{r}) z \psi_1\left(\vec{r}\right) \dif \vec{r} +\end{align} +We can now transform z into the spherical coordinates $z= r \cos(\theta) = r\sqrt{\frac{4\pi}{3}}Y_{10}(\theta, \phi)$. We can further separate out the angular part of the integral to obtain: +\begin{align} +\bra{2}\vec{D}\cdot\vec{e}_z\ket{1} \propto e \int \sin(\theta) d\theta d\varphi Y_{l',m'}(\theta, \varphi) Y_{10}(\theta, \phi) Y_{l,m}(\theta, \varphi) +\end{align} +This element is only non-zero if $m = m'$ (see appendix C of \cite{Hertel_2015} for all the gorious details).\selectlanguage{english} +\begin{figure}[h!] +\begin{center} +\includegraphics[width=0.70\columnwidth]{figures/Bildschirmfoto-2018-10-30-um-13-22-49/Bildschirmfoto-2018-10-30-um-13-22-49} +\caption{{Dipole selection rules for different polarizations of light. +{\label{852353}}% +}} +\end{center} +\end{figure} + +\subsection{Circularly polarized light} +Light has not just linear polarization, but it might also have some circular polarization. In this case we can write: +\begin{align} +\vec{E}(t) &= \frac{E_0}{\sqrt{2}} \left(\cos(\omega t +\varphi)\vec{e}_x + \sin(\omega t +\varphi)\vec{e}_y\right)\\ +\vec{E}(t) &= \text{Re}\left(\vec{e}_+ E_0 e^{-i\omega t +\phi}\right)\\ +\vec{e}_\pm &= \frac{\vec{e}_x\pm i\vec{e}_y}{\sqrt{2}} +\end{align} +So light with polarization $\vec{\epsilon} = \vec{e}_+$ is called right-hand circular ($\sigma^+$) and $\vec{\epsilon} = \vec{e}_-$ is called left-hand circular ($\sigma^-$). Let us now evaluate the transition elements here. The dipole operator element boils now down to the evaluation of the integral: +\begin{align} +\bra{l',m',n'}x+iy\ket{l,m,n} +\end{align} +As previously we can express the coupling term in spherical coordinates: +\begin{align} +\frac{x+iy}{\sqrt{2}} = -r \sqrt{\frac{4\pi}{3}}Y_{11}(\theta, \varphi) +\end{align} +Evaluation of the integrals lead now to the rule the projection of the quantum number has to change $m' = m+1$. In a similiar fashion we find for left-hand circular light the selection rule $m' = m - 1$. All the results are summed up in Fig. \ref{852353}. + +In the next lecture~\cite{atoma} we will investigate the influence +of perturbative effects and see how the fine structure arises. + +\selectlanguage{english} +\FloatBarrier +\nocite{*} + +\bibliographystyle{plainnat} +\bibliography{bibliography/converted_to_latex.bib% +} + +\end{document} + diff --git a/amo/tex_files/Lecture 7 - Beyond the boring hydrogen atom.tex b/amo/tex_files/Lecture 7 - Beyond the boring hydrogen atom.tex new file mode 100644 index 0000000..239ba9a --- /dev/null +++ b/amo/tex_files/Lecture 7 - Beyond the boring hydrogen atom.tex @@ -0,0 +1,345 @@ +\documentclass[10pt]{article} + +\usepackage{fullpage} +\usepackage{setspace} +\usepackage{parskip} +\usepackage{titlesec} +\usepackage[section]{placeins} +\usepackage{xcolor} +\usepackage{breakcites} +\usepackage{lineno} +\usepackage{hyphenat} + + + + + +\PassOptionsToPackage{hyphens}{url} +\usepackage[colorlinks = true, + linkcolor = blue, + urlcolor = blue, + citecolor = blue, + anchorcolor = blue]{hyperref} +\usepackage{etoolbox} +\makeatletter +\patchcmd\@combinedblfloats{\box\@outputbox}{\unvbox\@outputbox}{}{% + \errmessage{\noexpand\@combinedblfloats could not be patched}% +}% +\makeatother + + +\usepackage[round]{natbib} +\let\cite\citep + + + + +\renewenvironment{abstract} + {{\bfseries\noindent{\abstractname}\par\nobreak}\footnotesize} + {\bigskip} + +\titlespacing{\section}{0pt}{*3}{*1} +\titlespacing{\subsection}{0pt}{*2}{*0.5} +\titlespacing{\subsubsection}{0pt}{*1.5}{0pt} + + +\usepackage{authblk} + + +\usepackage{graphicx} +\usepackage[space]{grffile} +\usepackage{latexsym} +\usepackage{textcomp} +\usepackage{longtable} +\usepackage{tabulary} +\usepackage{booktabs,array,multirow} +\usepackage{amsfonts,amsmath,amssymb} +\providecommand\citet{\cite} +\providecommand\citep{\cite} +\providecommand\citealt{\cite} +% You can conditionalize code for latexml or normal latex using this. +\newif\iflatexml\latexmlfalse +\AtBeginDocument{\DeclareGraphicsExtensions{.pdf,.PDF,.eps,.EPS,.png,.PNG,.tif,.TIF,.jpg,.JPG,.jpeg,.JPEG}} + +\usepackage[utf8]{inputenc} +\usepackage[ngerman,english]{babel} + + + + + + + + +\usepackage{siunitx} +\usepackage{amsmath} +\newcommand{\bra}[1]{\ensuremath{\left\langle#1\right|}} +\newcommand{\ket}[1]{\ensuremath{\left|#1\right\rangle}} +\newcommand{\braket}[1]{\ensuremath{\left\langle#1\right\rangle}} +\newcommand{\rhohat}{\hat{\rho}} +\newcommand{\tr}[1]{\mathrm{tr}(#1)} +\newcommand{\trarb}[2]{\mathrm{tr}_{#1}(#2)} +\newcommand{\vv}[1]{\mathbf{#1}} +\newcommand*\dif{\mathop{}\!\mathrm{d}} +\newcommand{\eexp}[1]{\mathrm{e}^{#1}} +\newcommand{\Hzero}{\hat{H}_0} +\newcommand{\Wop}{\hat{W}} + +\begin{document} + +\title{Lecture 7 - Beyond the 'boring' hydrogen atom} + + + +\author[1]{Fred Jendrzejewski}% +\author[2]{Selim Jochim}% +\affil[1]{Kirchhoff-Institut für Physik}% +\affil[2]{Physikalisches Institut der Universität Heidelberg}% + + +\vspace{-1em} + + + + + \date{January 04, 2025} + + +\begingroup +\let\center\flushleft +\let\endcenter\endflushleft +\maketitle +\endgroup + + + + + +\selectlanguage{english} +\begin{abstract} +In this lecture we will use the hydrogen atom to study static perturbations in form of external magnetic fields and relativistic effects, leading to the fine structure splitting.% +\end{abstract}% + + + +\sloppy + + +We spend quite some time on the properties of the hydrogen atom in the previous lectures \cite{Jendrzejewski,atom}. However, we completely neglected any effects of quantum-electrodynamics and relativistic physics. In this lecture we will study, why this is a good approximation for the hydrogen atom and then investigate in a perturbative fashion the terms. Most importantly, we will introduce that coupling between the orbital angular momentum and the spin of the electron, which leads to the fine splitting. + +\section{Perturbation theory} +Up to now have studied the hydrogen atom to find its eigensystem and then studied how it evolves under the presence of oscillating electric fields. This allowed us to understand in more detail the idea of eigenstates and then of time-dependent perturbation theory. However, one of the most important concepts that can be introduced very nicely on the hydrogen atom is stationnary perturbation theory in form of external magnetic fields or relativistic corrections. We will remind you of perturbation theory here and then apply it to some simple cases. + +We can now simply write down the problem as: +\begin{eqnarray} +\left(\Hzero +\lambda \Wop\right)\ket{\psi_m} = E_m\ket{\psi_m} +\end{eqnarray} +$\lambda$ is a very small parameter and $\Hzero$ is describing the hydrogen atom system. We will note the eigenvalues and eigenstates of this system as: +\begin{align}\label{Eq:EigsUnperturb} +\Hzero \ket{\varphi_n} = \epsilon_n \ket{\varphi_n} +\end{align} +While, we do not know the exact solution of $\ket{\psi_m}$ and the energy $E_m$, we decide to decompose them in the following expansion of the small parameter $\lambda$: +\begin{align} +\ket{\psi_m} &= \ket{\psi_m^{(0)}} + \lambda\ket{\psi_m^{(1)}}+\lambda^2\ket{\psi_m^{(2)}}+O(\lambda^3)\\ +E_m &= E_m^{(0)} +\lambda E_m^{(1)} + \lambda^2 E_m^{(2)}+O(\lambda^3)\, +\end{align} +To zeroth order in $\lambda$ we obtain: +\begin{eqnarray} +\Hzero \ket{\psi_m^{(0)}} = E_m^{(0)}\ket{\psi_m^{(0)}} +\end{eqnarray} +So it is just the unperturbed system and we can identify: +\begin{eqnarray} +\ket{\psi_m^{(0)}} = \ket{\varphi_m}~~E_m^{(0)} = \epsilon_m +\end{eqnarray} +For the first order we have to solve +\begin{eqnarray}\label{Eq:FirstOrder} +(\Hzero-E_m^{(0)}) \ket{\psi_m^{(1)}} + (\Wop-E_m^{(1)})\ket{\psi_m^{(0)}}= 0\\ +(\Hzero-\epsilon_m) \ket{\psi_m^{(1)}} + (\Wop-E_m^{(1)})\ket{\varphi_m}= 0 +\end{eqnarray} +We can multiply the whole equation by $\bra{\varphi_m}$ from the right. As $\bra{\varphi_m}\Hzero = \epsilon_m\bra{\varphi_m}$, the first term cancels out. Hence, we obtain: +\begin{eqnarray}\label{Eq:PerturbFirstOrder} +\boxed{E_m^{(1)} = \bra{\varphi_m}\Wop\ket{\varphi_m}} +\end{eqnarray} +We now also need to obtain the correction to the eigenstate. For that, we put \eqref{Eq:PerturbFirstOrder} into \eqref{Eq:FirstOrder}: +\begin{eqnarray} +(\Hzero-\epsilon_m) \ket{\psi_m^{(1)}} + (\Wop\ket{\varphi_m}-\ket{\varphi_m}\bra{\varphi_m}\Wop\ket{\varphi_m})= 0 +\end{eqnarray} +We can now multiply the whole equation by $\bra{\varphi_i}$ from the right and obtain: +\begin{eqnarray} +(\epsilon_i-\epsilon_m)\bra{\varphi_i}\ket{\psi_m^{(1)}}+\bra{\varphi_i}\Wop\ket{\varphi_m} &=& 0 +\end{eqnarray} +By rewriting the above equation, this directly gives us the decompositon of the $\ket{\psi_m^{(1)}}$ onto the original eigenstates and have: +\begin{eqnarray}\label{Eq:FirstOrderState} +\boxed{\ket{\psi_m^{(1)}} = \sum_{i\neq m} \frac{\bra{\varphi_i}\Wop\ket{\varphi_m}}{(\epsilon_m-\epsilon_i)}\ket{\varphi_i}} +\end{eqnarray} +And we end the calculation with second order pertubation in $\lambda$ +\begin{eqnarray} +(\Hzero-E_m^{(0)}) \ket{\psi_m^{(2)}} + (\Wop-E_m^{(1)})\ket{\psi_m^{(1)}}-E_m^{(2)} \ket{\psi_m^{(0)}}= 0\\ +(\Hzero-\epsilon_m) \ket{\psi_m^{(2)}} + (\Wop-E_m^{(1)})\ket{\psi_m^{(1)}}-E_m^{(2)} \ket{\varphi_m}= 0\\ +\end{eqnarray} +We can multiply once again whole equation by $\bra{\varphi_m}$ from the right, which directly drops the first term. The term $E_m^{(1)}\bra{\varphi_m}\ket{\psi_m^{(1)}}$ drops out as the first order perturbation does not contain a projection onto the initial state. So we can write: +\begin{eqnarray} +E_m^{(2)}= \bra{\varphi_m}\Wop\ket{\psi_m^{(1)}} +\end{eqnarray} +Plugging in our solution \eqref{Eq:FirstOrderState}, we obtain: +\begin{equation}\label{Eq:PerturbSecOrder} +\boxed{E_m^{(2)} = \sum_{i\neq m} \frac{|\bra{\varphi_i}\Wop\ket{\varphi_m}|^2}{(\epsilon_m-\epsilon_i)}} +\end{equation} + +\section{Static external magnetic fields} + +A first beautiful application of perturbation theory is the study of static magnetic fields (see Ch 1.9 and Ch. 2.7.1 of \cite{Hertel_2015} for more details). +The motion of the electron around the nucleus implies a magnetic current +\begin{align} +I = \frac{e}{t} = \frac{ev}{2\pi r} +\end{align} +and this implies a magnetic moment $M = I A$, with the enclosed surface $A=\pi r^2$. It may be rewritten as: +\begin{align} +\vec{M}_L &= -\frac{e}{2m_e}\vec{L} &=-\frac{\mu_B}{\hbar} \vec{L} \\ +\mu_B &= \frac{\hbar e}{2m_e} +\end{align} +where $\mu_B$ is the \textbf{Bohr magneton}. Its potential energy in a magnetic field $\vec{B} = B_0 \vec{e}_z$ is then: +\begin{align} +V_B &= -\vec{M}_L\cdot \vec{B}\\ +&= \frac{\mu_B}{\hbar} L_z B_0 +\end{align} +Its contribution is directly evaluated from Eq. \eqref{Eq:PerturbFirstOrder} to be: +\begin{align} +E_{Zeeman} = \mu_B m B_0 +\end{align} +This is the Zeeman splitting of the different magnetic substates. It is visualized in Fig. \ref{982283}.\selectlanguage{english} +\begin{figure}[h!] +\begin{center} +\includegraphics[width=0.70\columnwidth]{figures/Bildschirmfoto-2018-10-31-um-08-07-29/Bildschirmfoto-2018-10-31-um-08-07-29} +\caption{{The Zeeman effect in the hydrogen atom. +{\label{982283}}% +}} +\end{center} +\end{figure} + + + + +\section{Trapping with electric or magnetic fields} +We have now investigated the structure of the hydrogen atom and seen how its energy gets shifted in external magnetic fields. We can combine this understanding to study conservative traps for atoms and ions. Neutral atoms experience the external field: +\begin{align} +E_{mag}(x,y) = \mu_B m B_0(x,y) +\end{align} +For ions on the other hand we have fully charged particles. So they simply experience the external electric field directly: +\begin{align} +E_{el}(x,y) = -q E(x,y) +\end{align} + +Trapping atoms and ions has to be done under very good vacuum such that they are well isolate from the enviromnent and high precision experiments can be performed. + +However, the trap construction is not trivial given Maxwells equation $\text{div} \vec{E} = 0$ and $\text{div} \vec{B} = 0$. So, the experimentalists have to play some tricks with oscillating fields. We will not derive in detail how a resulting \textbf{Paul trap} works, but the \href{https://youtu.be/Xb-zpM0UOzk}{linked video} gives a very nice impression of the idea behind it. A sketch is presented in Fig. \ref{149591}.\selectlanguage{english} +\begin{figure}[h!] +\begin{center} +\includegraphics[width=0.70\columnwidth]{figures/IonTraps-01/IonTraps-01} +\caption{{The upper stage shows the phases of The two phases of the oscillating +electric field of a Paul trap. Taken +from~\href{https://en.wikipedia.org/wiki/Quadrupole_ion_trap}{wikipedia}. +Below we can see a linear ion (Paul) trap containing six calcium 40 +ions. Taken +from~\href{https://quantumoptics.at/en/research/lintrap.html}{here} . +{\label{149591}}% +}} +\end{center} +\end{figure} + + +This work on trapping ions dates back to the middle of the last century (!!!) and was recognized by the\href{https://www.nobelprize.org/prizes/physics/1989/summary/}{ Nobel prize in 1989} for Wolfgang Paul \cite{Paul_1990} and Hans Dehmelt \cite{Dehmelt_1990}. They shared the prize with Norman Ramsey, who developped extremely precise spectroscopic methods, now known as Ramsey spectroscopy \cite{Ramsey_1990}. + +For atoms we can play similiar games with magnetic traps. Again we have to solve the problem of the zero magnetic fields. Widely used configurations are the Ioffe-Pritchard trap, where quadrupole fields are superposed with a bias field \cite{Pritchard_1983}, or TOP-traps \cite{Petrich_1995}. + +Ion traps are now the basis of ionic quantum computers \cite{ions} and magnetic traps paved the way for quantum simulators with cold atoms \cite{Jendrzejewskia}. + + + +\section{What we missed from the Dirac equation} +Until now we have completely neglected relativistic effects, i.e. we should have really solved the Dirac equation instead of the Schr\selectlanguage{ngerman}ödinger equation. However, this is is major task, which we will not undertake here. But what were the main approximations ? +\begin{enumerate} +\item We neglected the existance of the electron spin. +\item We did not take into account the relativistic effects. +\end{enumerate} + +So, how does relativity affect the hydrogen spectrum? In a first step, we should actually introduce the magnetic moment of the spin: +\begin{align} +\vec{M}_S = -g_e \mu_B \frac{\vec{S}}{\hbar} +\end{align} +The spin of the electron is $1/2$, making it a fermion and the \textit{g factor of the electron} reads +\begin{align} +g_e \approx 2.0023 +\end{align} +Further discussions of the g-factor might be found in Chapter 6.6 of \cite{Hertel_2015}. + +\subsection{Amplitude of the relativistic effects} + +We saw in lecture 5 \cite{Jendrzejewski} and 6 \cite{Jendrzejewskib}, that the energy levels of hydrogenlike atoms are given by: +\begin{align}\label{Eq:EnergyHydrogen} +E_n &= \frac{Z^2 R_{y,\infty}}{n^2}\\ +R_{y,\infty} &= \frac{m_e e^4}{32 \pi^2 \epsilon_0^2 \hbar^2} +\end{align} +We can now use the fine-structure constant, which measures the coupling strength of the electric charges to the electromagnetic field: +\begin{align}\label{Eq:FS} +\alpha &= \frac{e^2}{4\pi\epsilon_0\hbar c}\\ +&= \frac{1}{137.035999139(31)} +\end{align} +We can now rewrite Eq. \eqref{Eq:EnergyHydrogen} as: +\begin{align} +E_n = \frac{1}{2} \underbrace{m_e c^2}_{\text{rest mass energy}} Z^2 \alpha^2 \frac{1}{n^2} +\end{align} +Here, $m_e c^2\approx \SI{511}{\kilo eV}$ is the rest mass energy of the electron. $E_n \approx \SI{10}{eV}$ on the other hand is the energy of the bound state and therefore in the order of the kinetic energy of the electron. As long as it is much smaller than the rest-mass of the electron, we can neglect the relativistic effects. A few observations: + +\begin{itemize} +\item Relativistic effects are most pronounced for deeply bound states of small quantum number $n$. +\item Relativistic effects effects will become important once $(Z\alpha)\approx 1$, so they will play a major role in heavy nuclei. +\end{itemize} + +For the hydrogen atom we can thus treat the relativistic effects in a perturbative approach.But the most important consequence of the relativistic terms is actually the existance of the electron spin. + +\subsection{The relativistic mass and Darwin term} + + +\begin{enumerate} +\item ``Relativistic mass'': +The relativistic relation between energy and momentum reads: +\begin{align} +E_\text{rel} &= \sqrt{(mc^2)^2+(\vec{p}c)^2}\\ +&\approx mc^2 + \frac{p^2}{2m}- \frac{\vec{p}^{\,4}}{8m^3c^2} + \cdots +\end{align} +The first two terms of the expansion are the nonrelativistic limit and the third term is the first correction. Therefore, the corresponding Hamiltonian is: +\begin{align} +\hat{H}_\text{rm} = - \frac{\hat{\vec{p}}^{\,4}}{8m^3c^2}. +\end{align} + +\item Darwin term: +If $r=0$, the potential $V(r)$ diverges to $-\infty$. We get: +\begin{align} +\hat{H}_\text{Darwin} = \frac{\pi \hbar^2}{2m^2c^2}\left( \frac{Ze^2}{4\pi\epsilon_0}\right) \delta(\hat{\vec{r}}) +\end{align} + +\end{enumerate} + +If we perform a first correction to the energy of the eigenstates $\braket{n,l,m}$ by calculating +% +\begin{align} +\braket{n,l,m|\hat{H}'|n,l,m}, +\end{align} +% +we find that it works perfectly for case (1) and (2) which is due to degeneracy. +$\hat{H}_\text{rm}$ and $\hat{H}_\text{Darwin}$ commute with all observables forming the complete set of commuting observables (CSCO) for $\hat{H}_0$ +\begin{align} +\hat{H}_0,\hat{\vec{L}}^2, \hat{L}_z, +\end{align} +% +with states described by $\ket{n,l,m}$. + +\selectlanguage{english} +\FloatBarrier +\bibliographystyle{plainnat} +\bibliography{bibliography/converted_to_latex.bib% +} + +\end{document} + diff --git a/amo/tex_files/Lecture 8 - The Helium atom.tex b/amo/tex_files/Lecture 8 - The Helium atom.tex new file mode 100644 index 0000000..8e52c08 --- /dev/null +++ b/amo/tex_files/Lecture 8 - The Helium atom.tex @@ -0,0 +1,414 @@ +\documentclass[10pt]{article} + +\usepackage{fullpage} +\usepackage{setspace} +\usepackage{parskip} +\usepackage{titlesec} +\usepackage[section]{placeins} +\usepackage{xcolor} +\usepackage{breakcites} +\usepackage{lineno} +\usepackage{hyphenat} + + + + + +\PassOptionsToPackage{hyphens}{url} +\usepackage[colorlinks = true, + linkcolor = blue, + urlcolor = blue, + citecolor = blue, + anchorcolor = blue]{hyperref} +\usepackage{etoolbox} +\makeatletter +\patchcmd\@combinedblfloats{\box\@outputbox}{\unvbox\@outputbox}{}{% + \errmessage{\noexpand\@combinedblfloats could not be patched}% +}% +\makeatother + + +\usepackage[round]{natbib} +\let\cite\citep + + + + +\renewenvironment{abstract} + {{\bfseries\noindent{\abstractname}\par\nobreak}\footnotesize} + {\bigskip} + +\titlespacing{\section}{0pt}{*3}{*1} +\titlespacing{\subsection}{0pt}{*2}{*0.5} +\titlespacing{\subsubsection}{0pt}{*1.5}{0pt} + + +\usepackage{authblk} + + +\usepackage{graphicx} +\usepackage[space]{grffile} +\usepackage{latexsym} +\usepackage{textcomp} +\usepackage{longtable} +\usepackage{tabulary} +\usepackage{booktabs,array,multirow} +\usepackage{amsfonts,amsmath,amssymb} +\providecommand\citet{\cite} +\providecommand\citep{\cite} +\providecommand\citealt{\cite} +% You can conditionalize code for latexml or normal latex using this. +\newif\iflatexml\latexmlfalse +\AtBeginDocument{\DeclareGraphicsExtensions{.pdf,.PDF,.eps,.EPS,.png,.PNG,.tif,.TIF,.jpg,.JPG,.jpeg,.JPEG}} + +\usepackage[utf8]{inputenc} +\usepackage[ngerman,english]{babel} + + + + + + + + +\usepackage{siunitx} +\usepackage{amsmath} +\newcommand{\bra}[1]{\ensuremath{\left\langle#1\right|}} +\newcommand{\ket}[1]{\ensuremath{\left|#1\right\rangle}} +\newcommand{\braket}[1]{\ensuremath{\left\langle#1\right\rangle}} +\newcommand{\rhohat}{\hat{\rho}} +\newcommand{\tr}[1]{\mathrm{tr}(#1)} +\newcommand{\trarb}[2]{\mathrm{tr}_{#1}(#2)} +\newcommand{\vv}[1]{\mathbf{#1}} +\newcommand*\dif{\mathop{}\!\mathrm{d}} +\newcommand{\eexp}[1]{\mathrm{e}^{#1}} +\newcommand{\Hzero}{\hat{H}_0} +\newcommand{\Wop}{\hat{W}} + +\begin{document} + +\title{Lecture 8 - The Helium atom} + + + +\author[1]{Fred Jendrzejewski}% +\author[2]{Selim Jochim}% +\affil[1]{Kirchhoff-Institut für Physik}% +\affil[2]{Physikalisches Institut der Universität Heidelberg}% + + +\vspace{-1em} + + + + + \date{January 04, 2025} + + +\begingroup +\let\center\flushleft +\let\endcenter\endflushleft +\maketitle +\endgroup + + + + + +\selectlanguage{english} +\begin{abstract} +In this lecture we will discuss some basic properties of the Helium atom. We will introduce first some useful notations for the specific Hamiltonian at hand. Then we will focus on the important consequences played by the electron-electron interaction on the spin structure and the level scheme of the system. Finally, we will introduce the variational method for the estimation of the ground state energy.