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\def\bop{\mathcal{B}}
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\def\ea{\end{eqnarray}}
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\def\ee{\end{equation}}
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\begin{document}

\title{Birefringence}
\maketitle

\section{Why Birefringence}

From WMAP Komatsu paper:
Since the temperature and E-mode polarization are
parity-even and the B-mode polarization is parity-odd, the TB and EB correlations should vanish in a universe that conserves parity. For this reason the TB and EB correlations are usually used to check for systematics, and not widely used as a cosmological probe.
Polarization of photons offers a powerful way of probing the cosmological parity violation, or the “cosmological birefringence”. Let us consider a parity-violating interaction term in the Lagrangian such as the Chern-Simons term ${\cal L}_{CS}= -\frac{1}{2}p_{\alpha}A_{\beta}{\tilde F}^{\alpha \beta}$ where $F^{\alpha \beta}$ and $A_{\beta}$ are the usual electromagnetic tensor and vector potential, ${\tilde F}^{\alpha \beta}$ is the dual tensor and $p_{\alpha}$ is an arbitrary timelike vector. This Chern-Simons term makes two polarization states of photons propagate with different group velocities, causing the polarization plane to rotate by an angle $\alpha$ (Carroll 1990). $p_{\alpha}$ can be sources by a scalar field. Such a field might have something to do with dark energy, for example. We are therefore looking at a potential parity-violating interaction between the visible section (i.e., photons) and dark sector (i.e., dark energy)

\section{Effects of Birefringence on the Power Spectra}

(from 1904.12440)
When polarisation angles are rotated uniformly over the sky by an angle $\alpha$, spherical harmonics coefficients of the observed E- and B-mode polarisation, denoted by "o", are related
to the intrinsic ones by
\ba
a_{\ell m}^{E, o} &=& a_{\ell m}^{E} \cos 2\alpha - a_{\ell m}^{B} \sin 2\alpha  \nonumber \\
a_{\ell m}^{B, o} &=& a_{\ell m}^{E} \sin 2\alpha + a_{\ell m}^{B} \cos 2\alpha
\ea
For the power spectra, we obtained
\ba
C^{EE,o}_{\ell} &=&  C^{EE}_{\ell} \cos^{2} 2\alpha + C^{BB}_{\ell} \sin^{2} 2\alpha -  C^{EB}_{\ell} \sin 4\alpha \nonumber \\
C^{BB,o}_{\ell} &=&  C^{EE}_{\ell} \sin^{2} 2\alpha + C^{BB}_{\ell} \cos^{2} 2\alpha + C^{EB}_{\ell} \sin 4\alpha \nonumber \\
C^{EB,o}_{\ell} &=&  C^{EE}_{\ell} \sin 2\alpha  \cos 2\alpha. - C^{BB}_{\ell} \sin 2\alpha  \cos 2\alpha + C^{EB}_{\ell}(\cos^{2} 2\alpha - \sin^{2} 2\alpha) \nonumber \\
C^{EB,o}_{\ell} &=&  \frac{\sin 4 \alpha}{2} (C^{EE}_{\ell} . - C^{BB}_{\ell} ) + \cos 4\alpha \ C^{EB}_{\ell}
\ea
We can relate the observed EB power spectrum to the observed EE and BB power spectrum
\ba
C^{EB,o}_{\ell} &=&  \frac{1}{2} ( C^{EE,o}_{\ell} - C^{BB,o}_{\ell} ) \tan 4\alpha  + \frac{ C^{EB}_{\ell}}{\cos 4 \alpha}
\ea
indeed
\ba
C^{EB,o}_{\ell} &=&  \frac{1}{2} ( C^{EE}_{\ell} \cos^{2} 2\alpha + C^{BB}_{\ell} \sin^{2} 2\alpha -  C^{EB}_{\ell} \sin 4\alpha - C^{EE}_{\ell} \sin^{2} 2\alpha - C^{BB}_{\ell} \cos^{2} 2\alpha - C^{EB}_{\ell} \sin 4\alpha \nonumber) \tan 4\alpha  \nonumber \\
 &+& \frac{ C^{EB}_{\ell}}{\cos 4 \alpha} \nonumber \\
&=&   \frac{1}{2} ( C^{EE}_{\ell}  - C^{BB}_{\ell} ) \sin 4\alpha + C^{EB}_{\ell}  (\frac{1 - \sin^{2}  4 \alpha}{\cos 4 \alpha} ) = \frac{1}{2} ( C^{EE}_{\ell}  - C^{BB}_{\ell} ) \sin 4\alpha + C^{EB}_{\ell}  \cos 4 \alpha
\ea

\section{Likelihood}

Assuming the intrinsic EB is equal to zero, we can therefore form two different $\chi^{2}$
\ba
\chi^{2} (\alpha)_{\rm data} &=& \sum_{bb'} \left[C^{EB,o}_{b} -  \frac{1}{2} ( C^{EE,o}_{b} - C^{BB,o}_{b} ) \tan 4\alpha \right] \Xi^{-1}_{bb'} \left[ C^{EB,o}_{b'} -  \frac{1}{2} ( C^{EE,o}_{b'} - C^{BB,o}_{b'} ) \tan 4\alpha \right] \\
\chi^{2} (\alpha)_{\rm th} &=& \sum_{bb'} \left[C^{EB, o}_{b} -  \frac{1}{2} ( C^{EE}_{b} - C^{BB}_{b} ) \sin 4\alpha \right] \Xi^{-1}_{bb'} \left[C^{EB, o}_{b} -  \frac{1}{2} ( C^{EE}_{b} - C^{BB}_{b} ) \sin 4\alpha \right]
\ea

\section{Fisher}
\ba
F_{\alpha \alpha} &=& \left \langle \frac{ \partial^{2} {\cal L} }{ \partial^{2}{\alpha} }\right \rangle \Bigr\rvert_{\alpha=0} = 4 \sum_{bb'}( C^{EE,o}_{b} - C^{BB,o}_{b} ) \Xi^{-1}_{bb'} ( C^{EE,o}_{b'} - C^{BB,o}_{b'} ) \\
\sigma_{\alpha} &=& \frac{1}{2} ( \sum_{bb'}( C^{EE,o}_{b} - C^{BB,o}_{b} ) \Xi^{-1}_{bb'} ( C^{EE,o}_{b'} - C^{BB,o}_{b'} ))^{-1/2}
\ea

\section{Separating CMB and foregrounds}

The foregrounds are sensitive to the instrumental miscalibrated angle but is not sensitive to the cosmic birefringence (because fg is at redshift zero) we can therefore separate CMB and fg.
Now let's call $\alpha$ the instrumental miscalibrated angle and $\beta$ the birefringence angle
\ba
a_{\ell m}^{E, o} &=& a^{\rm fg, E}_{\ell m} \cos (2\alpha) - a^{\rm fg, B}_{\ell m} \sin (2\alpha)  + a^{\rm CMB, E}_{\ell m} \cos (2\alpha + 2\beta) - a^{\rm CMB, B}_{\ell m} \sin (2\alpha + 2\beta) + a_{\ell m}^{E, N} \nonumber \\
a_{\ell m}^{B, o} &=& a^{\rm fg, E}_{\ell m} \sin (2\alpha) + a^{\rm fg, B}_{\ell m} \cos (2\alpha) + a^{\rm CMB, E}_{\ell m} \sin (2\alpha + 2\beta ) + a^{\rm CMB, B}_{\ell m} \cos (2\alpha + 2\beta) + a_{\ell m}^{B, N}
\ea
Assuming we form cross spectra
\ba
C^{EE,o}_{\ell} &=& \langle( a^{\rm fg, E}_{\ell m} \cos (2\alpha) - a^{\rm fg, B}_{\ell m} \sin (2\alpha)  + a^{\rm CMB, E}_{\ell m} \cos (2\alpha + 2\beta) - a^{\rm CMB, B}_{\ell m} \sin (2\alpha + 2\beta)) \nonumber \\
&\times& (a^{\rm fg, E}_{\ell m} \cos (2\alpha) - a^{\rm fg, B}_{\ell m} \sin (2\alpha)  + a^{\rm CMB, E}_{\ell m} \cos (2\alpha + 2\beta) - a^{\rm CMB, B}_{\ell m} \sin (2\alpha + 2\beta)) \rangle \nonumber \\
&=& C^{\rm EE, fg}_{\ell} \cos^{2} (2\alpha) + C^{\rm BB, fg}_{\ell} \sin^{2} (2\alpha) + C^{\rm EE, CMB}_{\ell} \cos^{2} (2\alpha + 2\beta)   + C^{\rm BB, CMB}_{\ell} \sin^{2} (2\alpha + 2\beta)  \nonumber \\
&-& 2 C^{\rm EB, fg}_{\ell} \cos (2\alpha)\sin (2\alpha) -  2 C^{\rm EB, CMB}_{\ell} \cos (2\alpha + 2\beta)\sin (2\alpha + 2\beta)  \nonumber \\
C^{BB,o}_{\ell}  &=& C^{\rm EE, fg}_{\ell} \sin^{2} (2\alpha) + C^{\rm BB, fg}_{\ell} \cos^{2} (2\alpha) + C^{\rm EE, CMB}_{\ell} \sin^{2} (2\alpha + 2\beta)   + C^{\rm BB, CMB}_{\ell} \cos^{2} (2\alpha + 2\beta)  \nonumber \\
&+& 2 C^{\rm EB, fg}_{\ell} \cos (2\alpha)\sin (2\alpha) +  2 C^{\rm EB, CMB}_{\ell} \cos (2\alpha + 2\beta)\sin (2\alpha + 2\beta)  \nonumber \\
\ea
And
\ba
C^{EB,o}_{\ell} &=&  \langle( a^{\rm fg, E}_{\ell m} \cos (2\alpha) - a^{\rm fg, B}_{\ell m} \sin (2\alpha)  + a^{\rm CMB, E}_{\ell m} \cos (2\alpha + 2\beta) - a^{\rm CMB, B}_{\ell m} \sin (2\alpha + 2\beta)) \nonumber \\
&\times& (a^{\rm fg, E}_{\ell m} \sin (2\alpha) + a^{\rm fg, B}_{\ell m} \cos (2\alpha)  + a^{\rm CMB, E}_{\ell m} \sin (2\alpha + 2\beta) + a^{\rm CMB, B}_{\ell m} \cos (2\alpha + 2\beta)) \rangle \nonumber \\
&=&  C^{\rm EE, fg}_{\ell} \cos (2\alpha) \sin (2\alpha)  - C^{\rm BB, fg}_{\ell} \sin (2\alpha) \cos (2\alpha) + C^{\rm EE, CMB}_{\ell} \cos (2\alpha +2 \beta) \sin (2\alpha +2 \beta)  \nonumber \\
&-& C^{\rm BB, CMB}_{\ell} \sin (2\alpha + 2\beta) \cos (2\alpha + 2\beta)  +  C^{\rm EB, fg}_{\ell} ( \cos^{2} (2\alpha) - \sin^{2} (2\alpha) )  \nonumber \\
&+&  C^{\rm EB, CMB}_{\ell} ( \cos^{2} (2\alpha + 2\beta) - \sin^{2} (2\alpha + 2 \beta)) \nonumber \\
C^{EB,o}_{\ell} &=& \frac{\sin(4\alpha)}{2}(C^{\rm EE, fg}_{\ell}  - C^{\rm BB, fg}_{\ell} ) + \frac{\sin(4\alpha + 4\beta)}{2}(C^{\rm EE, CMB}_{\ell}  - C^{\rm BB, CMB}_{\ell} )  \nonumber \\
&+& \cos(4\alpha) C^{\rm EB, fg}_{\ell} + \cos(4\alpha + 4\beta) C^{\rm EB, CMB}_{\ell}
\ea
Let's demonstrate that we can refactorize $C^{EB,o}_{\ell}$ as
\ba
C^{EB,o}_{\ell} &=& \frac{\tan(4\alpha)}{2} \left( C^{EB,o}_{\ell} - C^{BB,o}_{\ell} \right) + \frac{ \sin(4\beta)}{2 \cos(4\alpha)}( C^{\rm EE, CMB}_{\ell} - C^{\rm BB, CMB}_{\ell}) \nonumber \\
&+& \frac{1}{\cos(4\alpha)} C^{\rm EB, fg}_{\ell}  + \frac{ \cos(4\beta)}{ \cos(4\alpha)} C^{\rm EB, CMB}_{\ell}
\ea
We can expand
\ba
 \left( C^{EE,o}_{\ell} - C^{BB,o}_{\ell} \right) &=&   C^{\rm EE, fg}_{\ell} \cos^{2} (2\alpha) + C^{\rm BB, fg}_{\ell} \sin^{2} (2\alpha) + C^{\rm EE, CMB}_{\ell} \cos^{2} (2\alpha + 2\beta)   \nonumber \\
 &+& C^{\rm BB, CMB}_{\ell} \sin^{2} (2\alpha + 2\beta)  - 2 C^{\rm EB, fg}_{\ell} \cos (2\alpha)\sin (2\alpha) -  2 C^{\rm EB, CMB}_{\ell} \cos (2\alpha + 2\beta)\sin (2\alpha + 2\beta)  \nonumber \\
&-& C^{\rm EE, fg}_{\ell} \sin^{2} (2\alpha) - C^{\rm BB, fg}_{\ell} \cos^{2} (2\alpha) - C^{\rm EE, CMB}_{\ell} \sin^{2} (2\alpha + 2\beta)   - C^{\rm BB, CMB}_{\ell} \cos^{2} (2\alpha + 2\beta)  \nonumber \\
&-& 2 C^{\rm EB, fg}_{\ell} \cos (2\alpha)\sin (2\alpha) -  2 C^{\rm EB, CMB}_{\ell} \cos (2\alpha + 2\beta)\sin (2\alpha + 2\beta) \nonumber \\
&=&   \cos (4\alpha) (C^{\rm EE, fg}_{\ell}  - C^{\rm BB, fg}_{\ell} ) + \cos (4\alpha + 4\beta) (C^{\rm EE, CMB}_{\ell}  - C^{\rm BB, CMB}_{\ell} ) \nonumber \\
&-& 2 C^{\rm EB, fg}_{\ell} \sin (4\alpha) - 2 C^{\rm EB, CMB}_{\ell} \sin (4\alpha + 4\beta)
\ea
so that
\ba
C^{EB,o}_{\ell} &=& \frac{\sin(4\alpha)}{2}  (C^{\rm EE, fg}_{\ell}  - C^{\rm BB, fg}_{\ell} ) +  \frac{\tan(4\alpha)}{2} \cos (4\alpha + 4\beta) (C^{\rm EE, CMB}_{\ell}  - C^{\rm BB, CMB}_{\ell} ) \nonumber \\
&-& C^{\rm EB, fg}_{\ell} \sin (4\alpha )\tan(4\alpha)  - C^{\rm EB, CMB}_{\ell} \sin (4\alpha + 4\beta)\tan(4\alpha)  + \frac{ \sin(4\beta)}{2 \cos(4\alpha)}( C^{\rm EE, CMB}_{\ell} - C^{\rm BB, CMB}_{\ell}) \nonumber \\
&+& \frac{1}{\cos(4\alpha)} C^{\rm EB, fg}_{\ell}  + \frac{ \cos(4\beta)}{ \cos(4\alpha)} C^{\rm EB, CMB}_{\ell}  \nonumber \\
&=& \frac{\sin(4\alpha)}{2}  (C^{\rm EE, fg}_{\ell}  - C^{\rm BB, fg}_{\ell} )  +  (C^{\rm EE, CMB}_{\ell}  - C^{\rm BB, CMB}_{\ell} )\left( \frac{\tan(4\alpha)}{2} \cos (4\alpha + 4\beta) + \frac{ \sin(4\beta)}{2 \cos(4\alpha)}\right) \nonumber \\
&+& \left( \frac{1}{\cos(4\alpha)} - \sin (4\alpha )\tan(4\alpha) \right) C^{\rm EB, fg}_{\ell} + \left(\frac{ \cos(4\beta)}{ \cos(4\alpha)} - \sin (4\alpha + 4\beta)\tan(4\alpha) \right) C^{\rm EB, CMB}_{\ell}
\ea
Let's do some trigometry
\ba
 \frac{1}{\cos(4\alpha)} - \sin (4\alpha )\tan(4\alpha)  &=& \cos(4\alpha) \nonumber \\
 \frac{ \cos(4\beta)}{ \cos(4\alpha)} - \sin (4\alpha + 4\beta)\tan(4\alpha)   &=&  \frac{ \cos(4\beta)}{ \cos(4\alpha)}  - (\sin (4\alpha) \cos(4\beta)+ \sin (4\beta) \cos(4\alpha) )\tan(4\alpha) \nonumber \\
 &=&  \frac{ \cos(4\beta)}{ \cos(4\alpha)}   - \left(\sin^{2} (4\alpha)\frac{ \cos(4\beta)}{ \cos(4\alpha)}+ \sin (4\beta) \sin(4\alpha) \right) \nonumber \\
 &=&  \frac{ \cos(4\beta)}{ \cos(4\alpha)}(1 - \sin^{2} (4\alpha)) - \sin (4\beta) \sin(4\alpha) )  \nonumber \\
 &=&  \cos(4\beta) \cos(4\alpha) -  \sin (4\beta) \sin(4\alpha)  = \cos(4\beta + 4\alpha) \nonumber \\
  \frac{\tan(4\alpha)}{2} \cos (4\alpha + 4\beta) + \frac{ \sin(4\beta)}{2 \cos(4\alpha)} &=&    \frac{\tan(4\alpha)}{2} [\cos (4\alpha) \cos( 4\beta)- \sin(4\alpha) \sin( 4\beta) ]+ \frac{ \sin(4\beta)}{2 \cos(4\alpha)} \nonumber \\
 &=& \frac{\sin (4\alpha) \cos( 4\beta)}{2} - \frac{ \sin^{2}(4\alpha) \sin( 4\beta) }{2 \cos(4 \alpha)} + \frac{ \sin(4\beta)}{2 \cos(4\alpha)} \nonumber \\
 &=& \frac{\sin (4\alpha) \cos( 4\beta)}{2}  -  \frac{ \sin(4\beta)}{2 \cos(4\alpha)}  ( \sin^{2}(4\alpha)-1)  \nonumber \\
 &=& \frac{\sin (4\alpha) \cos( 4\beta)}{2}  + \frac{\cos (4\alpha) \sin( 4\beta)}{2} = \frac{ \sin(4\alpha + 4\beta)}{2}
\ea
With this, we demonstrate that
\ba
C^{EB,o}_{\ell} &=& \frac{\tan(4\alpha)}{2} \left( C^{EB,o}_{\ell} - C^{BB,o}_{\ell} \right) + \frac{ \sin(4\beta)}{2 \cos(4\alpha)}( C^{\rm EE, CMB}_{\ell} - C^{\rm BB, CMB}_{\ell}) \nonumber \\
&+& \frac{1}{\cos(4\alpha)} C^{\rm EB, fg}_{\ell}  + \frac{ \cos(4\beta)}{ \cos(4\alpha)} C^{\rm EB, CMB}_{\ell}
\ea
is indeed equal to
\ba
C^{EB,o}_{\ell} &=& \frac{\sin(4\alpha)}{2}(C^{\rm EE, fg}_{\ell}  - C^{\rm BB, fg}_{\ell} ) + \frac{\sin(4\alpha + 4\beta)}{2}(C^{\rm EE, CMB}_{\ell}  - C^{\rm BB, CMB}_{\ell} )  \nonumber \\
&+& \cos(4\alpha) C^{\rm EB, fg}_{\ell} + \cos(4\alpha + 4\beta) C^{\rm EB, CMB}_{\ell}
\ea
which complete the proof.

