  \documentclass[12pt]{article}

  \textwidth=17cm
\textheight=22.5cm

\topmargin -1.5cm
\oddsidemargin -0.3cm
%\evensidemargin -0.8cm

\parskip 0.3cm

\pagestyle{plain}




  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  %% Indranil's LaTeX macros
  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  \def\ek{\epsilon_{k}}
  \def\xk{\xi_k} 
  \def\la{{\langle\, }}
  \def\ra{{\,\rangle }}
  \def\p{\,|\,}
  \def\infintd3k{ \int_{-\infty}^\infty d^3r\,}

  % Equation environments
  \def\bea{\begin{eqnarray}}
  \def\eea{\end{eqnarray}}
  \def\ben{\begin{equation}}
  \def\een{\end{equation}}
  \def\benu{\begin{enumerate}}
  \def\enu{\end{enumerate}}

  
  
   \begin{document}

  
   
   \begin{center}
   Solution to Problem Set 2
   \end{center}
        
A system of fermions in the presence of impurities with concentration
$n_i = N/V$ is described by the Hamiltonian,
\ben
\mathcal{H} = \sum_k \ek a^{\dagger}_k a_k + \int dx U(x)\psi^{\dagger}(x)
\psi(x)
\een
where $U(x) = \sum_i V(x-X_i)$ is the array of potentials due to impurities 
located at random positions $X_i$. The Fourier transform of the field operator
is,
\[
\psi(x) = \frac{1}{ \sqrt{V} } \sum_k e^{ikx} a_k
\]
and that of the impurity potential $V(x)$ is,
\[
V(q) = \int dx e^{-iqx}V(x)
\]
The Hamiltonian can be rewritten as
\ben
\mathcal{H} = \sum_k \ek a^{\dagger}_k a_k + \frac{1}{V} \sum_q V(q)
\rho_q \sum_p a^{\dagger}_{p+q}a_p
\een
where $\rho_q = \sum_i e^{-iqX_i}$, is a function of the random variables
$X_i$. We have to perform ensemble average over the random
positions. This makes the Hamiltonian translationally invariant and
therefore tractable. For a typical sample the ensemble average is justified 
only if the probability distribution over the ensembles is sharply peaked.
This condition is satisfied in the limit of weak disorder. The simplest  
ensemble average (with uniform weight) is defined as,
\ben
\overline{F} = \prod_{i=1}^N \frac{1}{V} \int d^3X_i F(X_1, \ldots , X_N)
\een
Then,
\ben
\overline{\rho_q} = \frac{N}{V} \int d^3X e^{-iqX} = N \delta_{q,0}
\een
This average will appear in the first order self energy diagram. For the  
second order term we have,
\[
\overline{\rho_{q_1}\rho_{q_2}} = N^2 \delta_{q_1,0}\delta_{q_2,0} +
N\delta_{q_1+q_2,0}
\]
where we have approximated $N(N-1)$ by $N^2$, assuming N to be
large. For the calculation of self energy we need,
\ben
\overline{\rho_q \rho_{-q}} = N^2 \delta_{q,0} + N
\een
The first term on the right side produce second order improper self energy
diagram, which can be disregarded. The second term will be relevant for 
second order proper self energy diagram.

