Merge pull request #13 from fretchen/AMO

Amo

amo/lecture3.mdx

@@ -0,0 +1,365 @@
+---
+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, 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
+    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.
+
+<img src="./lecture3_pic1.png" width="90%">
+
+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.
+
+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., $NH_3$
+
+- 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:
+
+$$
+
+\hat{H} = \frac{\hbar}{2}\left( \begin{array}{cc} \Delta  & \Omega_x - i\Omega_y\\ \Omega_x +i\Omega_y & -\Delta \end{array} \right)
+$$
+
+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:
+
+$$
+
+\hat{H} = \mathbf{B}\cdot\hat{\mathbf{s}}\text{ with }\mathbf{B} = (\Omega_x, \Omega_y, \Delta)
+$$
+
+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 the Hamiltonian we obtain:
+
+$$
+ E_\pm = \pm \frac{\hbar}{2} \sqrt{\Delta^2+|\Omega|^2}
+$$
+
+The energies can be nicely summarized as in Fig.
+
+<img src="./lecture3_pic2.png" width="90%">
+
+The Eigenstates then read:
+
+$$
+\left|\psi_+\right\rangle=\cos\left(\frac{\theta}{2}\right) \mathrm{e}^{-i{\varphi}/{2}}\left|0\right\rangle+\sin\left(\frac{\theta}{2}\right) \mathrm{e}^{i{\varphi}/{2}}\left|1\right\rangle,
+$$
+
+$$
+\left|\psi_-\right\rangle=-\sin\left(\frac{\theta}{2}\right) \mathrm{e}^{-i{\varphi}/{2}}\left|0\right\rangle+\cos\left(\frac{\theta}{2}\right) \mathrm{e}^{i{\varphi}/{2}}\left|1\right\rangle,
+$$
+
+where
+
+$$
+
+\tan(\theta) = \frac{|\Omega|}{\Delta}
+$$
+
+## 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 below
+
+<img src="./lecture3_pic3.png" width="90%">
+
+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:
+
+$$
+i\hbar \frac{d}{dt}\left|\psi(t)\right\rangle=\hat{H}\left|\psi(t)\right\rangle,\\
+i \frac{d}{dt}\left(\begin{array}{c} c_0(t) \\ c_1(t) \end{array}\right) = \frac{1}{2}\left( \begin{array}{cc} \Delta & \Omega \\ \Omega^* & -\Delta \end{array} \right) \left(\begin{array}{c} c_0(t) \\ c_1(t) \end{array} \right).
+$$
+
+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
+
+$$
+ \left|\psi(t)\right\rangle=\lambda \mathrm{e}^{-i{E_+}t/{\hbar}} \left|\psi_+\right\rangle + \mu \mathrm{e}^{-i{E_-}t/{\hbar}} \left|\psi_-\right\rangle
+$$
+
+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 and express
+$\left|\varphi\right\rangle$ in terms of $|\psi_+$ and $|\psi_-$:
+
+$$
+\left|\psi(0)\right\rangle \overset{!}{=} \left|0\right\rangle\\
+  = \mathrm{e}^{i{\varphi}/{2}} \left[ \cos\left( \frac{\theta}{2}\right) \left|\psi_+\right\rangle-\sin\left(\frac{\theta}{2}\right)\left|\psi_-\right\rangle\right]
+$$
+
+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:
+
+$$
+ P_1(t)=\frac{1}{1+\left(\frac{\Delta}{|\Omega|}\right)^2}\sin^2\left(\sqrt{|\Omega|^2+\Delta^2}\frac{t}{2}\right)
+$$
+
+<img src="./lecture3_pic4.png" width="90%">
+
+### 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 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
+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 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. Wiley-VCH, 2006.