\section{A Multifrequency likelihood}
We can rewrite
\ba
C^{EE,o}_{\ell} &=& C^{\rm EE, fg}_{\ell} \cos^{2} (2\alpha) + C^{\rm BB, fg}_{\ell} \sin^{2} (2\alpha) + C^{\rm EE, CMB}_{\ell} \cos^{2} (2\alpha + 2\beta)   + C^{\rm BB, CMB}_{\ell} \sin^{2} (2\alpha + 2\beta)  \nonumber \\
&-& 2 C^{\rm EB, fg}_{\ell} \cos (2\alpha)\sin (2\alpha) -  2 C^{\rm EB, CMB}_{\ell} \cos (2\alpha + 2\beta)\sin (2\alpha + 2\beta)  \nonumber \\
C^{BB,o}_{\ell}  &=& C^{\rm EE, fg}_{\ell} \sin^{2} (2\alpha) + C^{\rm BB, fg}_{\ell} \cos^{2} (2\alpha) + C^{\rm EE, CMB}_{\ell} \sin^{2} (2\alpha + 2\beta)   + C^{\rm BB, CMB}_{\ell} \cos^{2} (2\alpha + 2\beta)  \nonumber \\
&+& 2 C^{\rm EB, fg}_{\ell} \cos (2\alpha)\sin (2\alpha) +  2 C^{\rm EB, CMB}_{\ell} \cos (2\alpha + 2\beta)\sin (2\alpha + 2\beta)  \nonumber \\
C^{EB,o}_{\ell} &=& \frac{\sin(4\alpha)}{2}(C^{\rm EE, fg}_{\ell}  - C^{\rm BB, fg}_{\ell} ) + \frac{\sin(4\alpha + 4\beta)}{2}(C^{\rm EE, CMB}_{\ell}  - C^{\rm BB, CMB}_{\ell} )  \nonumber \\
&+& \cos(4\alpha) C^{\rm EB, fg}_{\ell} + \cos(4\alpha + 4\beta) C^{\rm EB, CMB}_{\ell}
\ea
In a matricial form, let's neglect intrinsic EB correlation
\ba
\begin{pmatrix}
C^{\rm E_{i}E_{j},o}_{\ell} \cr
C^{\rm B_{i}B_{j},o}_{\ell}
\end{pmatrix}
&=&
{\bm R}(\alpha_{i}, \alpha_{j})
\begin{pmatrix}
C^{\rm E_{i}E_{j},fg}_{\ell} \cr
C^{\rm B_{i}B_{j},fg}_{\ell}
\end{pmatrix}
+
{\bm R}(\alpha_{i} + \beta, \alpha_{j}+ \beta)
\begin{pmatrix}
C^{\rm E_{i}E_{j},CMB}_{\ell} \cr
C^{\rm B_{i}B_{j},CMB}_{\ell}
\end{pmatrix} \nonumber \\
C^{\rm E_{i}B_{j},o}_{\ell} &=& \vec{R}^{T} (\alpha_{i}, \alpha_{j})
\begin{pmatrix}
C^{\rm E_{i}E_{j},fg}_{\ell} \cr
C^{\rm B_{i}B_{j},fg}_{\ell}
\end{pmatrix}
+
\vec{R}^{T} (\alpha_{i} + \beta, \alpha_{j}+ \beta )
\begin{pmatrix}
C^{\rm E_{i}E_{j},CMB}_{\ell} \cr
C^{\rm B_{i}B_{j},CMB}_{\ell}
\end{pmatrix}
\ea
With  ${\bm R}$ a rotation matrix and $\vec{R}$ a rotation vector
\ba
{\bm R}(\alpha_{i}, \alpha_{j})  =
\begin{pmatrix}
\cos(2\alpha_{i}) \cos(2\alpha_{j}) &
\sin(2\alpha_{i}) \sin(2\alpha_{j}) \cr
\sin(2\alpha_{i}) \sin(2\alpha_{j}) &
\cos(2\alpha_{i}) \cos(2\alpha_{j})
\end{pmatrix}
\ea
\ba
\vec{R} (\alpha_{i}, \alpha_{j}) =
\begin{pmatrix}
\cos(2\alpha_{i}) \sin(2\alpha_{j}) \cr
-\sin(2\alpha_{i}) \cos(2\alpha_{j})
\end{pmatrix}
\ea
We can simply re-express $C^{EB,o}_{\ell} $ as a function of $C^{EE,o}_{\ell} $ and $C^{BB,o}_{\ell} $ using the equation above
\ba
C^{\rm E_{i}B_{j},o}_{\ell} &=& \vec{R}^{T} (\alpha_{i}, \alpha_{j}) \left[ {\bm R}^{-1}(\alpha_{i}, \alpha_{j})
\begin{pmatrix}
C^{\rm E_{i}E_{j},o}_{\ell} \cr
C^{\rm B_{i}B_{j},o}_{\ell}
\end{pmatrix}
- {\bm R}^{-1}(\alpha_{i}, \alpha_{j})
{\bm R}(\alpha_{i} + \beta, \alpha_{j}+ \beta)
\begin{pmatrix}
C^{\rm E_{i}E_{j},CMB}_{\ell} \cr
C^{\rm B_{i}B_{j},CMB}_{\ell}
\end{pmatrix}  \right] \nonumber \\
&+&
\vec{R}^{T} (\alpha_{i} + \beta, \alpha_{j}+ \beta )
\begin{pmatrix}
C^{\rm E_{i}E_{j},CMB}_{\ell} \cr
C^{\rm B_{i}B_{j},CMB}_{\ell}
\end{pmatrix} \nonumber \\
&=& \vec{R}^{T} (\alpha_{i}, \alpha_{j})  {\bm R}^{-1}(\alpha_{i}, \alpha_{j})
\begin{pmatrix}
C^{\rm E_{i}E_{j},o}_{\ell} \cr
C^{\rm B_{i}B_{j},o}_{\ell}
\end{pmatrix}  \nonumber \\
&+& \left[  \vec{R}^{T} (\alpha_{i} + \beta, \alpha_{j}+ \beta ) -  \vec{R}^{T} (\alpha_{i}, \alpha_{j})  {\bm R}^{-1}(\alpha_{i}, \alpha_{j})  {\bm R}(\alpha_{i} + \beta, \alpha_{j}+ \beta) \right]
\begin{pmatrix}
C^{\rm E_{i}E_{j},CMB}_{\ell} \cr
C^{\rm B_{i}B_{j},CMB}_{\ell}
\end{pmatrix}
\ea
This can finally be rewritten
\ba
C^{\rm E_{i}B_{j},o}_{\ell} - {\bm A}
\begin{pmatrix}
C^{\rm E_{i}E_{j},o}_{\ell} \cr
C^{\rm B_{i}B_{j},o}_{\ell}
\end{pmatrix}
-   {\bm B}
\begin{pmatrix}
C^{\rm E_{i}E_{j},CMB}_{\ell} \cr
C^{\rm B_{i}B_{j},CMB}_{\ell}
\end{pmatrix}
= 0
\ea
with
\ba
{\bm A} =
\begin{pmatrix}
 \vec{R}^{T} (\alpha_{i}, \alpha_{j})  {\bm R}^{-1}(\alpha_{i}, \alpha_{j})
\end{pmatrix}
\ea
and
\ba
{\bm B} = \vec{R}^{T} (\alpha_{i} + \beta, \alpha_{j}+ \beta ) -  \vec{R}^{T} (\alpha_{i}, \alpha_{j})  {\bm R}^{-1}(\alpha_{i}, \alpha_{j})  {\bm R}(\alpha_{i} + \beta, \alpha_{j}+ \beta)
\ea
%A multifrequency likelihood can be written
%\ba
%-2\ln {\cal L} &=& \sum_{\ell}  ({\bm A}  \vec{C_{\ell}}^{\rm o}  - {\bm B}\vec{C_{\ell}}^{\rm CMB} )^{T} {\bm C}^{-1}({\bm A}  \vec{C_{\ell}}^{\rm o}  - {\bm B}\vec{C_{\ell}}^{\rm CMB} ) \nonumber \\
%{\bm C} &=& {\bm A} {\rm Cov}( \vec{C_{\ell}}^{\rm o}, \vec{C_{\ell}}^{\rm o}) {\bm A}^{T}
%\ea

\end{document}


\documentclass{article}

\usepackage{pl-syntax}

\title{Example Usage of \texttt{pl-syntax} Package}
\author{Andrew Hirsch \and Josh Acay}

\begin{document}
\maketitle

\section{The Simply Typed Lambda Calculus}

Read the comments in the \texttt{.tex} source for an explanation of how
the following is generated.

\begin{figure}[h]
  \begin{syntax}
    % Define a syntactic category with no right-hand side.
    \abstractCategory[Variables]{x, y, z}

    % Define a syntactic category from a set
    \categoryFromSet[Naturals]{n}{\mathbf{N}}

    % Start a category definition. Optional input is the category
    % description, the other declares the meta variable(s) you will
    % use to refer to this category.
    \category[Expressions]{e}
    % Each alternative is introduced in its own line.
    \alternative{x}
    \alternative{n}
    \alternative{\lambda x . e}
    \alternative{e_1 e_2}

    \category[Values]{v}
    \alternative{\lambda x . e}

    % You can add vertical spacing between categories to visually group them.
    \separate
    % You can pass the amount of space explicitly if you want to manually control it:
    % \separate[5ex]

    \category[Types]{\tau}
    \alternativeLine{\mathbf{Nat}}
    % You can place alternatives on a new line.
    \alternativeLine{\tau_1 \rightarrow \tau_2}

    \category[Contexts]{\Gamma}
    \alternative{x_1 : \tau_1, \ldots, x_n : \tau_n}
    % You can use words to describe more complicated properties of
    % definitions if you don't want to use BNF.
    \note{no repeats}
    \note{unordered}

    % Alternative layout if you want more space.
    \category[Expressions]{e}
    \groupleft{
      \alternative{x}
      \alternative{n}
    }
    \groupleft{
      \alternative{\lambda x . e}
      \alternative{e_1 e_2}
    }
  \end{syntax}

  \caption{Terms and Types of the Simply Typed Lambda Calculus}
  \label{fig:abstract-syntax}
\end{figure}

\end{document}


%\documentclass[UTF8]{ctexart} % use larger type; default would be 10pt
\documentclass[a4paper]{article}
\usepackage{../mqc}

% TikZFeynman
\newcommand{\qr}{quarter right}
\newcommand{\ql}{quarter left}
\newcommand{\hfr}{half right}
\newcommand{\hfl}{half left}
\newcommand{\el}{edge label}

%\usepackage{tikz}
%\usepackage[european]{circuitikz}
\usepackage{tikz-feynman}
%\usetikzlibrary{arrows.meta, bending, positioning}

\title{\textbf{Modern Quantum Chemistry, Szabo \& Ostlund}\\HW}
\author{wsr
\vspace{5pt}\\
}
\date{\today} % Activate to display a given date or no date (if empty),
         % otherwise the current date is printed

\begin{document}
% \boldmath

\maketitle

\tableofcontents

\newpage

\setcounter{section}{5}

\section{Many-body Perturbation Theory}
\subsection{RS Perturbation Theory}

\subsection{Diagrammatic Representation of RS Perturbation Theory}
\subsubsection{Diagrammatic Perturbation Theory for Two States}
\ex{6.1}
~\\
\tikzfeynmanset{
every vertex={black,dot},
}

\begin{minipage}{0.3\linewidth}
	\hfill
	\begin{tikzpicture}
	\begin{feynman}[small]
	\vertex (a);
	\vertex [below=of a] (b);
	\vertex [below=of b] (c);
	\vertex [below=of c] (d);
	\vertex [below=of d] (e);
	\diagram[inline=(c)]{
		(a) -- [fermion, quarter right, edge label'=1] (b)
		-- [fermion, quarter right, edge label'=1] (c)
		-- [fermion, quarter right, edge label'=1] (d)
		-- [fermion, quarter right, edge label'=1] (e)
		-- [fermion, quarter right, edge label'=2] (a),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
\begin{minipage}{0.7\linewidth}
	$ = (-1)^5 \dfrac{V_{12}V_{21}V_{11}^3}{(E_1^{(0)} - E_2^{(0)})^4} = -\dfrac{V_{12}V_{21}V_{11}^3}{(E_1^{(0)} - E_2^{(0)})^4} $\\
\end{minipage}
~\vspace{15pt}\\
\begin{minipage}{0.3\linewidth}
	\hfill
	\begin{tikzpicture}
	\begin{feynman}[small]
	\vertex (a);
	\vertex [below=of a] (b);
	\vertex [below=of b] (c);
	\vertex [below=of c] (d);
	\vertex [below=of d] (e);
	\diagram[inline=(c)]{
		(e) -- [fermion, quarter left, edge label=2] (d)
		-- [fermion, quarter left, edge label=2] (c)
		-- [fermion, quarter left, edge label=2] (b)
		-- [fermion, quarter left, edge label=2] (a)
		-- [fermion, quarter left, edge label=1] (e),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
\begin{minipage}{0.7\linewidth}
	$ = (-1)^2 \dfrac{V_{12}V_{21}V_{22}^3}{(E_1^{(0)} - E_2^{(0)})^4} = \dfrac{V_{12}V_{21}V_{22}^3}{(E_1^{(0)} - E_2^{(0)})^4} $\\
\end{minipage}

~\vspace{15pt}\\
\begin{minipage}{0.3\linewidth}
	\hfill
	\begin{tikzpicture}
	\begin{feynman}[small]
	\vertex (a);
	\vertex [below=of a] (b);
	\vertex [below=of b] (c);
	\vertex [below=of c] (d);
	\vertex [below=of d] (e);
	\diagram[inline=(c)]{
		(a) -- [fermion, quarter right, edge label'=1] (b)
		-- [fermion, quarter right, edge label'=1] (d)
		-- [fermion, quarter right, edge label'=1] (e)
		-- [fermion, quarter right, edge label'=2] (c)
		-- [fermion, quarter right, edge label'=2] (a),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
\begin{minipage}{0.7\linewidth}
	$ = (-1)^4 \dfrac{V_{12}V_{21}V_{11}^2 V_{22}}{(E_1^{(0)} - E_2^{(0)})^4} = \dfrac{V_{12}V_{21}V_{11}^2 V_{22}}{(E_1^{(0)} - E_2^{(0)})^4} $\\
\end{minipage}