In the Keldysh formalism Green's function G and self energy $\Sigma$
are matrices defined as,
\[
G =\left[  \begin{array}{cc}
           G^t & G^< \\
           G^> & G^{\tilde{t}}
           \end{array} \right],
 \hspace{.2in}and \hspace{.2in} \Sigma = \left[  \begin{array}{cc}
                      \Sigma^t & \Sigma^<  \\
                      \Sigma^> & \Sigma^{\tilde{t}}
                      \end{array} \right]
\] 
Dyson's equation in frequency-momentum space takes the form,
\ben
G = G_0 + G_0 \sigma_3 \Sigma \sigma_3 G
\een 
where $\sigma_3$ is the Pauli matrix $\left(  \begin{array}{cc}
                                               1 & 0 \\
                                               0 & -1
                                               \end{array} \right)$.
Defining $\tilde{G} = G \sigma_3$, we can rewrite the above equation
as 
\ben
\tilde{G} = \tilde{G_0} +\tilde{G_0} \tilde{\Sigma} \tilde{G}
\een
So, the Feynmn rules to evaluate the diagrams are best written in terms
of $\tilde{G}$ and $\tilde{\Sigma}$. To understand what the Feynman rules
are, let us look at the object $ \la T \psi_{\alpha}\psi^{\dagger}_{\beta}
(\sigma_3)_{\beta\mu} \ra $. In perturbation expansion we have,
\ben
\la T \psi_{\alpha}\psi^{\dagger}_{\beta}(\sigma_3)_{\beta\mu} \ra
= \la T exp[-i\int_{\mathcal{C}}d\tau V(\tau)\psi^{\dagger}(\tau)\psi(\tau)]
  \psi_{\alpha}\psi^{\dagger}_{\beta}(\sigma_3)_{\beta\mu} \ra_0
\een
where the subscript zero indicates average with respect to the free 
Hamiltonian, and $\mathcal{C}$ indicates time integration over the Keldysh
contour. The lower line of the Keldysh contour that runs from $+\infty$ to
$-\infty$ can be reversed by introducing a negative sign. Then the expansion
of the S-matrix looks like,
\bea
 exp[-i\int_{\mathcal{C}}d\tau V(\tau)\psi^{\dagger}(\tau)\psi(\tau)]
= &1&  + (\sigma_3)_{\alpha\beta}\int_{-\infty}^{+\infty}d \tau (-i)
 V(\tau)\psi^{\dagger}_{\alpha}(\tau)\psi_{\beta}(\tau) \nonumber \\
&+& \frac{1}{2!}\left\{(\sigma_3)_{\alpha\beta}\int_{-\infty}^{+\infty}d
\tau (-i) V(\tau)\psi^{\dagger}_{\alpha}(\tau)\psi_{\beta}(\tau)\right\}^2
+ \cdots \nonumber
\eea
Using this expansion in eqn(8), and writing schematically we get,
\bea
\la \psi_{\alpha}\psi^{\dagger}_{\beta}(\sigma_3)_{\beta\mu} \ra
&=& \la \psi_{\alpha}\psi^{\dagger}_{\beta}(\sigma_3)_{\beta\mu} \ra_0
+ \la \psi_{\alpha}\psi^{\dagger}_a (\sigma_3)_{ab}\ra_0
\int_{-\infty}^{+\infty}d \tau (-i)V(\tau) 
\la \psi_{a}\psi^{\dagger}_{\beta}(\sigma_3)_{\beta\mu} \ra_0 \nonumber \\
&+&  \la \psi_{\alpha}\psi^{\dagger}_a (\sigma_3)_{ab}\ra_0
\int_{-\infty}^{+\infty}d \tau (-i)V(\tau) 
\la \psi_b \psi^{\dagger}_c (\sigma_3)_{cd}\ra_0
\int_{-\infty}^{+\infty}d \tau (-i)V(\tau) 
\la \psi_d \psi^{\dagger}_{\beta} (\sigma_3)_{\beta\mu}\ra_0 \nonumber \\
&+& \cdots
\eea
Now comparing eqn(7) and eqn(9) we arrive at the rules that each contraction
gives a single line that denotes the 
free propagator $i\tilde{G_0}$. The scattering vertex is represented
in momentum space by 
$\frac{-iV(q)\rho_q}{V}$, which has to be averaged according to equation (3).
The self energy diagram represents $-i \tilde{\Sigma}$.
Then, the first order self energy diagram, after the impurity average, yields,
\bea
-i \tilde{\Sigma}(k) &=& -i \frac{N}{V} V(q=0) \nonumber \\
 \Sigma^t(k) &=& n_i \int d^3x V(x) \nonumber \\
\Sigma^{\tilde{t}}(k) &=& - n_i \int d^3x V(x)
\eea
To first order there is no contribution to $\Sigma^<$ and $\Sigma^>$. Then,
$\Sigma^A = \Sigma^t - \Sigma^> = \Sigma^t$, and $\Sigma^R = \Sigma^t
- \Sigma^< = \Sigma^t$. Also $\Sigma^K = \Sigma^< + \Sigma^> = 0$. Also to 
be noted is that the first order contribution to self energy is
constant, and it leads to a simple shift in the energy levels of the
free theory. Therefore, this contribution is of no physical interest.