% +\end{abstract}% + + + +\sloppy + + +In todays lecture, we will see how the electron spin couples to the orbital angular momentum and how this creates spin-orbit coupling. We will then start out with the discussion of the Helium atom. + +\section{Spin-orbit coupling} + +The third term, which arises from the Dirac equation is the spin-orbit coupling. We will give here a common hand-waving explanation in a similiar spirit to the discussion of the magnetic moment for given angular momentum \cite{Demtr_der_2010}. Please, be aware that it misses a factor of 2. The electron has a spin 1/2 and hence a magnetic moment $\vec{M}_S = -g_e \mu_B \frac{\vec{S}}{\hbar}$. This magnetic moment experiences a magnetic field, simply due to the motion of the electron charge itself. Assuming a circular motion of the electron, we obtain the magnetic field amplitude: +\begin{align} +B &= \frac{\mu_0 i}{2r}\\ +B &= \frac{\mu_0 ev}{4\pi r^2}\\ +B &= \frac{\mu_0 e}{4\pi m_e r^3}L\\ +\end{align} +Through the coupling with the spin and introducing a fudge factor of 2\footnote{It's proper derivation is left to quantum field theory lectures}, we obtain the Hamiltonian: +\begin{align}\label{Eq:HamLS} +\hat{H}_{LS} = \frac{g_e}{4\pi \epsilon_0}\frac{e^2}{2m_e^2c^2 r^3} \hat{\vec{L}}\cdot \hat{\vec{S}} +\end{align} + +How does it act on a state $\ket{\psi}$? For the example +\begin{align} +\ket{\psi} = \ket{m_l} \otimes \ket{m_s} +\end{align} +we get: +\begin{align} +\hat{L}_z \cdot \hat{S}_z \left( \ket{m_l} \otimes \ket{m_s} \right) += \hbar^2 m_l \cdot m_s (\ket{m_l} \otimes \ket{m_s}) +\end{align} +The states +% +\begin{align} +\ket{n,l,m_l} \otimes \ket{s,m_s}. +\end{align} +% + +span the complete Hilbert space. Any state of the atom can be represented by: +% +\begin{align} +\ket{\psi} = \sum_{\{n,l,m_l,m_s\}} c_{n,l,m_l,m_s} \ket{n,l,m_l,m_s}. +\end{align} +% +As usual we can massively simplify the problem by using the appropiate conserved quantities. + + + +\subsection{Conservation of total angular momentum} + +We can look into it a bit further into the details and see that the Hamiltonian $\hat{H}_\textrm{LS}$ does not commute with $\hat{L}_z$: +\begin{align} +[L_z, \vec{L}\cdot \vec{S}] &= [L_z, L_x S_x + L_y S_y + L_z S_z]\\ +[L_z, \vec{L}\cdot \vec{S}] &= [L_z, L_x ]S_x + [L_z, L_y ]S_y\\ +[L_z, \vec{L}\cdot \vec{S}] &= i\hbar L_y S_x -i\hbar L_x S_y\neq 0 +\end{align} +This suggests that $L_z$ is not a good quantum number anymore. We have to include the spin degree of freedom into the description. Let us repeat the same procedure for the spin projection: +\begin{align} +[S_z, \vec{L}\cdot \vec{S}] &= [S_z, L_x S_x + L_y S_y + L_z S_z]\\ +[S_z, \vec{L}\cdot \vec{S}] &= L_x [S_z, S_x] + L_y [S_z, S_y]\\ +[S_z, \vec{L}\cdot \vec{S}] &= i\hbar L_x S_y -i\hbar L_y S_x\neq 0 +\end{align} +This implies that the spin projection is not a conserved quantity either. However, the sum of spin and orbital angular momentum will commute $[L_z + S_z, \vec{L}\vec{S}] =0$ according to the above calculations. Similiar calculations hold for the other components, indicating that the \textit{total angular momentum} is conserved \footnote{It should be as there is no external torque acting on the atom}: +\begin{align} +\vec{J} = \vec{L} + \vec{S} +\end{align} +We can now rewrite eq. \eqref{Eq:HamLS} in terms of the conserved quantities through the following following little trick: +\begin{align} +\hat{\vec{J}}^2 &= \left( \hat{\vec{L}} + \hat{\vec{S}} \right) ^2 = \hat{\vec{L}}^2 + 2 \hat{\vec{L}} \cdot \hat{\vec{S}} + \hat{\vec{S}}^2\\ +\hat{\vec{L}} \cdot \hat{\vec{S}} &= \frac{1}{2} \left( \hat{\vec{J}}^2 - \hat{\vec{L}}^2 - \hat{\vec{S}}^2 \right) +\end{align} + + + + +This directly implies that $\hat{J}^2$, $\hat{L}^2$ and $\hat{S}^2$ are new conserved quantities of the system. If we call $\hat{H}_0$ the Hamiltonian of the hydrogen atom, we previously used the complete set of commuting observables \footnote{see lecture 2 for a few words on the definition of such a set }: +\begin{align} +\left\{ \hat{H}_0, \hat{\vec{L}}^2, \hat{L}_z,\hat{\vec{S}}^2, \hat{S}_z \right\} +\end{align} + +We now use the complete set of commuting observables: +\begin{align} +\left\{ \hat{H}_0 + \hat{H}_{LS}, \hat{\vec{L}}^2,\hat{\vec{S}}^2, \hat{\vec{J}}^2, \hat{J}_z \right\}. +\end{align} +The corresponding basis states $\ket{n,l,j,m_j}$ are given by: +% +\begin{align} +\ket{n,l,j,m_j} = \sum_{m_l,m_s} \ket{n, l, m_l, m_s} \underbrace{\braket{n, l, m_l, m_s | n, l, j, m_j}}_{\text{Clebsch-Gordan coefficients}} +\end{align} +% +Here, the Clebsch-Gordan coefficients (cf. \cite{Olive_2014}, p. 557, or \url{http://pdg.lbl.gov/2002/clebrpp.pdf}) describe the coupling of angular momentum states. + + + +\textbf{Example: $l=1$ and $s=1/2$.} + +With the Clebsch-Gordan coefficients, the following example states---given by $Jj$ and $m_j$---can be expressed by linear combinations of states defined by $m_l$ and $m_s$: +% +\begin{align} +\ket{j=\frac{3}{2}, m_j = \frac{3}{2}} &=&& \ket{m_l=1, m_s = +\frac{1}{2}}\\ +\ket{j=\frac{3}{2}, m_j = \frac{1}{2}} &= &\sqrt{\frac{1}{3}} &\ket{m_l=1, m_s = -\frac{1}{2}} +\sqrt{\frac{2}{3}} &\ket{m_l = 0, m_s = +\frac{1}{2}} +\end{align} + +\subsection{Summary of the relativistic shifts} +We can now proceed to a summary of the relativistic effects in the hydrogen atom as presented in Fig. \ref{391959}. + +\begin{itemize} +\item The states should be characterized by angular momentum anymore, but by the total angular momentum $J$ and the orbital angular momentum. We introduce the notation: +\begin{align} +nl_{j} +\end{align} +\item All shifts are on the order of $\alpha^2$ and hence pertubative. +\item Some levels remain degenerate in relativistic theory, most importantly the $2s_{1/2}$ and the $2p_{1/2}$ state. +\end{itemize}\selectlanguage{english} +\begin{figure}[h!] +\begin{center} +\includegraphics[width=0.70\columnwidth]{figures/Bildschirmfoto-2018-11-05-um-08-30-48/Bildschirmfoto-2018-11-05-um-08-30-48} +\caption{{Fine structure of the Hydrogen atom. Adapted from~\protect\cite{Demtr_der_2010}~ +Fig. 5.33 +{\label{391959}}% +}} +\end{center} +\end{figure} + +\section{The Lamb shift} + +The previous discussions studied the effects of the Dirac equation onto our understanding of the Hydrogen atom. Most importantly, we saw that we can test those predictions quite well through the shifts in the level scheme. It is possible to push this analysis even further. One particularly important candidate here are the degenerate levels $2s_{1/2}$ and $2p_{1/2}$. Being able to see any splitting here, will be proof physics beyond the Dirac equation. And it is a relative measurement, for which it therefore not necessary to have insane absolute precisions. It is exactly this measurement that Lamb and Retherford undertook in 1947 \cite{Lamb_1947}. They observed actually a splitting of roughly $1$GHz, which they drove through direct rf-transitions. The observed shift was immediately explained by Bethe \cite{Bethe_1947} through the idea of QED a concept that we will come back to later in this lecture in a much simpler context of cavity QED. + +We would simply like to add here that the long story of the hydrogen atom and the Lamb shift is far from over as open questions remained until September 2019. Basically, a group of people measured the radius in some 'heavy' muonic hydrogen very precisely in 2010 \cite{Pohl_2010}. They could only explain them by changing the size of the proton radius, which was previously assumed to be well measured. It was only this year the another team reperformed a similiar measurement on electronic hydrogen (the normal one), obtaining consistent results \cite{Bezginov_2019}. A nice summary of the "proton radius puzzle" can be found \href{https://www.quantamagazine.org/physicists-finally-nail-the-protons-size-and-hope-dies-20190911/}{here}. + +\section{The helium problem} + +In this lecture we will discuss the Helium atom and what makes it so interesting in the laboratory. We will most importantly see that you cannot solve the problem exactly. This makes it a great historical example where a simple system was used to test state-of-the-art theories. An extensive discussion can be found in Chapter 7 of \cite{bransden2003physics} or Chapter 6 of \cite{Demtr_der_2010}. Even nowadays, the system continues to be a nice test-bed of many-body theories \cite{Combescot_2017, Ott_2019}. + +The Helium atom describes a two electron system as shown in the figure below.\selectlanguage{english} +\begin{figure}[h!] +\begin{center} +\includegraphics[width=0.70\columnwidth]{figures/HeliumSketch/HeliumSketch} +\caption{{The helium atom describes two electrons coupled to the nucleus of charge +Z=2.~~ +{\label{982117}}% +}} +\end{center} +\end{figure} + + + +In the reference frame of center-of-mass we obtain the following Hamiltonian: +\begin{equation} +H = -\frac{\hbar^2}{2\mu}\nabla_{r_1}^2 -\frac{\hbar^2}{2\mu}\nabla_{r_2}^2-\frac{\hbar^2}{M}\nabla_{r_1}\cdot\nabla_{r_2}+\frac{e^2}{4\pi \epsilon_0}\left(-\frac{Z}{r_1}-\frac{Z}{r_2}+\frac{1}{r_{12}}\right) +\end{equation} + +The term in the middle is the mass polarization term. We further introduced the reduced mass +\begin{equation} +\mu = \frac{m_eM}{m_e + M} +\end{equation} +For the very large mass differences $M= 7300 m_e \gg m_e$, we can do two simplifications: +\begin{itemize} +\item Omit the term on the mass polarization. +\item Set the reduced mass to the mass of the electron. +\end{itemize} + +So we obtain the simplified Hamiltonian +\begin{equation} +H = -\frac{\hbar^2}{2m_e}\nabla_{r_1}^2 -\frac{\hbar^2}{2m_e}\nabla_{r_2}^2+\frac{e^2}{4\pi \epsilon_0}\left(-\frac{Z}{r_1}-\frac{Z}{r_2}+\frac{1}{r_{12}}\right) +\end{equation} + + + +\section{Natural units} +For simplicity it is actually nice to work in the so-called \textbf{natural units}, where we measure all energies and distance on typical scales. We will start out by measuring all distances in units of $a_0$, which is defined as: +\begin{equation} +a_0 = \frac{4\pi \epsilon_0 \hbar^2}{me^2} = \SI{0.5}{angstrom} +\end{equation} +So we can introduce the replacement: +\begin{equation} +\mathbf{r} = \mathbf{\tilde{r}}a_0 +\end{equation} +So the Hamiltonian reads: +\begin{eqnarray} +H &= -\frac{\hbar^2}{2m_ea_0^2}\nabla_{\tilde{r}_1}^2 -\frac{\hbar^2}{2m_ea_0^2}\nabla_{\tilde{r}_2}^2+\frac{e^2}{4\pi \epsilon_0 a_0}\left(-\frac{Z}{\tilde{r}_1}-\frac{Z}{\tilde{r}_2}+\frac{1}{\tilde{r}_{12}}\right)\\ +H &= -\frac{e^4 m}{2(4\pi\epsilon_0)^2 \hbar^2}\nabla_{\tilde{r}_1}^2 -\frac{e^4 m}{2(4\pi\epsilon_0)^2 \hbar^2}\nabla_{\tilde{r}_2}^2+\frac{e^4 m}{(4\pi \epsilon_0)^2\hbar^2}\left(-\frac{Z}{\tilde{r}_1}-\frac{Z}{\tilde{r}_2}+\frac{1}{\tilde{r}_{12}}\right) +\end{eqnarray} +And finally we can measure all energies in units of +\begin{equation} +E_0 = \frac{e^4 m}{(4\pi\epsilon_0)^2\hbar^2} = \SI{1}{hartree} = \SI{27.2}{eV} +\end{equation} +So the Hamiltonian reads in these natural units: +\begin{equation} +\label{eq:HeliumReduced} +\tilde{H} = -\frac{1}{2}\nabla_{\tilde{r}_1}^2 -\frac{1}{2}\nabla_{\tilde{r}_2}^2+\left(-\frac{Z}{\tilde{r}_1}-\frac{Z}{\tilde{r}_2}+\frac{1}{\tilde{r}_{12}}\right) +\end{equation} +Another, more common, way of introducing this is to define: +\begin{eqnarray} +m &=& \hbar = e = 4\pi \epsilon_0 \equiv 1\\ +\alpha &=& \frac{e^2}{(4\pi \epsilon_0) \hbar c}= \frac{1}{137}\\ +\Rightarrow c &=& \frac{1}{\alpha} +\end{eqnarray} + +Within these units we have for the hydrogen atom: +\begin{equation} +E_n = \frac{Z^2}{2}\frac{1}{n^2}E_0 +\end{equation} + +\textbf{For the remainder of this lecture we will assume that we are working in natural units and just omit the tildas.} + + + +\section{Electron-electron interaction} + +Now we can decompose the Hamiltonian in the following fashion: +\begin{equation} +H = H_1 + H_2 + H_{12} +\end{equation} +So without the coupling term between the electrons we would just have once again two hydrogen atoms. The whole crux is now that the term $H_{12}$ is actually coupling or \textbf{entangling} the two electrons. + + + +\section{Symmetries} + +The \textbf{exchange} operator is defined as: +\begin{align} +P_{12}\psi(r_1,r_2) = \psi(r_2, r_1) +\end{align} +We directly see for \eqref{eq:HeliumReduced} that the exchange operator commutes with the Hamiltonian, $[H,P_{12}] = 0$. This implies directly that the parity is a conserved quantity of the system and that we have a set of Eigenstates associated with the parity. + +We can now apply the operator twice: +\begin{align} +P_{12}^2\psi(r_1,r_2) = \lambda^2 \psi(r_1, r_2) = \psi(r_1, r_2) +\end{align} + +So we can see that there are two sets of eigenvalues with $\lambda = \pm 1$. +\begin{align} +P_{12}\psi_\pm = \pm \psi_\pm +\end{align} + +We will call: +\begin{itemize} +\item $\psi_+$ are para-states +\item $\psi_-$ are ortho-states +\end{itemize} + +This symmetry is a really strong one and it was only recently that direct transitions between ortho and para-states were observed \cite{Kanamori_2017}. Interestingly, we did not need to look into the spin and the Pauli principle for this discussion at all. This will happen in the next step. + +\section{Spin and Pauli principle} + +We have seen that the Hamiltonian \eqref{eq:HeliumReduced} does not contain the spin degree of freedom. So we can decompose the total wave function as: +\begin{equation} +\overline{\psi} = \psi(\mathbf{r}_1, \mathbf{r}_2) \cdot \chi(1,2) +\end{equation} + +\subsection{Spin degree of freedom} + +Given that the electron is $s=\frac{1}{2}$, we can decompose each wavefunction as: +\begin{equation} +\chi = \alpha |\uparrow\rangle + \beta |\downarrow\rangle +\end{equation} +So if the two spins were \textit{not} correlated, we could just write the spin wavefunction as: +\begin{equation} +\chi(1,2) = \chi_\mathrm{1}\cdot\chi_\mathrm{2} +\end{equation} +However, the electron-electron interaction entangles the atoms. An example would be the singlet state: +\begin{equation} +\chi(1,2) = \frac{1}{\sqrt{2}}\left(|\uparrow \downarrow\rangle - |\downarrow\uparrow \rangle\right) +\end{equation} + +To construct the full wave function we need to take into account the \textit{Pauli} principle, which telles us for Fermions that the \textit{full} wavefunction should anti-sysmmetrc under exchange of particles: +\begin{equation} +\overline{\psi}(q_1, q_2, \cdots, q_i,\cdots, q_j, \cdots) = +-\overline{\psi}(q_1, q_2, \cdots, q_j,\cdots, q_i, \cdots) +\end{equation} +This tells us that each quantum state can be only occupied by a single electron at maximum. + +Now we can come back to the full wavefunction using the results of the previous section. We have: +\begin{equation} +\overline{\psi}(1,2) = \psi_{\pm}(r_1,r_2)\chi_\mp(1,2) +\end{equation} +with $P_{12}\chi_\pm = \pm \chi_\pm$. Now can once again look for good solutions to this problem. It is basically the total spin $\mathbf{S} = \mathbf{S}_1 + \mathbf{S}_2$, or better $\mathbf{S}^2$. This commutes with both the Hamiltonian and the parity operator, so it is a conserved quantity. Sorting out the solutions we have +\begin{align} +\chi_- &= \frac{1}{\sqrt{2}}\left(|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle\right)\\ +\chi_{+,1} &= |\uparrow\uparrow\rangle \\ +\chi_{+,1} &= \frac{1}{\sqrt{2}}\left(|\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle\right) \\ +\chi_{+,-1} &= |\downarrow\downarrow\rangle \\ +\end{align} +So $\chi_+$ is associated with spin 1 and $\chi_-$ is associated with spin 0. + +\selectlanguage{english} +\FloatBarrier +\bibliographystyle{plainnat} +\bibliography{bibliography/converted_to_latex.bib% +} + +\end{document} + diff --git a/utils/cleanMd.ts b/utils/cleanMd.ts index 5751df7..5871133 100644 --- a/utils/cleanMd.ts +++ b/utils/cleanMd.ts @@ -34,6 +34,13 @@ export const removeMath = (fileDirectory: string = "./blog") => { // replace any =& with = blogContent = blogContent.replace(/=&/g, "="); + // replace any line ending on ` $$` with and empty line in the middle + blogContent = blogContent.replace(/ \$\$\n/g, "\n\n$$$\n"); + + // if a line starts with $$, add two empty lines behind it + blogContent = blogContent.replace(/\n\$\$ /g, "\n$$$\n\n"); + + // write the new content to the file fs.writeFileSync(`${fileDirectory}/${file}`, blogContent); }); From e4a55f5d73de35a28f10b7534617d0303cd61a76 Mon Sep 17 00:00:00 2001 From: fretchen Date: Sat, 4 Jan 2025 11:21:18 +0100 Subject: [PATCH 02/13] Clean the equations --- amo/lecture3.md | 20 ++++++++++---------- utils/cleanMd.ts | 2 ++ 2 files changed, 12 insertions(+), 10 deletions(-) diff --git a/amo/lecture3.md b/amo/lecture3.md index eabc632..8477549 100644 --- a/amo/lecture3.md +++ b/amo/lecture3.md @@ -80,7 +80,7 @@ rather boring. However, we typically have the ability to change the Hamiltonian by switching on and off laser or microwave fields [^1]. We can then write the Hamiltonian in its most general form as: $$ -\label{Eq:TwoLevelGeneral} + \hat{H} = \frac{\hbar}{2}\left( \begin{array}{cc} \Delta & \Omega_x - i\Omega_y\\ \Omega_x +i\Omega_y & -\Delta \end{array} \right) $$ @@ -114,7 +114,7 @@ $$ We then obtain: $$ -\label{Eq:HamSpin} + \hat{H} = \mathbf{B}\cdot\hat{\mathbf{s}}\text{ with }\mathbf{B} = (\Omega_x, \Omega_y, \Delta) $$ @@ -152,7 +152,7 @@ system anymore. How should the system be described now ? We can once again diagonalize the system and write $$ -\hat{H}\left|\varphi_{\pm\right\rangle} = E_{\pm}\left|\varphi_\pm\right\rangle\\ +\hat{H}\left|\varphi_{\pm}\right\rangle = E_{\pm}\left|\varphi_\pm\right\rangle\\ E_{\pm} = \pm\frac{\hbar}{2}\Omega\\ \left|\varphi_\pm\right\rangle = \frac{\left|0\right\rangle\pm\left|1\right\rangle}{\sqrt{2}} $$ @@ -176,7 +176,7 @@ case. By diagonalizing Eq. reference="Eq:TwoLevelGeneral"} we obtain: $$ -\label{eq:Epm} + E_\pm = \pm \frac{\hbar}{2} \sqrt{\Delta^2+|\Omega|^2} $$ @@ -192,17 +192,17 @@ width="0.70\\columnwidth"} The Eigenstates then read: $$ -\left|\psi_+\right\rangle=\cos\left(\frac{\theta}{2}\right) \mathrm{e}^{-i{\varphi}/{2}}\left|0\right\rangle+\sin\left(\frac{\theta}{2}\right) \mathrm{e}^{i{\varphi}/{2}}\left|1\right\rangle, \label{eq:staticpsiplus} +\left|\psi_+\right\rangle=\cos\left(\frac{\theta}{2}\right) \mathrm{e}^{-i{\varphi}/{2}}\left|0\right\rangle+\sin\left(\frac{\theta}{2}\right) \mathrm{e}^{i{\varphi}/{2}}\left|1\right\rangle, $$ $$ -\left|\psi_-\right\rangle=-\sin\left(\frac{\theta}{2}\right) \mathrm{e}^{-i{\varphi}/{2}}\left|0\right\rangle+\cos\left(\frac{\theta}{2}\right) \mathrm{e}^{i{\varphi}/{2}}\left|1\right\rangle, \label{eq:staticpsiminus} +\left|\psi_-\right\rangle=-\sin\left(\frac{\theta}{2}\right) \mathrm{e}^{-i{\varphi}/{2}}\left|0\right\rangle+\cos\left(\frac{\theta}{2}\right) \mathrm{e}^{i{\varphi}/{2}}\left|1\right\rangle, $$ where $$ - \label{eq:parameters} + \tan(\theta) = \frac{|\Omega|}{\Delta} $$ @@ -245,7 +245,7 @@ previous section. For the state $\left|\psi(t)\right\rangle$ we get $$ - \left|\psi(t)\right\rangle=\lambda \mathrm{e}^{-i{E_+}t/{\hbar}} \left|\psi_+\right\rangle + \mu \mathrm{e}^{-i{E_-}t/{\hbar}} \left|\psi_-\right\rangle \label{eq:psitimeevolution} + \left|\psi(t)\right\rangle=\lambda \mathrm{e}^{-i{E_+}t/{\hbar}} \left|\psi_+\right\rangle + \mu \mathrm{e}^{-i{E_-}t/{\hbar}} \left|\psi_-\right\rangle $$ with the factors $\lambda$ and $\mu$, which are defined @@ -270,8 +270,8 @@ reference="eq:staticpsiplus"} and reference="eq:staticpsiminus"}: $$ -\left|\psi(0)\right\rangle \overset{!}{=}& \left|0\right\rangle\\ - = & \mathrm{e}^{i{\varphi}/{2}} \left[ \cos\left( \frac{\theta}{2}\right) \left|\psi_+\right\rangle-\sin\left(\frac{\theta}{2}\right)\left|\psi_-\right\rangle\right] +\left|\psi(0)\right\rangle \overset{!}{=} \left|0\right\rangle\\ + = \mathrm{e}^{i{\varphi}/{2}} \left[ \cos\left( \frac{\theta}{2}\right) \left|\psi_+\right\rangle-\sin\left(\frac{\theta}{2}\right)\left|\psi_-\right\rangle\right] $$ By equating the coefficients we get for $\lambda$ and diff --git a/utils/cleanMd.ts b/utils/cleanMd.ts index 5871133..10ed016 100644 --- a/utils/cleanMd.ts +++ b/utils/cleanMd.ts @@ -40,6 +40,8 @@ export const removeMath = (fileDirectory: string = "./blog") => { // if a line starts with $$, add two empty lines behind it blogContent = blogContent.replace(/\n\$\$ /g, "\n$$$\n\n"); + // remove the \label{} with the text in it from the content + blogContent = blogContent.replace(/\\label{([^}]*)}/g, ""); // write the new content to the file fs.writeFileSync(`${fileDirectory}/${file}`, blogContent); From 8eb76ec257f3cd732fceca2044eef8651e049b2e Mon Sep 17 00:00:00 2001 From: fretchen Date: Sat, 4 Jan 2025 11:22:10 +0100 Subject: [PATCH 03/13] Delete blogs.json --- blog/blogs.json | 12 ------------ 1 file changed, 12 deletions(-) delete mode 100644 blog/blogs.json diff --git a/blog/blogs.json b/blog/blogs.json deleted file mode 100644 index ee4d866..0000000 --- a/blog/blogs.json +++ /dev/null @@ -1,12 +0,0 @@ -[ - { - "title": "Ideas behind the blog stack", - "content": "\nOver the last few years I kept writing notes and code in all kind of different ways. Sometimes I would use Wordpress, personal notes, markdown or Jupyter notebooks. They get saved in some repo and there you go. But these days I would like to bring them slowly together into some more common structure, i.e. on one common website.\n\nMy rather heavy reliance on Jupyter notebooks and markdown really mostly ruled out Wordpress. I also really like the ideas behind static site generators. They are simple, fast and can be version controlled. Then I had to choose the appropiate stack. The first logical idea would have been [mkdocs-material](https://squidfunk.github.io/mkdocs-material/). I have made great experiences with it in the past. It is super simple to set up, very configurable and it looks great. However, I recently started to have a deeper look into proper web tech of the type of React and it is simply sooo much more natural to work with those components etc.\n\nHaving settled on *React*, I first thought that it is totally enough to work with [create-react-app](https://create-react-app.dev/). However, you soon realize that there have been no releases over the last few years and that the project is not really maintained anymore. A cute little solution was then [nano-react-app](ttps://github.com/nano-react-app/nano-react-app). It is a super minimalistic setup and worked well as I started to play around.\n\nThis got me far enough with a single webpage. But as I wanted to have a blog with multiple posts, I had to think about how to structure the whole thing. And this is the moment where you need some kind of routers. And this is the moment, where I had to learn what react meant with the following statement [on their website](https://react.dev/learn/start-a-new-react-project):\n\n> If you want to build a new app or a new website fully with React, we recommend picking one of the React-powered frameworks popular in the community.