~\vspace{15pt}\\
\begin{minipage}{0.3\linewidth}
	\hfill
	\begin{tikzpicture}
	\begin{feynman}[small]
	\vertex (a);
	\vertex [below=of a] (b);
	\vertex [below=of b] (c);
	\vertex [below=of c] (d);
	\vertex [below=of d] (e);
	\diagram[inline=(c)]{
		(a) -- [fermion, quarter left, edge label=1] (c)
		-- [fermion, quarter left, edge label=1] (e)
		-- [fermion, quarter left, edge label=2] (d)
		-- [fermion, quarter left, edge label=2] (b)
		-- [fermion, quarter left, edge label=2] (a),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
\begin{minipage}{0.7\linewidth}
	$ = (-1)^3 \dfrac{V_{12}V_{21}V_{11} V_{22}^2}{(E_1^{(0)} - E_2^{(0)})^4} = -\dfrac{V_{12}V_{21}V_{11} V_{22}^2}{(E_1^{(0)} - E_2^{(0)})^4} $\\
\end{minipage}

\newpage
Similarly,\\
%~\vspace{15pt}\\
\begin{minipage}{0.2\linewidth}
	\hfill
	\begin{tikzpicture}
	\begin{feynman}[small]
	\vertex (a);
	\vertex [below=of a] (b);
	\vertex [below=of b] (c);
	\vertex [below=of c] (d);
	\vertex [below=of d] (e);
	\diagram[inline=(c)]{
		(a) -- [fermion, quarter right, edge label'=1] (c)
		-- [fermion, quarter right, edge label'=1] (d)
		-- [fermion, quarter right, edge label'=1] (e)
		-- [fermion, quarter right, edge label'=2] (b)
		-- [fermion, quarter right, edge label'=2] (a),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
,
\begin{minipage}{0.15\linewidth}
	%\hfill
	\begin{tikzpicture}
	\begin{feynman}[small]
	\vertex (a);
	\vertex [below=of a] (b);
	\vertex [below=of b] (c);
	\vertex [below=of c] (d);
	\vertex [below=of d] (e);
	\diagram[inline=(c)]{
		(a) -- [fermion, quarter right, edge label'=1] (b)
		-- [fermion, quarter right, edge label'=1] (c)
		-- [fermion, quarter right, edge label'=1] (e)
		-- [fermion, quarter right, edge label'=2] (d)
		-- [fermion, quarter right, edge label'=2] (a),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
\begin{minipage}{0.6\linewidth}
	$ = \dfrac{V_{12}V_{21}V_{11}^2 V_{22}}{(E_1^{(0)} - E_2^{(0)})^4} $\\
\end{minipage}

~\vspace{15pt}\\
\begin{minipage}{0.2\linewidth}
	\hfill
	\begin{tikzpicture}
	\begin{feynman}[small]
	\vertex (a);
	\vertex [below=of a] (b);
	\vertex [below=of b] (c);
	\vertex [below=of c] (d);
	\vertex [below=of d] (e);
	\diagram[inline=(c)]{
		(a) -- [fermion, quarter left, edge label=1] (b)
		-- [fermion, quarter left, edge label=1] (e)
		-- [fermion, quarter left, edge label=2] (d)
		-- [fermion, quarter left, edge label=2] (c)
		-- [fermion, quarter left, edge label=2] (a),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
,
\begin{minipage}{0.15\linewidth}
	\begin{tikzpicture}
	\begin{feynman}[small]
	\vertex (a);
	\vertex [below=of a] (b);
	\vertex [below=of b] (c);
	\vertex [below=of c] (d);
	\vertex [below=of d] (e);
	\diagram[inline=(c)]{
		(a) -- [fermion, quarter left, edge label=1] (d)
		-- [fermion, quarter left, edge label=1] (e)
		-- [fermion, quarter left, edge label=2] (c)
		-- [fermion, quarter left, edge label=2] (b)
		-- [fermion, quarter left, edge label=2] (a),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
\begin{minipage}{0.6\linewidth}
	$ = -\dfrac{V_{12}V_{21}V_{11} V_{22}^2}{(E_1^{(0)} - E_2^{(0)})^4} $\\
\end{minipage}
~\vspace{10pt}\\
thus, the sum of above terms is
\begin{align}
\dfrac{V_{12}V_{21}(V_{22}^3 - V_{11}^3)}{(E_1^{(0)} - E_2^{(0)})^4} + 3\times\dfrac{V_{12}V_{21}(V_{11}^2 V_{22} - V_{11}V_{22}^2)}{(E_1^{(0)} - E_2^{(0)})^4}
= \dfrac{V_{12}V_{21}(V_{22} - V_{11})^3}{(E_1^{(0)} - E_2^{(0)})^4}
\end{align}

\subsubsection{Diagrammatic Perturbation Theory for $ N $ States}
\ex{6.2}
The 4th-order perturbation energy of state $ i $ can be expressed as
\begin{align}
&\sum_{k,n,m\neq i} \dfrac{V_{ki}V_{nk}V_{mn}V_{im}}{(E_i^{(0)}-E_k^{(0)})(E_i^{(0)}-E_n^{(0)})(E_i^{(0)}-E_m^{(0)})} + \sum_{n\neq i} \dfrac{V_{ii}^2 V_{ni}V_{in}}{(E_i^{(0)}-E_n^{(0)})^3} - \sum_{m,n\neq i} \dfrac{V_{ii}V_{mi}V_{in}V_{nm}}{(E_i^{(0)}-E_m^{(0)})^2(E_i^{(0)}-E_n^{(0)})} \notag\\
&- \sum_{m,n\neq i} \dfrac{V_{ii}V_{ni}V_{im}V_{mn}}{(E_i^{(0)}-E_m^{(0)})^2(E_i^{(0)}-E_n^{(0)})} - \sum_{m,n\neq i} \dfrac{V_{mi}V_{im}V_{in}V_{ni}}{(E_i^{(0)}-E_m^{(0)})(E_i^{(0)}-E_n^{(0)})(2E_i^{(0)}-E_n^{(0)}-E_m^{(0)})} \notag\\
&- \sum_{m,n\neq i} \dfrac{V_{mi}V_{im}V_{in}V_{ni}}{(E_i^{(0)}-E_n^{(0)})^2(2E_i^{(0)}-E_n^{(0)}-E_m^{(0)})} \notag\\
= &\sum_{k,n,m\neq i} \dfrac{V_{ki}V_{nk}V_{mn}V_{im}}{(E_i^{(0)}-E_k^{(0)})(E_i^{(0)}-E_n^{(0)})(E_i^{(0)}-E_m^{(0)})} + \sum_{n\neq i} \dfrac{V_{ii}^2 V_{ni}V_{in}}{(E_i^{(0)}-E_n^{(0)})^3} - 2\sum_{m,n\neq i} \dfrac{V_{ii}V_{mi}V_{in}V_{nm}}{(E_i^{(0)}-E_m^{(0)})^2(E_i^{(0)}-E_n^{(0)})} \notag\\
&%- \sum_{m,n\neq i} \dfrac{V_{ii}V_{ni}V_{im}V_{mn}}{(E_i^{(0)}-E_m^{(0)})^2(E_i^{(0)}-E_n^{(0)})}
- \sum_{m,n\neq i} \dfrac{V_{mi}V_{im}V_{in}V_{ni}}{(E_i^{(0)}-E_m^{(0)})(E_i^{(0)}-E_n^{(0)})^2}
\end{align}
while
\begin{equation}\label{key}
\Braket{n | \sH | \Psi_i^{(3)}} + \Braket{n | \mathscr{V} | \Psi_i^{(2)}} = E_i^{(0)} \Braket{n|\Psi_i^{(3)}} + E_i^{(1)} \Braket{n|\Psi_i^{(2)}} + E_i^{(2)} \Braket{n|\Psi_i^{(1)}}
\end{equation}
\begin{align}\label{key}
(E_i^{(0)} - E_n^{(0)})\Braket{n | \Psi_i^{(3)}}
&= \Braket{n | \mathscr{V} | \Psi_i^{(2)}} - E_i^{(1)} \Braket{n|\Psi_i^{(2)}} - E_i^{(2)} \Braket{n|\Psi_i^{(1)}} \notag\\
&= \Braket{n | \mathscr{V} | \Psi_i^{(2)}}
- E_i^{(1)} \dfrac{\Braket{n | \mathscr{V} | \Psi_i^{(1)}} - E_i^{(1)}\Braket{n|\Psi_i^{(1)}}}{E_i^{(0)} - E_n^{(0)}}
- E_i^{(2)} \Braket{n|\Psi_i^{(1)}} \notag\\
&= \Braket{n | \mathscr{V} | \Psi_i^{(2)}}
- E_i^{(1)} \dfrac{\Braket{n | \mathscr{V} | \Psi_i^{(1)}}}{E_i^{(0)} - E_n^{(0)}}
+ \qty[E_i^{(1)}]^2 \dfrac{\Braket{n|\mathscr{V}|i}}{\qty[E_i^{(0)} - E_n^{(0)}]^2}
- E_i^{(2)} \dfrac{\Braket{n|\mathscr{V}|i}}{E_i^{(0)} - E_n^{(0)}}
\end{align}
\begin{align}
E_i^{(4)} &= \Braket{i | \mathscr{V} | \Psi_i^{(3)}} \notag\\
&= \sum_{n\neq i} \dfrac{\Braket{i | \mathscr{V} | n}}{E_i^{(0)} - E_n^{(0)}} \qty{\Braket{n | \mathscr{V} | \Psi_i^{(2)}}
- E_i^{(1)} \dfrac{\Braket{n | \mathscr{V} | \Psi_i^{(1)}}}{E_i^{(0)} - E_n^{(0)}}
+ \qty[E_i^{(1)}]^2 \dfrac{\Braket{n|\mathscr{V}|i}}{\qty[E_i^{(0)} - E_n^{(0)}]^2}
- E_i^{(2)} \dfrac{\Braket{n|\mathscr{V}|i}}{E_i^{(0)} - E_n^{(0)}} }\notag\\
&= \sum_{n\neq i} \dfrac{\Braket{i | \mathscr{V} | n}}{E_i^{(0)} - E_n^{(0)}} \Braket{n | \mathscr{V} | \Psi_i^{(2)}}
- E_i^{(1)} \sum_{n\neq i} \dfrac{\Braket{i | \mathscr{V} | n}}{\qty[E_i^{(0)} - E_n^{(0)}]^2} \Braket{n | \mathscr{V} | \Psi_i^{(1)}} \notag\\
&{}\quad +  \qty[E_i^{(1)}]^2 \sum_{n\neq i} \dfrac{V_{in}V_{ni}}{\qty[E_i^{(0)} - E_n^{(0)}]^3}
-  E_i^{(2)} \sum_{n\neq i} \dfrac{V_{in}V_{ni}}{\qty[E_i^{(0)} - E_n^{(0)}]^2} \notag\\
&= \sum_{n,m\neq i} \dfrac{\Braket{i | \mathscr{V} | n}}{E_i^{(0)} - E_n^{(0)}} \Braket{n | \mathscr{V} | m} \Braket{m |\Psi_i^{(2)}}
- E_i^{(1)} \sum_{n,m\neq i} \dfrac{\Braket{i | \mathscr{V} | n}\Braket{n | \mathscr{V} | m}\Braket{m | \mathscr{V} | i} }{\qty[E_i^{(0)} - E_n^{(0)}]^2\qty[E_i^{(0)} - E_m^{(0)}]}  \notag\\
&{}\quad +  \qty[E_i^{(1)}]^2 \sum_{n\neq i} \dfrac{V_{in}V_{ni}}{\qty[E_i^{(0)} - E_n^{(0)}]^3}
-  E_i^{(2)} \sum_{n\neq i} \dfrac{V_{in}V_{ni}}{\qty[E_i^{(0)} - E_n^{(0)}]^2} \notag\\
&= \sum_{n,m\neq i} \dfrac{V_{in} V_{nm}}{E_i^{(0)} - E_n^{(0)}} \dfrac{\Braket{m | \mathscr{V} | \Psi_i^{(1)}} - E_i^{(1)}\Braket{m|\Psi_i^{(1)}}}{E_i^{(0)} - E_m^{(0)}}
- E_i^{(1)} \sum_{n,m\neq i} \dfrac{V_{in}V_{nm}V_{mi} }{\qty[E_i^{(0)} - E_n^{(0)}]^2\qty[E_i^{(0)} - E_m^{(0)}]}  \notag\\
&{}\quad +  \qty[E_i^{(1)}]^2 \sum_{n\neq i} \dfrac{V_{in}V_{ni}}{\qty[E_i^{(0)} - E_n^{(0)}]^3}
-  E_i^{(2)} \sum_{n\neq i} \dfrac{V_{in}V_{ni}}{\qty[E_i^{(0)} - E_n^{(0)}]^2} \notag\\
&= \sum_{n,m,k\neq i} \dfrac{V_{in} V_{nm}}{E_i^{(0)} - E_n^{(0)}} \dfrac{\Braket{m | \mathscr{V} | k} \Braket{k| \mathscr{V} | i} }{\qty[E_i^{(0)} - E_m^{(0)}]\qty[E_i^{(0)} - E_k^{(0)}]}
- E_i^{(1)}\sum_{n,m\neq i} \dfrac{V_{in} V_{nm}}{E_i^{(0)} - E_n^{(0)}} \dfrac{ \Braket{m| \mathscr{V} |i}}{\qty[E_i^{(0)} - E_m^{(0)}]^2} \notag\\
&{}\quad- E_i^{(1)} \sum_{n,m\neq i} \dfrac{V_{in}V_{nm}V_{mi} }{\qty[E_i^{(0)} - E_n^{(0)}]^2\qty[E_i^{(0)} - E_m^{(0)}]}
 +  \qty[E_i^{(1)}]^2 \sum_{n\neq i} \dfrac{V_{in}V_{ni}}{\qty[E_i^{(0)} - E_n^{(0)}]^3}
-  E_i^{(2)} \sum_{n\neq i} \dfrac{V_{in}V_{ni}}{\qty[E_i^{(0)} - E_n^{(0)}]^2} \notag\\
&= \sum_{n,m,k\neq i} \dfrac{V_{in} V_{nm}V_{mk}V_{ki}}{\qty[E_i^{(0)} - E_n^{(0)}]\qty[E_i^{(0)} - E_m^{(0)}]\qty[E_i^{(0)} - E_k^{(0)}]}
- 2V_{ii}\sum_{n,m\neq i} \dfrac{V_{in} V_{nm} V_{mi}}{\qty[E_i^{(0)} - E_n^{(0)}]\qty[E_i^{(0)} - E_m^{(0)}]^2}  \notag\\
&{}\quad %- V_{ii} \sum_{n,m\neq i} \dfrac{V_{in}V_{nm}V_{mi} }{\qty[E_i^{(0)} - E_n^{(0)}]^2\qty[E_i^{(0)} - E_m^{(0)}]}
+  V_{ii}^2 \sum_{n\neq i} \dfrac{V_{in}V_{ni}}{\qty[E_i^{(0)} - E_n^{(0)}]^3}
-  \sum_{m\neq i} \dfrac{V_{mi}V_{im}}{\qty[E_i^{(0)} - E_m^{(0)}]} \sum_{n\neq i} \dfrac{V_{in}V_{ni}}{\qty[E_i^{(0)} - E_n^{(0)}]^2}
\end{align}
which agrees with diagrammatic results above.