The second order proper self energy diagram, after impurity agerage, gives,
\ben
-i \tilde{\Sigma}(k) = \frac{N}{V^2}\sum_p (-iV(p-k))(-iV(k-p))(i\tilde{G}(p))
\een
To get a closed form expression let us assume a short range potential,
namely, $V(x) = V_0 \delta(x)$. Since,
\bea
G^t(k,\omega) &=& \frac{1-f(\ek)}{\omega - \ek + i\eta} + \frac{f(\ek)}
{\omega - \ek - i\eta} \nonumber \\
&=& P \frac{1}{\omega - \ek} - i \pi \delta(\omega - \ek)(1 - 2f(\ek))
\nonumber
\eea
from equation (9) we get,
\[
\Sigma^t(k,\omega) = \frac{n_i V_0^2}{V} \sum_p \left\{ P \frac{1}{\omega 
 - \epsilon_p} - i \pi \delta(\omega - \epsilon_p)(1 - 2f(\epsilon_p) \right\}
\]
We define density of states as,
\ben
\rho(\epsilon) = \frac{1}{V} \sum_p \delta(\epsilon - \epsilon_p)
\een
then, the above equation can be rewritten as,
\ben
\Sigma^t(k,\omega) = n_i V_0^2 \left\{ P \int_{-\infty}^{\infty}
\frac{\rho(\epsilon)d\epsilon}{\omega - \epsilon} - i \pi \rho(\omega)
(1 - 2f(\omega)) \right\}
\een
Evaluation of the remaining self energies from equation (9) is on
similar lines, and we get,
\bea
\Sigma^{\tilde{t}}(k, \omega) &=& -n_i V_0^2 \left\{ P \int_{-\infty}^{\infty}
\frac{\rho(\epsilon)d\epsilon}{\omega - \epsilon} + i \pi \rho(\omega)
(1 - 2f(\omega)) \right\} \nonumber \\
\Sigma^>(k, \omega) &=& -i (1 - f(\omega))2\pi \rho(\omega)n_i V_0^2
\nonumber \\
\Sigma^<(k, \omega) &=& if(\omega)2\pi \rho(\omega)n_i V_0^2
\eea
And from the definition of $\Sigma^R$ and $\Sigma^A$ we get,
\bea
\Sigma^R &=& n_i V_0^2 \left\{ P \int_{-\infty}^{\infty}\frac{\rho(\epsilon)
d\epsilon}{\omega - \epsilon} - i \pi\rho(\omega)\right\} \nonumber \\
\Sigma^A &=& n_i V_0^2 \left\{ P \int_{-\infty}^{\infty}\frac{\rho(\epsilon)
d\epsilon}{\omega - \epsilon} + i \pi\rho(\omega)\right\}
\eea
Also
\ben
\Sigma^K = -i (1 - 2f(\omega))2\pi \rho(\omega)n_i V_0^2
\een


\end{document}From kotliar@electron.rutgers.edu Thu Mar  4 20:23:49 1999
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Date: Thu, 4 Mar 1999 20:23:45 -0500 (EST)
From: Gabi Kotliar <kotliar@physics.rutgers.edu>
Message-Id: <199903050123.UAA11332@electron.rutgers.edu>
To: kotliar@physics.rutgers.edu
Subject: solnw.tex
Content-Length: 8328

  \documentclass[12pt]{article}

  \textwidth=17cm
\textheight=22.5cm

\topmargin -1.5cm
\oddsidemargin -0.3cm
%\evensidemargin -0.8cm

\parskip 0.3cm

\pagestyle{plain}




  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  %% Indranil's LaTeX macros
  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  \def\ek{\epsilon_{k}}
  \def\xk{\xi_k} 
  \def\la{{\langle\, }}
  \def\ra{{\,\rangle }}
  \def\p{\,|\,}
  \def\infintd3k{ \int_{-\infty}^\infty d^3r\,}

  % Equation environments
  \def\bea{\begin{eqnarray}}
  \def\eea{\end{eqnarray}}
  \def\ben{\begin{equation}}
  \def\een{\end{equation}}
  \def\benu{\begin{enumerate}}
  \def\enu{\end{enumerate}}

  
  
   \begin{document}

  
   
   \begin{center}
   Solution to Problem Set 2
   \end{center}
        
A system of fermions in the presence of impurities with concentration
$n_i = N/V$ is described by the Hamiltonian,
\ben
\mathcal{H} = \sum_k \ek a^{\dagger}_k a_k + \int dx U(x)\psi^{\dagger}(x)
\psi(x)
\een
where $U(x) = \sum_i V(x-X_i)$ is the array of potentials due to impurities 
located at random positions $X_i$. The Fourier transform of the field operator
is,
\[
\psi(x) = \frac{1}{ \sqrt{V} } \sum_k e^{ikx} a_k
\]
and that of the impurity potential $V(x)$ is,
\[
V(q) = \int dx e^{-iqx}V(x)
\]
The Hamiltonian can be rewritten as
\ben
\mathcal{H} = \sum_k \ek a^{\dagger}_k a_k + \frac{1}{V} \sum_q V(q)
\rho_q \sum_p a^{\dagger}_{p+q}a_p
\een
where $\rho_q = \sum_i e^{-iqX_i}$, is a function of the random variables
$X_i$. We have to perform ensemble average over the random
positions. This makes the Hamiltonian translationally invariant and
therefore tractable. For a typical sample the ensemble average is justified 
only if the probability distribution over the ensembles is sharply peaked.
This condition is satisfied in the limit of weak disorder. The simplest  
ensemble average (with uniform weight) is defined as,
\ben
\overline{F} = \prod_{i=1}^N \frac{1}{V} \int d^3X_i F(X_1, \ldots , X_N)
\een
Then,
\ben
\overline{\rho_q} = \frac{N}{V} \int d^3X e^{-iqX} = N \delta_{q,0}
\een
This average will appear in the first order self energy diagram. For the  
second order term we have,
\[
\overline{\rho_{q_1}\rho_{q_2}} = N^2 \delta_{q_1,0}\delta_{q_2,0} +
N\delta_{q_1+q_2,0}
\]
where we have approximated $N(N-1)$ by $N^2$, assuming N to be
large. For the calculation of self energy we need,
\ben
\overline{\rho_q \rho_{-q}} = N^2 \delta_{q,0} + N
\een
The first term on the right side produce second order improper self energy
diagram, which can be disregarded. The second term will be relevant for 
second order proper self energy diagram.