\n>\n> You can use React without a framework, however we’ve found that most apps and sites eventually build solutions to common problems such as code-splitting, routing, data fetching, and generating HTML. These problems are common to all UI libraries, not just React.\n>\n> By starting with a framework, you can get started with React quickly, and avoid essentially building your own framework later.\n\nI really wanted to avoid this blow at the beginning but with the need for multiple pages I had to dive into this. After some research, I settled on [vike](https://vike.dev/). It provides everything I need and super flexible. It also has a bit of an indie vibe, which made it more sympathic. Finally setting it up is made quite easy with [create-bati](https://batijs.dev).\n\nSo here we are. I have a blog stack that I can work with. It is not perfect but it is a start. I will keep you updated on how it goes.", - "publishing_date": "2025-01-06" - }, - { - "title": "Hello World", - "content": "\n\nThis is the first post. It shows that the whole thing is working out nicely.", - "publishing_date": "2024-12-02" - } -] \ No newline at end of file From 30d33eff65fd3772e058cccc962cc5fbe365a5cf Mon Sep 17 00:00:00 2001 From: fretchen Date: Sat, 4 Jan 2025 11:39:49 +0100 Subject: [PATCH 04/13] Update lecture3.md --- amo/lecture3.md | 57 +++++++++++++++++++++++++++---------------------- 1 file changed, 31 insertions(+), 26 deletions(-) diff --git a/amo/lecture3.md b/amo/lecture3.md index 8477549..8c244b9 100644 --- a/amo/lecture3.md +++ b/amo/lecture3.md @@ -1,9 +1,9 @@ --- author: -- Fred Jendrzejewski -- Selim Jochim + - Fred Jendrzejewski + - Selim Jochim bibliography: -- bibliography/converted_to_latex.bib + - bibliography/converted_to_latex.bib date: January 04, 2025 title: Lecture 3 - The two-level system --- @@ -39,18 +39,18 @@ width="0.70\\columnwidth"} Some of the many examples for two-level systems that can be found in nature: -- Spin of the electron: Up vs. down state +- Spin of the electron: Up vs. down state -- Two-level atom with one electron (simplified): Excited vs. ground - state +- Two-level atom with one electron (simplified): Excited vs. ground + state -- Structures of molecules, e.g., [NH~3~](#fig:twostate) +- Structures of molecules, e.g., [NH~3~](#fig:twostate) -- Occupation of mesoscopic capacitors in nanodevices. +- Occupation of mesoscopic capacitors in nanodevices. -- Current states in superconducting loops. +- Current states in superconducting loops. -- Nitrogen-vacancy centers in diamond. +- Nitrogen-vacancy centers in diamond. # Hamiltonian, Eigenstates and Matrix Notation @@ -71,6 +71,7 @@ $$ $$ so that $\hat{H}_0$ be written as a diagonal matrix + $$ \hat{H}_0 = \left(\begin{array}{cc} E_0 & 0 \\ 0 & E_1 \end{array}\right). $$ @@ -79,12 +80,14 @@ If we would only prepare eigenstates the system would be rather boring. However, we typically have the ability to change the Hamiltonian by switching on and off laser or microwave fields [^1]. We can then write the Hamiltonian in its most general form as: + $$ \hat{H} = \frac{\hbar}{2}\left( \begin{array}{cc} \Delta & \Omega_x - i\Omega_y\\ \Omega_x +i\Omega_y & -\Delta \end{array} \right) $$ Sometimes we will also chose the definition: + $$ \Omega = |\Omega| e^{i\varphi}=\Omega_x + i\Omega_y $$ @@ -192,18 +195,18 @@ width="0.70\\columnwidth"} The Eigenstates then read: $$ -\left|\psi_+\right\rangle=\cos\left(\frac{\theta}{2}\right) \mathrm{e}^{-i{\varphi}/{2}}\left|0\right\rangle+\sin\left(\frac{\theta}{2}\right) \mathrm{e}^{i{\varphi}/{2}}\left|1\right\rangle, +\left|\psi_+\right\rangle=\cos\left(\frac{\theta}{2}\right) \mathrm{e}^{-i{\varphi}/{2}}\left|0\right\rangle+\sin\left(\frac{\theta}{2}\right) \mathrm{e}^{i{\varphi}/{2}}\left|1\right\rangle, $$ $$ -\left|\psi_-\right\rangle=-\sin\left(\frac{\theta}{2}\right) \mathrm{e}^{-i{\varphi}/{2}}\left|0\right\rangle+\cos\left(\frac{\theta}{2}\right) \mathrm{e}^{i{\varphi}/{2}}\left|1\right\rangle, +\left|\psi_-\right\rangle=-\sin\left(\frac{\theta}{2}\right) \mathrm{e}^{-i{\varphi}/{2}}\left|0\right\rangle+\cos\left(\frac{\theta}{2}\right) \mathrm{e}^{i{\varphi}/{2}}\left|1\right\rangle, $$ where $$ - -\tan(\theta) = \frac{|\Omega|}{\Delta} + +\tan(\theta) = \frac{|\Omega|}{\Delta} $$ # The Bloch sphere @@ -245,7 +248,7 @@ previous section. For the state $\left|\psi(t)\right\rangle$ we get $$ - \left|\psi(t)\right\rangle=\lambda \mathrm{e}^{-i{E_+}t/{\hbar}} \left|\psi_+\right\rangle + \mu \mathrm{e}^{-i{E_-}t/{\hbar}} \left|\psi_-\right\rangle + \left|\psi(t)\right\rangle=\lambda \mathrm{e}^{-i{E_+}t/{\hbar}} \left|\psi_+\right\rangle + \mu \mathrm{e}^{-i{E_-}t/{\hbar}} \left|\psi_-\right\rangle $$ with the factors $\lambda$ and $\mu$, which are defined @@ -320,7 +323,8 @@ picture we have: $$ \frac{d}{dt} \hat{s}_i = \frac{i}{\hbar}\left[\hat{H},\hat{s}_i\right]\\ \frac{d}{dt} \hat{s}_i = \frac{i}{\hbar}\sum_j B_j \left[\hat{s}_j,\hat{s}_i\right]\\ - + + $$ So to understand we time evolution, we only need to @@ -337,7 +341,8 @@ $$ \frac{d}{dt} \hat{s}_x = 0\\ \frac{d}{dt} \hat{s}_y = -\Omega \hat{s}_z\\ \frac{d}{dt} \hat{s}_z = \Omega \hat{s}_y - + + $$ So applying a field in x-direction leads to a rotation of the spin @@ -353,20 +358,20 @@ experience](https://quantum-computing.ibm.com/). However, you will typically not find Pauli matrices etc within these systems. The typical notation there is: -- $R_x(\phi)$ is a rotation around the x-axis for an angle $\phi$. +- $R_x(\phi)$ is a rotation around the x-axis for an angle $\phi$. -- Same holds for $R_y$ and $R_z$. +- Same holds for $R_y$ and $R_z$. -- $X$ denotes the rotation around the x axis for an angle $\pi$. So it - transforms $\left|1\right\rangle$ into - $\left|0\right\rangle$ and vise versa. +- $X$ denotes the rotation around the x axis for an angle $\pi$. So it + transforms $\left|1\right\rangle$ into + $\left|0\right\rangle$ and vise versa. -- $Z$ denotes the rotation around the x axis for an angle $\pi$. So it - transforms $\left|+\right\rangle$ into - $\left|-\right\rangle$ and vise versa. +- $Z$ denotes the rotation around the x axis for an angle $\pi$. So it + transforms $\left|+\right\rangle$ into + $\left|-\right\rangle$ and vise versa. The most commonly used gate is actually one that we did not talk about -at all, it is the *Hadamard* gate, which transforms +at all, it is the _Hadamard_ gate, which transforms $\left|1\right\rangle$ into $\left|-\right\rangle$ and $\left|0\right\rangle$ into From a6ad2a05e57eb5e38849976e39b0304b27135625 Mon Sep 17 00:00:00 2001 From: fretchen Date: Sat, 4 Jan 2025 12:34:00 +0100 Subject: [PATCH 05/13] Add the third lecture --- amo/{lecture3.md => lecture3.mdx} | 62 ++++++++------------------ amo/lecture3_pic1.png | Bin 0 -> 62926 bytes amo/lecture3_pic2.png | Bin 0 -> 90510 bytes amo/lecture3_pic3.png | Bin 0 -> 168156 bytes amo/lecture3_pic4.png | Bin 0 -> 12968 bytes components/Post.tsx | 4 +- package-lock.json | 70 ++++++++++++++++++++++++++++++ package.json | 1 + 8 files changed, 92 insertions(+), 45 deletions(-) rename amo/{lecture3.md => lecture3.mdx} (84%) create mode 100644 amo/lecture3_pic1.png create mode 100644 amo/lecture3_pic2.png create mode 100644 amo/lecture3_pic3.png create mode 100644 amo/lecture3_pic4.png diff --git a/amo/lecture3.md b/amo/lecture3.mdx similarity index 84% rename from amo/lecture3.md rename to amo/lecture3.mdx index 8c244b9..b6f0408 100644 --- a/amo/lecture3.md +++ b/amo/lecture3.mdx @@ -13,9 +13,8 @@ like level splitting at avoided crossings and dynamical properties like Rabi oscillations. After the previous discussions of some basic cooking recipes to quantum -mechanics in last weeks lectures [@Jendrzejewskia] and [@Jendrzejewski], -we will use them to understand the two-level system. A very detailled -discussion can be found in chapter 4 of Ref. [@1]. The importance of the +mechanics in last weeks lectures, we will use them to understand the two-level system. A very detailled +discussion can be found in chapter 4 of Ref. [^CT1]. The importance of the two-level system is at least three-fold: 1. It is the simplest system of quantum mechanics as it spans a Hilbert @@ -28,13 +27,13 @@ two-level system is at least three-fold: fundamental building block of the exploding field of quantum computing and quantum information science. -![Examples for two-state systems. a) Benzene: In the ground state, the + + +Examples for two-state systems. a) Benzene: In the ground state, the electrons are delocalized. b) Ammonia: The nitrogen atom is either found above or below the hydrogen triangle. The state changes when the nitrogen atom tunnels. c) Molecular ion : The electron is either localized near proton 1 or 2. -](figures/Bildschirmfoto-2018-09-28-um-14-39-281/Bildschirmfoto-2018-09-28-um-14-39-281){#217687 -width="0.70\\columnwidth"} Some of the many examples for two-level systems that can be found in nature: @@ -44,7 +43,7 @@ nature: - Two-level atom with one electron (simplified): Excited vs. ground state -- Structures of molecules, e.g., [NH~3~](#fig:twostate) +- Structures of molecules, e.g., $NH_3$ - Occupation of mesoscopic capacitors in nanodevices. @@ -174,23 +173,15 @@ $\Omega = 0$ and the 'dressed' states for the coupled system. ## General case Quite importantly we can solve the system completely even in the general -case. By diagonalizing Eq. -[\[Eq:TwoLevelGeneral\]](#Eq:TwoLevelGeneral){reference-type="eqref" -reference="Eq:TwoLevelGeneral"} we obtain: +case. By diagonalizing the Hamiltonian we obtain: $$ - E_\pm = \pm \frac{\hbar}{2} \sqrt{\Delta^2+|\Omega|^2} $$ The energies can be nicely summarized as in Fig. -[2](#326199){reference-type="ref" reference="326199"} - -![Anticrossing of energy levels. -](figures/Bildschirmfoto-2018-09-28-um-14-35-34/AvoidedCrossing){#326199 -width="0.70\\columnwidth"} -  + The Eigenstates then read: @@ -216,12 +207,9 @@ abstract, it is extremely helpful to visualize the problem on the Bloch sphere. The idea of the Bloch sphere is that the we have a complex wave function of well defined norm and two free parameters. So it seems quite natural to look for a good representation of it. And this is the Bloch -sphere as drawn in Fig. [3](#613576){reference-type="ref" -reference="613576"}. +sphere as drawn below -![The presentation of the eigenstate on the Bloch sphere. -](figures/BlochSphereWithVectorForLecture/BlochSphereWithVectorForLecture){#613576 -width="0.70\\columnwidth"} + We will see especially its usefulness especially as we discuss the dynamics of the two-state system. @@ -233,9 +221,7 @@ dynamics of the two-state system. After the static case we now want to investigate the dynamical properties of the two-state system. We calculate the time evolution of $\left|\psi(t)\right\rangle = c_0(t)\left|0\right\rangle + c_1(t)\left|1\right\rangle$ -with the Schrödinger equation and the perturbed Hamiltonian -[\[Eq:TwoLevelGeneral\]](#Eq:TwoLevelGeneral){reference-type="eqref" -reference="Eq:TwoLevelGeneral"}: +with the Schrödinger equation and the perturbed Hamiltonian: $$ i\hbar \frac{d}{dt}\left|\psi(t)\right\rangle=\hat{H}\left|\psi(t)\right\rangle,\\ @@ -263,14 +249,8 @@ P_1(t)=\left|\left\langle 1|\psi(t)\right\rangle\right|^2. $$ As a first step, we have to apply the initial condition -to -[\[eq:psitimeevolution\]](#eq:psitimeevolution){reference-type="eqref" -reference="eq:psitimeevolution"} and express -$\left|\varphi\right\rangle$ in terms of -[\[eq:staticpsiplus\]](#eq:staticpsiplus){reference-type="eqref" -reference="eq:staticpsiplus"} and -[\[eq:staticpsiminus\]](#eq:staticpsiminus){reference-type="eqref" -reference="eq:staticpsiminus"}: +to and express +$\left|\varphi\right\rangle$ in terms of $|\psi_+$ and $|\psi_-$: $$ \left|\psi(0)\right\rangle \overset{!}{=} \left|0\right\rangle\\ @@ -299,9 +279,7 @@ $$ P_1(t)=\frac{1}{1+\left(\frac{\Delta}{|\Omega|}\right)^2}\sin^2\left(\sqrt{|\Omega|^2+\Delta^2}\frac{t}{2}\right) $$ -![Rabi oscillations -](figures/Bildschirmfoto-2018-09-28-um-14-43-51/RabiOscillation){#833990 -width="0.70\\columnwidth"} + ## Visualization of the dynamics in the spin picture @@ -309,9 +287,7 @@ While the previous derivation might be the standard one, which certainly leads to the right results it might not be the most intuitive way of thinking about the dynamics. They become actually quite transparent in the spin language and on the Bloch sphere. So let us go back to the -formulation of the Hamiltonian in terms of spins as in Eq. -[\[Eq:HamSpin\]](#Eq:HamSpin){reference-type="eqref" -reference="Eq:HamSpin"}. +formulation of the Hamiltonian in terms of spins as at the beginning of the lecture. How would the question of the time evolution from $0$ to $1$ and back go now ? Basically, we would assume that the spin has been initialize into @@ -382,9 +358,9 @@ $$ \hat{H}\left|-\right\rangle = \left|1\right\rangle ~ \hat{H}\left|+\right\rangle = \left|0\right\rangle $$ -In the [forth -lecture](https://www.authorea.com/326506/emMDRkXxtm44IKqpCtDi6g) we will -see how it is that a time-dependent field can actually couple two atomic -states, which are normally of very different energies. +In the forth lecture we will see how it is that a time-dependent field can actually couple two atomic states, which are normally of very different energies. [^1]: See the discussions of the next lecture + + +[^CT1]: Quantum Mechanics, Volume 1. Cohen-Tannoudji, Diu, Laloe. 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{title}

{publishing_date &&

Published on: {publishing_date}

} - + {content} diff --git a/package-lock.json b/package-lock.json index ce2ce45..7da6596 100644 --- a/package-lock.json +++ b/package-lock.json @@ -14,6 +14,7 @@ "react-dom": "^19.0.0", "react-markdown": "^9.0.1", "rehype-katex": "^7.0.1", + "rehype-raw": "^7.0.0", "remark-gfm": "^4.0.0", "remark-math": "^6.0.0", "tsx": "^4.19.2", @@ -4027,6 +4028,31 @@ "url": "https://opencollective.com/unified" } }, + "node_modules/hast-util-raw": { + "version": "9.1.0", + "resolved": "https://registry.npmjs.org/hast-util-raw/-/hast-util-raw-9.1.0.tgz", + "integrity": "sha512-Y8/SBAHkZGoNkpzqqfCldijcuUKh7/su31kEBp67cFY09Wy0mTRgtsLYsiIxMJxlu0f6AA5SUTbDR8K0rxnbUw==", + "license": "MIT", + "dependencies": { + "@types/hast": "^3.0.0", + "@types/unist": "^3.0.0", + "@ungap/structured-clone": "^1.0.0", + "hast-util-from-parse5": "^8.0.0", + "hast-util-to-parse5": "^8.0.0", + "html-void-elements": "^3.0.0", + "mdast-util-to-hast": "^13.0.0", + "parse5": "^7.0.0", + "unist-util-position": "^5.0.0", + "unist-util-visit": "^5.0.0", + "vfile": "^6.0.0", + "web-namespaces": "^2.0.0", + "zwitch": "^2.0.0" + }, + "funding": { + "type": "opencollective", + "url": "https://opencollective.com/unified" + } + }, "node_modules/hast-util-to-estree": { "version": "3.1.0", "resolved": "https://registry.npmjs.org/hast-util-to-estree/-/hast-util-to-estree-3.1.0.tgz", @@ -4097,6 +4123,25 @@ "url": "https://opencollective.com/unified" } }, + "node_modules/hast-util-to-parse5": { + "version": "8.0.0", + "resolved": "https://registry.npmjs.org/hast-util-to-parse5/-/hast-util-to-parse5-8.0.0.tgz", + "integrity": "sha512-3KKrV5ZVI8if87DVSi1vDeByYrkGzg4mEfeu4alwgmmIeARiBLKCZS2uw5Gb6nU9x9Yufyj3iudm6i7nl52PFw==", + "license": "MIT", + "dependencies": { + "@types/hast": "^3.0.0", + "comma-separated-tokens": "^2.0.0", + "devlop": "^1.0.0", + "property-information": "^6.0.0", + "space-separated-tokens": "^2.0.0", + "web-namespaces": "^2.0.0", + "zwitch": "^2.0.0" + }, + "funding": { + "type": "opencollective", + "url": "https://opencollective.com/unified" + } + }, "node_modules/hast-util-to-text": { "version": "4.0.2", "resolved": "https://registry.npmjs.org/hast-util-to-text/-/hast-util-to-text-4.0.2.tgz", @@ -4162,6 +4207,16 @@ "url": "https://opencollective.com/unified" } }, + "node_modules/html-void-elements": { + "version": "3.0.0", + "resolved": "https://registry.npmjs.org/html-void-elements/-/html-void-elements-3.0.0.tgz", + "integrity": "sha512-bEqo66MRXsUGxWHV5IP0PUiAWwoEjba4VCzg0LjFJBpchPaTfyfCKTG6bc5F8ucKec3q5y6qOdGyYTSBEvhCrg==", + "license": "MIT", + "funding": { + "type": "github", + "url": "https://github.com/sponsors/wooorm" + } + }, "node_modules/ignore": { "version": "5.3.2", "resolved": "https://registry.npmjs.org/ignore/-/ignore-5.3.2.tgz", @@ -7018,6 +7073,21 @@ "url": "https://opencollective.com/unified" } }, + "node_modules/rehype-raw": { + "version": "7.0.0", + "resolved": "https://registry.npmjs.org/rehype-raw/-/rehype-raw-7.0.0.tgz", + "integrity": "sha512-/aE8hCfKlQeA8LmyeyQvQF3eBiLRGNlfBJEvWH7ivp9sBqs7TNqBL5X3v157rM4IFETqDnIOO+z5M/biZbo9Ww==", + "license": "MIT", + "dependencies": { + "@types/hast": "^3.0.0", + "hast-util-raw": "^9.0.0", + "vfile": "^6.0.0" + }, + "funding": { + "type": "opencollective", + "url": "https://opencollective.com/unified" + } + }, "node_modules/rehype-recma": { "version": "1.0.0", "resolved": "https://registry.npmjs.org/rehype-recma/-/rehype-recma-1.0.0.tgz", diff --git a/package.json b/package.json index 2ab7d08..985e438 100644 --- a/package.json +++ b/package.json @@ -19,6 +19,7 @@ "react-dom": "^19.0.0", "react-markdown": "^9.0.1", "rehype-katex": "^7.0.1", + "rehype-raw": "^7.0.0", "remark-gfm": "^4.0.0", "remark-math": "^6.0.0", "tsx": "^4.19.2", From 4953f2ac96614f158de3edf219230467df90591f Mon Sep 17 00:00:00 2001 From: fretchen Date: Sat, 4 Jan 2025 12:35:39 +0100 Subject: [PATCH 06/13] Create lecture4.md --- amo/lecture4.md | 245 ++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 245 insertions(+) create mode 100644 amo/lecture4.md diff --git a/amo/lecture4.md b/amo/lecture4.md new file mode 100644 index 0000000..66f6467 --- /dev/null +++ b/amo/lecture4.md @@ -0,0 +1,245 @@ +--- +author: +- Fred Jendrzejewski +- Selim Jochim +bibliography: +- bibliography/converted_to_latex.bib +date: January 04, 2025 +nocite: "[@*]" +title: Lecture 4 - Atoms in oscillating fields +--- + +In the lecture, we will see how a time dependent coupling allows us to +engineer a new Hamiltonian. Most importantly, we will discuss the +resonant coupling of two levels and the decay of a single level to a +continuum. + +In the last lecture [@Jendrzejewski], we discussed the properties of two +coupled levels. However, we did not elaborate at any stage how such a +system might emerge in a true atom. Two fundamental questions come to +mind: + +1. How is it that a laser allows to treat two atomic levels of very + different energies as if they were degenerate ? + +2. An atom has many energy levels $E_n$ and most of them are not + degenerate. How can we reduce this complicated structure to a + two-level system? + +The solution is to resonantly couple two of the atom's levels by +applying an external, oscillatory field, which is very nicely discussed +in chapter 12 of Ref. [@2002] [@Cohen_Tannoudji_1998]. We will discuss +important and fundamental properties of systems with a time-dependent +Hamiltonian. + +We will discuss a simple model for the atom in the oscillatory field. We +can write down the Hamiltonian: + +$$\begin{aligned} + \hat{H} = \hat{H}_0 + \hat{V}(t). +\end{aligned}$$ Here, $\hat{H}_0$ belongs to the atom and $V(t)$ +describes the time-dependent field and its interaction with the atom. We +assume that $\ensuremath{\left|n\right\rangle}$ is an eigenstate of +$\hat{H}_0$ and write: $$\begin{aligned} +\hat{H}_0\ensuremath{\left|n\right\rangle} = E_n \ensuremath{\left|n\right\rangle}. +\end{aligned}$$ + +If the system is initially prepared in the state +$\ensuremath{\left|i\right\rangle}$, so that $$\begin{aligned} +\ensuremath{\left|\psi(t=0)\right\rangle} = \ensuremath{\left|i\right\rangle}, +\end{aligned}$$ what is the probability $$\begin{aligned} +P_m(t) = \left|\ensuremath{\left\langle m|\psi(t)\right\rangle}\right|^2 +\end{aligned}$$ to find the system in the state +$\ensuremath{\left|m\right\rangle}$ at the time $t$? + +# Evolution Equation + +The system $\ensuremath{\left|\psi(t)\right\rangle}$ can be expressed as +follows: $$\begin{aligned} +\ensuremath{\left|\psi(t)\right\rangle} = \sum_n \gamma_n(t) \mathrm{e}^{-i{E_n}t/{\hbar}} \ensuremath{\left|n\right\rangle}, +\end{aligned}$$ where the exponential is the time evolution for +$\hat{H}_1 =~0$. We plug this equation in the Schrödinger equation and +get: $$\begin{aligned} +i\hbar \sum_n\left(\dot{\gamma}_n(t)-i\frac{E_n}{\hbar}\gamma_n(t)\right)\mathrm{e}^{-i{E_n}t/{\hbar}}\ensuremath{\left|n\right\rangle} = \sum_n \gamma_n(t) \mathrm{e}^{-i{E_n}t/{\hbar}}\left(\hat{H}_0 + \hat{V}\right) \ensuremath{\left|n\right\rangle}\label{eq:timeev}\\ +\Longleftrightarrow i\hbar\sum_n \dot{\gamma}_n(t) \mathrm{e}^{-i{E_n}t/{\hbar}} \ensuremath{\left|n\right\rangle} + = \sum_n \gamma_n(t) \mathrm{e}^{-i{E_n}t/{\hbar}} \hat{V} \ensuremath{\left|n\right\rangle} +\end{aligned}$$ If we multiply +[\[eq:timeev\]](#eq:timeev){reference-type="eqref" +reference="eq:timeev"} with $\ensuremath{\left\langle k\right|}$ we +obtain a set of coupled differential equations $$\begin{aligned} +i\hbar \dot{\gamma}_k \mathrm{e}^{-i{E_k}t/{\hbar}} &= \sum_n \gamma_n \mathrm{e}^{-{E_n}t/{\hbar}}\ensuremath{\left\langle k\right|}\hat{V}\ensuremath{\left|n\right\rangle},\\ +i\hbar \dot{\gamma}_k &= \sum_n \gamma_n \mathrm{e}^{-i {(E_n-E_k)}t/{\hbar}} \ensuremath{\left\langle k\right|} \hat{V}\ensuremath{\left|n\right\rangle} +\end{aligned}$$ with initial conditions +$\ensuremath{\left|\psi(t=0)\right\rangle}$. They determine the full +time evolution. + +The solution of this set of equations depends on the details of the +system. However, there are a few important points: + +- For short enough times, the dynamics are driving by the coupling + strength + $\ensuremath{\left\langle k\right|}\hat{V} \ensuremath{\left|n\right\rangle}$. + +- The right-hand sight will oscillate on time scales of $E_n-E_k$ and + typically average to zero for long times. + +- If the coupling element is an oscillating field + $\propto e^{i\omega_L t}$, it might put certain times on resonance + and allow us to avoid the averaging effect. It is exactly this + effect, which allows us to isolate specific transitions to a very + high degree [^1] + +We will now see how the two-state system emerges from these +approximations and then set-up the perturbative treatment step-by-step. + +# Rotating wave approximation + +We will now assume that the coupling term in indeed an oscillating field +with frequency $\omega_L$, so it reads: $$\begin{aligned} +\hat{V} = \hat{V}_0 \cos(\omega_Lt) = \frac{\hat{V}_0}{2} \left(e^{i\omega_lt}+e^{-i\omega_lt}\right) +\end{aligned}$$ We will further assume the we would like use it to +isolate the transition $i\rightarrow f$, which is of frequency +$\hbar \omega_0 = E_f - E_i$. The relevant quantity is then the detuning +$\delta = \omega_0 - \omega_L$. If it is much smaller than any other +energy difference $E_n-E_i$, we directly reduce the system to the +following closed system: + +$$\begin{aligned} +i\dot{\gamma}_i &= \gamma_f \mathrm{e}^{-i \delta t} \Omega\\ +i\dot{\gamma}_f &= \gamma_i \mathrm{e}^{i \delta t}\Omega^* +\end{aligned}$$ Here we defined +$\Omega = \ensuremath{\left\langle i\right|} \frac{\hat{V_0}}{2\hbar}\ensuremath{\left|f\right\rangle}$. +And to make it really a time-of the same form as the two-level system +from the last lecture, we perform the transformation +$\gamma_f = \tilde{\gamma}_f e^{i\delta t}$, which reduces the system +too: $$\begin{aligned} +i \dot{\gamma}_i &= \Omega \tilde{\gamma}_f \\ +i\dot{\tilde{\gamma}}_f &= \delta \tilde{\gamma}_f + \Omega^* \gamma_i +\end{aligned}$$ This has exactly the form of the two-level system that +we studied previously. + +## Adiabatic elimination + +We can now proceed to the quite important case of far detuning, where +$\delta \gg \Omega$. In this case, the final state +$\ensuremath{\left|f\right\rangle}$ gets barely populated and the time +evolution can be approximated to to be zero [@lukin]. $$\begin{aligned} +\dot{\tilde{\gamma}}_f = 0 +\end{aligned}$$ We can use this equation to eliminate $\gamma$ from the +time evolution of the ground state. This approximation is known as +*adiabatic elimination*: $$\begin{aligned} +\tilde{\gamma}_f &= \frac{\Omega^*}{\delta}\gamma_i\\ +\Rightarrow i\hbar \dot{\gamma}_i &= \frac{|\Omega|^2}{\delta} \tilde{\gamma}_i +\end{aligned}$$ The last equation described the evolution of the initial +state with an energy $E_i = \frac{|\Omega|^2}{\delta}$. If the Rabi +coupling is created through an oscillating electric field, i.e. a laser, +this is know as the **light shift** or the **optical dipole potential**. +It is this concept that underlies the optical tweezer for which Arthur +Ashkin got the nobel prize in the 2018 [@2018]. + +## Example: Atomic clocks in optical tweezers + +A neat example that ties the previous concepts together is the recent +paper [@readout]. The experimental setup is visualized in Fig. +[1](#870855){reference-type="ref" reference="870855"}. + +![Experimental setup of an atomic array optical clock as taken from +[@readout]. +](figures/Bildschirmfoto-2019-10-23-um-11-27-17/Bildschirmfoto-2019-10-23-um-11-27-17){#870855 +width="0.70\\columnwidth"} + +While nice examples these clocks are still far away from the best clocks +out there, which are based on optical lattice clocks and ions +[@Ludlow_2015]. + +# Perturbative Solution + +The more formal student might wonder at which points all these rather +hefty approximation are actually valid, which is obviously a very +substantial question. So, we will now try to isolate the most important +contributions to the complicated system through perturbation theory. For +that we will assume that we can write: $$\begin{aligned} +\hat{V}(t) =\lambda \hat{H}_1(t) +\end{aligned}$$ , where $\lambda$ is a small parameter. In other words +we assume that the initial system $\hat{H}_0$ is only weakly perturbed. +Having identified the small parameter $\lambda$, we make the +*perturbative ansatz* $$\begin{aligned} + \gamma_n(t) = \gamma_n^{(0)} + \lambda \gamma_n^{(1)} + \lambda^2 \gamma_n^{(2)} + \cdots +\end{aligned}$$ and plug this ansatz in the evolution equations and sort +them by terms of equal power in $\lambda$. + +The $0$th order reads $$\begin{aligned} + i\hbar \dot{\gamma}_k^{(0)} = 0. +\end{aligned}$$ The $0$th order does not have a time evolution since we +prepared it in an eigenstate of $\hat{H}_0$. Any evolution arises due +the coupling, which is at least of order $\lambda$. + +So, for the $1$st order we get $$\begin{aligned} + \label{eq:1storderapprox} +i\hbar \dot{\gamma}_k^{(1)} = \sum_n \gamma_n^{(0)} \mathrm{e}^{-i(E_n-E_k)t/{\hbar}}\ensuremath{\left\langle k\right|}\hat{H}_1\ensuremath{\left|n\right\rangle}. +\end{aligned}$$ + +## First Order Solution (Born Approximation) + +For the initial conditions $\psi(t=0)=\ensuremath{\left|i\right\rangle}$ +we get $$\begin{aligned} +\gamma_k^{(0)}(t) = \delta_{ik}. +\end{aligned}$$ We plug this in the $1$st order approximation +[\[eq:1storderapprox\]](#eq:1storderapprox){reference-type="eqref" +reference="eq:1storderapprox"} and obtain the rate for the system to go +to the final state $\ensuremath{\left|f\right\rangle}$: +$$\begin{aligned} +i \hbar\dot{\gamma}^{(1)} = \mathrm{e}^{i(E_f-E_i)t/{\hbar}} \ensuremath{\left\langle f\right|}\hat{H}_1 \ensuremath{\left|i\right\rangle} +\end{aligned}$$ Integration with $\gamma_f^{(1)}(t=0) = 0$ yields +$$\begin{aligned} +\label{eq:gammaf1} +\gamma_f^{(1)} = \frac{1}{i\hbar}\int\limits_0^t \mathrm{e}^{i(E_f-E_i)t'/{\hbar}} \ensuremath{\left\langle f\right|} \hat{H}_1(t')\ensuremath{\left|i\right\rangle} \mathop{}\!\mathrm{d}t', +\end{aligned}$$ so that we obtain the probability for ending up in the +final state: $$\begin{aligned} +P_{i\to f}(t) = \lambda^2\left| \gamma_f^{(1)}(t)\right|^2. +\end{aligned}$$ Note that $P_{i\to f}(t) \ll 1$ is the condition for +this approximation to be valid! + +**Example 1: Constant Perturbation.** + +![Sketch of a constant perturbation +](figures/Bildschirmfoto-2018-09-28-um-15-46-04/Bildschirmfoto-2018-09-28-um-15-46-04){#723552 +width="0.70\\columnwidth"} + +We apply a constant perturbation in the time interval +$\left[0,T\right]$, as shown in [2](#723552){reference-type="ref" +reference="723552"}. If we use +[\[eq:gammaf1\]](#eq:gammaf1){reference-type="eqref" +reference="eq:gammaf1"} and set $\hbar \omega_0 = E_f-E_i$, we get +$$\begin{aligned} +\gamma_f^{(1)}(t\geq T) = \frac{1}{i \hbar} \ensuremath{\left\langle f\right|}\hat{H}_1\ensuremath{\left|i\right\rangle} \frac{\mathrm{e}^{i\omega_0 T}-1}{i\omega_0}, +\end{aligned}$$ and therefore $$\begin{aligned} +P_{i\to f} = \frac{1}{\hbar^2}\left|\ensuremath{\left\langle f\right|}\hat{V}\ensuremath{\left|i\right\rangle}\right|^2 \underbrace{\frac{\sin^2\left(\omega_0\frac{T}{2}\right)}{\left(\frac{\omega_0}{2}\right)^2}}_{\mathrm{y}(\omega_0,T)}. +\end{aligned}$$ A sketch of $\mathrm{y}(\omega_0,T)$ is shown in +[3](#615128){reference-type="ref" reference="615128"}. + +![A sketch of y +](figures/Bildschirmfoto-2018-09-28-um-15-54-58/Bildschirmfoto-2018-09-28-um-15-54-58){#615128 +width="0.70\\columnwidth"} + +We can push this calculation to the extreme case of +$T\rightarrow \infty$. This results in a delta function, which is peaked +round $\omega_0 = 0$ and we can write: $$\begin{aligned} +P_{i\to f} = T\frac{2\pi}{\hbar^2}\left|\ensuremath{\left\langle f\right|}\hat{V}\ensuremath{\left|i\right\rangle}\right|^2\delta(\omega_0) +\end{aligned}$$ This is the celebrated **Fermi's golden rule**. + +**Example 2: Sinusoidal Perturbation.** For the perturbation +$$\begin{aligned} +\hat{H}_1(t) = \left\{ \begin{array}{ccl} \hat{H}_1\mathrm{e}^{-i\omega t} && \text{for}\; 0 < t < T \\ 0 &&\text{otherwise}\end{array} \right. +\end{aligned}$$ we obtain the probability $$\begin{aligned} +P_{i\to f} (t \geq T) = \frac{1}{\hbar^2} \left|\ensuremath{\left\langle f\right|}\hat{V}\ensuremath{\left|i\right\rangle}\right|^2 \mathrm{y}(\omega_0 - \omega, T). +\end{aligned}$$ + +At $\omega = \left|E_f - E_i\right|/\hbar$ we are on resonance. + +In the [fifth +lecture](https://www.authorea.com/users/143341/articles/326514-lecture-5-the-hydrogen-atom), +we will start to dive into the hydrogen atom. + +[^1]: This is the idea behind atomic and optical clocks, which work + nowadays at $10^{-18}$. From e8ef0f5b1cf5489ac6d05e48b48b1b9fddddd4d2 Mon Sep 17 00:00:00 2001 From: fretchen Date: Sat, 4 Jan 2025 12:37:15 +0100 Subject: [PATCH 07/13] Update lecture4.md --- amo/lecture4.md | 250 ++++++++++++++++++++++++++++++++---------------- 1 file changed, 165 insertions(+), 85 deletions(-) diff --git a/amo/lecture4.md b/amo/lecture4.md index 66f6467..5dbc61f 100644 --- a/amo/lecture4.md +++ b/amo/lecture4.md @@ -14,7 +14,7 @@ engineer a new Hamiltonian. Most importantly, we will discuss the resonant coupling of two levels and the decay of a single level to a continuum. -In the last lecture [@Jendrzejewski], we discussed the properties of two +In the last lecture, we discussed the properties of two coupled levels. However, we did not elaborate at any stage how such a system might emerge in a true atom. Two fundamental questions come to mind: @@ -35,42 +35,66 @@ Hamiltonian. We will discuss a simple model for the atom in the oscillatory field. We can write down the Hamiltonian: -$$\begin{aligned} +$$ \hat{H} = \hat{H}_0 + \hat{V}(t). -\end{aligned}$$ Here, $\hat{H}_0$ belongs to the atom and $V(t)$ +$$ + +Here, $\hat{H}_0$ belongs to the atom and $V(t)$ describes the time-dependent field and its interaction with the atom. We -assume that $\ensuremath{\left|n\right\rangle}$ is an eigenstate of -$\hat{H}_0$ and write: $$\begin{aligned} -\hat{H}_0\ensuremath{\left|n\right\rangle} = E_n \ensuremath{\left|n\right\rangle}. -\end{aligned}$$ +assume that $\left|n\right\rangle$ is an eigenstate of +$\hat{H}_0$ and write: + +$$ +\hat{H}_0\left|n\right\rangle = E_n \left|n\right\rangle. +$$ If the system is initially prepared in the state -$\ensuremath{\left|i\right\rangle}$, so that $$\begin{aligned} -\ensuremath{\left|\psi(t=0)\right\rangle} = \ensuremath{\left|i\right\rangle}, -\end{aligned}$$ what is the probability $$\begin{aligned} -P_m(t) = \left|\ensuremath{\left\langle m|\psi(t)\right\rangle}\right|^2 -\end{aligned}$$ to find the system in the state -$\ensuremath{\left|m\right\rangle}$ at the time $t$? +$\left|i\right\rangle$, so that + +$$ +\left|\psi(t=0)\right\rangle = \left|i\right\rangle, +$$ + +what is the probability + +$$ +P_m(t) = \left|\left\langle m|\psi(t)\right\rangle\right|^2 +$$ + +to find the system in the state +$\left|m\right\rangle$ at the time $t$? # Evolution Equation -The system $\ensuremath{\left|\psi(t)\right\rangle}$ can be expressed as -follows: $$\begin{aligned} -\ensuremath{\left|\psi(t)\right\rangle} = \sum_n \gamma_n(t) \mathrm{e}^{-i{E_n}t/{\hbar}} \ensuremath{\left|n\right\rangle}, -\end{aligned}$$ where the exponential is the time evolution for +The system $\left|\psi(t)\right\rangle$ can be expressed as +follows: + +$$ +\left|\psi(t)\right\rangle = \sum_n \gamma_n(t) \mathrm{e}^{-i{E_n}t/{\hbar}} \left|n\right\rangle, +$$ + +where the exponential is the time evolution for $\hat{H}_1 =~0$. We plug this equation in the Schrödinger equation and -get: $$\begin{aligned} -i\hbar \sum_n\left(\dot{\gamma}_n(t)-i\frac{E_n}{\hbar}\gamma_n(t)\right)\mathrm{e}^{-i{E_n}t/{\hbar}}\ensuremath{\left|n\right\rangle} = \sum_n \gamma_n(t) \mathrm{e}^{-i{E_n}t/{\hbar}}\left(\hat{H}_0 + \hat{V}\right) \ensuremath{\left|n\right\rangle}\label{eq:timeev}\\ -\Longleftrightarrow i\hbar\sum_n \dot{\gamma}_n(t) \mathrm{e}^{-i{E_n}t/{\hbar}} \ensuremath{\left|n\right\rangle} - = \sum_n \gamma_n(t) \mathrm{e}^{-i{E_n}t/{\hbar}} \hat{V} \ensuremath{\left|n\right\rangle} -\end{aligned}$$ If we multiply +get: + +$$ +i\hbar \sum_n\left(\dot{\gamma}_n(t)-i\frac{E_n}{\hbar}\gamma_n(t)\right)\mathrm{e}^{-i{E_n}t/{\hbar}}\left|n\right\rangle = \sum_n \gamma_n(t) \mathrm{e}^{-i{E_n}t/{\hbar}}\left(\hat{H}_0 + \hat{V}\right) \left|n\right\rangle\\ +\Longleftrightarrow i\hbar\sum_n \dot{\gamma}_n(t) \mathrm{e}^{-i{E_n}t/{\hbar}} \left|n\right\rangle + = \sum_n \gamma_n(t) \mathrm{e}^{-i{E_n}t/{\hbar}} \hat{V} \left|n\right\rangle +$$ + +If we multiply [\[eq:timeev\]](#eq:timeev){reference-type="eqref" -reference="eq:timeev"} with $\ensuremath{\left\langle k\right|}$ we -obtain a set of coupled differential equations $$\begin{aligned} -i\hbar \dot{\gamma}_k \mathrm{e}^{-i{E_k}t/{\hbar}} &= \sum_n \gamma_n \mathrm{e}^{-{E_n}t/{\hbar}}\ensuremath{\left\langle k\right|}\hat{V}\ensuremath{\left|n\right\rangle},\\ -i\hbar \dot{\gamma}_k &= \sum_n \gamma_n \mathrm{e}^{-i {(E_n-E_k)}t/{\hbar}} \ensuremath{\left\langle k\right|} \hat{V}\ensuremath{\left|n\right\rangle} -\end{aligned}$$ with initial conditions -$\ensuremath{\left|\psi(t=0)\right\rangle}$. They determine the full +reference="eq:timeev"} with $\left\langle k\right|$ we +obtain a set of coupled differential equations + +$$ +i\hbar \dot{\gamma}_k \mathrm{e}^{-i{E_k}t/{\hbar}} = \sum_n \gamma_n \mathrm{e}^{-{E_n}t/{\hbar}}\left\langle k\right|\hat{V}\left|n\right\rangle,\\ +i\hbar \dot{\gamma}_k = \sum_n \gamma_n \mathrm{e}^{-i {(E_n-E_k)}t/{\hbar}} \left\langle k\right| \hat{V}\left|n\right\rangle +$$ + +with initial conditions +$\left|\psi(t=0)\right\rangle$. They determine the full time evolution. The solution of this set of equations depends on the details of the @@ -78,7 +102,7 @@ system. However, there are a few important points: - For short enough times, the dynamics are driving by the coupling strength - $\ensuremath{\left\langle k\right|}\hat{V} \ensuremath{\left|n\right\rangle}$. + $\left\langle k\right|\hat{V} \left|n\right\rangle$. - The right-hand sight will oscillate on time scales of $E_n-E_k$ and typically average to zero for long times. @@ -95,42 +119,60 @@ approximations and then set-up the perturbative treatment step-by-step. # Rotating wave approximation We will now assume that the coupling term in indeed an oscillating field -with frequency $\omega_L$, so it reads: $$\begin{aligned} +with frequency $\omega_L$, so it reads: + +$$ \hat{V} = \hat{V}_0 \cos(\omega_Lt) = \frac{\hat{V}_0}{2} \left(e^{i\omega_lt}+e^{-i\omega_lt}\right) -\end{aligned}$$ We will further assume the we would like use it to +$$ + +We will further assume the we would like use it to isolate the transition $i\rightarrow f$, which is of frequency $\hbar \omega_0 = E_f - E_i$. The relevant quantity is then the detuning $\delta = \omega_0 - \omega_L$. If it is much smaller than any other energy difference $E_n-E_i$, we directly reduce the system to the following closed system: -$$\begin{aligned} -i\dot{\gamma}_i &= \gamma_f \mathrm{e}^{-i \delta t} \Omega\\ -i\dot{\gamma}_f &= \gamma_i \mathrm{e}^{i \delta t}\Omega^* -\end{aligned}$$ Here we defined -$\Omega = \ensuremath{\left\langle i\right|} \frac{\hat{V_0}}{2\hbar}\ensuremath{\left|f\right\rangle}$. +$$ +i\dot{\gamma}_i = \gamma_f \mathrm{e}^{-i \delta t} \Omega\\ +i\dot{\gamma}_f = \gamma_i \mathrm{e}^{i \delta t}\Omega^* +$$ + +Here we defined +$\Omega = \left\langle i\right| \frac{\hat{V_0}}{2\hbar}\left|f\right\rangle$. And to make it really a time-of the same form as the two-level system from the last lecture, we perform the transformation $\gamma_f = \tilde{\gamma}_f e^{i\delta t}$, which reduces the system -too: $$\begin{aligned} -i \dot{\gamma}_i &= \Omega \tilde{\gamma}_f \\ -i\dot{\tilde{\gamma}}_f &= \delta \tilde{\gamma}_f + \Omega^* \gamma_i -\end{aligned}$$ This has exactly the form of the two-level system that +too: + +$$ +i \dot{\gamma}_i = \Omega \tilde{\gamma}_f \\ +i\dot{\tilde{\gamma}}_f = \delta \tilde{\gamma}_f + \Omega^* \gamma_i +$$ + +This has exactly the form of the two-level system that we studied previously. ## Adiabatic elimination We can now proceed to the quite important case of far detuning, where $\delta \gg \Omega$. In this case, the final state -$\ensuremath{\left|f\right\rangle}$ gets barely populated and the time -evolution can be approximated to to be zero [@lukin]. $$\begin{aligned} +$\left|f\right\rangle$ gets barely populated and the time +evolution can be approximated to to be zero [@lukin]. + +$$ \dot{\tilde{\gamma}}_f = 0 -\end{aligned}$$ We can use this equation to eliminate $\gamma$ from the +$$ + +We can use this equation to eliminate $\gamma$ from the time evolution of the ground state. This approximation is known as -*adiabatic elimination*: $$\begin{aligned} -\tilde{\gamma}_f &= \frac{\Omega^*}{\delta}\gamma_i\\ -\Rightarrow i\hbar \dot{\gamma}_i &= \frac{|\Omega|^2}{\delta} \tilde{\gamma}_i -\end{aligned}$$ The last equation described the evolution of the initial +*adiabatic elimination*: + +$$ +\tilde{\gamma}_f = \frac{\Omega^*}{\delta}\gamma_i\\ +\Rightarrow i\hbar \dot{\gamma}_i = \frac{|\Omega|^2}{\delta} \tilde{\gamma}_i +$$ + +The last equation described the evolution of the initial state with an energy $E_i = \frac{|\Omega|^2}{\delta}$. If the Rabi coupling is created through an oscillating electric field, i.e. a laser, this is know as the **light shift** or the **optical dipole potential**. @@ -158,46 +200,72 @@ The more formal student might wonder at which points all these rather hefty approximation are actually valid, which is obviously a very substantial question. So, we will now try to isolate the most important contributions to the complicated system through perturbation theory. For -that we will assume that we can write: $$\begin{aligned} +that we will assume that we can write: + +$$ \hat{V}(t) =\lambda \hat{H}_1(t) -\end{aligned}$$ , where $\lambda$ is a small parameter. In other words +$$ + +, where $\lambda$ is a small parameter. In other words we assume that the initial system $\hat{H}_0$ is only weakly perturbed. Having identified the small parameter $\lambda$, we make the -*perturbative ansatz* $$\begin{aligned} +*perturbative ansatz* + +$$ \gamma_n(t) = \gamma_n^{(0)} + \lambda \gamma_n^{(1)} + \lambda^2 \gamma_n^{(2)} + \cdots -\end{aligned}$$ and plug this ansatz in the evolution equations and sort +$$ + +and plug this ansatz in the evolution equations and sort them by terms of equal power in $\lambda$. -The $0$th order reads $$\begin{aligned} +The $0$th order reads + +$$ i\hbar \dot{\gamma}_k^{(0)} = 0. -\end{aligned}$$ The $0$th order does not have a time evolution since we +$$ + +The $0$th order does not have a time evolution since we prepared it in an eigenstate of $\hat{H}_0$. Any evolution arises due the coupling, which is at least of order $\lambda$. -So, for the $1$st order we get $$\begin{aligned} - \label{eq:1storderapprox} -i\hbar \dot{\gamma}_k^{(1)} = \sum_n \gamma_n^{(0)} \mathrm{e}^{-i(E_n-E_k)t/{\hbar}}\ensuremath{\left\langle k\right|}\hat{H}_1\ensuremath{\left|n\right\rangle}. -\end{aligned}$$ +So, for the $1$st order we get + +$$ + +i\hbar \dot{\gamma}_k^{(1)} = \sum_n \gamma_n^{(0)} \mathrm{e}^{-i(E_n-E_k)t/{\hbar}}\left\langle k\right|\hat{H}_1\left|n\right\rangle. +$$ ## First Order Solution (Born Approximation) -For the initial conditions $\psi(t=0)=\ensuremath{\left|i\right\rangle}$ -we get $$\begin{aligned} +For the initial conditions $\psi(t=0)=\left|i\right\rangle$ +we get + +$$ \gamma_k^{(0)}(t) = \delta_{ik}. -\end{aligned}$$ We plug this in the $1$st order approximation +$$ + +We plug this in the $1$st order approximation [\[eq:1storderapprox\]](#eq:1storderapprox){reference-type="eqref" reference="eq:1storderapprox"} and obtain the rate for the system to go -to the final state $\ensuremath{\left|f\right\rangle}$: -$$\begin{aligned} -i \hbar\dot{\gamma}^{(1)} = \mathrm{e}^{i(E_f-E_i)t/{\hbar}} \ensuremath{\left\langle f\right|}\hat{H}_1 \ensuremath{\left|i\right\rangle} -\end{aligned}$$ Integration with $\gamma_f^{(1)}(t=0) = 0$ yields -$$\begin{aligned} -\label{eq:gammaf1} -\gamma_f^{(1)} = \frac{1}{i\hbar}\int\limits_0^t \mathrm{e}^{i(E_f-E_i)t'/{\hbar}} \ensuremath{\left\langle f\right|} \hat{H}_1(t')\ensuremath{\left|i\right\rangle} \mathop{}\!\mathrm{d}t', -\end{aligned}$$ so that we obtain the probability for ending up in the -final state: $$\begin{aligned} +to the final state $\left|f\right\rangle$: +$$ +i \hbar\dot{\gamma}^{(1)} = \mathrm{e}^{i(E_f-E_i)t/{\hbar}} \left\langle f\right|\hat{H}_1 \left|i\right\rangle +$$ + +Integration with $\gamma_f^{(1)}(t=0) = 0$ yields +$$ + +\gamma_f^{(1)} = \frac{1}{i\hbar}\int\limits_0^t \mathrm{e}^{i(E_f-E_i)t'/{\hbar}} \left\langle f\right| \hat{H}_1(t')\left|i\right\rangle \mathop{}\!\mathrm{d}t', +$$ + +so that we obtain the probability for ending up in the +final state: + +$$ P_{i\to f}(t) = \lambda^2\left| \gamma_f^{(1)}(t)\right|^2. -\end{aligned}$$ Note that $P_{i\to f}(t) \ll 1$ is the condition for +$$ + +Note that $P_{i\to f}(t) \ll 1$ is the condition for this approximation to be valid! **Example 1: Constant Perturbation.** @@ -211,11 +279,17 @@ $\left[0,T\right]$, as shown in [2](#723552){reference-type="ref" reference="723552"}. If we use [\[eq:gammaf1\]](#eq:gammaf1){reference-type="eqref" reference="eq:gammaf1"} and set $\hbar \omega_0 = E_f-E_i$, we get -$$\begin{aligned} -\gamma_f^{(1)}(t\geq T) = \frac{1}{i \hbar} \ensuremath{\left\langle f\right|}\hat{H}_1\ensuremath{\left|i\right\rangle} \frac{\mathrm{e}^{i\omega_0 T}-1}{i\omega_0}, -\end{aligned}$$ and therefore $$\begin{aligned} -P_{i\to f} = \frac{1}{\hbar^2}\left|\ensuremath{\left\langle f\right|}\hat{V}\ensuremath{\left|i\right\rangle}\right|^2 \underbrace{\frac{\sin^2\left(\omega_0\frac{T}{2}\right)}{\left(\frac{\omega_0}{2}\right)^2}}_{\mathrm{y}(\omega_0,T)}. -\end{aligned}$$ A sketch of $\mathrm{y}(\omega_0,T)$ is shown in +$$ +\gamma_f^{(1)}(t\geq T) = \frac{1}{i \hbar} \left\langle f\right|\hat{H}_1\left|i\right\rangle \frac{\mathrm{e}^{i\omega_0 T}-1}{i\omega_0}, +$$ + +and therefore + +$$ +P_{i\to f} = \frac{1}{\hbar^2}\left|\left\langle f\right|\hat{V}\left|i\right\rangle\right|^2 \underbrace{\frac{\sin^2\left(\omega_0\frac{T}{2}\right)}{\left(\frac{\omega_0}{2}\right)^2}}_{\mathrm{y}(\omega_0,T)}. +$$ + +A sketch of $\mathrm{y}(\omega_0,T)$ is shown in [3](#615128){reference-type="ref" reference="615128"}. ![A sketch of y @@ -224,22 +298,28 @@ width="0.70\\columnwidth"} We can push this calculation to the extreme case of $T\rightarrow \infty$. This results in a delta function, which is peaked -round $\omega_0 = 0$ and we can write: $$\begin{aligned} -P_{i\to f} = T\frac{2\pi}{\hbar^2}\left|\ensuremath{\left\langle f\right|}\hat{V}\ensuremath{\left|i\right\rangle}\right|^2\delta(\omega_0) -\end{aligned}$$ This is the celebrated **Fermi's golden rule**. +round $\omega_0 = 0$ and we can write: + +$$ +P_{i\to f} = T\frac{2\pi}{\hbar^2}\left|\left\langle f\right|\hat{V}\left|i\right\rangle\right|^2\delta(\omega_0) +$$ + +This is the celebrated **Fermi's golden rule**. **Example 2: Sinusoidal Perturbation.