\subsubsection{Summation of Diagrams}

\newpage
\subsection{Orbital Perturbation Theory: One-Particle Perturbations}
\ex{6.3}
Since $ n\neq 0 $ and $ v(i) $ is one-particle operator, $ n $ must be single-excited, i.e. $ \ket{\Psi_a^r} $. Thus,
\begin{align}
E_0^{(2)} &= \sum_{a,r} \dfrac{\abs{\Braket{\Psi_0 | \sum_i v(i) | \Psi_a^r}}^2}{\Braket{\Psi_0 | \sH | \Psi_0} - \Braket{\Psi_a^r| \sH | \Psi_a^r}} \notag\\
&= \sum_{a,r} \dfrac{v_{ar} v_{ra}}{\sum_b \varepsilon_b^{(0)} - \qty(\sum_{b\neq a}\varepsilon_b^{(0)} + \varepsilon_r^{(0)})} \notag\\
&= \sum_{a,r} \dfrac{v_{ar} v_{ra}}{\varepsilon_a^{(0)} - \varepsilon_r^{(0)}}
\end{align}

\ex{6.4}
Eq 6.15 in textbook gives
\begin{align}
E_i^{(3)} &= \sum_{n,m\neq i} \dfrac{\Braket{i|\mathscr{V}|n} \Braket{n|\mathscr{V}|m} \Braket{m|\mathscr{V}|i}}{(E_i^{(0)}-E_n^{(0)}) (E_i^{(0)}-E_m^{(0)})}
- E_i^{(1)}\sum_{n\neq i} \dfrac{\abs{\Braket{i|\mathscr{V}|n}}^2 }{(E_i^{(0)}-E_n^{(0)})^2} \notag\\
&= A_i^{(3)} + B_i^{(3)}
\end{align}
\subex{a.}
\begin{align}
B_0^{(3)} &= - E_0^{(1)}\sum_{n\neq 0} \dfrac{\abs{\Braket{\Psi_0|\mathscr{V}|n}}^2 }{(E_0^{(0)}-E_n^{(0)})^2} \notag\\
&= -\sum_b v_{bb} \sum_{a,r}\dfrac{v_{ar}v_{ra}}{(\varepsilon_a^{(0)} - \varepsilon_r^{(0)})^2} \notag\\
&= - \sum_{a,b,r}\dfrac{v_{aa} v_{br}v_{rb}}{(\varepsilon_b^{(0)} - \varepsilon_r^{(0)})^2}
\end{align}
\subex{b.}
\begin{align}
A_0^{(3)} &= \sum_{n,m\neq 0} \dfrac{\Braket{\Psi_0|\mathscr{V}|n} \Braket{n|\mathscr{V}|m} \Braket{m|\mathscr{V}|\Psi_0}}{(E_0^{(0)}-E_n^{(0)}) (E_0^{(0)}-E_m^{(0)})} \notag\\
&= \sum_{a,r,b,s} \dfrac{\Braket{\Psi_0|\mathscr{V}|\Psi_a^r} \Braket{\Psi_a^r|\mathscr{V}|\Psi_b^s} \Braket{\Psi_b^s|\mathscr{V}|\Psi_0}}{(\varepsilon_a^{(0)} - \varepsilon_r^{(0)}) (\varepsilon_b^{(0)} - \varepsilon_s^{(0)})} \notag\\
&= \sum_{a,r,b,s} \dfrac{v_{ar}v_{sb} \Braket{\Psi_a^r|\mathscr{V}|\Psi_b^s} }{(\varepsilon_a^{(0)} - \varepsilon_r^{(0)}) (\varepsilon_b^{(0)} - \varepsilon_s^{(0)})}
\end{align}
\subex{c.}
Clearly, if $ a\neq b, r\neq s $
\begin{align}
\Braket{\Psi_a^r | \mathscr{V} | \Psi_b^s} = 0
\end{align}
If $ a=b, r\neq s $,
\begin{align}
\Braket{\Psi_a^r | \mathscr{V} | \Psi_b^s}
&= \Braket{r| v |s} \notag\\
&= v_{rs}
\end{align}
If $ a\neq b, r= s $,
\begin{align}
\Braket{\Psi_a^r | \mathscr{V} | \Psi_b^s}
&= \Braket{\Psi_a^r | \mathscr{V} | \Psi_b^r} \notag\\
&= \Braket{\Psi_a^r | \mathscr{V} | -\Psi_{ab}^{ra}} \notag\\
&= -\Braket{b| v |a} \notag\\
&= -v_{ba}
\end{align}
If $ a= b, r= s $,
\begin{align}
\Braket{\Psi_a^r | \mathscr{V} | \Psi_b^s}
&= \Braket{\Psi_a^r | \mathscr{V} | \Psi_a^r} \notag\\
&= \sum_c v_{cc} - v_{aa} + v_{rr}
\end{align}

\subex{d.}
\begin{align}
E_0^{(3)} &= A_0^{(3)} + B_0^{(3)} \notag\\
&= \sum_{a,r,b,s}
\dfrac{v_{ar}v_{sb} \Braket{\Psi_a^r|\mathscr{V}|\Psi_b^s} }{(\varepsilon_a^{(0)} - \varepsilon_r^{(0)}) (\varepsilon_b^{(0)} - \varepsilon_s^{(0)})}
- \sum_{a,b,r}\dfrac{v_{aa} v_{br}v_{rb}}{(\varepsilon_b - \varepsilon_r)^2}
\notag\\
&=  \sum_{a,r\neq s}
\dfrac{v_{ar}v_{sa} v_{rs} }{(\varepsilon_a^{(0)} - \varepsilon_r^{(0)}) (\varepsilon_a^{(0)} - \varepsilon_s^{(0)})}
+ \sum_{a\neq b,r}
\dfrac{v_{ar}v_{rb}(-v_{ba}) }{(\varepsilon_a^{(0)} - \varepsilon_r^{(0)}) (\varepsilon_b^{(0)} - \varepsilon_r^{(0)})}
\notag\\
&\quad {} + \sum_{a,r}
\dfrac{v_{ar}v_{ra} (\sum_c v_{cc} - v_{aa} + v_{rr}) }{(\varepsilon_a^{(0)} - \varepsilon_r^{(0)})^2}
- \sum_{a,b,r}\dfrac{v_{aa} v_{br}v_{rb}}{(\varepsilon_b^{(0)} - \varepsilon_r^{(0)})^2}
\notag\\
&=  \sum_{a,r\neq s}
\dfrac{v_{ar}v_{sa} v_{rs} }{(\varepsilon_a^{(0)} - \varepsilon_r^{(0)}) (\varepsilon_a^{(0)} - \varepsilon_s^{(0)})}
+ \sum_{a\neq b,r}
\dfrac{v_{ar}v_{rb}(-v_{ba}) }{(\varepsilon_a^{(0)} - \varepsilon_r^{(0)}) (\varepsilon_b^{(0)} - \varepsilon_r^{(0)})}
\notag\\
&\quad {} + \sum_{a,r}
\dfrac{v_{ar}v_{ra} (\sum_c v_{cc} - v_{aa} + v_{rr}) }{(\varepsilon_a^{(0)} - \varepsilon_r^{(0)})^2}
- \sum_{a,r}\dfrac{\sum_c v_{cc} v_{ar}v_{ra}}{(\varepsilon_a^{(0)} - \varepsilon_r^{(0)})^2}
\notag\\
&=  \sum_{a,r\neq s}
\dfrac{v_{ar}v_{sa} v_{rs} }{(\varepsilon_a^{(0)} - \varepsilon_r^{(0)}) (\varepsilon_a^{(0)} - \varepsilon_s^{(0)})}
+ \sum_{a\neq b,r}
\dfrac{v_{ar}v_{rb}(-v_{ba}) }{(\varepsilon_a^{(0)} - \varepsilon_r^{(0)}) (\varepsilon_b^{(0)} - \varepsilon_r^{(0)})}
+ \sum_{a,r}
\dfrac{v_{ar}v_{ra} ( - v_{aa} + v_{rr}) }{(\varepsilon_a^{(0)} - \varepsilon_r^{(0)})^2}
\notag\\
&=  \sum_{a,r,s}
\dfrac{v_{ar}v_{sa} v_{rs} }{(\varepsilon_a^{(0)} - \varepsilon_r^{(0)}) (\varepsilon_a^{(0)} - \varepsilon_s^{(0)})}
- \sum_{a,b,r}
\dfrac{v_{ar}v_{rb}v_{ba} }{(\varepsilon_a^{(0)} - \varepsilon_r^{(0)}) (\varepsilon_b^{(0)} - \varepsilon_r^{(0)})}
\end{align}

\subex{e.}
That's obvious.

\ex{6.5}
Since $ a,b $ run over all $ n $ occupied orbitals $ i,j $ and $ r $ runs over all $ n $ unoccupied orbitals $ k^* $, we have
\begin{align}
-2\sum_{a,b,r}^{N/2} \dfrac{v_{ra}v_{ab}v_{br}}{(\varepsilon_a^{(0)} - \varepsilon_r^{(0)}) (\varepsilon_b^{(0)} - \varepsilon_r^{(0)})}
&= -\dfrac{2}{(2\beta)^2} \sum_i^n \sum_j^n \sum_k^n \Braket{i|v|j} \Braket{j|v|k^*} \Braket{k^*|v|i} \notag\\
&= -\dfrac{2}{(2\beta)^2} \sum_i^3
\mqty[\Braket{i|v|i+1} \Braket{i+1|v|(i+2)^*} \Braket{(i+2)^*|v|i} \\
+ \Braket{i|v|i+2} \Braket{i+2|v|(i+1)^*} \Braket{(i+1)^*|v|i} ] \notag\\
&= -\dfrac{2}{(2\beta)^2} \sum_i^3 \qty[(\beta/2)(\beta/2)(-\beta/2) + (\beta/2)(-\beta/2)(\beta/2)] \notag\\
&= -\dfrac{2}{(2\beta)^2} \times 3\times (-\beta^3/4) \notag\\
&= 3\beta/8
\end{align}

\ex{6.6}
\subex{a.}
Using the general expression, we get
\begin{align}
\mathscr{E}_0 &= 6\alpha - 2\sum_{j=-1}^1 (\beta_1^2 + \beta_2^2 + 2\beta_1\beta_2\cos\dfrac{2j\pi}{3})^{1/2} \notag\\
&= 6\alpha - 2(\beta_1^2 + \beta_2^2 + 2\beta_1\beta_2\cos\dfrac{-2\pi}{3})^{1/2}
- 2(\beta_1^2 + \beta_2^2 + 2\beta_1\beta_2\cos 0)^{1/2}
- 2(\beta_1^2 + \beta_2^2 + 2\beta_1\beta_2\cos\dfrac{2\pi}{3})^{1/2}
\notag\\
&= 6\alpha - 2(\beta_1^2 + \beta_2^2 - \beta_1\beta_2)^{1/2}
- 2(\beta_1^2 + \beta_2^2 + 2\beta_1\beta_2)^{1/2}
- 2(\beta_1^2 + \beta_2^2 - \beta_1\beta_2)^{1/2} \notag\\
&= 6\alpha
- 2\abs{\beta_1 + \beta_2}
- 4(\beta_1^2 + \beta_2^2 - \beta_1\beta_2)^{1/2} \notag\\
&= 6\alpha + 2(\beta_1 + \beta_2)
- 4(\beta_1^2 + \beta_2^2 - \beta_1\beta_2)^{1/2}
\end{align}

Using H\"uckel matrix:
\begin{align}
\vb{H} = \mqty(
\alpha  & \beta_1 & 0       & 0       & 0       & \beta_2\\
\beta_1 & \alpha  & \beta_2 & 0       & 0       & 0      \\
0       & \beta_2 & \alpha  & \beta_1 & 0       & 0      \\
0       & 0       & \beta_1 & \alpha  & \beta_2 & 0      \\
0       & 0       & 0       & \beta_2 & \alpha  & \beta_1\\
\beta_2 & 0       & 0       & 0       & \beta_1 & \alpha
)
\end{align}
%Diagonalize $ \vb{H} - \bm\varepsilon $, we get
Eigenvalues of $ \vb{H} $ are
\begin{align}\label{key}
&\alpha + (\beta_1 + \beta_2), \notag\\
&\alpha - \sqrt{\beta_1^2 + \beta_2^2 - \beta_1\beta_2} \;\text{ (2-fold)}, \notag\\
&\alpha + \sqrt{\beta_1^2 + \beta_2^2 - \beta_1\beta_2} \;\text{ (2-fold)}, \notag\\
&\alpha - (\beta_1 + \beta_2),
\end{align}
thus
\begin{align}
\mathscr{E}_0 &= 2\qty[\alpha + (\beta_1 + \beta_2)] + 4\qty[\alpha - \sqrt{\beta_1^2 + \beta_2^2 - \beta_1\beta_2}] \notag\\
&= 6\alpha + 2(\beta_1 + \beta_2) - 4\sqrt{\beta_1^2 + \beta_2^2 - \beta_1\beta_2}
\end{align}

\subex{b.}
\begin{align}
E_R &= \mathscr{E}_0 - (N\alpha + N\beta) \notag\\
&= 6\alpha + 2(\beta_1 + \beta_2) - 4\sqrt{\beta_1^2 + \beta_2^2 - \beta_1\beta_2} - (6\alpha + 6\beta) \notag\\
&= -4\beta_1 + 2\beta_2 - 4\sqrt{\beta_1^2 + \beta_2^2 - \beta_1\beta_2} \notag\\
&= 4\beta\qty(-1 + \dfrac{1}{2}x + \sqrt{1 + x^2 - x})
\end{align}

\subex{c.}
\begin{align}
E_R &= 4\beta\qty(-1 + \dfrac{1}{2}x + \sqrt{1 + x^2 - x}) \notag\\
&= 4\beta\qty[ -1 + \dfrac{1}{2}x
+ 1 + \dfrac{1}{2}(x^2-x) - \dfrac{1}{8}(x^2-x)^2 + \dfrac{1}{16}(x^2-x)^3 - \dfrac{5}{128}(x^2-x)^4 ] \notag\\
&= 4\beta\qty[
\dfrac{1}{2}x^2 - \dfrac{1}{8}(x^4 + x^2 - 2x^3)
+ \dfrac{1}{16}(-x^3 + 3x^4) - \dfrac{5}{128}x^4 + \cdots ] \notag\\
&= 4\beta\qty[
\dfrac{3}{8}x^2 + \dfrac{3}{16}x^3 + \dfrac{3}{128}x^4 + \cdots ] \notag\\
&= \beta\qty[
\dfrac{3}{2}x^2 + \dfrac{3}{4}x^3 + \dfrac{3}{32}x^4 + \cdots ]
\end{align}

\newpage
\subsection{Diagrammatic Representation of Orbital Perturbation Theory}
\ex{6.7}
\subex{a.}
~\\
\begin{minipage}{0.2\linewidth}
		\hfill
	\begin{tikzpicture}
	\begin{feynman}[small]
	\vertex (a);
	\vertex [below=of a] (b);
	\vertex [below=of b] (c);
	\vertex [below=of c] (d);
	%\vertex [below=of d] (e);
	\diagram[inline=(c)]{
		(a) -- [fermion, quarter right, edge label'=a] (b)
		-- [fermion, quarter right, edge label'=b] (d)
		-- [fermion, quarter right, edge label'=s] (c)
		-- [fermion, quarter right, edge label'=r] (a)
		%-- [fermion, quarter right, edge label'=2] (a),
	};
	\end{feynman}
	\end{tikzpicture}
	\\
	\vspace{25pt}\\
\end{minipage}
\begin{minipage}{0.8\linewidth}
	\vspace{25pt}~\\
	\begin{align}
	&= - \sum_{a,b,r,s} \dfrac{v_{ab}v_{bs}v_{sr}v_{ra}}{(\varepsilon_a^{(0)} - \varepsilon_r^{(0)})(\varepsilon_b^{(0)} - \varepsilon_r^{(0)})(\varepsilon_b^{(0)} - \varepsilon_b^{(0)})}  &\notag\\
	&= -\dfrac{1}{(2\beta)^3} \sum_{i,j,k,l} \Braket{i|v|j}\Braket{j|v|k^*}\Braket{k^*|v|l^*}\Braket{l^*|v|i} & \notag\\
	&= -\dfrac{2}{(2\beta)^3} \sum_i^{N/2} [-1 + 1 -1 -1 +1 -1]\times (\beta/2)^4  & \notag\\
	&= \dfrac{N\beta}{64}  & \notag\\
	\end{align}
\end{minipage}
%\tikzfeynmanset{
%}
The pictorial representation of the summation are as follows\\
\begin{center}
	\begin{tikzpicture}
	\begin{feynman}[nodes=circle, large]
	\vertex (a) {$ (i) $};
	\vertex [below right=of a] (b1) {$ (i-1) $};
	\vertex [above right=of a] (b2) {$ (i+1) $};
	\vertex [below right=of b1] (c1) {$ (i-2)^* $};
	\vertex [above right=of b1] (c2) {$ (i)^* $};
	\vertex [above right=of b2] (c3) {$ (i+2)^* $};
	\vertex [above right=of c1] (d1) {$ (i-1)^* $};
	\vertex [above right=of c2] (d2) {$ (i+1)^* $};
	\vertex [above right=of d1] (e) {$ (i) $};
	\diagram[inline=(c)]{
		(a) -- [plain, edge label'=+] (b1);
		(a) -- [plain, edge label=+] (b2);
		(b1) -- [plain, edge label'=--] (c1);
		(b1) -- [plain, edge label'=+] (c2);
		(b2) -- [plain, edge label=--] (c2);
		(b2) -- [plain, edge label=+] (c3);
		(c1) -- [plain, edge label'=--] (d1);
		(c2) -- [plain, edge label'=--] (d1);
		(c2) -- [plain, edge label=--] (d2);
		(c3) -- [plain, edge label=--] (d2);
		(d1) -- [plain, edge label'=--] (e);
		(d2) -- [plain, edge label=+] (e);
		%-- [fermion, quarter right, edge label'=2] (a),
	};
	\end{feynman}
	\end{tikzpicture}
\end{center}