In the Keldysh formalism Green's function G and self energy $\Sigma$
are matrices defined as,
\[
G =\left[  \begin{array}{cc}
           G^t & G^< \\
           G^> & G^{\tilde{t}}
           \end{array} \right],
 \hspace{.2in}and \hspace{.2in} \Sigma = \left[  \begin{array}{cc}
                      \Sigma^t & \Sigma^<  \\
                      \Sigma^> & \Sigma^{\tilde{t}}
                      \end{array} \right]
\] 
Dyson's equation in frequency-momentum space takes the form,
\ben
G = G_0 + G_0 \sigma_3 \Sigma \sigma_3 G
\een 
where $\sigma_3$ is the Pauli matrix $\left(  \begin{array}{cc}
                                               1 & 0 \\
                                               0 & -1
                                               \end{array} \right)$.
Defining $\tilde{G} = G \sigma_3$, we can rewrite the above equation
as 
\ben
\tilde{G} = \tilde{G_0} +\tilde{G_0} \tilde{\Sigma} \tilde{G}
\een
So, the Feynmn rules to evaluate the diagrams are best written in terms
of $\tilde{G}$ and $\tilde{\Sigma}$. To understand what the Feynman rules
are, let us look at the object $ \la T \psi_{\alpha}\psi^{\dagger}_{\beta}
(\sigma_3)_{\beta\mu} \ra $. In perturbation expansion we have,
\ben
\la T \psi_{\alpha}\psi^{\dagger}_{\beta}(\sigma_3)_{\beta\mu} \ra
= \la T exp[-i\int_{\mathcal{C}}d\tau V(\tau)\psi^{\dagger}(\tau)\psi(\tau)]
  \psi_{\alpha}\psi^{\dagger}_{\beta}(\sigma_3)_{\beta\mu} \ra_0
\een
where the subscript zero indicates average with respect to the free 
Hamiltonian, and $\mathcal{C}$ indicates time integration over the Keldysh
contour. The lower line of the Keldysh contour that runs from $+\infty$ to
$-\infty$ can be reversed by introducing a negative sign. Then the expansion
of the S-matrix looks like,
\bea
 exp[-i\int_{\mathcal{C}}d\tau V(\tau)\psi^{\dagger}(\tau)\psi(\tau)]
= &1&  + (\sigma_3)_{\alpha\beta}\int_{-\infty}^{+\infty}d \tau (-i)
 V(\tau)\psi^{\dagger}_{\alpha}(\tau)\psi_{\beta}(\tau) \nonumber \\
&+& \frac{1}{2!}\left\{(\sigma_3)_{\alpha\beta}\int_{-\infty}^{+\infty}d
\tau (-i) V(\tau)\psi^{\dagger}_{\alpha}(\tau)\psi_{\beta}(\tau)\right\}^2
+ \cdots \nonumber
\eea
Using this expansion in eqn(8), and writing schematically we get,
\bea
\la \psi_{\alpha}\psi^{\dagger}_{\beta}(\sigma_3)_{\beta\mu} \ra
&=& \la \psi_{\alpha}\psi^{\dagger}_{\beta}(\sigma_3)_{\beta\mu} \ra_0
+ \la \psi_{\alpha}\psi^{\dagger}_a (\sigma_3)_{ab}\ra_0
\int_{-\infty}^{+\infty}d \tau (-i)V(\tau) 
\la \psi_{a}\psi^{\dagger}_{\beta}(\sigma_3)_{\beta\mu} \ra_0 \nonumber \\
&+&  \la \psi_{\alpha}\psi^{\dagger}_a (\sigma_3)_{ab}\ra_0
\int_{-\infty}^{+\infty}d \tau (-i)V(\tau) 
\la \psi_b \psi^{\dagger}_c (\sigma_3)_{cd}\ra_0
\int_{-\infty}^{+\infty}d \tau (-i)V(\tau) 
\la \psi_d \psi^{\dagger}_{\beta} (\sigma_3)_{\beta\mu}\ra_0 \nonumber \\
&+& \cdots
\eea
Now comparing eqn(7) and eqn(9) we arrive at the rules that each contraction
gives a single line that denotes the 
free propagator $i\tilde{G_0}$. The scattering vertex is represented
in momentum space by 
$\frac{-iV(q)\rho_q}{V}$, which has to be averaged according to equation (3).
The self energy diagram represents $-i \tilde{\Sigma}$.
Then, the first order self energy diagram, after the impurity average, yields,
\bea
-i \tilde{\Sigma}(k) &=& -i \frac{N}{V} V(q=0) \nonumber \\
 \Sigma^t(k) &=& n_i \int d^3x V(x) \nonumber \\
\Sigma^{\tilde{t}}(k) &=& - n_i \int d^3x V(x)
\eea
To first order there is no contribution to $\Sigma^<$ and $\Sigma^>$. Then,
$\Sigma^A = \Sigma^t - \Sigma^> = \Sigma^t$, and $\Sigma^R = \Sigma^t
- \Sigma^< = \Sigma^t$. Also $\Sigma^K = \Sigma^< + \Sigma^> = 0$. Also to 
be noted is that the first order contribution to self energy is
constant, and it leads to a simple shift in the energy levels of the
free theory. Therefore, this contribution is of no physical interest.