** For the perturbation -$$\begin{aligned} +$$ \hat{H}_1(t) = \left\{ \begin{array}{ccl} \hat{H}_1\mathrm{e}^{-i\omega t} && \text{for}\; 0 < t < T \\ 0 &&\text{otherwise}\end{array} \right. -\end{aligned}$$ we obtain the probability $$\begin{aligned} -P_{i\to f} (t \geq T) = \frac{1}{\hbar^2} \left|\ensuremath{\left\langle f\right|}\hat{V}\ensuremath{\left|i\right\rangle}\right|^2 \mathrm{y}(\omega_0 - \omega, T). -\end{aligned}$$ +$$ + +we obtain the probability + +$$ +P_{i\to f} (t \geq T) = \frac{1}{\hbar^2} \left|\left\langle f\right|\hat{V}\left|i\right\rangle\right|^2 \mathrm{y}(\omega_0 - \omega, T). +$$ At $\omega = \left|E_f - E_i\right|/\hbar$ we are on resonance. -In the [fifth -lecture](https://www.authorea.com/users/143341/articles/326514-lecture-5-the-hydrogen-atom), -we will start to dive into the hydrogen atom. +In the fifth lecture, we will start to dive into the hydrogen atom. [^1]: This is the idea behind atomic and optical clocks, which work nowadays at $10^{-18}$. From 8c1b280a0670c2c97228fdfaf281f463c0509c60 Mon Sep 17 00:00:00 2001 From: fretchen Date: Sat, 4 Jan 2025 12:51:07 +0100 Subject: [PATCH 08/13] Adding lecture 4 --- amo/lecture3.mdx | 21 ++++----- amo/lecture4.md | 104 ++++++++++++++++++++---------------------- amo/lecture4_pic1.png | Bin 0 -> 316033 bytes amo/lecture4_pic2.png | Bin 0 -> 8715 bytes amo/lecture4_pic3.png | Bin 0 -> 23964 bytes 5 files changed, 60 insertions(+), 65 deletions(-) create mode 100644 amo/lecture4_pic1.png create mode 100644 amo/lecture4_pic2.png create mode 100644 amo/lecture4_pic3.png diff --git a/amo/lecture3.mdx b/amo/lecture3.mdx index b6f0408..8ba2cdf 100644 --- a/amo/lecture3.mdx +++ b/amo/lecture3.mdx @@ -51,7 +51,7 @@ nature: - Nitrogen-vacancy centers in diamond. -# Hamiltonian, Eigenstates and Matrix Notation +## Hamiltonian, Eigenstates and Matrix Notation To start out, we will consider two eigenstates $\left|0\right\rangle$, $\left|1\right\rangle$ @@ -123,7 +123,7 @@ $$ You will go through this calculation in the excercise of this week. -## Case of no perturbation $\Omega = 0$ +### Case of no perturbation $\Omega = 0$ This is exactly the case of no applied laser fields that we discussed previously. We simply removed the energy offset @@ -138,7 +138,7 @@ $$ We typically call $\Delta$ the energy difference between the levels or the **detuning**. -## Case of no detuning $\Delta = 0$ +### Case of no detuning $\Delta = 0$ Let us suppose that the diagonal elements are exactly zero. And for simplicity we will also keep $\Omega_y =0$ as it simply complicates the @@ -170,7 +170,7 @@ result: This is also a motivation the formulation of the 'bare' system for $\Omega = 0$ and the 'dressed' states for the coupled system. -## General case +### General case Quite importantly we can solve the system completely even in the general case. By diagonalizing the Hamiltonian we obtain: @@ -200,7 +200,7 @@ $$ \tan(\theta) = \frac{|\Omega|}{\Delta} $$ -# The Bloch sphere +## The Bloch sphere While we could just discuss the details of the above state in the abstract, it is extremely helpful to visualize the problem on the Bloch @@ -214,9 +214,9 @@ sphere as drawn below We will see especially its usefulness especially as we discuss the dynamics of the two-state system. -# Dynamical Aspects +## Dynamical Aspects -## Time Evolution of $\left|\psi(t)\right\rangle$ +### Time Evolution of $\left|\psi(t)\right\rangle$ After the static case we now want to investigate the dynamical properties of the two-state system. We calculate the time evolution of @@ -281,7 +281,7 @@ $$ -## Visualization of the dynamics in the spin picture +### Visualization of the dynamics in the spin picture While the previous derivation might be the standard one, which certainly leads to the right results it might not be the most intuitive way of @@ -326,7 +326,7 @@ around the $x$ axis with velocity $\Omega$. We can now use this general picture to understand the dynamics as rotations around an axis, which is defined by the different components of the magnetic field. -# A few words on the quantum information notation +## A few words on the quantum information notation The qubit is THE basic ingredient of quantum computers. A nice way to play around with them is actually the [IBM Quantum @@ -362,5 +362,4 @@ In the forth lecture we will see how it is that a time-dependent field can actua [^1]: See the discussions of the next lecture - -[^CT1]: Quantum Mechanics, Volume 1. Cohen-Tannoudji, Diu, Laloe. Wiley-VCH, 2006. \ No newline at end of file +[^CT1]: Quantum Mechanics, Volume 1. Cohen-Tannoudji, Diu, Laloe. Wiley-VCH, 2006. diff --git a/amo/lecture4.md b/amo/lecture4.md index 5dbc61f..1f3d7d1 100644 --- a/amo/lecture4.md +++ b/amo/lecture4.md @@ -1,9 +1,9 @@ --- author: -- Fred Jendrzejewski -- Selim Jochim + - Fred Jendrzejewski + - Selim Jochim bibliography: -- bibliography/converted_to_latex.bib + - bibliography/converted_to_latex.bib date: January 04, 2025 nocite: "[@*]" title: Lecture 4 - Atoms in oscillating fields @@ -28,7 +28,7 @@ mind: The solution is to resonantly couple two of the atom's levels by applying an external, oscillatory field, which is very nicely discussed -in chapter 12 of Ref. [@2002] [@Cohen_Tannoudji_1998]. We will discuss +in chapter 12 of Ref. [^2002] [^Cohen_Tannoudji_1998]. We will discuss important and fundamental properties of systems with a time-dependent Hamiltonian. @@ -64,7 +64,7 @@ $$ to find the system in the state $\left|m\right\rangle$ at the time $t$? -# Evolution Equation +## Evolution Equation The system $\left|\psi(t)\right\rangle$ can be expressed as follows: @@ -83,9 +83,7 @@ i\hbar \sum_n\left(\dot{\gamma}_n(t)-i\frac{E_n}{\hbar}\gamma_n(t)\right)\mathrm = \sum_n \gamma_n(t) \mathrm{e}^{-i{E_n}t/{\hbar}} \hat{V} \left|n\right\rangle $$ -If we multiply -[\[eq:timeev\]](#eq:timeev){reference-type="eqref" -reference="eq:timeev"} with $\left\langle k\right|$ we +If we multiply the equation with $\left\langle k\right|$ we obtain a set of coupled differential equations $$ @@ -100,29 +98,29 @@ time evolution. The solution of this set of equations depends on the details of the system. However, there are a few important points: -- For short enough times, the dynamics are driving by the coupling - strength - $\left\langle k\right|\hat{V} \left|n\right\rangle$. +- For short enough times, the dynamics are driving by the coupling + strength + $\left\langle k\right|\hat{V} \left|n\right\rangle$. -- The right-hand sight will oscillate on time scales of $E_n-E_k$ and - typically average to zero for long times. +- The right-hand sight will oscillate on time scales of $E_n-E_k$ and + typically average to zero for long times. -- If the coupling element is an oscillating field - $\propto e^{i\omega_L t}$, it might put certain times on resonance - and allow us to avoid the averaging effect. It is exactly this - effect, which allows us to isolate specific transitions to a very - high degree [^1] +- If the coupling element is an oscillating field + $\propto e^{i\omega_L t}$, it might put certain times on resonance + and allow us to avoid the averaging effect. It is exactly this + effect, which allows us to isolate specific transitions to a very + high degree [^1] We will now see how the two-state system emerges from these approximations and then set-up the perturbative treatment step-by-step. -# Rotating wave approximation +## Rotating wave approximation We will now assume that the coupling term in indeed an oscillating field with frequency $\omega_L$, so it reads: $$ -\hat{V} = \hat{V}_0 \cos(\omega_Lt) = \frac{\hat{V}_0}{2} \left(e^{i\omega_lt}+e^{-i\omega_lt}\right) +\hat{V} = \hat{V}_0 \cos(\omega_Lt) = \frac{\hat{V}_0}{2} \left(e^{i\omega_lt}+e^{-i\omega_lt}\right) $$ We will further assume the we would like use it to @@ -146,13 +144,13 @@ too: $$ i \dot{\gamma}_i = \Omega \tilde{\gamma}_f \\ -i\dot{\tilde{\gamma}}_f = \delta \tilde{\gamma}_f + \Omega^* \gamma_i +i\dot{\tilde{\gamma}}_f = \delta \tilde{\gamma}_f + \Omega^* \gamma_i $$ This has exactly the form of the two-level system that we studied previously. -## Adiabatic elimination +### Adiabatic elimination We can now proceed to the quite important case of far detuning, where $\delta \gg \Omega$. In this case, the final state @@ -165,7 +163,7 @@ $$ We can use this equation to eliminate $\gamma$ from the time evolution of the ground state. This approximation is known as -*adiabatic elimination*: +_adiabatic elimination_: $$ \tilde{\gamma}_f = \frac{\Omega^*}{\delta}\gamma_i\\ @@ -177,24 +175,19 @@ state with an energy $E_i = \frac{|\Omega|^2}{\delta}$. If the Rabi coupling is created through an oscillating electric field, i.e. a laser, this is know as the **light shift** or the **optical dipole potential**. It is this concept that underlies the optical tweezer for which Arthur -Ashkin got the nobel prize in the 2018 [@2018]. +Ashkin got the [nobel prize in the 2018](https://www.nobelprize.org/uploads/2018/10/advanced-physicsprize2018.pdf). -## Example: Atomic clocks in optical tweezers +### Example: Atomic clocks in optical tweezers -A neat example that ties the previous concepts together is the recent -paper [@readout]. The experimental setup is visualized in Fig. -[1](#870855){reference-type="ref" reference="870855"}. +A neat example that ties the previous concepts together is [the recent +paper](https://arxiv.org/abs/1908.05619v2). The experimental setup is visualized in the figure below. -![Experimental setup of an atomic array optical clock as taken from -[@readout]. -](figures/Bildschirmfoto-2019-10-23-um-11-27-17/Bildschirmfoto-2019-10-23-um-11-27-17){#870855 -width="0.70\\columnwidth"} + While nice examples these clocks are still far away from the best clocks -out there, which are based on optical lattice clocks and ions -[@Ludlow_2015]. +out there, which are based on [optical lattice clocks and ions](http://dx.doi.org/10.1103/revmodphys.87.637). -# Perturbative Solution +## Perturbative Solution The more formal student might wonder at which points all these rather hefty approximation are actually valid, which is obviously a very @@ -209,7 +202,7 @@ $$ , where $\lambda$ is a small parameter. In other words we assume that the initial system $\hat{H}_0$ is only weakly perturbed. Having identified the small parameter $\lambda$, we make the -*perturbative ansatz* +_perturbative ansatz_ $$ \gamma_n(t) = \gamma_n^{(0)} + \lambda \gamma_n^{(1)} + \lambda^2 \gamma_n^{(2)} + \cdots @@ -231,11 +224,11 @@ the coupling, which is at least of order $\lambda$. So, for the $1$st order we get $$ - + i\hbar \dot{\gamma}_k^{(1)} = \sum_n \gamma_n^{(0)} \mathrm{e}^{-i(E_n-E_k)t/{\hbar}}\left\langle k\right|\hat{H}_1\left|n\right\rangle. $$ -## First Order Solution (Born Approximation) +### First Order Solution (Born Approximation) For the initial conditions $\psi(t=0)=\left|i\right\rangle$ we get @@ -244,15 +237,15 @@ $$ \gamma_k^{(0)}(t) = \delta_{ik}. $$ -We plug this in the $1$st order approximation -[\[eq:1storderapprox\]](#eq:1storderapprox){reference-type="eqref" -reference="eq:1storderapprox"} and obtain the rate for the system to go +We plug this in the $1$st order approximation and obtain the rate for the system to go to the final state $\left|f\right\rangle$: + $$ i \hbar\dot{\gamma}^{(1)} = \mathrm{e}^{i(E_f-E_i)t/{\hbar}} \left\langle f\right|\hat{H}_1 \left|i\right\rangle $$ Integration with $\gamma_f^{(1)}(t=0) = 0$ yields + $$ \gamma_f^{(1)} = \frac{1}{i\hbar}\int\limits_0^t \mathrm{e}^{i(E_f-E_i)t'/{\hbar}} \left\langle f\right| \hat{H}_1(t')\left|i\right\rangle \mathop{}\!\mathrm{d}t', @@ -270,15 +263,13 @@ this approximation to be valid! **Example 1: Constant Perturbation.** -![Sketch of a constant perturbation -](figures/Bildschirmfoto-2018-09-28-um-15-46-04/Bildschirmfoto-2018-09-28-um-15-46-04){#723552 -width="0.70\\columnwidth"} + + +Sketch of a constant perturbation. We apply a constant perturbation in the time interval -$\left[0,T\right]$, as shown in [2](#723552){reference-type="ref" -reference="723552"}. If we use -[\[eq:gammaf1\]](#eq:gammaf1){reference-type="eqref" -reference="eq:gammaf1"} and set $\hbar \omega_0 = E_f-E_i$, we get +$\left[0,T\right]$, as shown in above. If we use the expression for $\gamma_f^{(1)}$ and set $\hbar \omega_0 = E_f-E_i$, we get + $$ \gamma_f^{(1)}(t\geq T) = \frac{1}{i \hbar} \left\langle f\right|\hat{H}_1\left|i\right\rangle \frac{\mathrm{e}^{i\omega_0 T}-1}{i\omega_0}, $$ @@ -289,12 +280,11 @@ $$ P_{i\to f} = \frac{1}{\hbar^2}\left|\left\langle f\right|\hat{V}\left|i\right\rangle\right|^2 \underbrace{\frac{\sin^2\left(\omega_0\frac{T}{2}\right)}{\left(\frac{\omega_0}{2}\right)^2}}_{\mathrm{y}(\omega_0,T)}. $$ -A sketch of $\mathrm{y}(\omega_0,T)$ is shown in -[3](#615128){reference-type="ref" reference="615128"}. +A sketch of $\mathrm{y}(\omega_0,T)$ is shown below -![A sketch of y -](figures/Bildschirmfoto-2018-09-28-um-15-54-58/Bildschirmfoto-2018-09-28-um-15-54-58){#615128 -width="0.70\\columnwidth"} + + +A sketch of y We can push this calculation to the extreme case of $T\rightarrow \infty$. This results in a delta function, which is peaked @@ -307,6 +297,7 @@ $$ This is the celebrated **Fermi's golden rule**. **Example 2: Sinusoidal Perturbation.** For the perturbation + $$ \hat{H}_1(t) = \left\{ \begin{array}{ccl} \hat{H}_1\mathrm{e}^{-i\omega t} && \text{for}\; 0 < t < T \\ 0 &&\text{otherwise}\end{array} \right. $$ @@ -321,5 +312,10 @@ At $\omega = \left|E_f - E_i\right|/\hbar$ we are on resonance. In the fifth lecture, we will start to dive into the hydrogen atom. -[^1]: This is the idea behind atomic and optical clocks, which work +[^1]: + This is the idea behind atomic and optical clocks, which work nowadays at $10^{-18}$. + +[^2002]: Jean Dalibard Jean-Louis Basdevant. Quantum Mechanics. Springer-Verlag, 2002. + +[^Cohen_Tannoudji_1998]: Claude Cohen-Tannoudji, Jacques Dupont-Roc, Gilbert Grynberg. Atom-Photon Interactions. 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Fred Jendrzejewski +- Selim Jochim +bibliography: +- bibliography/converted_to_latex.bib +date: January 04, 2025 +nocite: "[@*]" +title: Lecture 5 - The Hydrogen Atom +--- + +In this lecture we will first discuss the diagonalization of the +harmonic oscillator and then discuss the main properties of the hydrogen +atom. + +In the previous lectures we have seen how to treat eigenstates of the +two-level system and then how we can derive its effective emergence from +some complex level structure if we [apply oscillating +fields](https://www.authorea.com/users/143341/articles/326506-lecture-4-atoms-in-oscillating-fields). + +Today, we will increase the complexity towards the harmonic oscillator +and the hydrogen atom. + +# The harmonic oscillator + +The harmonic oscillator is another great toy model to understand certain +properties of quantum mechanical systems. Most importantly, it is a +great introduction into the properties of bound systems and ladder +operators. The basic Hamiltonian comes along in a rather innocent +fashion, namely: + +$$\begin{aligned} +\label{Eq:HamHO} +\hat{H} &= \frac{\hat{p}^2}{2m}+ \frac{m\omega^2}{2}\hat{x}^2 +\end{aligned}$$ The two variables $\hat{p}$ and $\hat{x}$ are +non-commuting $[\hat{x}, \hat{p}] = i\hbar$, so they cannot be measured +at the same time. We would now like to put the operator into a diagonal +form such that it reads something like: $$\begin{aligned} +\label{Eq:HamHO} +\hat{H} &= \sum_n \epsilon_n \ensuremath{\left|n\right\rangle}\ensuremath{\left\langle n\right|} +\end{aligned}$$ + +We will follow he quite closely the discussion of Ref. [@interactions]. + +## The ladder operators + +We would like to get the spectrum first. So make the equation look a bit +nicer we will define $\hat{p} = \hat{P} \sqrt{m\omega}$ and +$\hat{x} = \frac{\hat{X}}{\sqrt{m\omega}}$ such that we have: +$$\begin{aligned} +\label{Eq:HamHO} +\hat{H} &= \frac{\omega}{2}\left(\hat{P}^2 + \hat{X}^2\right) +\end{aligned}$$ [^1] The next step is then to define the ladder +operators: $$\begin{aligned} +\hat{a} = \frac{1}{\sqrt{2\hbar}}\left(\hat{X}+i\hat{P}\right)\\ +\hat{a}^\dag = \frac{1}{\sqrt{2\hbar}}\left(\hat{X}-i\hat{P}\right)\\ +\end{aligned}$$ At this stage we can just try to rewrite the Hamiltonian +in terms of the operators, such that: $$\begin{aligned} +\hat{a}^\dag \hat{a} &= \frac{1}{2\hbar}(\hat{X}-i\hat{P})(\hat{X}+i\hat{P})\\ +&= \frac{1}{2\hbar}(\hat{X}^2 +\hat{P}^2 -\hbar)\\ + \frac{1}{2}(X^2 +\hat{P}^2 ) &= \hbar \left(\hat{a}^\dag \hat{a}-\frac{1}{2}\right) +\end{aligned}$$ So the Hamiltonian can now be written as: +$$\begin{aligned} +\hat{H} &= \hbar \omega \left(\hat{N} + \frac{1}{2}\right)\text{ with } \hat{N} = a^\dag a +\end{aligned}$$ At this stage we have diagonalized the Hamiltonian, what +remains to be understood is the the values that $\hat{a}^\dag a$ can +take. + +## Action of the ladder operators in the Fock basis + +We would like to understand the basis, which is defined by: +$$\begin{aligned} +\hat{N} \ensuremath{\left|n\right\rangle} = n \ensuremath{\left|n\right\rangle} +\end{aligned}$$ The non-commutation between $\hat{X}$ and $\hat{P}$ is +translated to the ladder operators as: $$\begin{aligned} +&= \frac{1}{2\hbar}[\hat{X}+iP,\hat{X}-i\hat{P}] = 1\\ +~[\hat{N}, a] &= -\hat{a}\\ +~[\hat{N}, a^\dag] &= a^\dag +\end{aligned}$$ From these relationship we can show then that: +$$\begin{aligned} +\hat{a}\ensuremath{\left|n\right\rangle} = \sqrt{n}\ensuremath{\left|n-1\right\rangle}\\ +\hat{a}^\dag \ensuremath{\left|n\right\rangle} = \sqrt{n+1}\ensuremath{\left|n+1\right\rangle}\\ +\end{aligned}$$ These relations are the motivation for the name ladder +operators as they connect the different eigenstates. And they are +raising/lowering the quantum number by one. Finally we have to find the +lower limit. And this is quite naturally 0 as +$n = \ensuremath{\left\langle n\right|}\hat{N}\ensuremath{\left|n\right\rangle} = \ensuremath{\left\langle\psi_1\right|}\ensuremath{\left|\psi_1\right\rangle}\geq 0$. +So we can construct the full basis by just defining the action of the +lowering operator on the zero element +$a\ensuremath{\left|0\right\rangle} = 0$ and the other operators are +then constructed as: $$\begin{aligned} +\ensuremath{\left|n\right\rangle} = \frac{(a^\dag)^n}{\sqrt{n!}}\ensuremath{\left|0\right\rangle} +\end{aligned}$$ + +## Spatial representation of the eigenstates + +While we now have the spectrum it would be really nice to obtain the +spatial properties of the different states. For that we have to project +them onto the x basis. Let us start out with the ground state for which +we have $\hat{a}\ensuremath{\left|0\right\rangle}= 0$: $$\begin{aligned} +\ensuremath{\left\langle x\right|}\frac{1}{\sqrt{2\hbar}}\left(\sqrt{m\omega}\hat{x} +i \frac{1}{\sqrt{m\omega}}\hat{p}\right)\ensuremath{\left|0\right\rangle}= 0\\ +\left(\sqrt{\frac{m\omega}{\hbar}}x + \sqrt{\frac{\hbar}{m\omega}}\partial_x\right)\psi_0(x)= 0\\ +\Rightarrow \psi_0(x) \propto e^{-\frac{x^2}{2a_{HO}^2}} +\end{aligned}$$ This also introduces the typical distance in the quantum +harmonic oscillator which is given by $a_{HO} =\sqrt{\hbar/m\omega}$. +The other states are solutions to the defining equations: +$$\begin{aligned} +\psi_n(x) = \frac{1}{\sqrt{n!}2^n}\left(\sqrt{m\omega}x - \frac{1}{\sqrt{m\omega}}\frac{d}{dx}\right)^n \psi_0(x)\\ +\psi_n(x) = \frac{1}{\sqrt{n!