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Merge branch 'main' of github.com:niclaurenti/QFT_exercises
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niclaurenti committed Jun 11, 2024
2 parents 2a6034e + 462e2e0 commit 75d3cb3
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42 changes: 42 additions & 0 deletions .github/workflows/upload.yml
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name: Deploy PDF

on:
push:
branches:
- main

jobs:
build:
runs-on: ubuntu-latest

steps:
- name: Checkout code
uses: actions/checkout@v2

- name: Install LaTeX
run: |
sudo apt-get update
sudo apt-get install -y texlive-full
- name: Compile LaTeX to PDF
run: make

- name: Clone niclaurenti.github.io
uses: actions/checkout@v2
with:
repository: niclaurenti/niclaurenti.github.io
token: ${{ secrets.NICLAURENTI_TOKEN }}
path: niclaurenti.github.io

- name: Copy main.pdf
run: |
cp main.pdf niclaurenti.github.io/files/qft_exercises.pdf
- name: Commit and Push Changes
run: |
cd niclaurenti.github.io
git config user.email "[email protected]"
git config user.name "niclaurenti"
git add files/qft_exercises.pdf
git commit -m "Automated update of qft_exercises.pdf: ${{ github.event.head_commit.message }}"
git push
1 change: 1 addition & 0 deletions Makefile
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default:
pdflatex main.tex
pdflatex main.tex
2 changes: 1 addition & 1 deletion exercises/Ex2.tex
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\begin{split}
\phi(t,\vx) & = \int \frac{\rmd^{3} \vp}{(2\pi)^{3}} \big( A(\vp) \, e^{-i \Ep t + i \vp \cdot \vx} + B(\vp) \, e^{i \Ep t + i \vp \cdot \vx} \big)
\\
& = \int \frac{d^{3} \vp}{(2\pi)^{3}} \big( A(\vp) \, e^{-i \Ep t + i \vp \cdot \vx} + B(-\vp) \, e^{i \Ep t - i \vp \cdot \vx} \big) \,,
& = \int \frac{\rmd^{3} \vp}{(2\pi)^{3}} \big( A(\vp) \, e^{-i \Ep t + i \vp \cdot \vx} + B(-\vp) \, e^{i \Ep t - i \vp \cdot \vx} \big) \,,
\end{split}
\end{equation}
where in the second line we replaced the variable of integration as $\vp \mapsto - \vp$.
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2 changes: 1 addition & 1 deletion exercises/Ex30.tex
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\end{tikzpicture} ~ = ~ i\mathcal{M}_s - i\mathcal{M}_u \, .
\end{align}
Observe that the two diagrams have a difference of a minus sign
since in Eq.~\eqref{u_contraction} we moved $\sbar{\psi}(x)$ of two spaces to the right, while $\psi(x)$ must be moved of one space to the right, so we are left with a minus sign. The reason why also $\psi(x)$ must be moved is that all the external state contractions must be done in the same order\footnote{this is related to the fact that we can define $|f(p_1), \sbar{f}(p_2)\rangle \sim \acon_{\vp_1,s} \bcon_{\vp_2,s}\ket{0}$ or $|f(p_1), \sbar{f}(p_2)\rangle \sim \bcon_{\vp_1,s} \acon_{\vp_2,s} \ket{0}$. Both definitions are possible but differ by a minus sign.}: in Eq.~\eqref{s_contraction} we applied to the final particles first $\sbar{\psi}$ and then $\psi$.
since in Eq.~\eqref{u_contraction} we moved $\sbar{\psi}(x)$ of two spaces to the right, while $\psi(x)$ must be moved of one space to the right, so we are left with a minus sign. The reason why also $\psi(x)$ must be moved is that all the external state contractions must be done in the same order\footnote{this is related to the fact that we can define $|f(p_1), \sbar{f}(p_2)\rangle \sim \acon_{\vp_1,s_1} \bcon_{\vp_2,s_2}\ket{0}$ or $|f(p_1), \sbar{f}(p_2)\rangle \sim \bcon_{\vp_2,s_2} \acon_{\vp_1,s_1} \ket{0}$. Both definitions are possible but differ by a minus sign.}: in Eq.~\eqref{s_contraction} we applied to the final particles first $\sbar{\psi}$ and then $\psi$.
However, in Eq.~\eqref{u_contraction} the fields are placed in the opposite order.
Therefore, in order to be consistent with the choice done in Eq.~\eqref{s_contraction}, we have to exchange $\psi(x)$ and $\sbar{\psi}(y)$, getting an extra minus sign.
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