The second order proper self energy diagram, after impurity agerage, gives,
\ben
-i \tilde{\Sigma}(k) = \frac{N}{V^2}\sum_p (-iV(p-k))(-iV(k-p))(i\tilde{G}(p))
\een
To get a closed form expression let us assume a short range potential,
namely, $V(x) = V_0 \delta(x)$. Since,
\bea
G^t(k,\omega) &=& \frac{1-f(\ek)}{\omega - \ek + i\eta} + \frac{f(\ek)}
{\omega - \ek - i\eta} \nonumber \\
&=& P \frac{1}{\omega - \ek} - i \pi \delta(\omega - \ek)(1 - 2f(\ek))
\nonumber
\eea
from equation (9) we get,
\[
\Sigma^t(k,\omega) = \frac{n_i V_0^2}{V} \sum_p \left\{ P \frac{1}{\omega 
 - \epsilon_p} - i \pi \delta(\omega - \epsilon_p)(1 - 2f(\epsilon_p) \right\}
\]
We define density of states as,
\ben
\rho(\epsilon) = \frac{1}{V} \sum_p \delta(\epsilon - \epsilon_p)
\een
then, the above equation can be rewritten as,
\ben
\Sigma^t(k,\omega) = n_i V_0^2 \left\{ P \int_{-\infty}^{\infty}
\frac{\rho(\epsilon)d\epsilon}{\omega - \epsilon} - i \pi \rho(\omega)
(1 - 2f(\omega)) \right\}
\een
Evaluation of the remaining self energies from equation (9) is on
similar lines, and we get,
\bea
\Sigma^{\tilde{t}}(k, \omega) &=& -n_i V_0^2 \left\{ P \int_{-\infty}^{\infty}
\frac{\rho(\epsilon)d\epsilon}{\omega - \epsilon} + i \pi \rho(\omega)
(1 - 2f(\omega)) \right\} \nonumber \\
\Sigma^>(k, \omega) &=& -i (1 - f(\omega))2\pi \rho(\omega)n_i V_0^2
\nonumber \\
\Sigma^<(k, \omega) &=& if(\omega)2\pi \rho(\omega)n_i V_0^2
\eea
And from the definition of $\Sigma^R$ and $\Sigma^A$ we get,
\bea
\Sigma^R &=& n_i V_0^2 \left\{ P \int_{-\infty}^{\infty}\frac{\rho(\epsilon)
d\epsilon}{\omega - \epsilon} - i \pi\rho(\omega)\right\} \nonumber \\
\Sigma^A &=& n_i V_0^2 \left\{ P \int_{-\infty}^{\infty}\frac{\rho(\epsilon)
d\epsilon}{\omega - \epsilon} + i \pi\rho(\omega)\right\}
\eea
Also
\ben
\Sigma^K = -i (1 - 2f(\omega))2\pi \rho(\omega)n_i V_0^2
\een


\end{document}