}2^n}H_n(x) \psi_0(x)\\ +\end{aligned}$$ where $H_n(x)$ are the Hermite polynoms. + +# The hamiltonian of the hydrogen atom + +The hydrogen atom plays at central role in atomic physics as it is *the* +basic ingredient of atomic structures. It describes a single *electron*, +which is bound to the nucleus of a single *proton*. As such it is the +simplest of all atoms and can be described analytically within high +precision. This has motivated an enormous body of literature on the +problem, which derives all imaginable properties in nauseating detail. +Therefore, we will focus here on the main properties and only sketch the +derivations, while we will reference to the more technical details. + +For the hydrogen atom as shown in [1](#261310){reference-type="ref" +reference="261310"}, we can write down the Hamiltonian $$\begin{aligned} +\hat{H}=\frac{{{\hat{\vec{p}}}^2_\text{p}}}{2m_\text{p}} + \frac{{\hat{\vec{p}}}^2_\text{e}}{2m_\text{e}} - \frac{Ze^2}{4\pi\epsilon_0 r}, +\end{aligned}$$ where $Ze$ is the nuclear charge. To solve the problem, +we have to find the right Hilbert space. We can not solve the problem of +the electron alone. If we do a separation of coordinates, i.e., we +separate the Hamiltonian into the the center of mass and the relative +motion, we get $$\begin{aligned} +\hat{H} = \underbrace{\frac{{\hat{\vec{p}}}^2_{\textrm{cm}}}{2M}}_{\hat{H}_{\textrm{cm}}} + \underbrace{\frac{{\hat{\vec{p}}}^2_\text{r}}{2\mu}- \frac{Ze^2}{4\pi\epsilon_0r}}_{\hat{H}_{\text{atom}}} \label{eq:hydrogencmatomsplit} +\end{aligned}$$ with the reduced mass $1/\mu=1/m_\text{e}+1/m_\text{p}$. +If the state of the hydrogen atom $\ensuremath{\left|\psi\right\rangle}$ +is an eigenstate of $\hat{H}$, we can write $$\begin{aligned} +\hat{H}\ensuremath{\left|\psi\right\rangle}=&\left( \hat{H}_\textrm{cm}+\hat{H}_{\text{atom}} \right)\ensuremath{\left|\psi_\textrm{cm}\right\rangle}\otimes \ensuremath{\left|\psi_\text{atom}\right\rangle} \label{eq:hydrogencmatom}\\ +=& \left( E_{\text{kin}} + E_\text{atom} \right) \ensuremath{\left|\psi\right\rangle}. +\end{aligned}$$ Both states in +[\[eq:hydrogencmatom\]](#eq:hydrogencmatom){reference-type="eqref" +reference="eq:hydrogencmatom"} are eigenstates of the system. The state +$\ensuremath{\left|\psi\right\rangle}$ can be split up as shown since +the two degrees of freedom are generally not entangled. + +![Sketch of the hydrogen atom with the relative coordinate and the +coordinates of the proton and the electron. +](figures/Bildschirmfoto-2018-09-28-um-16-07-07/Bildschirmfoto-2018-09-28-um-16-07-07){#261310 +width="0.70\\columnwidth"} + +The wave function of the system then reads: $$\begin{aligned} +\psi(\vec{R},\vec{r}) =& \left( \ensuremath{\left\langle R\right|} \otimes \ensuremath{\left\langle r\right|}\right)\left( \ensuremath{\left|\psi_\textrm{cm}\right\rangle} \otimes \ensuremath{\left|\psi_{\text{atom}}\right\rangle}\right)\\ +=& \psi(\vec{R}) \cdot \psi (\vec{r}) +\end{aligned}$$ Our goal is now to find the eigenfunctions and +eigenenergies of $\hat{H}_\text{atom}$. In order to further divide the +Hilbert space, we can use the symmetries. + +# Conservation of orbital angular momentum + +$\hat{H}_\text{atom}$ possesses spherical symmetry, which implies that +**orbital angular momentum** $\hat{\vec{L}}$ is conserved. It is defined +as: $$\begin{aligned} +\hat{\vec{L}}=\hat{\vec{r}} \times \hat{\vec{p}} +\end{aligned}$$ In other words, we have: $$\begin{aligned} += 0 +\end{aligned}$$ Let us show first that the kinetic term commutes with +the angular momentum operator, We will employ the commutator +relationships for position and momentum $[x_i, p_j]=i\hbar$ and the +relationship $[A,BC] = [A,B]C+B[A,C]$ and +$[f(x), p_x] = [x,p_x]\frac{\partial f(x)}{\partial x}$. So we obtain: +$$\begin{aligned} +&= [p_x^2,xp_y]-[p_y^2,yp_x] \\ + &= [p_x^2,x]p_y-[p_y^2,y] p_x\\ + &=i\hbar 2 p_xp_y-2i\hbar p_y p_x\\ + &= 0 +\end{aligned}$$ Analog calculations show that $L_y$ and $L_z$ commute. +In a similiar fashion we can verify that the potential term commutes +with the different components of $\hat{\vec{L}}$ $$\begin{aligned} +&= [\frac{1}{r}, xp_y]-[\frac{1}{r}, yp_x]\\ +&= x[\frac{1}{r}, p_y]-y[\frac{1}{r}, p_x]\\ +&= -x \frac{yi\hbar}{2r^{3/2}}+y\frac{xi\hbar}{2r^{3/2}}\\ +&=0 +\end{aligned}$$ We can therefore decompose the eigenfunctions of the +hydrogen atom over the eigenbasis of the angular momentum operator. A +detailled discussion of the properties of $\vec{L}$ can be found in +Appendix B of [@Hertel_2015]. To find the eigenbasis, we first need to +identify the commutation relationships between the components of +$\hat{\vec{L}}$. We can calculate them following commutation +relationships: $$\begin{aligned} +&= [yp_z - zp_y, zp_x - xp_z]\\ +&=[yp_z, zp_x]-[yp_z,xp_z]- [zp_y, zp_x] + [zp_y,xp_z]\\ +&=[yp_z, zp_x] + [zp_y,xp_z]\\ +&=[yp_z, z]p_x +x[zp_y,p_z]\\ +&=-i\hbar yp_x +i\hbar xp_y\\ +&= i\hbar L_z +\end{aligned}$$ This relationship holds for all the other components too +and we have in general: $$\begin{aligned} += i\hbar \epsilon_{ijk}L_k +\end{aligned}$$ The orbital angular momentum is therefore part of the +large family of angular momentum operators, which also comprises spin +etc. In particular the different components are not independent, and +therefore we cannot form a basis out the three components. A suitable +choice is actually to use the following combinations: $$\begin{aligned} +\hat{\vec{L}}^2\ensuremath{\left|l,m_l\right\rangle} =& \hbar^2 l (l+1)\ensuremath{\left|l,m_l\right\rangle}\\ +\hat{L}_z\ensuremath{\left|l,m_l\right\rangle} =& \hbar m_l \ensuremath{\left|l,m_l\right\rangle} +\end{aligned}$$ + +- $l$ is a non-negative integer and it is called the **orbital angular + momentum quantum number**. + +- $m_l$ takes values $-l, -l+1, ..., l-1, l$ and it is sometimes + called the **projection of the angular momentum**. + +## Eigenfunction of the angular momentum operators + +Having identified the relevant operators it would be nice to obtain a +space representation of them. This works especially nicely in spherical +coordinates. There, we get $$\begin{aligned} +\hat{L}_z&= - i \hbar \partial_{\phi}\\ +\hat{\vec{L}}^2 &= - \hbar^2 \left[\frac{1}{\sin(\theta)}\partial_{\theta} \left( \sin(\theta) \partial_\theta\right) + \frac{1}{\sin^2(\theta)} \partial_{\phi\phi} \right]. +\end{aligned}$$ The corresponding wave functions are $$\begin{aligned} +\ensuremath{\left\langle\theta, \phi | l,m_l\right\rangle} = Y_{lm}(\theta,\phi). +\end{aligned}$$ + +Where $Y_{lm}(\theta, \phi)$ are the **spherical harmonics**. + +# The radial wave equation + +Given that we now know that the angular momentum is conserved for the +hydrogen atom, we can actually rewrite the Hamltonian +[\[eq:hydrogencmatomsplit\]](#eq:hydrogencmatomsplit){reference-type="ref" +reference="eq:hydrogencmatomsplit"} in terms of the angular momentum as +we find: $$\begin{aligned} +\hat{H}_\text{atom} = \hat{H}_r + \frac{\hat{L}}{2\mu r^2}+V(r) \\ +\hat{H}_r = -\frac{\hbar^2}{2\mu}\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) +\end{aligned}$$ We can now separate out the angular part and decompose +it over the eigenfunctions of $\hat{\vec{L}}$, such that we make the +ansatz [^2]: $$\begin{aligned} +\psi (r,\theta,\phi) = R(r) Y_{lm}(\theta,\phi) +\end{aligned}$$ + +We can plug this separated ansatz in the Schrödinger equation. We +already solved the angular in the discussion of the angular momentum and +for the radial part we obtain: $$\begin{aligned} +-\frac{\hbar^2}{2\mu}\frac{1}{r}\frac{d^2(rR(r))}{dr^2} - \frac{Ze^2}{4\pi\epsilon_0 r} R(r) + \frac{\hbar^2}{2\mu}\frac{l(l+1)}{r^2}R(r) = ER(r) +\end{aligned}$$ Substituting $R(r)=u(r)/r$ leads to $$\begin{aligned} +-\frac{\hbar^2}{2\mu}\frac{d^2}{dr^2}u(r) +\underbrace{ \left( -\frac{Ze^2}{4\pi\epsilon_0 r} + \frac{\hbar^2}{2\mu} \frac{l(l+1)}{r^2} \right)}_{V_{\text{eff}}} u(r) = E \, u(r), +\end{aligned}$$ which is known as the "radial wave equation". It is a +very general result for *any* central potential. It can also be used to +describe unbound states ($E>0$) that occur during scattering. + +In the [next +lecture](https://www.authorea.com/users/143341/articles/326674-lecture-6-the-dipole-approximation-in-the-hydrogen-atom) +we will look into the energy scales of the hydrogen atom and then start +coupling different levels. + +[^1]: The commutator between $\hat{X}$ and $\hat{P}$ is still as for $x$ + and $p$. + +[^2]: Only if the system is in a well-defined angular momentum state, we + can write it down like this. From ea42bccf0aa657845fc8f916e62a7dbf6787295e Mon Sep 17 00:00:00 2001 From: fretchen Date: Sat, 4 Jan 2025 13:06:21 +0100 Subject: [PATCH 10/13] Add lecture 5 --- amo/lecture5.md | 352 +++++++++++++++++++++++++++--------------- amo/lecture5_pic1.png | Bin 0 -> 10579 bytes 2 files changed, 225 insertions(+), 127 deletions(-) create mode 100644 amo/lecture5_pic1.png diff --git a/amo/lecture5.md b/amo/lecture5.md index 3c8285b..259a3e8 100644 --- a/amo/lecture5.md +++ b/amo/lecture5.md @@ -1,9 +1,9 @@ --- author: -- Fred Jendrzejewski -- Selim Jochim + - Fred Jendrzejewski + - Selim Jochim bibliography: -- bibliography/converted_to_latex.bib + - bibliography/converted_to_latex.bib date: January 04, 2025 nocite: "[@*]" title: Lecture 5 - The Hydrogen Atom @@ -15,8 +15,8 @@ atom. In the previous lectures we have seen how to treat eigenstates of the two-level system and then how we can derive its effective emergence from -some complex level structure if we [apply oscillating -fields](https://www.authorea.com/users/143341/articles/326506-lecture-4-atoms-in-oscillating-fields). +some complex level structure if we apply oscillating +fields. Today, we will increase the complexity towards the harmonic oscillator and the hydrogen atom. @@ -29,115 +29,166 @@ great introduction into the properties of bound systems and ladder operators. The basic Hamiltonian comes along in a rather innocent fashion, namely: -$$\begin{aligned} -\label{Eq:HamHO} -\hat{H} &= \frac{\hat{p}^2}{2m}+ \frac{m\omega^2}{2}\hat{x}^2 -\end{aligned}$$ The two variables $\hat{p}$ and $\hat{x}$ are +$$ + +\hat{H} = \frac{\hat{p}^2}{2m}+ \frac{m\omega^2}{2}\hat{x}^2 +$$ + +The two variables $\hat{p}$ and $\hat{x}$ are non-commuting $[\hat{x}, \hat{p}] = i\hbar$, so they cannot be measured at the same time. We would now like to put the operator into a diagonal -form such that it reads something like: $$\begin{aligned} -\label{Eq:HamHO} -\hat{H} &= \sum_n \epsilon_n \ensuremath{\left|n\right\rangle}\ensuremath{\left\langle n\right|} -\end{aligned}$$ +form such that it reads something like: + +$$ + +\hat{H} = \sum_n \epsilon_n \left|n\right\rangle\left\langle n\right| +$$ -We will follow he quite closely the discussion of Ref. [@interactions]. +We will follow he quite closely [this discussion](https://ocw.mit.edu/courses/nuclear-engineering/22-51-quantum-theory-of-radiation-interactions-fall-2012/lecture-notes/MIT22_51F12_Ch9.pdf). ## The ladder operators We would like to get the spectrum first. So make the equation look a bit nicer we will define $\hat{p} = \hat{P} \sqrt{m\omega}$ and $\hat{x} = \frac{\hat{X}}{\sqrt{m\omega}}$ such that we have: -$$\begin{aligned} -\label{Eq:HamHO} -\hat{H} &= \frac{\omega}{2}\left(\hat{P}^2 + \hat{X}^2\right) -\end{aligned}$$ [^1] The next step is then to define the ladder -operators: $$\begin{aligned} + +$$ + +\hat{H} = \frac{\omega}{2}\left(\hat{P}^2 + \hat{X}^2\right) +$$ + +[^1] The next step is then to define the ladder +operators: + +$$ \hat{a} = \frac{1}{\sqrt{2\hbar}}\left(\hat{X}+i\hat{P}\right)\\ \hat{a}^\dag = \frac{1}{\sqrt{2\hbar}}\left(\hat{X}-i\hat{P}\right)\\ -\end{aligned}$$ At this stage we can just try to rewrite the Hamiltonian -in terms of the operators, such that: $$\begin{aligned} -\hat{a}^\dag \hat{a} &= \frac{1}{2\hbar}(\hat{X}-i\hat{P})(\hat{X}+i\hat{P})\\ -&= \frac{1}{2\hbar}(\hat{X}^2 +\hat{P}^2 -\hbar)\\ - \frac{1}{2}(X^2 +\hat{P}^2 ) &= \hbar \left(\hat{a}^\dag \hat{a}-\frac{1}{2}\right) -\end{aligned}$$ So the Hamiltonian can now be written as: -$$\begin{aligned} -\hat{H} &= \hbar \omega \left(\hat{N} + \frac{1}{2}\right)\text{ with } \hat{N} = a^\dag a -\end{aligned}$$ At this stage we have diagonalized the Hamiltonian, what +$$ + +At this stage we can just try to rewrite the Hamiltonian +in terms of the operators, such that: + +$$ +\hat{a}^\dag \hat{a} = \frac{1}{2\hbar}(\hat{X}-i\hat{P})(\hat{X}+i\hat{P})\\ += \frac{1}{2\hbar}(\hat{X}^2 +\hat{P}^2 -\hbar)\\ + \frac{1}{2}(X^2 +\hat{P}^2 ) = \hbar \left(\hat{a}^\dag \hat{a}-\frac{1}{2}\right) +$$ + +So the Hamiltonian can now be written as: + +$$ +\hat{H} = \hbar \omega \left(\hat{N} + \frac{1}{2}\right)\text{ with } \hat{N} = a^\dag a +$$ + +At this stage we have diagonalized the Hamiltonian, what remains to be understood is the the values that $\hat{a}^\dag a$ can take. ## Action of the ladder operators in the Fock basis We would like to understand the basis, which is defined by: -$$\begin{aligned} -\hat{N} \ensuremath{\left|n\right\rangle} = n \ensuremath{\left|n\right\rangle} -\end{aligned}$$ The non-commutation between $\hat{X}$ and $\hat{P}$ is -translated to the ladder operators as: $$\begin{aligned} -&= \frac{1}{2\hbar}[\hat{X}+iP,\hat{X}-i\hat{P}] = 1\\ -~[\hat{N}, a] &= -\hat{a}\\ -~[\hat{N}, a^\dag] &= a^\dag -\end{aligned}$$ From these relationship we can show then that: -$$\begin{aligned} -\hat{a}\ensuremath{\left|n\right\rangle} = \sqrt{n}\ensuremath{\left|n-1\right\rangle}\\ -\hat{a}^\dag \ensuremath{\left|n\right\rangle} = \sqrt{n+1}\ensuremath{\left|n+1\right\rangle}\\ -\end{aligned}$$ These relations are the motivation for the name ladder + +$$ +\hat{N} \left|n\right\rangle = n \left|n\right\rangle +$$ + +The non-commutation between $\hat{X}$ and $\hat{P}$ is +translated to the ladder operators as: + +$$ += \frac{1}{2\hbar}[\hat{X}+iP,\hat{X}-i\hat{P}] = 1\\ +~[\hat{N}, a] = -\hat{a}\\ +~[\hat{N}, a^\dag] = a^\dag +$$ + +From these relationship we can show then that: + +$$ +\hat{a}\left|n\right\rangle = \sqrt{n}\left|n-1\right\rangle\\ +\hat{a}^\dag \left|n\right\rangle = \sqrt{n+1}\left|n+1\right\rangle\\ +$$ + +These relations are the motivation for the name ladder operators as they connect the different eigenstates. And they are raising/lowering the quantum number by one. Finally we have to find the lower limit. And this is quite naturally 0 as -$n = \ensuremath{\left\langle n\right|}\hat{N}\ensuremath{\left|n\right\rangle} = \ensuremath{\left\langle\psi_1\right|}\ensuremath{\left|\psi_1\right\rangle}\geq 0$. +$n = \left\langle n\right|\hat{N}\left|n\right\rangle = \left\langle\psi_1\right|\left|\psi_1\right\rangle\geq 0$. So we can construct the full basis by just defining the action of the lowering operator on the zero element -$a\ensuremath{\left|0\right\rangle} = 0$ and the other operators are -then constructed as: $$\begin{aligned} -\ensuremath{\left|n\right\rangle} = \frac{(a^\dag)^n}{\sqrt{n!}}\ensuremath{\left|0\right\rangle} -\end{aligned}$$ +$a\left|0\right\rangle = 0$ and the other operators are +then constructed as: + +$$ +\left|n\right\rangle = \frac{(a^\dag)^n}{\sqrt{n!}}\left|0\right\rangle +$$ ## Spatial representation of the eigenstates While we now have the spectrum it would be really nice to obtain the spatial properties of the different states. For that we have to project them onto the x basis. Let us start out with the ground state for which -we have $\hat{a}\ensuremath{\left|0\right\rangle}= 0$: $$\begin{aligned} -\ensuremath{\left\langle x\right|}\frac{1}{\sqrt{2\hbar}}\left(\sqrt{m\omega}\hat{x} +i \frac{1}{\sqrt{m\omega}}\hat{p}\right)\ensuremath{\left|0\right\rangle}= 0\\ +we have $\hat{a}\left|0\right\rangle= 0$: + +$$ +\left\langle x\right|\frac{1}{\sqrt{2\hbar}}\left(\sqrt{m\omega}\hat{x} +i \frac{1}{\sqrt{m\omega}}\hat{p}\right)\left|0\right\rangle= 0\\ \left(\sqrt{\frac{m\omega}{\hbar}}x + \sqrt{\frac{\hbar}{m\omega}}\partial_x\right)\psi_0(x)= 0\\ \Rightarrow \psi_0(x) \propto e^{-\frac{x^2}{2a_{HO}^2}} -\end{aligned}$$ This also introduces the typical distance in the quantum +$$ + +This also introduces the typical distance in the quantum harmonic oscillator which is given by $a_{HO} =\sqrt{\hbar/m\omega}$. The other states are solutions to the defining equations: -$$\begin{aligned} + +$$ \psi_n(x) = \frac{1}{\sqrt{n!}2^n}\left(\sqrt{m\omega}x - \frac{1}{\sqrt{m\omega}}\frac{d}{dx}\right)^n \psi_0(x)\\ \psi_n(x) = \frac{1}{\sqrt{n!}2^n}H_n(x) \psi_0(x)\\ -\end{aligned}$$ where $H_n(x)$ are the Hermite polynoms. +$$ + +where $H_n(x)$ are the Hermite polynoms. # The hamiltonian of the hydrogen atom -The hydrogen atom plays at central role in atomic physics as it is *the* -basic ingredient of atomic structures. It describes a single *electron*, -which is bound to the nucleus of a single *proton*. As such it is the +The hydrogen atom plays at central role in atomic physics as it is _the_ +basic ingredient of atomic structures. It describes a single _electron_, +which is bound to the nucleus of a single _proton_. As such it is the simplest of all atoms and can be described analytically within high precision. This has motivated an enormous body of literature on the problem, which derives all imaginable properties in nauseating detail. Therefore, we will focus here on the main properties and only sketch the derivations, while we will reference to the more technical details. -For the hydrogen atom as shown in [1](#261310){reference-type="ref" -reference="261310"}, we can write down the Hamiltonian $$\begin{aligned} + + +Sketch of the hydrogen atom with the relative coordinate and the +coordinates of the proton and the electron. + +For the hydrogen atom as shown in above, we can write down the Hamiltonian + +$$ \hat{H}=\frac{{{\hat{\vec{p}}}^2_\text{p}}}{2m_\text{p}} + \frac{{\hat{\vec{p}}}^2_\text{e}}{2m_\text{e}} - \frac{Ze^2}{4\pi\epsilon_0 r}, -\end{aligned}$$ where $Ze$ is the nuclear charge. To solve the problem, +$$ + +where $Ze$ is the nuclear charge. To solve the problem, we have to find the right Hilbert space. We can not solve the problem of the electron alone. If we do a separation of coordinates, i.e., we separate the Hamiltonian into the the center of mass and the relative -motion, we get $$\begin{aligned} -\hat{H} = \underbrace{\frac{{\hat{\vec{p}}}^2_{\textrm{cm}}}{2M}}_{\hat{H}_{\textrm{cm}}} + \underbrace{\frac{{\hat{\vec{p}}}^2_\text{r}}{2\mu}- \frac{Ze^2}{4\pi\epsilon_0r}}_{\hat{H}_{\text{atom}}} \label{eq:hydrogencmatomsplit} -\end{aligned}$$ with the reduced mass $1/\mu=1/m_\text{e}+1/m_\text{p}$. -If the state of the hydrogen atom $\ensuremath{\left|\psi\right\rangle}$ -is an eigenstate of $\hat{H}$, we can write $$\begin{aligned} -\hat{H}\ensuremath{\left|\psi\right\rangle}=&\left( \hat{H}_\textrm{cm}+\hat{H}_{\text{atom}} \right)\ensuremath{\left|\psi_\textrm{cm}\right\rangle}\otimes \ensuremath{\left|\psi_\text{atom}\right\rangle} \label{eq:hydrogencmatom}\\ -=& \left( E_{\text{kin}} + E_\text{atom} \right) \ensuremath{\left|\psi\right\rangle}. -\end{aligned}$$ Both states in -[\[eq:hydrogencmatom\]](#eq:hydrogencmatom){reference-type="eqref" -reference="eq:hydrogencmatom"} are eigenstates of the system. The state -$\ensuremath{\left|\psi\right\rangle}$ can be split up as shown since +motion, we get + +$$ +\hat{H} = \underbrace{\frac{{\hat{\vec{p}}}^2_{\textrm{cm}}}{2M}}_{\hat{H}_{\textrm{cm}}} + \underbrace{\frac{{\hat{\vec{p}}}^2_\text{r}}{2\mu}- \frac{Ze^2}{4\pi\epsilon_0r}}_{\hat{H}_{\text{atom}}} +$$ + +with the reduced mass $1/\mu=1/m_\text{e}+1/m_\text{p}$. +If the state of the hydrogen atom $\left|\psi\right\rangle$ +is an eigenstate of $\hat{H}$, we can write + +$$ +\hat{H}\left|\psi\right\rangle=\left(\hat{H}_\textrm{cm}+\hat{H}_{\text{atom}} \right)\left|\psi_\textrm{cm}\right\rangle\otimes \left|\psi_\text{atom}\right\rangle \\ += \left( E_{\text{kin}} + E_\text{atom} \right) \left|\psi\right\rangle. +$$ + +Both states are eigenstates of the system. The state +$\left|\psi\right\rangle$ can be split up as shown since the two degrees of freedom are generally not entangled. ![Sketch of the hydrogen atom with the relative coordinate and the @@ -145,10 +196,14 @@ coordinates of the proton and the electron. ](figures/Bildschirmfoto-2018-09-28-um-16-07-07/Bildschirmfoto-2018-09-28-um-16-07-07){#261310 width="0.70\\columnwidth"} -The wave function of the system then reads: $$\begin{aligned} -\psi(\vec{R},\vec{r}) =& \left( \ensuremath{\left\langle R\right|} \otimes \ensuremath{\left\langle r\right|}\right)\left( \ensuremath{\left|\psi_\textrm{cm}\right\rangle} \otimes \ensuremath{\left|\psi_{\text{atom}}\right\rangle}\right)\\ -=& \psi(\vec{R}) \cdot \psi (\vec{r}) -\end{aligned}$$ Our goal is now to find the eigenfunctions and +The wave function of the system then reads: + +$$ +\psi(\vec{R},\vec{r}) = \left( \left\langle R\right| \otimes \left\langle r\right|\right)\left( \left|\psi_\textrm{cm}\right\rangle \otimes \left|\psi_{\text{atom}}\right\rangle\right)\\ += \psi(\vec{R}) \cdot \psi (\vec{r}) +$$ + +Our goal is now to find the eigenfunctions and eigenenergies of $\hat{H}_\text{atom}$. In order to further divide the Hilbert space, we can use the symmetries. @@ -156,103 +211,146 @@ Hilbert space, we can use the symmetries. $\hat{H}_\text{atom}$ possesses spherical symmetry, which implies that **orbital angular momentum** $\hat{\vec{L}}$ is conserved. It is defined -as: $$\begin{aligned} +as: + +$$ \hat{\vec{L}}=\hat{\vec{r}} \times \hat{\vec{p}} -\end{aligned}$$ In other words, we have: $$\begin{aligned} +$$ + +In other words, we have: + +$$ = 0 -\end{aligned}$$ Let us show first that the kinetic term commutes with +$$ + +Let us show first that the kinetic term commutes with the angular momentum operator, We will employ the commutator relationships for position and momentum $[x_i, p_j]=i\hbar$ and the relationship $[A,BC] = [A,B]C+B[A,C]$ and $[f(x), p_x] = [x,p_x]\frac{\partial f(x)}{\partial x}$. So we obtain: -$$\begin{aligned} -&= [p_x^2,xp_y]-[p_y^2,yp_x] \\ - &= [p_x^2,x]p_y-[p_y^2,y] p_x\\ - &=i\hbar 2 p_xp_y-2i\hbar p_y p_x\\ - &= 0 -\end{aligned}$$ Analog calculations show that $L_y$ and $L_z$ commute. + +$$ += [p_x^2,xp_y]-[p_y^2,yp_x] \\ + = [p_x^2,x]p_y-[p_y^2,y] p_x\\ + =i\hbar 2 p_xp_y-2i\hbar p_y p_x\\ + = 0 +$$ + +Analog calculations show that $L_y$ and $L_z$ commute. In a similiar fashion we can verify that the potential term commutes -with the different components of $\hat{\vec{L}}$ $$\begin{aligned} -&= [\frac{1}{r}, xp_y]-[\frac{1}{r}, yp_x]\\ -&= x[\frac{1}{r}, p_y]-y[\frac{1}{r}, p_x]\\ -&= -x \frac{yi\hbar}{2r^{3/2}}+y\frac{xi\hbar}{2r^{3/2}}\\ -&=0 -\end{aligned}$$ We can therefore decompose the eigenfunctions of the +with the different components of $\hat{\vec{L}}$ + +$$ += [\frac{1}{r}, xp_y]-[\frac{1}{r}, yp_x]\\ += x[\frac{1}{r}, p_y]-y[\frac{1}{r}, p_x]\\ += -x \frac{yi\hbar}{2r^{3/2}}+y\frac{xi\hbar}{2r^{3/2}}\\ +=0 +$$ + +We can therefore decompose the eigenfunctions of the hydrogen atom over the eigenbasis of the angular momentum operator. A detailled discussion of the properties of $\vec{L}$ can be found in -Appendix B of [@Hertel_2015]. To find the eigenbasis, we first need to +[Appendix B of Hertel](http://dx.doi.org/10.1007/978-3-642-54322-7). To find the eigenbasis, we first need to identify the commutation relationships between the components of $\hat{\vec{L}}$. We can calculate them following commutation -relationships: $$\begin{aligned} -&= [yp_z - zp_y, zp_x - xp_z]\\ -&=[yp_z, zp_x]-[yp_z,xp_z]- [zp_y, zp_x] + [zp_y,xp_z]\\ -&=[yp_z, zp_x] + [zp_y,xp_z]\\ -&=[yp_z, z]p_x +x[zp_y,p_z]\\ -&=-i\hbar yp_x +i\hbar xp_y\\ -&= i\hbar L_z -\end{aligned}$$ This relationship holds for all the other components too -and we have in general: $$\begin{aligned} +relationships: + +$$ += [yp_z - zp_y, zp_x - xp_z]\\ +=[yp_z, zp_x]-[yp_z,xp_z]- [zp_y, zp_x] + [zp_y,xp_z]\\ +=[yp_z, zp_x] + [zp_y,xp_z]\\ +=[yp_z, z]p_x +x[zp_y,p_z]\\ +=-i\hbar yp_x +i\hbar xp_y\\ += i\hbar L_z +$$ + +This relationship holds for all the other components too +and we have in general: + +$$ = i\hbar \epsilon_{ijk}L_k -\end{aligned}$$ The orbital angular momentum is therefore part of the +$$ + +The orbital angular momentum is therefore part of the large family of angular momentum operators, which also comprises spin etc. In particular the different components are not independent, and therefore we cannot form a basis out the three components. A suitable -choice is actually to use the following combinations: $$\begin{aligned} -\hat{\vec{L}}^2\ensuremath{\left|l,m_l\right\rangle} =& \hbar^2 l (l+1)\ensuremath{\left|l,m_l\right\rangle}\\ -\hat{L}_z\ensuremath{\left|l,m_l\right\rangle} =& \hbar m_l \ensuremath{\left|l,m_l\right\rangle} -\end{aligned}$$ +choice is actually to use the following combinations: + +$$ +\hat{\vec{L}}^2\left|l,m_l\right\rangle = \hbar^2 l (l+1)\left|l,m_l\right\rangle\\ +\hat{L}_z\left|l,m_l\right\rangle = \hbar m_l \left|l,m_l\right\rangle +$$ -- $l$ is a non-negative integer and it is called the **orbital angular - momentum quantum number**. +- $l$ is a non-negative integer and it is called the **orbital angular + momentum quantum number**. -- $m_l$ takes values $-l, -l+1, ..., l-1, l$ and it is sometimes - called the **projection of the angular momentum**. +- $m_l$ takes values $-l, -l+1, ..., l-1, l$ and it is sometimes + called the **projection of the angular momentum**. ## Eigenfunction of the angular momentum operators Having identified the relevant operators it would be nice to obtain a space representation of them. This works especially nicely in spherical -coordinates. There, we get $$\begin{aligned} -\hat{L}_z&= - i \hbar \partial_{\phi}\\ -\hat{\vec{L}}^2 &= - \hbar^2 \left[\frac{1}{\sin(\theta)}\partial_{\theta} \left( \sin(\theta) \partial_\theta\right) + \frac{1}{\sin^2(\theta)} \partial_{\phi\phi} \right]. -\end{aligned}$$ The corresponding wave functions are $$\begin{aligned} -\ensuremath{\left\langle\theta, \phi | l,m_l\right\rangle} = Y_{lm}(\theta,\phi). -\end{aligned}$$ +coordinates. There, we get + +$$ +\hat{L}_z= - i \hbar \partial_{\phi}\\ +\hat{\vec{L}}^2 = - \hbar^2 \left[\frac{1}{\sin(\theta)}\partial_{\theta} \left( \sin(\theta) \partial_\theta\right) + \frac{1}{\sin^2(\theta)} \partial_{\phi\phi} \right]. +$$ + +The corresponding wave functions are + +$$ +\left\langle\theta, \phi | l,m_l\right\rangle = Y_{lm}(\theta,\phi). +$$ Where $Y_{lm}(\theta, \phi)$ are the **spherical harmonics**. # The radial wave equation Given that we now know that the angular momentum is conserved for the -hydrogen atom, we can actually rewrite the Hamltonian -[\[eq:hydrogencmatomsplit\]](#eq:hydrogencmatomsplit){reference-type="ref" -reference="eq:hydrogencmatomsplit"} in terms of the angular momentum as -we find: $$\begin{aligned} +hydrogen atom, we can actually rewrite the Hamltonian in terms of the angular momentum as +we find: + +$$ \hat{H}_\text{atom} = \hat{H}_r + \frac{\hat{L}}{2\mu r^2}+V(r) \\ \hat{H}_r = -\frac{\hbar^2}{2\mu}\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) -\end{aligned}$$ We can now separate out the angular part and decompose +$$ + +We can now separate out the angular part and decompose it over the eigenfunctions of $\hat{\vec{L}}$, such that we make the -ansatz [^2]: $$\begin{aligned} +ansatz [^2]: + +$$ \psi (r,\theta,\phi) = R(r) Y_{lm}(\theta,\phi) -\end{aligned}$$ +$$ We can plug this separated ansatz in the Schrödinger equation. We already solved the angular in the discussion of the angular momentum and -for the radial part we obtain: $$\begin{aligned} +for the radial part we obtain: + +$$ -\frac{\hbar^2}{2\mu}\frac{1}{r}\frac{d^2(rR(r))}{dr^2} - \frac{Ze^2}{4\pi\epsilon_0 r} R(r) + \frac{\hbar^2}{2\mu}\frac{l(l+1)}{r^2}R(r) = ER(r) -\end{aligned}$$ Substituting $R(r)=u(r)/r$ leads to $$\begin{aligned} +$$ + +Substituting $R(r)=u(r)/r$ leads to + +$$ -\frac{\hbar^2}{2\mu}\frac{d^2}{dr^2}u(r) +\underbrace{ \left( -\frac{Ze^2}{4\pi\epsilon_0 r} + \frac{\hbar^2}{2\mu} \frac{l(l+1)}{r^2} \right)}_{V_{\text{eff}}} u(r) = E \, u(r), -\end{aligned}$$ which is known as the "radial wave equation". It is a -very general result for *any* central potential. It can also be used to +$$ + +which is known as the "radial wave equation". It is a +very general result for _any_ central potential. It can also be used to describe unbound states ($E>0$) that occur during scattering. -In the [next -lecture](https://www.authorea.com/users/143341/articles/326674-lecture-6-the-dipole-approximation-in-the-hydrogen-atom) -we will look into the energy scales of the hydrogen atom and then start +In the next lecture we will look into the energy scales of the hydrogen atom and then start coupling different levels. -[^1]: The commutator between $\hat{X}$ and $\hat{P}$ is still as for $x$ +[^1]: + The commutator between $\hat{X}$ and $\hat{P}$ is still as for $x$ and $p$. -[^2]: Only if the system is in a well-defined angular momentum state, we +[^2]: + Only if the system is in a well-defined angular momentum state, we can write it down like this. diff --git a/amo/lecture5_pic1.png b/amo/lecture5_pic1.png new file mode 100644 index 0000000000000000000000000000000000000000..008f58e9692230bac5b010728fe30dea3a49f7db GIT binary patch literal 10579 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z&zhNJVMPEQ+`oBDOr4FLEFD}d?d?F~md*|~hMu?tNzDKW-rpp)hIX!IhQ=y(T|E94wF}3-7@H#US3=D!~B;>`bL=8Uv5BJl{@Bjb+ literal 0 HcmV?d00001 From 6aa802fee63d916943b695db7c6ef7f650a1a0a1 Mon Sep 17 00:00:00 2001 From: fretchen Date: Sat, 4 Jan 2025 13:07:32 +0100 Subject: [PATCH 11/13] Create lecture6.md --- amo/lecture6.md | 276 ++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 276 insertions(+) create mode 100644 amo/lecture6.md diff --git a/amo/lecture6.md b/amo/lecture6.md new file mode 100644 index 0000000..48815e9 --- /dev/null +++ b/amo/lecture6.md @@ -0,0 +1,276 @@ +--- +author: +- Fred Jendrzejewski +- Selim Jochim +bibliography: +- bibliography/converted_to_latex.bib +date: January 04, 2025 +nocite: "[@*]" +title: Lecture 6 - The dipole approximation in the hydrogen atom +--- + +We will continue with some properties of the hydrogen atom. First +compare it to the harmonic oscillator, then look into dipole transitions +and end with the coupling to static magnetic fields. + +In the last lecture [@atom] we discussed the basic properties of the +hydrogen atom and found its eigenstates. We will now summarize the most +important properties and look into its orbitals. From that we will +understand the understand the interaction with electromagnetic waves and +introduce the selection rules for dipole transitions. + +# The energies of Hydrogen and its wavefunctions + +In the last lecture, we looked into hydrogen and saw that we could write +it's Hamiltonian as: $$\begin{aligned} +\hat{H}_\text{atom} = \hat{H}_r + \frac{\hat{L}}{2\mu r^2}+V(r) \\ +\hat{H}_r = -\frac{\hbar^2}{2\mu}\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) +\end{aligned}$$ We could then separate out the angular part and +decompose it as: $$\begin{aligned} +\psi (r,\theta,\phi) = \frac{u(r)}{r} Y_{lm}(\theta,\phi) +\end{aligned}$$ The radial wave equation reads then: $$\begin{aligned} +\label{Eq:RadWF} +-\frac{\hbar^2}{2\mu}\frac{d^2}{dr^2}u(r) +\underbrace{ \left( -\frac{Ze^2}{4\pi\epsilon_0 r} + \frac{\hbar^2}{2\mu} \frac{l(l+1)}{r^2} \right)}_{V_{\text{eff}}} u(r) = E \, u(r), +\end{aligned}$$ + +## Energy scales + +We can now make [\[Eq:RadWF\]](#Eq:RadWF){reference-type="eqref" +reference="Eq:RadWF"} dimensionless, by rewriting: $$\begin{aligned} +r = \rho \tilde{a}_{0} +\end{aligned}$$ So we rewrite: $$\begin{aligned} +-\frac{\hbar^2}{2\mu \tilde{a}_{0}^2}\frac{d^2}{d\rho^2}u(r) + \left( -\frac{Ze^2}{4\pi\epsilon_0\tilde{a}_{0}}\frac{1}{\rho} + \frac{\hbar^2}{2\mu \tilde{a}_{0}^2} \frac{l(l+1)}{\rho^2} \right) u(r) = E \, u(r), +\end{aligned}$$ This allows us to measure energies in units of: +$$\begin{aligned} +E &= \epsilon R_{y,\textrm{m}}\\ +R_{y,\textrm{m}} &= -\frac{\hbar^2}{2\mu \tilde{a}_{0}^2} +\end{aligned}$$ The equation reads then: $$\begin{aligned} +\frac{d^2}{d\rho^2}u(\rho) + \left( \frac{\mu Ze^2 \tilde{a}_{0}}{\hbar^2 4\pi\epsilon_0}\frac{2}{\rho} - \frac{l(l+1)}{\rho^2} \right) u(\rho) = \epsilon u(\rho), +\end{aligned}$$ If we finally set $$\begin{aligned} +\tilde{a}_{0}&=\frac{4\pi\epsilon_0 \hbar^2}{\mu Z e^2} +\end{aligned}$$ We obtain the especially elegant formulation: +$$\begin{aligned} +\frac{d^2}{d\rho^2}u(\rho) + \left( \frac{2}{\rho} - \frac{l(l+1)}{\rho^2} \right) u(\rho) = \epsilon u(\rho), +\end{aligned}$$ We typically call $\tilde{a}_{0}$ the **Bohr radius** +for an atom with reduced mass $\mu$ and with a nucleus with charge +number $Z$. $R_{y,\textrm{m}}$ is the **Rydberg energy** of such an +atom. + +The universal constant is defined for the infinite mass limit +$\mu \approx m_e$ and for $Z=1$. As a length scale we introduce the Bohr +radius for infinite nuclear mass $$\begin{aligned} +a_0 &= \frac{4\pi\epsilon_0\hbar^2}{m_e e^2} = \text{\num{0.5} \text{angstrom}} = \text{\SI{0.05}{\nano\meter}}. +\end{aligned}$$ The energy scale reads: $$\begin{aligned} +R_{y,\infty} &= \frac{m_e e^4}{32 \pi^2 \epsilon_0^2 \hbar^2}\\ +&\approx \SI{2.179e-18}{J}\\ +& \approx e \times\SI{13.6}{eV}\\ +&\approx h \times\SI{3289}{T\hertz} +\end{aligned}$$ So if we excite the hydrogen atom for time scales of a +few attoseconds, we will coherently create superposition states of all +existing levels. But which ones ? And at which frequency ? + +## Solution of the radial wave equation + +At this stage we can have a look into the energy landscape: + +![Energy potential of the hydrogen atom +](figures/Bildschirmfoto-2018-09-28-um-16-21-46/Bildschirmfoto-2018-09-28-um-16-21-46){#951159 +width="0.70\\columnwidth"} + +The energies read then $$\begin{aligned} +E_n = -\frac{R_{y,\textrm{m}}}{n^2} \qquad \text{with} \qquad n=1,2,3,\cdots +\end{aligned}$$ for $l=0$ and $$\begin{aligned} +E_n = -\frac{R_{y,\textrm{m}}}{n^2} \qquad \text{with} \qquad n=2,3,4,\cdots +\end{aligned}$$ for $l=1$. Despite the different effective potentials +(see [1](#951159){reference-type="ref" reference="951159"}), we get the +same eigenstates. This looks like an accidental degeneracy. Actually, +there is a hidden symmetry which comes from the so-called "Runge-Lenz" +vector. It only occurs in an attractive $1/r$-potential [@atom]. This +vector reads: $$\mathbf{A} =\mathbf{p}\times\mathbf{L}-\mathbf{r}$$ + +Finally, we can also visualize the radial wavefunctions for the hydrogen +atom as shown in Fig. [2](#785001){reference-type="ref" +reference="785001"}. + +![Radial wavefunctions +](figures/Bildschirmfoto-2018-10-29-um-08-04-45/Bildschirmfoto-2018-10-29-um-08-04-45){#785001 +width="0.70\\columnwidth"} + +Associated with these radial wavefunctions, we also have the angular +profiles. Where $Y_{lm}(\theta, \phi)$ are the **spherical harmonics** +as shown in Fig. [3](#175742){reference-type="ref" reference="175742"}. + +![The spherical harmonics. Fig is taken from Ref. [@Demtr_der_2018] +](figures/Bildschirmfoto-2018-10-18-um-09-04-42/Bildschirmfoto-2018-10-18-um-09-04-42){#175742 +width="0.70\\columnwidth"} + +Their shape is especially important for understanding the possibility of +coupling different orbits through electromagnetic waves. + +# The electric dipole approximation + +![Interaction between an atom and an electromagnetic wave $\vec{E}$ with +wave vector $\vec{k}$. The states $\text{|g>}$ and $\text{|e>}$ stand +for the ground and excited state and $\hbar\omega_0$ is the energy of +the resonant transition between the states. +](figures/Bildschirmfoto-2018-09-29-um-21-46-38/Bildschirmfoto-2018-09-29-um-21-46-38){#823292 +width="0.70\\columnwidth"} + +We consider an atom which is located in a radiation field. By resonant +coupling with the frequency $\omega_0$, it can go from the ground state +$\ensuremath{\left|g\right\rangle}$ to the excited state +$\ensuremath{\left|e\right\rangle}$ (see +[4](#823292){reference-type="ref" reference="823292"}). + +The potential energy of a charge distribution in a homogeneous +electromagnetic field $\vec{E}$ is: $$\begin{aligned} +E_\text{pot} = \sum_i q_i \vec{r}_i\cdot \vec{E}. +\end{aligned}$$ If the upper limit of the sum is 2, we obtain the dipole +moment $$\begin{aligned} +\vec{D} = e \vec{r}. +\end{aligned}$$ For the hydrogen atom, the distance corresponds to the +Bohr radius. + +![A charge distribution in an electromagnetic field \$\\vec{E}\$ +](figures/Bildschirmfoto-2018-09-29-um-21-49-06/Bildschirmfoto-2018-09-29-um-21-49-06){#241421 +width="0.70\\columnwidth"} + +**Note.** Apart from the monopole, the dipole potential is the lowest +order term of the multipole expansion of the scalar potential $\phi$: +$$\begin{aligned} +\phi \left( \vec{r} \right) =& \frac{1}{4\pi\epsilon_0}\frac{\vec{D}\cdot\vec{r}}{|\vec{r}|^3}\\ +\vec{E}(\vec{r})=& \vec{\nabla}\phi(\vec{r}) = \frac{ 3 \left(\vec{D}\cdot \vec{r}\right) \vec{r}/{|\vec{r}|^2}- \vec{D}}{4\pi\epsilon_0|\vec{r}|^3}. +\end{aligned}$$ + +For the dipole approximation we consider the size of the atom and +compare it to the wavelength $\lambda$ of the electromagnetic field: +$$\begin{aligned} +\ensuremath{\left\langle|r|\right\rangle} \sim 1\text{angstrom}\ll \lambda \sim 10^3\text{angstrom} +\end{aligned}$$ + +- Therefore, we assume that the field is homogeneous in space and omit + the spatial dependence: $$\begin{aligned} + E(r,t) \approx E(t) + \end{aligned}$$ + +- The correction term resulting from the semi-classical dipole + approximation then is $$\begin{aligned} + \hat{H}_1(t)=-e\hat{\vec{r}} \cdot \vec{E}(t) = -\hat{\vec{D}} \cdot \vec{E}(t) + \end{aligned}$$ + +- Why can the magnetic field be ignored in this approximation? The + velocity of an electron is $\sim \alpha c$. The hydrogen atom only + has small relativistic corrections. If we compare the modulus of the + magnetic and the electric field, we get: $$\begin{aligned} + \left| \vec{B} \right| = \frac{|\vec{E}|}{c} + \end{aligned}$$ The electric field contribution thus dominates. + +Now we choose $$\begin{aligned} +\vec{E} = E_0 \vec{\epsilon} \cos \left(\omega t - \vec{k} \cdot \vec{r}\right) +\end{aligned}$$ and do time-dependent perturbation theory (see +[@Jendrzejewski]): + +$$\begin{aligned} +\ensuremath{\left|\psi(t)\right\rangle} = \gamma_1(t) \mathrm{e}^{-iE_1t/\hbar} \ensuremath{\left|1\right\rangle} +& \gamma_2(t) \mathrm{e}^{-iE_2t/\hbar} \ensuremath{\left|2\right\rangle}\\ ++&\sum_{n=3}^\infty \gamma_n \mathrm{e}^{-iE_nt/\hbar} \ensuremath{\left|n\right\rangle} +\end{aligned}$$ As initial condition we choose $$\begin{aligned} + \gamma_i(0) = \left\{ \begin{array}{ccc} 1 &\text{for}& i=1 \\ 0 &\text{for}& i>1 \end{array} \right. +\end{aligned}$$ We write $\omega_0 = (E_2-E_1)/\hbar$ and get to first +order $\hat{vec{D}}$: + +$$\begin{aligned} +\gamma_2(t) = \overbrace{\frac{E_0}{2\hbar} \ensuremath{\left\langle 2|\hat{\vec{D}}\cdot \vec{\epsilon}\,|1\right\rangle}}^{\text{Rabi frequency }\Omega} \underbrace{\left(\frac{\mathrm{e}^{i(\omega_0 + \omega)t}-1}{\omega_0 + \omega} + \frac{\mathrm{e}^{i(\omega_0 - \omega)t}-1}{\omega_0 - \omega}\right)}_{\text{time evolution of the system}} +\end{aligned}$$ The term before the round brackets is called dipole +matrix element: $$\begin{aligned} +\label{Eq:DipOp} +\ensuremath{\left\langle 2|\hat{\vec{D}}\cdot \vec{\epsilon}\,|1\right\rangle} =e \int \psi_2\left(\vec{r}\right) \cdot \vec{r} \cdot \vec{\epsilon} \cdot \psi_1\left(\vec{r}\right) \mathop{}\!\mathrm{d}\vec{r}. +\end{aligned}$$ + +![Coupling of different orbitals. +](figures/Bildschirmfoto-2018-10-31-um-11-54-46/SelectionRules){#708926 +width="0.70\\columnwidth"} + +# Selection rules + +We can now look into the allowed transition in the atom as they are what +we will typically observe within experiments. + +## Change of parity + +The parity operator is defined as: $$\begin{aligned} +\hat{P}\psi(\vec{r}) = \psi(-\vec{r}) +\end{aligned}$$ For the eigenfunction we have: $$\begin{aligned} +\hat{P} \psi(\vec{r}) = \lambda \psi(\vec{r})\\ +\lambda = \pm 1 +\end{aligned}$$ The eigenvalues are called *odd* and *even*. From the +definition of the dipole operator we can see that it is of odd parity. +What about the parity of the states that it is coupling ? If they have +both the same parity than the whole integral will disappear and no +dipole transition can appear. + +We can become more concrete for the given eigenfunctions as we have +within spherical coordinates: $$\begin{aligned} +(r, \theta, \phi) \rightarrow (r, \pi -\theta, \phi+\pi) +\end{aligned}$$ For the orbitals of the hydrogen atom we then have +explicitly: $$\begin{aligned} +\hat{P}\psi_{nlm}(r, \theta, \phi) &= R_{nl}(r)Y_{lm}(\pi -\theta, \phi+\pi)\\ +&= (-1)^l R_{nl}(r)Y_{lm}(, \theta, \phi) +\end{aligned}$$ This gives us the first selection rule that the +**orbital angular momentum has to change for dipole transitions** +$\Delta l = \pm 1$. + +- $s$ orbitals are only coupled to $p$ orbitals through dipole + transitions. + +- $p$ orbitals are only coupled to $s$ and $d$ orbitals through dipole + transitions. + +## Coupling for linearly polarized light + +Having established the need for parity change, we also need to +investigate the influence of the polarization of the light, which enters +the dipole operator through the vector $\epsilon$. In the simplest case +the light has linear polarization ($\pi$ polarized) and we can write: +$$\begin{aligned} +\vec{E}(t) = \vec{e}_zE_0 \cos(\omega t +\varphi) +\end{aligned}$$ This means that the dipole transition element +[\[Eq:DipOp\]](#Eq:DipOp){reference-type="eqref" reference="Eq:DipOp"} +is now given by: $$\begin{aligned} +\ensuremath{\left\langle 2\right|}\vec{D}\cdot\vec{e}_z\ensuremath{\left|1\right\rangle} = e \int \psi_2(\vec{r}) z \psi_1\left(\vec{r}\right) \mathop{}\!\mathrm{d}\vec{r} +\end{aligned}$$ We can now transform z into the spherical coordinates +$z= r \cos(\theta) = r\sqrt{\frac{4\pi}{3}}Y_{10}(\theta, \phi)$. We can +further separate out the angular part of the integral to obtain: +$$\begin{aligned} +\ensuremath{\left\langle 2\right|}\vec{D}\cdot\vec{e}_z\ensuremath{\left|1\right\rangle} \propto e \int \sin(\theta) d\theta d\varphi Y_{l',m'}(\theta, \varphi) Y_{10}(\theta, \phi) Y_{l,m}(\theta, \varphi) +\end{aligned}$$ This element is only non-zero if $m = m'$ (see appendix +C of [@Hertel_2015] for all the gorious details). + +![Dipole selection rules for different polarizations of light. +](figures/Bildschirmfoto-2018-10-30-um-13-22-49/Bildschirmfoto-2018-10-30-um-13-22-49){#852353 +width="0.70\\columnwidth"} + +## Circularly polarized light + +Light has not just linear polarization, but it might also have some +circular polarization. In this case we can write: $$\begin{aligned} +\vec{E}(t) &= \frac{E_0}{\sqrt{2}} \left(\cos(\omega t +\varphi)\vec{e}_x + \sin(\omega t +\varphi)\vec{e}_y\right)\\ +\vec{E}(t) &= \text{Re}\left(\vec{e}_+ E_0 e^{-i\omega t +\phi}\right)\\ +\vec{e}_\pm &= \frac{\vec{e}_x\pm i\vec{e}_y}{\sqrt{2}} +\end{aligned}$$ So light with polarization $\vec{\epsilon} = \vec{e}_+$ +is called right-hand circular ($\sigma^+$) and +$\vec{\epsilon} = \vec{e}_-$ is called left-hand circular ($\sigma^-$). +Let us now evaluate the transition elements here. The dipole operator +element boils now down to the evaluation of the integral: +$$\begin{aligned} +\ensuremath{\left\langle l',m',n'\right|}x+iy\ensuremath{\left|l,m,n\right\rangle} +\end{aligned}$$ As previously we can express the coupling term in +spherical coordinates: $$\begin{aligned} +\frac{x+iy}{\sqrt{2}} = -r \sqrt{\frac{4\pi}{3}}Y_{11}(\theta, \varphi) +\end{aligned}$$ Evaluation of the integrals lead now to the rule the +projection of the quantum number has to change $m' = m+1$. In a similiar +fashion we find for left-hand circular light the selection rule +$m' = m - 1$. All the results are summed up in Fig. +[7](#852353){reference-type="ref" reference="852353"}. + +In the next lecture [@atoma] we will investigate the influence of +perturbative effects and see how the fine structure arises. From db192ae077c895d2093144785a129ce6962e4364 Mon Sep 17 00:00:00 2001 From: fretchen Date: Sat, 4 Jan 2025 13:24:57 +0100 Subject: [PATCH 12/13] Add lecture 6 --- amo/lecture6.md | 379 +++++++++++++++++++++++++++--------------- amo/lecture6_pic1.png | Bin 0 -> 62537 bytes amo/lecture6_pic2.png | Bin 0 -> 70079 bytes amo/lecture6_pic3.png | Bin 0 -> 197672 bytes amo/lecture6_pic4.png | Bin 0 -> 35099 bytes amo/lecture6_pic5.png | Bin 0 -> 17935 bytes amo/lecture6_pic6.svg | 233 ++++++++++++++++++++++++++ amo/lecture6_pic7.svg | Bin 0 -> 56866 bytes 8 files changed, 475 insertions(+), 137 deletions(-) create mode 100644 amo/lecture6_pic1.png create mode 100644 amo/lecture6_pic2.png create mode 100644 amo/lecture6_pic3.png create mode 100644 amo/lecture6_pic4.png create mode 100644 amo/lecture6_pic5.png create mode 100644 amo/lecture6_pic6.svg create mode 100644 amo/lecture6_pic7.svg diff --git a/amo/lecture6.md b/amo/lecture6.md index 48815e9..0854098 100644 --- a/amo/lecture6.md +++ b/amo/lecture6.md @@ -1,9 +1,9 @@ --- author: -- Fred Jendrzejewski -- Selim Jochim + - Fred Jendrzejewski + - Selim Jochim bibliography: -- bibliography/converted_to_latex.bib + - bibliography/converted_to_latex.bib date: January 04, 2025 nocite: "[@*]" title: Lecture 6 - The dipole approximation in the hydrogen atom @@ -13,7 +13,7 @@ We will continue with some properties of the hydrogen atom. First compare it to the harmonic oscillator, then look into dipole transitions and end with the coupling to static magnetic fields. -In the last lecture [@atom] we discussed the basic properties of the +In the last lecture, we discussed the basic properties of the hydrogen atom and found its eigenstates. We will now summarize the most important properties and look into its orbitals. From that we will understand the understand the interaction with electromagnetic waves and @@ -22,50 +22,89 @@ introduce the selection rules for dipole transitions. # The energies of Hydrogen and its wavefunctions In the last lecture, we looked into hydrogen and saw that we could write -it's Hamiltonian as: $$\begin{aligned} +it's Hamiltonian as: + +$$ \hat{H}_\text{atom} = \hat{H}_r + \frac{\hat{L}}{2\mu r^2}+V(r) \\ \hat{H}_r = -\frac{\hbar^2}{2\mu}\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) -\end{aligned}$$ We could then separate out the angular part and -decompose it as: $$\begin{aligned} +$$ + +We could then separate out the angular part and +decompose it as: + +$$ \psi (r,\theta,\phi) = \frac{u(r)}{r} Y_{lm}(\theta,\phi) -\end{aligned}$$ The radial wave equation reads then: $$\begin{aligned} -\label{Eq:RadWF} +$$ + +The radial wave equation reads then: + +$$ + -\frac{\hbar^2}{2\mu}\frac{d^2}{dr^2}u(r) +\underbrace{ \left( -\frac{Ze^2}{4\pi\epsilon_0 r} + \frac{\hbar^2}{2\mu} \frac{l(l+1)}{r^2} \right)}_{V_{\text{eff}}} u(r) = E \, u(r), -\end{aligned}$$ +$$ ## Energy scales -We can now make [\[Eq:RadWF\]](#Eq:RadWF){reference-type="eqref" -reference="Eq:RadWF"} dimensionless, by rewriting: $$\begin{aligned} +We can now make the last equation dimensionless, by rewriting: + +$$ r = \rho \tilde{a}_{0} -\end{aligned}$$ So we rewrite: $$\begin{aligned} +$$ + +So we rewrite: + +$$ -\frac{\hbar^2}{2\mu \tilde{a}_{0}^2}\frac{d^2}{d\rho^2}u(r) + \left( -\frac{Ze^2}{4\pi\epsilon_0\tilde{a}_{0}}\frac{1}{\rho} + \frac{\hbar^2}{2\mu \tilde{a}_{0}^2} \frac{l(l+1)}{\rho^2} \right) u(r) = E \, u(r), -\end{aligned}$$ This allows us to measure energies in units of: -$$\begin{aligned} -E &= \epsilon R_{y,\textrm{m}}\\ -R_{y,\textrm{m}} &= -\frac{\hbar^2}{2\mu \tilde{a}_{0}^2} -\end{aligned}$$ The equation reads then: $$\begin{aligned} +$$ + +This allows us to measure energies in units of: + +$$ +E = \epsilon R_{y,\textrm{m}}\\ +R_{y,\textrm{m}} = -\frac{\hbar^2}{2\mu \tilde{a}_{0}^2} +$$ + +The equation reads then: + +$$ \frac{d^2}{d\rho^2}u(\rho) + \left( \frac{\mu Ze^2 \tilde{a}_{0}}{\hbar^2 4\pi\epsilon_0}\frac{2}{\rho} - \frac{l(l+1)}{\rho^2} \right) u(\rho) = \epsilon u(\rho), -\end{aligned}$$ If we finally set $$\begin{aligned} -\tilde{a}_{0}&=\frac{4\pi\epsilon_0 \hbar^2}{\mu Z e^2} -\end{aligned}$$ We obtain the especially elegant formulation: -$$\begin{aligned} +$$ + +If we finally set + +$$ +\tilde{a}_{0}=\frac{4\pi\epsilon_0 \hbar^2}{\mu Z e^2} +$$ + +We obtain the especially elegant formulation: + +$$ \frac{d^2}{d\rho^2}u(\rho) + \left( \frac{2}{\rho} - \frac{l(l+1)}{\rho^2} \right) u(\rho) = \epsilon u(\rho), -\end{aligned}$$ We typically call $\tilde{a}_{0}$ the **Bohr radius** +$$ + +We typically call $\tilde{a}_{0}$ the **Bohr radius** for an atom with reduced mass $\mu$ and with a nucleus with charge number $Z$. $R_{y,\textrm{m}}$ is the **Rydberg energy** of such an atom. The universal constant is defined for the infinite mass limit $\mu \approx m_e$ and for $Z=1$. As a length scale we introduce the Bohr -radius for infinite nuclear mass $$\begin{aligned} -a_0 &= \frac{4\pi\epsilon_0\hbar^2}{m_e e^2} = \text{\num{0.5} \text{angstrom}} = \text{\SI{0.05}{\nano\meter}}. -\end{aligned}$$ The energy scale reads: $$\begin{aligned} -R_{y,\infty} &= \frac{m_e e^4}{32 \pi^2 \epsilon_0^2 \hbar^2}\\ -&\approx \SI{2.179e-18}{J}\\ -& \approx e \times\SI{13.6}{eV}\\ -&\approx h \times\SI{3289}{T\hertz} -\end{aligned}$$ So if we excite the hydrogen atom for time scales of a +radius for infinite nuclear mass + +$$ +a_0 = \frac{4\pi\epsilon_0\hbar^2}{m_e e^2} = 0.5\text{angstrom} = 0.05 \text{nm}. +$$ + +The energy scale reads: + +$$ +R_{y,\infty} = \frac{m_e e^4}{32 \pi^2 \epsilon_0^2 \hbar^2}\\ +\approx 2.179e-18\textrm{J}\\ + \approx e \times 13.6\textrm{eV}\\ +\approx h \times 3289\textrm{THz} +$$ + +So if we excite the hydrogen atom for time scales of a few attoseconds, we will coherently create superposition states of all existing levels. But which ones ? And at which frequency ? @@ -73,122 +112,154 @@ existing levels. But which ones ? And at which frequency ? At this stage we can have a look into the energy landscape: -![Energy potential of the hydrogen atom -](figures/Bildschirmfoto-2018-09-28-um-16-21-46/Bildschirmfoto-2018-09-28-um-16-21-46){#951159 -width="0.70\\columnwidth"} + + +Energy potential of the hydrogen atom -The energies read then $$\begin{aligned} +The energies read then + +$$ E_n = -\frac{R_{y,\textrm{m}}}{n^2} \qquad \text{with} \qquad n=1,2,3,\cdots -\end{aligned}$$ for $l=0$ and $$\begin{aligned} +$$ + +for $l=0$ and + +$$ E_n = -\frac{R_{y,\textrm{m}}}{n^2} \qquad \text{with} \qquad n=2,3,4,\cdots -\end{aligned}$$ for $l=1$. Despite the different effective potentials -(see [1](#951159){reference-type="ref" reference="951159"}), we get the +$$ + +for $l=1$. Despite the different effective potentials, we get the same eigenstates. This looks like an accidental degeneracy. Actually, there is a hidden symmetry which comes from the so-called "Runge-Lenz" -vector. It only occurs in an attractive $1/r$-potential [@atom]. This +vector. It only occurs in an attractive $1/r$-potential . This vector reads: $$\mathbf{A} =\mathbf{p}\times\mathbf{L}-\mathbf{r}$$ Finally, we can also visualize the radial wavefunctions for the hydrogen -atom as shown in Fig. [2](#785001){reference-type="ref" -reference="785001"}. +atom as shown below -![Radial wavefunctions -](figures/Bildschirmfoto-2018-10-29-um-08-04-45/Bildschirmfoto-2018-10-29-um-08-04-45){#785001 -width="0.70\\columnwidth"} + Associated with these radial wavefunctions, we also have the angular profiles. Where $Y_{lm}(\theta, \phi)$ are the **spherical harmonics** -as shown in Fig. [3](#175742){reference-type="ref" reference="175742"}. +as shown below -![The spherical harmonics. Fig is taken from Ref. [@Demtr_der_2018] -](figures/Bildschirmfoto-2018-10-18-um-09-04-42/Bildschirmfoto-2018-10-18-um-09-04-42){#175742 -width="0.70\\columnwidth"} + Their shape is especially important for understanding the possibility of coupling different orbits through electromagnetic waves. # The electric dipole approximation -![Interaction between an atom and an electromagnetic wave $\vec{E}$ with +Below you see the interaction between an atom and an electromagnetic wave $\vec{E}$ with wave vector $\vec{k}$. The states $\text{|g>}$ and $\text{|e>}$ stand for the ground and excited state and $\hbar\omega_0$ is the energy of the resonant transition between the states. -](figures/Bildschirmfoto-2018-09-29-um-21-46-38/Bildschirmfoto-2018-09-29-um-21-46-38){#823292 -width="0.70\\columnwidth"} + + We consider an atom which is located in a radiation field. By resonant coupling with the frequency $\omega_0$, it can go from the ground state -$\ensuremath{\left|g\right\rangle}$ to the excited state -$\ensuremath{\left|e\right\rangle}$ (see -[4](#823292){reference-type="ref" reference="823292"}). +$\left|g\right\rangle$ to the excited state +$\left|e\right\rangle$. The potential energy of a charge distribution in a homogeneous -electromagnetic field $\vec{E}$ is: $$\begin{aligned} +electromagnetic field $\vec{E}$ is: + +$$ E_\text{pot} = \sum_i q_i \vec{r}_i\cdot \vec{E}. -\end{aligned}$$ If the upper limit of the sum is 2, we obtain the dipole -moment $$\begin{aligned} +$$ + +If the upper limit of the sum is 2, we obtain the dipole +moment + +$$ \vec{D} = e \vec{r}. -\end{aligned}$$ For the hydrogen atom, the distance corresponds to the +$$ + +For the hydrogen atom, the distance corresponds to the Bohr radius. -![A charge distribution in an electromagnetic field \$\\vec{E}\$ -](figures/Bildschirmfoto-2018-09-29-um-21-49-06/Bildschirmfoto-2018-09-29-um-21-49-06){#241421 -width="0.70\\columnwidth"} + **Note.** Apart from the monopole, the dipole potential is the lowest order term of the multipole expansion of the scalar potential $\phi$: -$$\begin{aligned} -\phi \left( \vec{r} \right) =& \frac{1}{4\pi\epsilon_0}\frac{\vec{D}\cdot\vec{r}}{|\vec{r}|^3}\\ -\vec{E}(\vec{r})=& \vec{\nabla}\phi(\vec{r}) = \frac{ 3 \left(\vec{D}\cdot \vec{r}\right) \vec{r}/{|\vec{r}|^2}- \vec{D}}{4\pi\epsilon_0|\vec{r}|^3}. -\end{aligned}$$ + +$$ +\phi \left( \vec{r} \right) = \frac{1}{4\pi\epsilon_0}\frac{\vec{D}\cdot\vec{r}}{|\vec{r}|^3}\\ +\vec{E}(\vec{r})= \vec{\nabla}\phi(\vec{r}) = \frac{ 3 \left(\vec{D}\cdot \vec{r}\right) \vec{r}/{|\vec{r}|^2}- \vec{D}}{4\pi\epsilon_0|\vec{r}|^3}. +$$ For the dipole approximation we consider the size of the atom and compare it to the wavelength $\lambda$ of the electromagnetic field: -$$\begin{aligned} -\ensuremath{\left\langle|r|\right\rangle} \sim 1\text{angstrom}\ll \lambda \sim 10^3\text{angstrom} -\end{aligned}$$ -- Therefore, we assume that the field is homogeneous in space and omit - the spatial dependence: $$\begin{aligned} +$$ +\left\langle|r|\right\rangle \sim 1\text{angstrom}\ll \lambda \sim 10^3\text{angstrom} +$$ + +- Therefore, we assume that the field is homogeneous in space and omit + the spatial dependence: + +$$ E(r,t) \approx E(t) - \end{aligned}$$ -- The correction term resulting from the semi-classical dipole - approximation then is $$\begin{aligned} + + +$$ + +- The correction term resulting from the semi-classical dipole + approximation then is + +$$ \hat{H}_1(t)=-e\hat{\vec{r}} \cdot \vec{E}(t) = -\hat{\vec{D}} \cdot \vec{E}(t) - \end{aligned}$$ -- Why can the magnetic field be ignored in this approximation? The - velocity of an electron is $\sim \alpha c$. The hydrogen atom only - has small relativistic corrections. If we compare the modulus of the - magnetic and the electric field, we get: $$\begin{aligned} + + +$$ + +- Why can the magnetic field be ignored in this approximation? The + velocity of an electron is $\sim \alpha c$. The hydrogen atom only + has small relativistic corrections. If we compare the modulus of the + magnetic and the electric field, we get: + +$$ \left| \vec{B} \right| = \frac{|\vec{E}|}{c} - \end{aligned}$$ The electric field contribution thus dominates. +$$ -Now we choose $$\begin{aligned} +The electric field contribution thus dominates. Now we choose + +$$ \vec{E} = E_0 \vec{\epsilon} \cos \left(\omega t - \vec{k} \cdot \vec{r}\right) -\end{aligned}$$ and do time-dependent perturbation theory (see -[@Jendrzejewski]): +$$ + +and do time-dependent perturbation theory: -$$\begin{aligned} -\ensuremath{\left|\psi(t)\right\rangle} = \gamma_1(t) \mathrm{e}^{-iE_1t/\hbar} \ensuremath{\left|1\right\rangle} +& \gamma_2(t) \mathrm{e}^{-iE_2t/\hbar} \ensuremath{\left|2\right\rangle}\\ -+&\sum_{n=3}^\infty \gamma_n \mathrm{e}^{-iE_nt/\hbar} \ensuremath{\left|n\right\rangle} -\end{aligned}$$ As initial condition we choose $$\begin{aligned} +$$ +\left|\psi(t)\right\rangle = \gamma_1(t) \mathrm{e}^{-iE_1t/\hbar} \left|1\right\rangle + \gamma_2(t) \mathrm{e}^{-iE_2t/\hbar} \left|2\right\rangle\\ ++\sum_{n=3}^\infty \gamma_n \mathrm{e}^{-iE_nt/\hbar} \left|n\right\rangle +$$ + +As initial condition we choose + +$$ \gamma_i(0) = \left\{ \begin{array}{ccc} 1 &\text{for}& i=1 \\ 0 &\text{for}& i>1 \end{array} \right. -\end{aligned}$$ We write $\omega_0 = (E_2-E_1)/\hbar$ and get to first -order $\hat{vec{D}}$: +$$ + +We write $\omega_0 = (E_2-E_1)/\hbar$ and get to first +order $\hat{\vec{D}}$: + +$$ +\gamma_2(t) = \overbrace{\frac{E_0}{2\hbar} \left\langle 2|\hat{\vec{D}}\cdot \vec{\epsilon}|1\right\rangle}^{\text{Rabi frequency }\Omega} \underbrace{\left(\frac{\mathrm{e}^{i(\omega_0 + \omega)t}-1}{\omega_0 + \omega} + \frac{\mathrm{e}^{i(\omega_0 - \omega)t}-1}{\omega_0 - \omega}\right)}_{\text{time evolution of the system}} +$$ + +The term before the round brackets is called dipole +matrix element: -$$\begin{aligned} -\gamma_2(t) = \overbrace{\frac{E_0}{2\hbar} \ensuremath{\left\langle 2|\hat{\vec{D}}\cdot \vec{\epsilon}\,|1\right\rangle}}^{\text{Rabi frequency }\Omega} \underbrace{\left(\frac{\mathrm{e}^{i(\omega_0 + \omega)t}-1}{\omega_0 + \omega} + \frac{\mathrm{e}^{i(\omega_0 - \omega)t}-1}{\omega_0 - \omega}\right)}_{\text{time evolution of the system}} -\end{aligned}$$ The term before the round brackets is called dipole -matrix element: $$\begin{aligned} -\label{Eq:DipOp} -\ensuremath{\left\langle 2|\hat{\vec{D}}\cdot \vec{\epsilon}\,|1\right\rangle} =e \int \psi_2\left(\vec{r}\right) \cdot \vec{r} \cdot \vec{\epsilon} \cdot \psi_1\left(\vec{r}\right) \mathop{}\!\mathrm{d}\vec{r}. -\end{aligned}$$ +$$ -![Coupling of different orbitals. -](figures/Bildschirmfoto-2018-10-31-um-11-54-46/SelectionRules){#708926 -width="0.70\\columnwidth"} +\left\langle 2|\hat{\vec{D}}\cdot \vec{\epsilon}\,|1\right\rangle =e \int \psi_2\left(\vec{r}\right) \cdot \vec{r} \cdot \vec{\epsilon} \cdot \psi_1\left(\vec{r}\right) \mathop{}\!\mathrm{d}\vec{r}. +$$ + + # Selection rules @@ -197,33 +268,49 @@ we will typically observe within experiments. ## Change of parity -The parity operator is defined as: $$\begin{aligned} +The parity operator is defined as: + +$$ \hat{P}\psi(\vec{r}) = \psi(-\vec{r}) -\end{aligned}$$ For the eigenfunction we have: $$\begin{aligned} +$$ + +For the eigenfunction we have: + +$$ \hat{P} \psi(\vec{r}) = \lambda \psi(\vec{r})\\ \lambda = \pm 1 -\end{aligned}$$ The eigenvalues are called *odd* and *even*. From the +$$ + +The eigenvalues are called _odd_ and _even_. From the definition of the dipole operator we can see that it is of odd parity. What about the parity of the states that it is coupling ? If they have both the same parity than the whole integral will disappear and no dipole transition can appear. We can become more concrete for the given eigenfunctions as we have -within spherical coordinates: $$\begin{aligned} +within spherical coordinates: + +$$ (r, \theta, \phi) \rightarrow (r, \pi -\theta, \phi+\pi) -\end{aligned}$$ For the orbitals of the hydrogen atom we then have -explicitly: $$\begin{aligned} -\hat{P}\psi_{nlm}(r, \theta, \phi) &= R_{nl}(r)Y_{lm}(\pi -\theta, \phi+\pi)\\ -&= (-1)^l R_{nl}(r)Y_{lm}(, \theta, \phi) -\end{aligned}$$ This gives us the first selection rule that the +$$ + +For the orbitals of the hydrogen atom we then have +explicitly: + +$$ +\hat{P}\psi_{nlm}(r, \theta, \phi) = R_{nl}(r)Y_{lm}(\pi -\theta, \phi+\pi)\\ += (-1)^l R_{nl}(r)Y_{lm}(, \theta, \phi) +$$ + +This gives us the first selection rule that the **orbital angular momentum has to change for dipole transitions** $\Delta l = \pm 1$. -- $s$ orbitals are only coupled to $p$ orbitals through dipole - transitions. +- $s$ orbitals are only coupled to $p$ orbitals through dipole + transitions. -- $p$ orbitals are only coupled to $s$ and $d$ orbitals through dipole - transitions. +- $p$ orbitals are only coupled to $s$ and $d$ orbitals through dipole + transitions. ## Coupling for linearly polarized light @@ -231,46 +318,64 @@ Having established the need for parity change, we also need to investigate the influence of the polarization of the light, which enters the dipole operator through the vector $\epsilon$. In the simplest case the light has linear polarization ($\pi$ polarized) and we can write: -$$\begin{aligned} + +$$ \vec{E}(t) = \vec{e}_zE_0 \cos(\omega t +\varphi) -\end{aligned}$$ This means that the dipole transition element -[\[Eq:DipOp\]](#Eq:DipOp){reference-type="eqref" reference="Eq:DipOp"} -is now given by: $$\begin{aligned} -\ensuremath{\left\langle 2\right|}\vec{D}\cdot\vec{e}_z\ensuremath{\left|1\right\rangle} = e \int \psi_2(\vec{r}) z \psi_1\left(\vec{r}\right) \mathop{}\!\mathrm{d}\vec{r} -\end{aligned}$$ We can now transform z into the spherical coordinates +$$ + +This means that the dipole transition element is now given by: + +$$ +\left\langle 2\right|\vec{D}\cdot\vec{e}_z\left|1\right\rangle = e \int \psi_2(\vec{r}) z \psi_1\left(\vec{r}\right) \mathop{}\!\mathrm{d}\vec{r} +$$ + +We can now transform z into the spherical coordinates $z= r \cos(\theta) = r\sqrt{\frac{4\pi}{3}}Y_{10}(\theta, \phi)$. We can further separate out the angular part of the integral to obtain: -$$\begin{aligned} -\ensuremath{\left\langle 2\right|}\vec{D}\cdot\vec{e}_z\ensuremath{\left|1\right\rangle} \propto e \int \sin(\theta) d\theta d\varphi Y_{l',m'}(\theta, \varphi) Y_{10}(\theta, \phi) Y_{l,m}(\theta, \varphi) -\end{aligned}$$ This element is only non-zero if $m = m'$ (see appendix -C of [@Hertel_2015] for all the gorious details). -![Dipole selection rules for different polarizations of light. -](figures/Bildschirmfoto-2018-10-30-um-13-22-49/Bildschirmfoto-2018-10-30-um-13-22-49){#852353 -width="0.70\\columnwidth"} +$$ +\left\langle 2\right|\vec{D}\cdot\vec{e}_z\left|1\right\rangle \propto e \int \sin(\theta) d\theta d\varphi Y_{l',m'}(\theta, \varphi) Y_{10}(\theta, \phi) Y_{l,m}(\theta, \varphi) +$$ + +This element is only non-zero if $m = m'$ (see [appendix +C of Hertel 2015](http://dx.doi.org/10.1007/978-3-642-54322-7) for all the gorious details). + + + +Above are the dipole selection rules for different polarizations of light. ## Circularly polarized light Light has not just linear polarization, but it might also have some -circular polarization. In this case we can write: $$\begin{aligned} -\vec{E}(t) &= \frac{E_0}{\sqrt{2}} \left(\cos(\omega t +\varphi)\vec{e}_x + \sin(\omega t +\varphi)\vec{e}_y\right)\\ -\vec{E}(t) &= \text{Re}\left(\vec{e}_+ E_0 e^{-i\omega t +\phi}\right)\\ -\vec{e}_\pm &= \frac{\vec{e}_x\pm i\vec{e}_y}{\sqrt{2}} -\end{aligned}$$ So light with polarization $\vec{\epsilon} = \vec{e}_+$ +circular polarization. In this case we can write: + +$$ +\vec{E}(t) = \frac{E_0}{\sqrt{2}} \left(\cos(\omega t +\varphi)\vec{e}_x + \sin(\omega t +\varphi)\vec{e}_y\right)\\ +\vec{E}(t) = \text{Re}\left(\vec{e}_+ E_0 e^{-i\omega t +\phi}\right)\\ +\vec{e}_\pm = \frac{\vec{e}_x\pm i\vec{e}_y}{\sqrt{2}} +$$ + +So light with polarization $\vec{\epsilon} = \vec{e}_+$ is called right-hand circular ($\sigma^+$) and $\vec{\epsilon} = \vec{e}_-$ is called left-hand circular ($\sigma^-$). Let us now evaluate the transition elements here. The dipole operator element boils now down to the evaluation of the integral: -$$\begin{aligned} -\ensuremath{\left\langle l',m',n'\right|}x+iy\ensuremath{\left|l,m,n\right\rangle} -\end{aligned}$$ As previously we can express the coupling term in -spherical coordinates: $$\begin{aligned} + +$$ +\left\langle l',m',n'\right|x+iy\left|l,m,n\right\rangle +$$ + +As previously we can express the coupling term in +spherical coordinates: + +$$ \frac{x+iy}{\sqrt{2}} = -r \sqrt{\frac{4\pi}{3}}Y_{11}(\theta, \varphi) -\end{aligned}$$ Evaluation of the integrals lead now to the rule the +$$ + +Evaluation of the integrals lead now to the rule the projection of the quantum number has to change $m' = m+1$. In a similiar fashion we find for left-hand circular light the selection rule -$m' = m - 1$. All the results are summed up in Fig. -[7](#852353){reference-type="ref" reference="852353"}. +$m' = m - 1$. -In the next lecture [@atoma] we will investigate the influence of +In the next lecture, we will investigate the influence of perturbative effects and see how the fine structure arises. diff --git a/amo/lecture6_pic1.png b/amo/lecture6_pic1.png new file mode 100644 index 0000000000000000000000000000000000000000..1054603d1eb56e01ebd0db844c8563e738b2cc24 GIT binary patch literal 62537 zcmbrmby(C}_%Dj0l+qnZZK|{O$&{0N4!_nFr4b5lzZJM`EQa{;;)kkHJ`p0()L+51KMvB77%CzoM(PKlB z9}tBgXbg2MI7>?{&glhZ_odS|21{c>xO%-ODKzoZX4~^&>nx9FIDiHxO@huAqr)$Hx zS+s~Z?HhSDcdy>R=^^j)L8If1A3fnS!&`cVw%{TjDS(eAjh~z~+?Vv2S(^K0EQu)A zl=Mdm_CFeKnFz*Y(fxN%`^wPb9ZjkP(9t|!n^y*yVbm9xVN7X$C6ATvSD@*=oKbX_9c)yk#B)_JHBr4T5*L}EdU9$NSzQPlp!kiq| z!)~Rw#OdtU6yg{^969>t4ZX8W2Q%u?Z_*1LR(zEQW7i*&-dPTz5u+u4Ud4DVBNY0S zc_yZkv%6!nyb zNqU+_cO>Ofgt%oux)X;`O0Vu&F{X$%UDgBMB^Gl6l1jwAM}J%82UaNEj$fh~GyA;k58cEFZvCLcs(C>@ 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zW#*)=4C>JH1&bjm&a?QS;|L>E>UqXI#heg9fnWIvP%XYB4D5QKE%H|N9n>iLi_7ziXzd zrb-SnMV}!e8b)t@I6U0!DmTu=ylhMmG!H9VAPsn64(ogF+__ISu&7l=zVPw?=ga-u z%?|oi&RI}Om?2Yu(8KAoG{%z7x#U}f%`}?9q4rTYoq#1=l}aCfqjWFY-E`9jjjLq#vN5%Bm|4Gr-L;B}Dp}O##S%|4J3PGcyGZn^^c;4-rCx_5?J+? zy*)SA`ughgxlzp~d3SEuZT~3+JQF$L5HRGxmGR7(`+6QedX1FwJw#$N_rjbmEzA)!~+IxN2S>(sJ$cUFR0NL(7sNl8gt3Lm@umdlB% ztC)QGfHGBwdLohUSr?)FK#IZ0z|cb1&_LJNAjHtf%E-XV#7Nt~z{Xn<_UOis2+&M&Ae%1qBFVQ_}A6dVf*a#9sM^OE&+ z6hJoX`>;9!^@zjtq@)%n7iAWdWaj5FxMUU=g7$K8bP?4{FKxjnB Date: Sat, 4 Jan 2025 22:27:22 +0100 Subject: [PATCH 13/13] Some better navigation --- layouts/LayoutDefault.tsx | 2 ++ pages/amo/+Page.tsx | 7 ++++++- 2 files changed, 8 insertions(+), 1 deletion(-) diff --git a/layouts/LayoutDefault.tsx b/layouts/LayoutDefault.tsx index 3be7f9b..18ca5e1 100644 --- a/layouts/LayoutDefault.tsx +++ b/layouts/LayoutDefault.tsx @@ -16,6 +16,8 @@ export default function LayoutDefault({ children }: { children: React.ReactNode

Website by fretchen

Welcome + Blog + AMO {""} {children} diff --git a/pages/amo/+Page.tsx b/pages/amo/+Page.tsx index 0a968b3..c7109e3 100644 --- a/pages/amo/+Page.tsx +++ b/pages/amo/+Page.tsx @@ -7,7 +7,12 @@ const App: React.FC = function () { console.log(blogs); return (
-

Welcome to my lecture notes on AMO!

+

AMO lecture notes

+

+ {" "} + Welcome, to my lecture notes on Atomic, Molecular and Optical physics that I prepared in my time in Heidelberg. + They consist of a total of 24 lectures, which I will recollect here again. +

);

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