\begin{minipage}{0.2\linewidth}
	\hfill
	\begin{tikzpicture}
	\begin{feynman}[small]
	\vertex (a);
	\vertex [below=of a] (b);
	\vertex [below=of b] (c);
	\vertex [below=of c] (d);
	%\vertex [below=of d] (e);
	\diagram[inline=(c)]{
		(a) -- [fermion, out=200, in=150, edge label'=a] (d)
		-- [fermion, out=45, in=330, edge label'=r] (b)
		-- [fermion, quarter left, edge label'=b] (c)
		-- [fermion, quarter left, edge label=s] (a)
		%-- [fermion, quarter right, edge label'=2] (a),
	};
	\end{feynman}
	\end{tikzpicture}
	\\
	\vspace{25pt}\\
\end{minipage}
\begin{minipage}{0.8\linewidth}
	\vspace{25pt}~\\
	\begin{align}
	&= - \sum_{a,r,b,s} \dfrac{v_{ar}v_{rb}v_{bs}v_{sa}}{(\varepsilon_a^{(0)} - \varepsilon_r^{(0)})(\varepsilon_a^{(0)} - \varepsilon_s^{(0)})(\varepsilon_a^{(0)} + \varepsilon_b^{(0)} - \varepsilon_r^{(0)} - \varepsilon_s^{(0)})}  &\notag\\
	&= -\dfrac{1}{(2\beta)^2\times 4\beta} \sum_{i,j,k,l} \Braket{i|v|j^*}\Braket{j^*|v|k}\Braket{k|v|l^*}\Braket{l^*|v|i} & \notag\\
	&= -\dfrac{2}{(2\beta)^2\times 4\beta} \sum_i^{N/2} 6\times (\beta/2)^4  & \notag\\
	&= -\dfrac{3N\beta}{128}  & \notag\\
	\end{align}
\end{minipage}
%\tikzfeynmanset{
%}
The pictorial representation of the summation are as follows\\
\begin{center}
	\begin{tikzpicture}
	\begin{feynman}[nodes=circle, large]
	\vertex (a) {$ (i) $};
	\vertex [below right=of a] (b1) {$ (i-1)^* $};
	\vertex [above right=of a] (b2) {$ (i+1)^* $};
	\vertex [below right=of b1] (c1) {$ (i-2) $};
	\vertex [above right=of b1] (c2) {$ (i) $};
	\vertex [above right=of b2] (c3) {$ (i+2) $};
	\vertex [above right=of c1] (d1) {$ (i-1)^* $};
	\vertex [above right=of c2] (d2) {$ (i+1)^* $};
	\vertex [above right=of d1] (e) {$ (i) $};
	\diagram[inline=(c)]{
		(a) -- [plain, edge label'=--] (b1);
		(a) -- [plain, edge label=+] (b2);
		(b1) -- [plain, edge label'=+] (c1);
		(b1) -- [plain, edge label'=--] (c2);
		(b2) -- [plain, edge label=+] (c2);
		(b2) -- [plain, edge label=--] (c3);
		(c1) -- [plain, edge label'=+] (d1);
		(c2) -- [plain, edge label'=--] (d1);
		(c2) -- [plain, edge label=+] (d2);
		(c3) -- [plain, edge label=--] (d2);
		(d1) -- [plain, edge label'=--] (e);
		(d2) -- [plain, edge label=+] (e);
		%-- [fermion, quarter right, edge label'=2] (a),
	};
	\end{feynman}
	\end{tikzpicture}
\end{center}
thus
\begin{equation}\label{key}
E_0^{(4)} = 4\times \dfrac{N\beta}{64} + 3\times\qty(-\dfrac{3N\beta}{128}) =  \dfrac{N\beta}{64}
\end{equation}

\subex{b.}
Let $ N=6 $, we get
\begin{equation}\label{key}
E_0^{(4)} = \dfrac{3\beta}{32}
\end{equation}
which agrees with the result in Ex 6.6.

\subsection{Perturbation Expansion of the Correlation Energy}
\ex{6.8}
\begin{align}
E_0^{(2)} &=
\dfrac{1}{4}\sum_{a,b,r,s} \dfrac{\abs{\Braket{ab||rs}}^2}{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s} \notag\\
&= \dfrac{1}{4}\sum_{a,b,r,s}
\dfrac{\qty(\Braket{ab|rs} - \Braket{ab|sr}) \qty(\Braket{rs|ab} - \Braket{sr|ab})}{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s} \notag\\
&= \dfrac{1}{4}\sum_{a,b,r,s}
\dfrac{\Braket{ab|rs}\Braket{rs|ab} - \Braket{ab|sr}\Braket{rs|ab}  - \Braket{ab|rs}\Braket{sr|ab} + \Braket{ab|sr}\Braket{sr|ab}}{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s} \notag\\
&= \dfrac{1}{4}\qty[\sum_{a,b,r,s}
\dfrac{\Braket{ab|rs}\Braket{rs|ab}}{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
- \sum_{a,b,r,s}
\dfrac{\Braket{ab|sr}\Braket{rs|ab}}{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
- \sum_{a,b,r,s}
\dfrac{ \Braket{ab|rs}\Braket{sr|ab}}{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
+ \sum_{a,b,r,s}
\dfrac{\Braket{ab|sr}\Braket{sr|ab}}{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
] \notag\\
&= \dfrac{1}{4}\qty[2\sum_{a,b,r,s}
\dfrac{\Braket{ab|rs}\Braket{rs|ab}}{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
- 2\sum_{a,b,r,s}
\dfrac{\Braket{ab|rs}\Braket{sr|ab}}{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
] \notag\\
&= \dfrac{1}{2}\sum_{a,b,r,s}
\dfrac{\Braket{ab|rs}\Braket{rs|ab}}{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
- \dfrac{1}{2}\sum_{a,b,r,s}
\dfrac{\Braket{ab|rs}\Braket{rs|ba}}{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
\end{align}
For a closed-shell system, the possible spin part of $ a,b,r,s $ of the non-zero terms are\\
first term: $ \alpha,\alpha,\alpha,\alpha $;\quad $ \alpha,\beta,\alpha,\beta $;\quad $ \beta,\alpha,\beta,\alpha $;\quad $ \beta,\beta,\beta,\beta $\\
second term: $ \alpha,\alpha,\alpha,\alpha $;\quad $ \beta,\beta,\beta,\beta $\\
thus
\begin{align}
E_0^{(2)} &= 2\sum_{a,b,r,s}^{N/2}
\dfrac{\Braket{ab|rs}\Braket{rs|ab}}{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
- \sum_{a,b,r,s}^{N/2}
\dfrac{\Braket{ab|rs}\Braket{rs|ba}}{\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_s}
\end{align}

\ex{6.9}
\begin{align}
E_{\corr} &= \Delta - (\Delta^2 + K_{12}^2)^{1/2} \notag\\
&= \Delta - \qty[\Delta + \dfrac{K_{12}^2}{2\Delta}] \notag\\
&= - \dfrac{K_{12}^2}{2\Delta} \notag\\
&= - \dfrac{K_{12}^2}{2(\varepsilon_2-\varepsilon_1) + J_{11} + J_{22} - 4J_{12} + 2K_{12}} \notag\\
&= - K_{12}^2 \qty(\dfrac{1}{2(\varepsilon_2-\varepsilon_1)} - \dfrac{ J_{11} + J_{22} - 4J_{12} + 2K_{12}}{4(\varepsilon_2-\varepsilon_1)^2}) \notag\\
&= \dfrac{K_{12}^2}{2(\varepsilon_1-\varepsilon_2)} + \dfrac{K_{12}^2 (J_{11} + J_{22} - 4J_{12} + 2K_{12})}{4(\varepsilon_1-\varepsilon_2)^2})
\end{align}

\subsection{The $ N $-dependence of the RS Perturbation Expansion}
\ex{6.10}
From Eq 6.68, we get
\begin{align}
E_0^{(1)} &= \Braket{\Psi_0 | \mathscr{V} | \Psi_0} = -\dfrac{1}{2}\sum_{ab}\Braket{ab||ab} \notag\\
&= -\dfrac{1}{2}\sum_{i=1}^N \qty[\Braket{1_i \bar{1}_i|| 1_i \bar{1}_i} + \Braket{\bar{1}_i 1_i || \bar{1}_i 1_i}] \notag\\
&= -\dfrac{1}{2}\sum_{i=1}^N
\qty[\Braket{1_i \bar{1}_i| 1_i \bar{1}_i}
- \Braket{1_i \bar{1}_i| \bar{1}_i 1_i}
+ \Braket{\bar{1}_i 1_i | \bar{1}_i 1_i}
- \Braket{\bar{1}_i 1_i | 1_i \bar{1}_i} ] \notag\\
&= -\dfrac{1}{2}\times 2N [1_i 1_i| 1_i 1_i] \notag\\
&= -N J_{11}
\end{align}
\begin{align}
\Braket{\Psi_{1_i\bar{1}_i}^{2_i\bar{2}_i} | \mathscr{V} | \Psi_{1_i\bar{1}_i}^{2_i\bar{2}_i}}
&= \Braket{\Psi_{1_i\bar{1}_i}^{2_i\bar{2}_i} | \sH | \Psi_{1_i\bar{1}_i}^{2_i\bar{2}_i}}
- \Braket{\Psi_{1_i\bar{1}_i}^{2_i\bar{2}_i} | \sH_0 | \Psi_{1_i\bar{1}_i}^{2_i\bar{2}_i}} \notag\\
&= (2N-2) h_{11} + 2h_{22} + (N-1)J_{11} + J_{22} - (2N-2)\varepsilon_1 - 2\varepsilon_2 \notag\\
&= (2N-2) h_{11} + 2h_{22} + (N-1)J_{11} + J_{22} - (2N-2)(h_{11} + J_{11}) - 2(h_{22} + 2J_{12} - K_{12}) \notag\\
&= -(N-1)J_{11} + J_{22} - 4J_{12} + 2K_{12}
\end{align}

\subsection{Diagrammatic Representation of the Perturbation Expansion of the Correlation Energy}

\subsubsection{Hugenholtz Diagrams}
\ex{6.11}
The numerator and denominator are obvious.\\
$ h=5 $, and $ l=2 $ since closed loops are $ r\ra a\ra d\ra t\ra e\ra r; \; s\ra c\ra b\ra s $. The number of quivalent line pairs is one ($ r,s $). Thus the pre-factor is $ -\dfrac{1}{2} $.

\subsubsection{Goldstone Diagrams}
\ex{6.12}
~\\
1
\begin{minipage}{0.4\linewidth}
	\centering
\begin{tikzpicture}
\begin{feynman}[large]
\vertex (a); \vertex [below=of a] (b);
\vertex [right=of a] (c); \vertex [right=of b] (d);

\diagram[%inline=(e.base)
]{
	(a) -- [fermion,  quarter left, edge label=r] (b)
	-- [fermion, quarter left, edge label=a] (a),
	(c)-- [fermion, quarter left, edge label=s] (d)
	-- [fermion, quarter left, edge label=b] (c),
	(a) -- [scalar] (c),
	(b) -- [scalar] (d),
	%-- [fermion, quarter right, edge label'=2] (a),
};
\end{feynman}
\end{tikzpicture}
\end{minipage}
$ \lra  $
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=of a] (b);
	%\vertex [right=of a] (c); %\vertex [right=of b] (d);
	\diagram[%inline=(e.base)
	]{
		(a) -- [fermion,  half left, edge label=r] (b)
		-- [fermion, half left, edge label=a] (a),
		(a)-- [fermion, quarter left, edge label=s] (b)
		-- [fermion, quarter left, edge label=b] (a),
		%(a) -- [scalar] (c),
		%(b) -- [scalar] (d),
		%-- [fermion, quarter right, edge label'=2] (a),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
~\\
~\vspace{10pt}\\
2
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=of a] (b);
	\vertex [right=of a] (c); \vertex [right=of b] (d);

	\diagram[%inline=(e.base)
	]{
		(a) -- [fermion, out=-60, in=150, edge label'=b] (d)
		-- [fermion, quarter right, edge label=s] (c),
		(c)-- [fermion, out=240, in=30, edge label=a] (b)
		-- [fermion, quarter left, edge label=r] (a),
		(a) -- [scalar] (c),
		(b) -- [scalar] (d),
		%-- [fermion, quarter right, edge label'=2] (a),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
$ \lra  $
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=of a] (b);
	%\vertex [right=of a] (c); %\vertex [right=of b] (d);
	\diagram[%inline=(e.base)
	]{
		(b) -- [fermion,  half right, edge label'=r] (a),
		(b) -- [fermion, half left, edge label'=s] (a),
		(a)-- [fermion, quarter left, edge label=a] (b),
		(a) -- [fermion, quarter right, edge label=b] (b),
		%(a) -- [scalar] (c),
		%(b) -- [scalar] (d),
		%-- [fermion, quarter right, edge label'=2] (a),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
~\\
~\vspace{10pt}\\
1
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=2.4cm of a] (b);
	\vertex [right=2.5cm of a] (c); \vertex [right=2.5cm of b] (d);
	\vertex at ($(a)!0.5!(b) + (0.7cm, 0)$) (e); \vertex at ($(c)!0.5!(d) + (-0.7cm, 0)$) (f);

	\diagram[%inline=(e.base)
	]{
		(a) -- [anti fermion,  quarter left, edge label=t] (e)
		-- [anti fermion,  quarter left, edge label=r] (b)
		-- [anti fermion, quarter left, edge label=a] (a),
		(c) -- [fermion, quarter left, edge label=b] (d)
		-- [fermion, quarter left, edge label=u] (f)
		-- [fermion, quarter left, edge label=s] (c),
		(a) -- [scalar] (c),
		(b) -- [scalar] (d),
		(e) -- [scalar] (f),
		%-- [fermion, quarter right, edge label'=2] (a),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
$ \lra  $
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=of a] (b);
	%\vertex [right=of a] (c); %\vertex [right=of b] (d);
	\vertex at ($(a)!0.5!(b) $) (e);
	\diagram[%inline=(e.base)
	]{
		(a) -- [anti fermion,  quarter left, edge label=t] (e)
		-- [anti fermion,  quarter left, edge label=r] (b)
		-- [anti fermion, half left, edge label=a] (a),
		(a) -- [fermion, half left, edge label=b] (b)
		-- [fermion, quarter left, edge label=u] (e)
		-- [fermion, quarter left, edge label=s] (a),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
~\\
~\vspace{10pt}\\
2
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=2.4cm of a] (b);
	\vertex [right=2.5cm of a] (c); \vertex [right=2.5cm of b] (d);
	\vertex at ($(a)!0.5!(b) + (0.7cm, 0)$) (e); \vertex at ($(c)!0.5!(d) + (-0.7cm, 0)$) (f);

	\diagram[%inline=(e.base)
	]{
		(a) -- [fermion,  quarter left, edge label=c] (e)
		-- [ fermion,  quarter left, edge label=a] (b)
		-- [ fermion, quarter left, edge label=r] (a),
		(c) -- [anti fermion, quarter left, edge label=s] (d)
		-- [anti fermion, quarter left, edge label=d] (f)
		-- [anti fermion, quarter left, edge label=b] (c),
		(a) -- [scalar] (c),
		(b) -- [scalar] (d),
		(e) -- [scalar] (f),
		%-- [fermion, quarter right, edge label'=2] (a),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
$ \lra  $
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=2.4cm of a] (b);
	%\vertex [right=2.5cm of a] (c); \vertex [right=2.5cm of b] (d);
	\vertex at ($(a)!0.5!(b) $) (e);
	%\vertex at ($(c)!0.5!(d) + (-0.7cm, 0)$) (f);

	\diagram[%inline=(e.base)
	]{
		(a) -- [fermion,  quarter left, edge label=c] (e)
		-- [ fermion,  quarter left, edge label=a] (b)
		-- [ fermion, half left, edge label=r] (a),
		(a) -- [anti fermion, half left, edge label=s] (b)
		-- [anti fermion, quarter left, edge label=d] (e)
		-- [anti fermion, quarter left, edge label=b] (a),
		%(a) -- [scalar] (c),
		%(b) -- [scalar] (d),
		%(e) -- [scalar] (f),
		%-- [fermion, quarter right, edge label'=2] (a),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
~\\
~\vspace{10pt}\\
3
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=2.4cm of a] (b);
	\vertex [right=2.5cm of a] (c); \vertex [right=2.5cm of b] (d);
	\vertex at ($(a)!0.5!(b) + (0.7cm, 0)$) (e); \vertex at ($(c)!0.5!(d) + (-0.7cm, 0)$) (f);

