%\documentclass[12pt,aps,floats,superscriptaddress]{revtex4}
\documentclass[12pt,aps]{revtex4}
\usepackage{amsmath}

\newcommand{\ci}{c_{i\sigma}}
\newcommand{\ck}{c_{k\sigma}}
\newcommand{\cid}{c_{i\sigma}^{\dagger}}
\newcommand{\ckd}{c_{k\sigma}^{\dagger}}
\newcommand{\cj}{c_{j\sigma}}
\newcommand{\cjd}{c_{j\sigma}^{\dagger}}
\newcommand{\cl}{c_{l\sigma}}
\newcommand{\cld}{c_{l\sigma}^{\dagger}}
\newcommand{\cm}{c_{m\sigma}}
\newcommand{\cmd}{c_{m\sigma}^{\dagger}}

\begin{document}

\begin{center}
{\Large Solution for Homework 1}
\end{center}

\vspace{.5cm}
\noindent
Problem 1.

\vspace{.5cm}
\noindent
a) The current and polarization operators are defined as:
$$
j_x=e\sum_{k\sigma}{\partial\epsilon_k\over\partial k_x}
c_{k\sigma}^{\dagger}c_{k\sigma},\qquad\qquad
P_x=e\sum_{i\sigma}
c_{i\sigma}^{\dagger}c_{i\sigma}
r^x_i
$$
Operators in momentum space $c_{k\sigma}$ are related to 
real space representation $c_{i\sigma}$ as:
$$
\ck={1\over V}\sum_ie^{-ikr_i}\ci,\qquad\qquad
\ckd={1\over V}\sum_ie^{ikr_i}\cid
$$
so their anticommutators are
$$
\{\ckd,\ci\}={1\over V}e^{ikr_i},\qquad\qquad
\{\ck,\cid\}={1\over V}e^{-ikr_i}
$$
We use a relationship
$$
[A,BC]=B[A,C]+[A,B]C=-B\{A,C\}+\{A,B\}C
$$
to find:
\begin{eqnarray}
[\ckd\ck,\cid\ci]&=&\cid[\ckd\ck,\ci]+[\ckd\ck,\cid]\ci=
-\cid\{\ckd,\ci\}\ck+\ckd\{\ck,\cid\}\ci\nonumber\\
&&=-{1\over V}e^{ikr_i}\cid\ck+{1\over V}e^{-ikr_i}\ckd\ci
\nonumber
\end{eqnarray}
Now we can evaluate the commutator
\begin{eqnarray}
[j_x,P_x]&=&e^2\sum_{ik\sigma}{\partial\epsilon_k\over\partial k_x}r^x_i
[\ckd\ck,\cid\ci]\nonumber\\
&=&ie^2\sum_{ik\sigma}{\partial\epsilon_k\over\partial k_x}{1\over V}
\left({\partial\over\partial k_x}(e^{ikr_i}\cid)\ck
+\ckd{\partial\over\partial k_x}(e^{-ikr_i}\ci)\right)\nonumber\\
&=&ie^2\sum_{k\sigma}{\partial\epsilon_k\over\partial k_x}
{\partial\over\partial k}\ckd\ck=
-ie^2\sum_{k\sigma}{\partial^2\epsilon_k\over\partial k^2}\ckd\ck
\label{jp}
\end{eqnarray}
To derive the f-sum rule we use the Kubo formula:
\begin{equation}
\sigma(\omega)={1\over\omega}\int_0^{+\infty}
dte^{i\omega t}\langle[j^{\dagger}(t),j(0)]\rangle+i{e^2\over\omega}
\sum_{k\sigma}{\partial^2\epsilon_k\over\partial k^2}
\langle\ckd\ck\rangle
\label{kubo}
\end{equation}
Noticing that $\partial P/\partial t=j$, integrating by parts 
Eq.(\ref{kubo}) and using the commutator Eq.(\ref{jp}) we get:
\begin{equation}
\sigma(\omega)=i\int_0^{+\infty}
dte^{i\omega t}\langle[j(0),P(t)]\rangle
\nonumber
\end{equation}
Integrating the conductivity other the frequency we derive the sum rule:
\begin{equation}
\int_{-\infty}^{+\infty}d\omega\sigma(\omega)=i2\pi\int_0^{+\infty}
dt\delta(t)\langle[j(0),P(t)]\rangle=i\pi\langle[j,P]\rangle
=\pi e^2\sum_{k\sigma}{\partial^2\epsilon_k\over\partial k^2}
\langle\ckd\ck\rangle
\nonumber
\end{equation}

\vspace{.5cm}
\noindent
b)
Integrating by parts a Fourier transform of a function we obtain
an expansion in $1/\omega$:
$$
f(\omega)=\int_{-\infty}^{+\infty}dt e^{i\omega t}f(t)=
-\sum_{k=0}^{\infty}(i\omega^{-1})^{k+1}[f^{(k)}(0)]
$$
where $[f^{(k)}(0)]$ is a jump of the function $f^{(k)}(t)$ at $t=0$:
$$
[f^{(k)}(0)]\equiv f^{(k)}(0^+)-f^{(k)}(0^-)
$$
For example, for a Green's function:
\begin{equation}
G_{ij}(i\omega)=-{1\over i\omega}[G_{ij}(0)]
+{1\over(i\omega)^2}[{\partial G_{ij}(0)\over\partial\tau}]
-{1\over(i\omega)^3}[{\partial^2 G_{ij}(0)\over\partial\tau^2}]+...
\label{gmoments}
\end{equation}
where 
$$
[G_{ij}(0)]=
-\langle c_i c_j^{\dagger}\rangle -\langle c_j^{\dagger}c_i\rangle
=-\langle\{c_i,c_j^{\dagger}\}\rangle=-\delta_{ij}
$$
$$
[{\partial G_{ij}(0)\over\partial\tau}]
=-\langle {\partial c_i\over\partial\tau} c_j^{\dagger}\rangle 
-\langle c_j^{\dagger}{\partial c_i\over\partial\tau}\rangle=
-\langle [H,c_i] c_j^{\dagger}\rangle
-\langle c_j^{\dagger}[H,c_i]\rangle
=-\langle\{[H,c_i],c_j^{\dagger}\}\rangle
$$
Similarly:
$$
[{\partial^2 G_{ij}(0)\over\partial\tau^2}]
=-\langle\{[H,c_i],[H,c_j^{\dagger}]\}\rangle
$$
Using $[\cid\cj,\ci]=-\cj$ we evaluate:
\begin{eqnarray}
[H,\ci]&=&-\sum_{lj}t_{ij}[\cld\cj,\ci]+
U\sum_l[n_{l\uparrow}n_{l\downarrow},\ci]+\mu[\cld\cl,\ci]\nonumber\\
&=&\sum_j(t_{ij}-\mu\delta_{ij})\cj-U\ci n_{i\bar\sigma}\nonumber\\
{[H,\cid]}&=& -[H,\ci]^{\dagger}=
-\sum_j(t_{ji}-\mu\delta_{ij})\cjd+U\cid n_{i\bar\sigma}\nonumber
\end{eqnarray}
From where:
\begin{equation}
\{[H,\ci],\cjd\}=(t_{ij}-\mu\delta_{ij})-U\delta_{ij}n_{i\bar\sigma}
\label{com1}
\end{equation}
\begin{equation}
\{[H,\ci],[H,\cjd]\}=
-\sum_l(t_{il}-\mu\delta_{il})(t_{lj}-\mu\delta_{lj})
+U(t_{ij}-\mu\delta_{ij})(n_{i\bar\sigma}+n_{j\bar\sigma})
-U^2\delta_{ij}n_{i\bar\sigma}n_{j\bar\sigma}
\label{com2}
\end{equation}
Assuming no symmetry breaking 
$\langle n_{i\sigma}\rangle=\langle n\rangle$ for 
$\forall\,\,\, i,\sigma$
we rewrite Eqs.(\ref{com1},\ref{com2}) in momentum space:
\begin{eqnarray}
{[G^{(1)}(0,k)]}&=&\epsilon_k+U\langle n\rangle\label{comk1}\\
{[G^{(2)}(0,k)]}&=&-\epsilon_k^2
-2U\epsilon_k\langle n\rangle+U^2\langle n^2\rangle\label{comk2}
\end{eqnarray}
With the self-energy $\Sigma(\omega,k)=A_1+A_2/(i\omega)$
expansion of the Green's function reads:
\begin{equation}
G(\omega,k)={1\over i\omega-\epsilon_k-\Sigma(\omega,k)}=
{1\over i\omega}+
{1\over (i\omega)^2}(\epsilon_k+A_1)+
{1\over (i\omega)^3}(A_2+(\epsilon_k+A_1)^2)
\label{g}
\end{equation}
From Eqs.(\ref{gmoments},\ref{comk1},\ref{comk2},\ref{g}) we
obtain:
\begin{eqnarray}
A_1&=&U\langle n\rangle\nonumber\\
A_2&=&U^2(\langle n^2\rangle-\langle n\rangle^2)=
U^2\langle n\rangle(1-\langle n\rangle)\nonumber
\end{eqnarray}

\end{document}