	\diagram[%inline=(e.base)
	]{
		(a) -- [anti fermion,  quarter left, edge label=s] (e)
		-- [anti fermion,  quarter left, edge label=r] (b)
		-- [anti fermion, quarter left, edge label=a] (a),
		(c) -- [anti fermion, quarter left, edge label=t] (d)
		-- [anti fermion, quarter left, edge label=c] (f)
		-- [anti fermion, quarter left, edge label=b] (c),
		(a) -- [scalar] (c),
		(b) -- [scalar] (d),
		(e) -- [scalar] (f),
		%-- [fermion, quarter right, edge label'=2] (a),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
$ \lra $
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=2.4cm of a] (b);
	%\vertex [right=2.5cm of a] (c); \vertex [right=2.5cm of b] (d);
	\vertex at ($(a)!0.5!(b) $) (e);
	%\vertex at ($(c)!0.5!(d) + (-0.7cm, 0)$) (f);

	\diagram[%inline=(e.base)
	]{
		(a) -- [anti fermion,  quarter left, edge label=s] (e)
		-- [anti fermion,  quarter left, edge label=r] (b)
		-- [anti fermion, half left, edge label=a] (a),
		(a) -- [anti fermion, half left, edge label=t] (b)
		-- [anti fermion, quarter left, edge label=c] (e)
		-- [anti fermion, quarter left, edge label=b] (a),
		%(a) -- [scalar] (c),
		%(b) -- [scalar] (d),
		%(e) -- [scalar] (f),
		%-- [fermion, quarter right, edge label'=2] (a),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
~\\
~\vspace{10pt}\\
4
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=2.4cm of a] (b);
	\vertex [right=2.5cm of a] (c); \vertex [right=2.5cm of b] (d);
	\vertex at ($(a)!0.5!(b) + (-0.7cm, 0)$) (e); \vertex at ($(a)!0.5!(b) + (0.7cm, 0)$) (f);

	\diagram[%inline=(e.base)
	]{
		(a) -- [fermion,  quarter left, edge label=a] (f)
		-- [ fermion,  quarter left, edge label=b] (b)
		-- [ fermion, quarter left, edge label=r] (e)
		-- [ fermion, quarter left, edge label=s] (a),
		(c) -- [fermion, quarter left, edge label=c] (d)
		%-- [anti fermion, quarter left, edge label=d] (f)
		-- [fermion, quarter left, edge label=t] (c),
		(a) -- [scalar] (c),
		(b) -- [scalar] (d),
		(e) -- [scalar] (f),
		%-- [fermion, quarter right, edge label'=2] (a),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
$ \lra  $
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=2.4cm of a] (b);
	%\vertex [right=2.5cm of a] (c); \vertex [right=2.5cm of b] (d);
	\vertex at ($(a)!0.5!(b) $) (e);
	%\vertex at ($(a)!0.5!(b) + (0.7cm, 0)$) (f);

	\diagram[%inline=(e.base)
	]{
		(a) -- [fermion,  quarter left, edge label=a] (e)
		-- [ fermion,  quarter left, edge label=b] (b)
		-- [ fermion, quarter left, edge label=r] (e)
		-- [ fermion, quarter left, edge label=s] (a),
		(a) -- [fermion, half left, edge label=c] (b)
		%-- [anti fermion, quarter left, edge label=d] (f)
		-- [fermion, half left, edge label=t] (a),
		%(a) -- [scalar] (c),
		%(b) -- [scalar] (d),
		%(e) -- [scalar] (f),
		%-- [fermion, quarter right, edge label'=2] (a),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
~\\
~\vspace{10pt}\\
5
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=2.4cm of a] (b);
	\vertex [right= of a] (c); \vertex [right=4cm of b] (d);
	\vertex [below=1.2cm of c] (e); \vertex [right= of e] (f);

	\diagram[%inline=(e.base)
	]{
		(a) %-- [fermion,  quarter left, edge label=a] (f)
		-- [ fermion,  quarter left, edge label=a] (b)
		%-- [ fermion, quarter left, edge label=r] (e)
		-- [ fermion, quarter left, edge label=r] (a),
		(c) -- [fermion, quarter left, edge label=b] (e)
		%-- [anti fermion, quarter left, edge label=d] (f)
		-- [fermion, quarter left, edge label=s] (c),
		(f) -- [fermion, quarter left, edge label=c] (d)
		%-- [anti fermion, quarter left, edge label=d] (f)
		-- [fermion, quarter left, edge label=t] (f),
		(a) -- [scalar] (c),
		(b) -- [scalar] (d),
		(e) -- [scalar] (f),
		%-- [fermion, quarter right, edge label'=2] (a),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
$ \lra  $
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=2.4cm of a] (b);
	%\vertex [right= of a] (c); \vertex [right=4cm of b] (d);
	\vertex [below=1.2cm of a] (e); %\vertex [right= of e] (f);

	\diagram[%inline=(e.base)
	]{
		(a) %-- [fermion,  quarter left, edge label=a] (f)
		-- [ fermion,  half left, edge label=a] (b)
		%-- [ fermion, quarter left, edge label=r] (e)
		-- [ fermion, half left, edge label=r] (a),
		(a) -- [fermion, quarter left, edge label=b] (e)
		%-- [anti fermion, quarter left, edge label=d] (f)
		-- [fermion, quarter left, edge label=s] (a),
		(e) -- [fermion, quarter left, edge label=c] (b)
		%-- [anti fermion, quarter left, edge label=d] (f)
		-- [fermion, quarter left, edge label=t] (e),
		%(a) -- [scalar] (c),
		%(b) -- [scalar] (d),
		%(e) -- [scalar] (f),
		%-- [fermion, quarter right, edge label'=2] (a),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}

~\\
~\vspace{10pt}\\
6
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=2.4cm of a] (b);
	\vertex [right=2.5cm of a] (c); \vertex [right=2.5cm of b] (d);
	\vertex at ($(a)!0.5!(b) + (0.6cm, 0)$) (e); \vertex at ($(c)!0.5!(d) + (-0.6cm, 0)$) (f);

	\diagram[%inline=(e.base)
	]{
		(a) -- [fermion,  quarter left, edge label=a] (e)
		-- [ fermion,  edge label=s] (c)
		-- [ fermion, quarter left, edge label=b] (d)
		-- [ fermion, quarter left, edge label=t] (f)
		-- [fermion, edge label=c] (b)
		%-- [anti fermion, quarter left, edge label=d] (f)
		-- [fermion, quarter left, edge label=r] (a),
		(a) -- [scalar] (c),
		(b) -- [scalar] (d),
		(e) -- [scalar] (f),
		%-- [fermion, quarter right, edge label'=2] (a),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
$ \lra  $
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=2.4cm of a] (b);
	%\vertex [right=2.5cm of a] (c); \vertex [right=2.5cm of b] (d);
	\vertex at ($(a)!0.5!(b)$) (e); %\vertex at ($(c)!0.5!(d) + (-0.6cm, 0)$) (f);

	\diagram[%inline=(e.base)
	]{
		(a) -- [fermion,  quarter left, edge label=a] (e)
		-- [ fermion,  quarter left, edge label=s] (a)
		-- [ fermion, half left, edge label=b] (b)
		-- [ fermion, quarter left, edge label=t] (e)
		-- [fermion, quarter left, edge label=c] (b)
		%-- [anti fermion, quarter left, edge label=d] (f)
		-- [fermion, half left, edge label=r] (a),
		%(a) -- [scalar] (c),
		%(b) -- [scalar] (d),
		%(e) -- [scalar] (f),
		%-- [fermion, quarter right, edge label'=2] (a),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
~\\
~\vspace{10pt}\\
7
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=2.4cm of a] (b);
	\vertex [right=2.5cm of a] (c); \vertex [right=2.5cm of b] (d);
	\vertex at ($(a)!0.5!(b) + (-0.6cm, 0)$) (e); \vertex at ($(c)!0.5!(d) + (0.6cm, 0)$) (f);

	\diagram[]{
		(a) -- [fermion,  quarter right, edge label'=d] (e)
		-- [ fermion,  quarter right,  edge label'=a] (b)
		-- [ fermion,out=20, in=-120,  edge label'=r] (c)
		-- [ fermion, quarter left, edge label=b] (f)
		-- [fermion, quarter left, edge label=c] (d)
		%-- [anti fermion, quarter left, edge label=d] (f)
		-- [fermion,out=160, in=-60, edge label=s] (a),
		(a) -- [scalar] (c),
		(b) -- [scalar] (d),
		(e) -- [scalar] (f),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
$ \lra  $
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=2.4cm of a] (b);
	%\vertex [right=2.5cm of a] (c); \vertex [right=2.5cm of b] (d);
	\vertex at ($(a)!0.5!(b) $) (e); %\vertex at ($(c)!0.5!(d) + (0.6cm, 0)$) (f);

	\diagram[]{
		(a) -- [fermion,  \qr, \el'=d] (e) -- [ fermion,  \qr,  \el'=a] (b)
		-- [ fermion, \hfr,  \el'=r] (a) -- [ fermion, \ql, \el=b] (e)
		-- [fermion, \ql, \el=c] (b)
		%-- [anti fermion, quarter left, edge label=d] (f)
		-- [fermion, \hfl, \el=s] (a),
		%(a) -- [scalar] (c),
		%(b) -- [scalar] (d),
		%(e) -- [scalar] (f),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
~\\
~\vspace{10pt}\\
8
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=2.4cm of a] (b);
	\vertex [right=2.5cm of a] (c); \vertex [right=2.5cm of b] (d);
	\vertex at ($(a)!0.5!(b) + (0.7cm, 0)$) (e); \vertex at ($(c)!0.5!(d) + (-0.7cm, 0)$) (f);
	\diagram[]{
		(a) -- [fermion,  \qr, \el'=a] (b) -- [ fermion, \el=r] (f)
		-- [ fermion,  \el'=s] (c) -- [ fermion, \ql, \el=b] (d)
		-- [fermion,  \el'=t] (e) -- [fermion, \el=u] (a),
		(a) -- [scalar] (c), 	(b) -- [scalar] (d), (e) -- [scalar] (f),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
$ \lra  $
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=2.4cm of a] (b);
	%\vertex [right=2.5cm of a] (c); \vertex [right=2.5cm of b] (d);
	\vertex at ($(a)!0.5!(b) $) (e); %\vertex at ($(c)!0.5!(d) + (-0.7cm, 0)$) (f);
	\diagram[]{
		(a) -- [fermion,  \hfr, \el'=a] (b) -- [ fermion,\qr, \el'=r] (e)
		-- [ fermion, \qr, \el'=s] (a) -- [ fermion, \hfl, \el=b] (b)
		-- [fermion, \ql, \el=t] (e) -- [fermion, \ql,\el=u] (a),
		%(a) -- [scalar] (c), 	(b) -- [scalar] (d), (e) -- [scalar] (f),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
~\\
~\vspace{10pt}\\
9
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=2.4cm of a] (b);
	\vertex [right=2.5cm of a] (c); \vertex [right=2.5cm of b] (d);
	\vertex at ($(a)!0.5!(b) + (-0.7cm, 0)$) (e); \vertex at ($(a)!0.5!(b) + (0.7cm, 0)$) (f);
	\diagram[]{
		(a) -- [fermion,  \qr, \el'=a] (e) -- [ fermion, \qr, \el'=b] (b)
		-- [ fermion,  \el'=r] (f) -- [ fermion, \el=s] (c)
		-- [fermion, \ql, \el'=c] (d) -- [fermion, \el'=t] (a),
		(a) -- [scalar] (c), 	(b) -- [scalar] (d), (e) -- [scalar] (f),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
$ \lra  $
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=2.4cm of a] (b);
	%\vertex [right=2.5cm of a] (c); \vertex [right=2.5cm of b] (d);
	\vertex at ($(a)!0.5!(b) $) (e); %\vertex at ($(a)!0.5!(b) + (0.7cm, 0)$) (f);
	\diagram[]{
		(a) -- [fermion,  \qr, \el'=a] (e) -- [ fermion, \qr, \el'=b] (b)
		-- [ fermion, \qr, \el'=r] (e) -- [ fermion, \qr, \el'=s] (a)
		-- [fermion, \hfl, \el=c] (b) -- [fermion, \hfl, \el=t] (a),
		%(a) -- [scalar] (c), 	(b) -- [scalar] (d), (e) -- [scalar] (f),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
~\\
~\vspace{10pt}\\
10
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=2.4cm of a] (b);
	\vertex [right=2.5cm of a] (c); \vertex [right=2.5cm of b] (d);
	\vertex at ($(a)!0.5!(b) + (-0.7cm, 0)$) (e); \vertex at ($(a)!0.5!(b) + (0.7cm, 0)$) (f);
	\diagram[]{
		(a) -- [fermion,  \el'=s] (f) -- [ fermion, \el'=r] (d)
		-- [ fermion, \qr,  \el'=c] (c) -- [ fermion, \el'=t] (b)
		-- [fermion, \ql, \el=b] (e) -- [fermion,\ql, \el=a] (a),
		(a) -- [scalar] (c), 	(b) -- [scalar] (d), (e) -- [scalar] (f),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
$ \lra  $
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=2.4cm of a] (b);
	%\vertex [right=2.5cm of a] (c); \vertex [right=2.5cm of b] (d);
	\vertex at ($(a)!0.5!(b) $) (e); %\vertex at ($(a)!0.5!(b) + (0.7cm, 0)$) (f);
	\diagram[]{
		(a) -- [fermion, \ql, \el=s] (e) -- [ fermion,\ql, \el=r] (b)
		-- [ fermion, \hfr,  \el'=c] (a) -- [ fermion, \hfr, \el'=t] (b)
		-- [fermion, \ql, \el=b] (e) -- [fermion,\ql, \el=a] (a),
		%(a) -- [scalar] (c), 	(b) -- [scalar] (d), (e) -- [scalar] (f),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
~\\
~\vspace{10pt}\\
11
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=2.4cm of a] (b);
	\vertex [right=3.5cm of a] (c); \vertex [right= of b] (d);
	\vertex [above=1.2cm of d] (e); \vertex [below=1.2cm of c] (f);
	\diagram[]{
		(a) -- [fermion,  \el'=b] (d) -- [ fermion, \qr, \el'=s] (e)
		-- [ fermion,  \el=a] (b) -- [ fermion, \ql, \el=r] (a),
		(c) -- [fermion, \ql, \el=c] (f) -- [fermion,\ql, \el=t] (c),
		(a) -- [scalar] (c), 	(b) -- [scalar] (d), (e) -- [scalar] (f),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
$ \lra  $
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=2.4cm of a] (b);
	%\vertex [right=3.5cm of a] (c); \vertex [right= of b] (d);
	\vertex at ($(a)!0.5!(b)$) (e); %\vertex [below=1.2cm of c] (f);
	\diagram[]{
		(a) -- [fermion, \hfl, \el=b] (b) -- [ fermion, \qr, \el'=s] (e)
		-- [ fermion, \qr, \el'=a] (b) -- [ fermion, \hfl, \el=r] (a),
		(a) -- [fermion, \ql, \el=c] (e) -- [fermion,\ql, \el=t] (a),
		%(a) -- [scalar] (c), 	(b) -- [scalar] (d), (e) -- [scalar] (f),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
~\\~\vspace{10pt}\\
12
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=2.4cm of a] (b);
	\vertex [right=of a] (c); \vertex [right=3.5cm of b] (d);
	\vertex [above=1.2cm of d] (f); \vertex [below=1.2cm of c] (e);
	\diagram[]{
		(a) -- [fermion,  \el'=a] (e) -- [ fermion, \qr, \el'=s] (c)
		-- [ fermion,  \el=b] (b) -- [ fermion, \ql, \el=r] (a),
		(f) -- [fermion, \ql, \el=c] (d) -- [fermion,\ql, \el=t] (f),
		(a) -- [scalar] (c), 	(b) -- [scalar] (d), (e) -- [scalar] (f),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
$ \lra  $
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=2.4cm of a] (b);
	%\vertex [right=of a] (c); \vertex [right=3.5cm of b] (d);
	%\vertex [above=1.2cm of d] (f);
	\vertex [below=1.2cm of a] (e);
	\diagram[]{
		(a) -- [fermion, \qr, \el'=a] (e) -- [ fermion, \qr, \el'=s] (a)
		-- [ fermion, \hfl, \el=b] (b) -- [ fermion, \hfl, \el=r] (a),
		(e) -- [fermion, \ql, \el=c] (b) -- [fermion,\ql, \el=t] (e),
		%(a) -- [scalar] (c), 	(b) -- [scalar] (d), (e) -- [scalar] (f),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
~\\
~\\~\\
For the Hugenholtz diagram provided, its value is\\
\begin{minipage}{0.3\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=of a] (b);
	%\vertex [right=of a] (c); %\vertex [right=of b] (d);
	\vertex at ($(a)!0.5!(b) $) (e);
	\diagram[%inline=(e.base)
	]{
		(a) -- [anti fermion,  quarter left, edge label=t] (e)
		-- [anti fermion,  quarter left, edge label=r] (b)
		-- [anti fermion, half left, edge label=a] (a),
		(a) -- [fermion, half left, edge label=b] (b)
		-- [fermion, quarter left, edge label=u] (e)
		-- [fermion, quarter left, edge label=s] (a),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
\begin{minipage}{0.6\linewidth}
	\vskip0.6cm
\begin{align}
&= \qty(\dfrac{1}{2})^3 (-1)^{2+2} \sum_{a,b,r,s,u,t} \dfrac{\Braket{ab||ru} \Braket{ru||ts}\Braket{ts||ab}}{(\varepsilon_a + \varepsilon_b - \varepsilon_u - \varepsilon_r)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)} &&& \notag\\
& =\dfrac{1}{8} \sum_{a,b,r,s,u,t} \dfrac{\Braket{ab||ru} \Braket{ru||ts}\Braket{ts||ab}}{(\varepsilon_a + \varepsilon_b - \varepsilon_u - \varepsilon_r)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)}  &&&
\notag
\end{align}
\end{minipage}
\begin{align}
&=\dfrac{1}{8} \sum_{a,b,r,s,u,t} \dfrac{(\Braket{ab|ru}-\Braket{ab|ur}) (\Braket{ru|ts}-\Braket{ru|st}) (\Braket{ts|ab}-\Braket{ts|ba})}{(\varepsilon_a + \varepsilon_b - \varepsilon_u - \varepsilon_r)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)}  \notag\\
&= \dfrac{1}{8} \sum_{a,b,r,s,u,t} \dfrac{\Braket{ab|ru} \Braket{ru|ts} \Braket{ts|ab}}{(\varepsilon_a + \varepsilon_b - \varepsilon_u - \varepsilon_r)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)}
- \dfrac{1}{8} \sum_{a,b,r,s,u,t} \dfrac{\Braket{ab|ur} \Braket{ru|ts} \Braket{ts|ab}}{(\varepsilon_a + \varepsilon_b - \varepsilon_u - \varepsilon_r)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)} \notag\\
&\quad{} - \dfrac{1}{8} \sum_{a,b,r,s,u,t} \dfrac{\Braket{ab|ru} \Braket{ru|st} \Braket{ts|ab}}{(\varepsilon_a + \varepsilon_b - \varepsilon_u - \varepsilon_r)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)}
+ \dfrac{1}{8} \sum_{a,b,r,s,u,t} \dfrac{\Braket{ab|ur} \Braket{ru|st} \Braket{ts|ab}}{(\varepsilon_a + \varepsilon_b - \varepsilon_u - \varepsilon_r)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)}
\notag\\
&\quad{} - \dfrac{1}{8} \sum_{a,b,r,s,u,t} \dfrac{\Braket{ab|ru} \Braket{ru|ts} \Braket{ts|ba}}{(\varepsilon_a + \varepsilon_b - \varepsilon_u - \varepsilon_r)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)}
+ \dfrac{1}{8} \sum_{a,b,r,s,u,t} + \dfrac{\Braket{ab|ur} \Braket{ru|ts} \Braket{ts|ba}}{(\varepsilon_a + \varepsilon_b - \varepsilon_u - \varepsilon_r)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)} \notag\\
&\quad{} + \dfrac{1}{8} \sum_{a,b,r,s,u,t} \dfrac{\Braket{ab|ru} \Braket{ru|st} \Braket{ts|ba}}{(\varepsilon_a + \varepsilon_b - \varepsilon_u - \varepsilon_r)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)}
- \dfrac{1}{8} \sum_{a,b,r,s,u,t} \dfrac{\Braket{ab|ur} \Braket{ru|st} \Braket{ts|ba}}{(\varepsilon_a + \varepsilon_b - \varepsilon_u - \varepsilon_r)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)}  \notag\\
&= \dfrac{1}{8} \sum_{a,b,r,s,u,t} \dfrac{\Braket{ab|ru} \Braket{ru|ts} \Braket{ts|ab}}{(\varepsilon_a + \varepsilon_b - \varepsilon_u - \varepsilon_r)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)}
- \dfrac{1}{8} \sum_{a,b,r,s,u,t} \dfrac{\Braket{ab|ur} \Braket{ru|ts} \Braket{ts|ab}}{(\varepsilon_a + \varepsilon_b - \varepsilon_u - \varepsilon_r)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)} \notag\\
&\quad{} - \dfrac{1}{8} \sum_{a,b,r,s,u,t} \dfrac{\Braket{ab|ur} \Braket{ur|st} \Braket{ts|ab}}{(\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_u)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)}
+ \dfrac{1}{8} \sum_{a,b,r,s,u,t} \dfrac{\Braket{ab|ru} \Braket{ur|st} \Braket{ts|ab}}{(\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_u)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)}
\notag\\
&\quad{} - \dfrac{1}{8} \sum_{a,b,r,s,u,t} \dfrac{\Braket{ab|ru} \Braket{ru|ts} \Braket{ts|ba}}{(\varepsilon_a + \varepsilon_b - \varepsilon_u - \varepsilon_r)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)}
+ \dfrac{1}{8} \sum_{a,b,r,s,u,t} + \dfrac{\Braket{ab|ur} \Braket{ru|ts} \Braket{ts|ba}}{(\varepsilon_a + \varepsilon_b - \varepsilon_u - \varepsilon_r)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)} \notag\\
&\quad{} + \dfrac{1}{8} \sum_{a,b,r,s,u,t} \dfrac{\Braket{ab|ur} \Braket{ur|st} \Braket{ts|ba}}{(\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_u)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)}
- \dfrac{1}{8} \sum_{a,b,r,s,u,t} \dfrac{\Braket{ab|ru} \Braket{ur|st} \Braket{ts|ba}}{(\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_u)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)}  \notag\\
&= \dfrac{1}{4} \sum_{a,b,r,s,u,t} \dfrac{\Braket{ab|ru} \Braket{ru|ts} \Braket{ts|ab}}{(\varepsilon_a + \varepsilon_b - \varepsilon_u - \varepsilon_r)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)}
- \dfrac{1}{4} \sum_{a,b,r,s,u,t} \dfrac{\Braket{ab|ur} \Braket{ru|ts} \Braket{ts|ab}}{(\varepsilon_a + \varepsilon_b - \varepsilon_u - \varepsilon_r)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)} \notag\\
&\quad{} - \dfrac{1}{4} \sum_{a,b,r,s,u,t} \dfrac{\Braket{ab|ru} \Braket{ru|ts} \Braket{ts|ba}}{(\varepsilon_a + \varepsilon_b - \varepsilon_u - \varepsilon_r)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)}
+ \dfrac{1}{4} \sum_{a,b,r,s,u,t} + \dfrac{\Braket{ab|ur} \Braket{ru|ts} \Braket{ts|ba}}{(\varepsilon_a + \varepsilon_b - \varepsilon_u - \varepsilon_r)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)} \notag\\
&= \dfrac{1}{4} \sum_{a,b,r,s,u,t} \dfrac{\Braket{ab|ru} \Braket{ru|ts} \Braket{ts|ab}}{(\varepsilon_a + \varepsilon_b - \varepsilon_u - \varepsilon_r)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)}
- \dfrac{1}{4} \sum_{a,b,r,s,u,t} \dfrac{\Braket{ab|ur} \Braket{ru|ts} \Braket{ts|ab}}{(\varepsilon_a + \varepsilon_b - \varepsilon_u - \varepsilon_r)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)} \notag\\
&\quad{} - \dfrac{1}{4} \sum_{a,b,r,s,u,t} \dfrac{\Braket{ba|ru} \Braket{ru|ts} \Braket{ts|ab}}{(\varepsilon_a + \varepsilon_b - \varepsilon_u - \varepsilon_r)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)}
+ \dfrac{1}{4} \sum_{a,b,r,s,u,t} + \dfrac{\Braket{ba|ur} \Braket{ru|ts} \Braket{ts|ab}}{(\varepsilon_a + \varepsilon_b - \varepsilon_u - \varepsilon_r)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)} \notag\\
&= \dfrac{1}{2} \sum_{a,b,r,s,u,t} \dfrac{\Braket{ab|ru} \Braket{ru|ts} \Braket{ts|ab}}{(\varepsilon_a + \varepsilon_b - \varepsilon_u - \varepsilon_r)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)}
- \dfrac{1}{2} \sum_{a,b,r,s,u,t} \dfrac{\Braket{ab|ur} \Braket{ru|ts} \Braket{ts|ab}}{(\varepsilon_a + \varepsilon_b - \varepsilon_u - \varepsilon_r)(\varepsilon_a + \varepsilon_b - \varepsilon_s - \varepsilon_t)}
\end{align}
=
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=2.4cm of a] (b);
	\vertex [right=2.5cm of a] (c); \vertex [right=2.5cm of b] (d);
	\vertex at ($(a)!0.5!(b) + (0.7cm, 0)$) (e); \vertex at ($(c)!0.5!(d) + (-0.7cm, 0)$) (f);

	\diagram[%inline=(e.base)
	]{
		(a) -- [anti fermion,  quarter left, edge label=t] (e)
		-- [anti fermion,  quarter left, edge label=r] (b)
		-- [anti fermion, quarter left, edge label=a] (a),
		(c) -- [fermion, quarter left, edge label=b] (d)
		-- [fermion, quarter left, edge label=u] (f)
		-- [fermion, quarter left, edge label=s] (c),
		(a) -- [scalar] (c),
		(b) -- [scalar] (d),
		(e) -- [scalar] (f),
		%-- [fermion, quarter right, edge label'=2] (a),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}
+
\begin{minipage}{0.4\linewidth}
	\centering
	\begin{tikzpicture}
	\begin{feynman}[large]
	\vertex (a); \vertex [below=2.4cm of a] (b);
	\vertex [right=2.5cm of a] (c); \vertex [right=2.5cm of b] (d);
	\vertex at ($(a)!0.5!(b) + (0.7cm, 0)$) (e); \vertex at ($(c)!0.5!(d) + (-0.7cm, 0)$) (f);
	\diagram[]{
		(a) -- [fermion,  \qr, \el'=a] (b) -- [ fermion, \el=u] (f)
		-- [ fermion,  \el'=s] (c) -- [ fermion, \ql, \el=b] (d)
		-- [fermion,  \el'=r] (e) -- [fermion, \el=t] (a),
		(a) -- [scalar] (c), 	(b) -- [scalar] (d), (e) -- [scalar] (f),
	};
	\end{feynman}
	\end{tikzpicture}
\end{minipage}

\subsubsection{Summation of Diagrams}

\subsubsection{What Is the Linked-Cluster Theorem?}
\ex{6.13}
For the 3rd-order Goldstone diagrams in Table 6.2,
\begin{equation}\label{key}
\text{diagram1} = (-1)^4 \qty(\dfrac{1}{2}) \sum_{ab}\sum_{rsut} \dfrac{\Braket{ab|ru}\Braket{ru|ts}\Braket{ts|ab} }{(\varepsilon_a + \varepsilon_b - \varepsilon_r - \varepsilon_u)(\varepsilon_a + \varepsilon_b - \varepsilon_t - \varepsilon_t)}
\end{equation}
$ a,b,r,s,u,t $ must come from 1 or 2 molecules. If they come from 2 molecules, $ \Braket{ru|ts} $ must be zero. Thus they only come from 1 molecule, i.e. the value of each Goldstone diagram is $ N $ times the result for a single molecule.

\subsection{Some Illustrative Calculations}

\end{document}


\section*{Reduction}

The
\emph{program reducer} $\Rrightarrow$ is a partial function from
programs to
programs
and $\Rrightarrow^*$ is its reflexive transitive closure.
Within the definition of $\Rrightarrow$, we make use of
the following two groups of analogous partial functions:
The statement
\emph{reducer} $\Rightarrow$ is a partial function from
statements to statements
and $\Rightarrow^*$ is its reflexive transitive closure.
The expression
\emph{reducer} $\rightarrow$ is a partial function from
expressions to expressions
and $\rightarrow^*$ is its reflexive transitive closure.

A \emph{program reduction} is a sequence of programs
$p_1 \Rrightarrow \cdots \Rrightarrow p_n$,
where $p_n$ is not reducible, i.e. there is no
program $q$ such that $p_n \Rightarrow q$.
Here, the program $p_n$ is called the \emph{result
of reducing} $p_1$.

A \emph{value} is a primitive number expression,
primitive boolean expression,
a primitive string expression, a function definition
expression or a function declaration statement.

The \emph{substitution} function
$p [ n \leftarrow v ]$ on programs/statements/expressions
replaces every free occurrence of the name $n$
in statement $p$ by value $v$. Care must be taken to introduce
and preserve
co-references in this process; substitution can introduce
cyclic references in the result of the substitution. For example,
$n$ may occur free in $v$, in which case
every occurrence of $n$ in $p$
will be replaced by $v$ such that $n$ in $v$ refers cyclically
to the node at which the replacement happens.

In summary:

\begin{enumerate}
\item $\Rrightarrow$: \emph{program reducer}
\item $\Rrightarrow^*$: transitive closure of \emph{program reducer}
\item $\Rightarrow$: \emph{statement reducer}
\item $\Rightarrow^*$: transitive closure of \emph{statement reducer}
\item $\rightarrow$: \emph{expression reducer}
\item $\rightarrow^*$: transitive closure of \emph{expression reducer}
\item $p [ n \leftarrow v ]$: substitution function
\end{enumerate}

\subsection*{How to read inference rules}

To read the rules below, think of each "fraction" as an
"if we know X then we can infer Y" implication statement.
For example,

\[
\frac{
  X = 5
}{
  X + 1 = 5 + 1
}
\]

corresponds to the statement
"If we know $X$ is equal to $5$,
then we can infer that $X + 1$ is equal to $5 + 1$".

\subsection*{Programs}

\textbf{First-statement}: In a sequence of statements, we can always
reduce the first one.
\[
\frac{
  \textit{statement}\ \Rightarrow\ \textit{statement}'
}{
  \textit{statement} \ldots
  \Rrightarrow
  \textit{statement}' \ldots
}
\]

\vspace{10mm}

\textbf{Eliminate-function-declaration}: Function declarations as first
statements are substituted in the remaining statements.
\[
\frac{
             f = \textbf{\texttt{function}}\  \textit{name} \
                 \textbf{\texttt{(}}\  \textit{parameters}
                 \ \textbf{\texttt{)}}\ \textit{block}
}{
f\ \textit{statement} \ldots\
  \Rrightarrow\
  \textit{statement} \ldots[\textit{name} \leftarrow f]
}
\]

\vspace{10mm}
\textbf{Eliminate-Values}: Values as first statemments are discarded, if
they are preceding one or more statements in a statement sequence.
\[
\frac{
v \ \mbox{is a value}
}{
v \textbf{\texttt{;}} \textit{statement}+\
   \Rrightarrow  \
  \textit{statement}+
}
\]

\subsection*{Statements: Expression statements}

\textbf{Expression-statement-reduce}: An expression statement
is reducible if its expression is reducible.
\[
\frac{
  e\ \rightarrow\ e'
}{
  e \textbf{\texttt{;}}
  \ \Rightarrow \
  e' \textbf{\texttt{;}}
}
\]

\subsection*{Expressions: Binary operators}

\textbf{Left-binary-reduce}: An expression with binary operator
can be reduced if its left sub-expression can be reduced.
\[
\frac{
  e_1 \ \rightarrow \ e_1'
}{
  e_1\  \textit{binary-operator} \ e_2
  \ \rightarrow \
  e_1'\  \textit{binary-operator} \ e_2
}
\]

\vspace{10mm}
\textbf{And-shortcut-false}: An expression with binary operator
$\textbf{\texttt{\&\&}}$ whose left sub-expression is
$\textbf{\texttt{false}}$ can be reduced to
$\textbf{\texttt{false}}$.
\[
\frac{
}{
  \textbf{\texttt{false}}\  \textbf{\texttt{\&\&}}\ e
  \ \rightarrow \
  \textbf{\texttt{false}}
}
\]

\vspace{10mm}
\textbf{And-shortcut-true}: An expression with binary operator
$\textbf{\texttt{\&\&}}$ whose left sub-expression is
$\textbf{\texttt{true}}$ can be reduced to
the right sub-expression.
\[
\frac{
}{
  \textbf{\texttt{true}}\  \textbf{\texttt{\&\&}}\ e
  \ \rightarrow \
  e
}
\]

\vspace{10mm}
\textbf{Or-shortcut-true}: An expression with binary operator
$\textbf{\texttt{||}}$ whose left sub-expression is
$\textbf{\texttt{true}}$ can be reduced to
$\textbf{\texttt{true}}$.
\[
\frac{
}{
  \textbf{\texttt{true}}\  \textbf{\texttt{||}}\ e
  \ \rightarrow \
  \textbf{\texttt{true}}
}
\]

\vspace{10mm}
\textbf{Or-shortcut-false}: An expression with binary operator
$\textbf{\texttt{||}}$ whose left sub-expression is
$\textbf{\texttt{false}}$ can be reduced to
the right sub-expression.
\[
\frac{
}{
  \textbf{\texttt{false}}\  \textbf{\texttt{||}}\ e
  \ \rightarrow \
  e
}
\]

\vspace{10mm}
\textbf{Right-binary-reduce}: An expression with binary operator
can be reduced if its left sub-expression is a value and its right
sub-expression can be reduced.
\[
\frac{
  e_2\ \rightarrow\ e_2', \textrm{and}\ \textit{binary-operator}
  \ \mbox{is not}\ \textbf{\texttt{\&\&}}\ \textrm{or}\ \texttt{\textbf{||}}
}{
  v\  \textit{binary-operator} \ e_2
  \ \rightarrow \
  v\  \textit{binary-operator} \ e_2'
}
\]

\vspace{10mm}
\textbf{Prim-binary-reduce}: An expression with binary operator
can be reduced if its left and right sub-expressions are values and
the corresponding function is defined for those values.
\[
\frac{
  v\ \mbox{is result of}\ v_1\  \textit{binary-operator} \ v_2
}{
  v_1\  \textit{binary-operator} \ v_2
  \ \rightarrow \
  v
}
\]

\subsection*{Expressions: Unary operators}

\textbf{Unary-reduce}: An expression with unary operator
can be reduced if its sub-expression can be reduced.
\[
\frac{
  e \ \rightarrow \ e'
}{
  \textit{unary-operator} \ e
  \ \rightarrow \
  \textit{unary-operator} \ e'
}
\]

\vspace{10mm}
\textbf{Prim-unary-reduce}: An expression with unary operator
can be reduced if its sub-expression is a value and
the corresponding function is defined for that value.
\[
\frac{
  v'\ \mbox{is result of}\ \textit{unary-operator} \ v
}{
  \textit{unary-operator} \ v
  \ \rightarrow \
  v'
}
\]

\subsection*{Expressions: conditionals}

\textbf{Conditional-predicate-reduce}: A conditional
expression can be reduced, if its predicate can be reduced.
\[
\frac{
  e_1 \ \rightarrow \ e_1'
}{
  e_1\  \textbf{\texttt{?}}\ e_2\ \textbf{\texttt{:}}\ e_3
  \ \rightarrow \
  e_1'\ \textbf{\texttt{?}}\ e_2\ \textbf{\texttt{:}}\ e_3
}
\]

\vspace{10mm}
\textbf{Conditional-true-reduce}: A conditional
expression whose predicate is the value
$\textbf{\texttt{true}}$
can be reduced to its consequent expression.
\[
\frac{
}{
  \textbf{\texttt{true}}\  \textbf{\texttt{?}}\ e_1\ \textbf{\texttt{:}}\ e_2
  \ \rightarrow \
  e_1
}
\]

\vspace{10mm}
\textbf{Conditional-false-reduce}: A conditional
expression whose predicate is the value
$\textbf{\texttt{false}}$
can be reduced to its alternative expression.
\[
\frac{
}{
  \textbf{\texttt{false}}\  \textbf{\texttt{?}}\ e_1\ \textbf{\texttt{:}}\ e_2
  \ \rightarrow \
  e_2
}
\]

\subsection*{Expressions: function application}

\textbf{Application-functor-reduce}: A function application
can be reduced if its functor expression can be reduced.
\[
\frac{
  e \ \rightarrow \ e'
}{
  e\  \textbf{\texttt{(}}\ \textit{expressions} \ \textbf{\texttt{)}}
  \ \rightarrow \
  e'\  \textbf{\texttt{(}}\ \textit{expressions} \ \textbf{\texttt{)}}
}
\]

\vspace{10mm}
\textbf{Application-argument-reduce}: A function application
can be reduced if one of its argument expressions can be reduced and all
preceding arguments are values.
\[
\frac{
  e \ \rightarrow \ e'
}{
  v\  \textbf{\texttt{(}}\ v_1 \ldots v_i \ e \ldots\ \textbf{\texttt{)}}
  \ \rightarrow \
  v\  \textbf{\texttt{(}}\ v_1 \ldots v_i \ e' \ldots\ \textbf{\texttt{)}}
}
\]

\vspace{10mm}
\textbf{Function-declaration-application-reduce}:
The application of a function declaration
can be reduced, if all
arguments are values.
\[
\frac{
  f = \textbf{\texttt{function}}\  \textit{n} \
                 \textbf{\texttt{(}}\  x_1 \ldots x_n
                 \ \textbf{\texttt{)\ \{\ return}}\ \textit{expression}
                 \ \textbf{\texttt{;\ \}}}
}{
  f\ \textbf{\texttt{(}}\ v_1 \ldots v_n\ \textbf{\texttt{)}}
  \ \rightarrow \
  \textit{expression} [x_1 \leftarrow v_1]\ldots[x_n \leftarrow v_n]
  [n \leftarrow f]
}
\]

    \end{document}


\subsection{Traffic Demand Per Subscriber}\label{subsec:behavior}

\begin{figure}[t]
\begin{minipage}{\linewidth}
\centering
%
\begin{subfigure}[b]{.49\linewidth}
\includegraphics[width=\linewidth]{figures/weekday_demand_mean.pdf}
               \caption{Weekday traffic demand.\label{fig:weekday-daily-usage}}
\end{subfigure}
%
\begin{subfigure}[b]{.49\linewidth}
\includegraphics[width=\linewidth]{figures/weekend_demand_mean.pdf}
               \caption{Weekend traffic demand.\label{fig:weekend-daily-usage}}
\end{subfigure}
%
\end{minipage}
\caption{Mean subscriber demand (bytes per 15-minute interval).}
\label{fig:traffic-demand-timeseries}
\end{figure}

\begin{table}[t]
\centering
\begin{tabular}{c c | c c c |}
\cline{3-5}
        &           & median & mean  & 95\%  \\ \hline
\multicolumn{1}{|c|}{\multirow{2}{*}{Weekday}} 	& treatment & 35.97  & 35.58 & 61.12 \\
\multicolumn{1}{|c|}{}					        & control   & 28.06  & 31.12 & 58.78 \\\hline
\multicolumn{1}{|c|}{\multirow{2}{*}{Weekend}}	& treatment & 45.27  & 40.10 & 64.27 \\
\multicolumn{1}{|c|}{} 					        & control   & 41.15  & 37.66 & 62.23 \\\hline
\end{tabular}
\caption{\green{Weekday and weekend traffic demands (MB) per measurement window.}}
\label{tab:traffic-demand-description}
\end{table}

We first explore how an upgrade to a higher service tier affected the
average traffic demand per subscriber, for different times of the day
and days of the week.  Figure~\ref{fig:traffic-demand-timeseries} shows
the average downlink traffic demand across subscribers for a week, for
both the treatment and control groups. We observe that subscriber
behavior differs significantly on weekdays and weekends.  The average
per subscriber demand over a weekday is 35.6 MB, and the 95th percentile
peak demand is 61.12 MB for subscribers in the treatment group
(Table~\ref{tab:traffic-demand-description}).  Over a weekend, the
average demand is 40.1 MB, and the 95th percentile demand is 64.3 MB for
treatment, but the median is 45.27 MB due to consistent use in the major
part of the day.  On weekdays, traffic demand increases monotonically
from morning until prime-time hours in the evening. On weekends, we
observed a sharp rise in demand in the early morning period, from 8:00
a.m. to 10:00 a.m. Then, the demand plateaued until the next rise
before evening prime-time hours. Previous reports indicate that
the aggregate traffic volume for US fixed access link providers usually
troughs during mid-afternoon hours (between
2:00~p.m.--6:00~p.m.)~\cite{sandvine20141h}. In contrast to these
  previous reports, we do not observe such troughs in
subscriber demand.

\begin{figure}[t]
\begin{minipage}{\linewidth}
\centering
\begin{subfigure}[b]{.49\linewidth}
\includegraphics[width=\linewidth]{figures/cdf_peak_demand-overall.pdf}
               \caption{\label{fig:CDF-data-rate-perc95}}
\end{subfigure}
\begin{subfigure}[b]{.49\linewidth}
% \includegraphics[width=\linewidth]{figures/diff_perc95_bytes_subsc-overall.pdf}	% 5 percentile
\includegraphics[width=\linewidth]{figures/diff_perc95_bytes_subsc-overall_01.pdf}		% 1 percentile
               \caption{\label{fig:diff-peak-overall}}
\end{subfigure}
%
\end{minipage}
\caption{95th percentile traffic demand (bytes per 15 minutes) per
  subscriber for the control and treatment groups over the three-month measurement period: (\protect\subref{fig:CDF-data-rate-perc95}) Peak (95\%) traffic demand per subscriber; (\protect\subref{fig:diff-peak-overall}) Change in overall peak (95\%) demand per subscriber. Subscribers were considered at every 5\% in each group. y-axis units are bytes transferred in the peak 15-minute interval, in MB. \label{fig:traffic-demand-overall}}
\end{figure}

% Figure \ref{CDF-data-rate} shows the data rate for each measurement period and every subscriber.
Figure \ref{fig:CDF-data-rate-perc95} shows the distribution of the the
95th percentile downlink traffic demand over the three-month measurement
period. The highest peak demand per 15-minute interval amongst
subscribers in the control group was 2.97 GB; in the treatment group,
the highest peak demand was 3.0 GB.
The average peak traffic demand was 169.8 MB for control and
186.6 MB for treatment. Given the 105 Mbps service-tier capacity,
this means that users rarely utilize their links,
even on averaging the 95th percentile demand (average utilization was
1.43\% for control and 1.5\% for treatment).

We suspected that the subscribers who downloaded most bytes in the
higher service tier would be the ones causing the largest difference in
mean demand, as previous studies have observed such a phenomenon. In
fact, we observed that the more moderate (median) subscribers actually seemed
to exhibit larger differences in traffic demand: The median peak demand
was 66.7 MB for the lower service tier, and 98.4 MB for the higher
tier.  \red{This result indicates that the more moderate subscribers who
received a service-tier upgrade significantly altered their peak
demand.}\sgfoot{Not sure if this needs change or not}
We also observed a significant difference in the mean peak
demand was present in the 50\% of subscribers in the control group
with the lowest traffic demand when compared to the same set of
subscribers of the treatment group. (This disparity appears as a large
gap under the 50\% tick in Figure~\ref{fig:CDF-data-rate-perc95}.)

\green{Figure~\ref{fig:diff-peak-overall} shows another way of looking at this
phenomenon: it explores the difference between the distribution of users
with particular traffic demands in the control and treatment groups.}
For each group, we sort the subscribers according to
increasing demand.  Then we compute the difference in peak demand for
each percentile in the group.  For example, the plot shows the median
user (50\% on the x-axis) increased their peak demand by about 25\% in
response to the service tier upgrade.  Comparing the 70\% subscribers of
both groups with the least demand, we see that peak demand in the
treatment group is higher than the peak demand in the control group,
\red{indicating that in fact even moderate users increase their demand as a
result of the service-tier increase, even though they are not using the
full capacity in either case.}
\sgfoot{Not sure if this needs change. It makes sense as a take-away message.}
 When we combine this analysis with that in
Figure~\ref{fig:CDF-data-rate-perc95}, we find that these subscribers
who respond with increased usage have a peak demand less than 200~MB.
\green{Naturally, the small number of users with the highest demand (closer to
100\%) also show a substantially larger usage for the higher service-tier.}

%\if 0
%\begin{figure}[t]
%\centering
%\includegraphics[width=.5\linewidth]{figures/cdf_diff_perc95_bytes_subsc-overall.pdf}
%               \caption{Distribution of the difference between \treatment{} and \control{}
%               peak traffic demands\label{fig:cdf-diff-perc95}}
%\end{figure}
%
%Figure~\ref{fig:cdf-diff-perc95} shows the distribution of the
%difference in demands of the two groups by percentile. There are 15\%
%subscribers in the \treatment{} that have peak demands have peak demands
%that are similar, or lower than the equivalent 15\% of subscribers in
%\control{}. Figure~\ref{fig:diff-peak-overall} shows that this group
%lies around between the 65 and 80 percentile
%demand. Figure~\ref{fig:CDF-data-rate-perc95} shows that between the
%65-80 percentile mark, subscribers had a demand between 200~MB to
%400~MB. For these subscribers in the treatment group, the equivalent
%subscribers in the control group had demands up to 10~MB higher. 20\% of the
%differences between the peak of the treatment group and the control
%group are higher than 30 MB. These subscribers lie between the between
%30\% and 50\% difference. Their peak demands are between 40--100
%MB as part of the higher tier, and 10--70~MB if they belong to the lower
%tier.  The subscribers with the highest traffic demand in the treatment
%group (beyond 800 MB peak demand) have demands 15--30~MB more than the
%highest peak demanding subscribers in the control group.  In the case of
%upstream traffic, 90\% of users in the treatment group consistently
%upload 2 MB more traffic than the control group. The top 10\% of the
%highest demanding subscribers in the treatment group uploaded between 2--10~MB
%more than the top 10\% of the control group.
%\fi

\begin{figure}[t]
\begin{minipage}{\linewidth}
\centering
\begin{subfigure}[b]{.49\linewidth}
 \includegraphics[width=\linewidth]{figures/cdf_peak_demand-daily.pdf}
                \caption{\label{fig:CDF-data-rate-daily-perc95}}
 \end{subfigure}
\begin{subfigure}[b]{.49\linewidth}
\includegraphics[width=\linewidth]{figures/diff_perc95_bytes_subsc-daily-overall_01.pdf}		%1 percentile
%\includegraphics[width=\linewidth]{figures/diff_perc95_bytes_subsc-daily-overall.pdf}		%5 percentile
                \caption{\label{fig:diff-peak-daily}}
\end{subfigure}
%
\end{minipage}
  \caption{\green{95th percentile traffic demand (bytes per 15 minutes) per day
  per subscriber for the control and treatment groups: (\protect\subref{fig:CDF-data-rate-daily-perc95}) Daily peak (95\%) traffic demand per subscriber; (\protect\subref{fig:diff-peak-daily}) Change in daily peak (95\%) demand per subscriber. Subscribers were considered at every 5\% in each group. y-axis units are bytes transferred in the peak 15-minute interval, in MB.\label{fig:traffic-demand-daily}}}
\end{figure}

\red{Further investigation revealed that users with moderate peak traffic
demands not only exhibit a large difference in their traffic demands in aggregate, but also
on a daily basis.} \origfoot{Further investigation revealed that users with moderate peak traffic
demands not only increase their traffic demands in aggregate, but also
on a daily basis.}  Figure~\ref{fig:traffic-demand-daily} shows that
when subscribers on the lower tier had a daily peak demand under 600~MB,
70\% of subscribers in the treatment group had 15-minute demands that
were 5--20~MB higher.  The ratio of the differences in demand across
percentiles also shows that the \red{40\% of subscribers with lowest peak
demands in the treatment group demonstrate more than
double the daily peak traffic demand of the control group.}
\origfoot{40\% of subscribers with lowest peak
demands in the control group more than
double their daily peak traffic demand in response to service-tier upgrades}

\red{One possible explanation for why moderate users increase their usage
in response to a service-tier upgrade is that the higher service tier
not only affords more capacity, but also a better user experience
(\eg, faster downloads).}\sgfoot{Don't this needs change in current context.}
\red{Thus, even though users may not be
exhausting the capacity of the higher service tier, they nonetheless
seem to respond to the service tier upgrade by using the Internet more
than they had before the service-tier upgrade.} \sgfoot{How to fix this one?}

%
%\if 0
%There could be many reasons for this increase in demand for subscribers with
%lower peak demands. One reason could be short term web activities (such as short videos,
%web browsing) that have a slightly higher traffic demand. Such an increase in demand would
%then be apparent during hours when users pursue such activities (mostly prime-time).
%Another reason could be an increase in background traffic (such as software updates).
%Studying the applications responsible for such behavioral changes in traffic demand is not
%in the scope of our current work. However, an interesting question that arises is: when
%does the demand increase the most throughout a day? Our analysis showed that the most
%subscribers reach within 95\% of daily maximum demand between 8:00 p.m.--12:00 a.m.
%\fi

% \begin{table}[h]
% \begin{tabular}{lllll}
%         &           & 1     & 2     & 3     \\
% Weekday & treatment & 21:45 & 00:00 & 23:30 \\
%         & control   & 22:30 & 22:15 & 00:00 \\
% Weekend & treatment & 23:30 & 23:45 & 14:30 \\
%         & control   & 22:30 & 21:30 &
% \end{tabular}
% \end{table}