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% 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}}
%
% Special symbols 
  \def\ua{\uparrow}
  \def\da{\downarrow}
  \def\dag{\dagger}
  \def\la{{\langle\, }}
  \def\ra{{\,\rangle }}
  \def\pm{\prime}
  \def\cd{c^{\dagger}}
%
% Special letters  
  \def\ek{\epsilon_{k}}
  \def\e {\epsilon} 
  \def\om{\omega}
  \def\b{\beta} 
  \def\r{\rho}  
  \def\s{\sigma}
  \def\t{\tau} 
  \def\d{\delta}
  \def\th{\theta}
  \def\f{\phi} 
  \def \vr{\vec{r}}
  \def\vk{\vec{k}}
  \def\vq{\vec{q}}
  \def\ham{\mathcal{H}}
%
% Particular Symbols
  \def\rq{\rho_q}
  \def\rdq{\rho_{-q}}
  \def\ddq{\Delta^{\dagger}_q}
  \def\dq{\Delta_q}
  \def\kp{k^{\prime}}
  \def\sp{\sigma^{\prime}}
  \def\kma{k_{{\rm max}}}
  \def\km{k_{{\rm  min}}}

\begin{document} 

\begin{center}
Solution to Problem Set 4
\end{center}

We are examining a system of free fermions characterized by the Hamiltonian,
\[
\ham_0 = \sum_{k, \s} \ek \cd_{k, \s} c_{k, \s},
\]
where $\ek = k^2/2m$.
The ground state of such a system is a filled Fermi sea. It can be represented
as 
\[
|\psi_0 \ra = \prod_{k < k_F} \cd_{k, \ua} \cd_{k, \da}.
\]

\mbox{{\bf Zero Temperature Calculation}}

We have to evaluate the correlation function,
\bea
G_{\r}(t,q) &=& (-i) \la T \left( \rq(t) \rdq(0) \right) \ra \nonumber \\
&=& (-i) \{ \th(t) \la \rq(t) \rdq(0) \ra + \th(-t) \la \rdq(0) \rq(t) \ra \},
\eea
where $\rq \equiv \sum_{k, \s} \cd_{k+q, \s} c_{k, \s}$. $\la \hat{O} \ra$  
denotes the ground state average $\la \psi_0 |\hat{O}
|\psi_0 \ra$, since we are evaluating matrix elements at zero temperature.
For a free theory the time evolution of the Heisenberg operators is simple. It
is easy to verify that 
\[
e^{i \ham_o t} \cd_k e^{-i \ham_o t} = e^{i \ek t} \cd_k,
\]
and 
\[
e^{i \ham_o t} c_k e^{-i \ham_o t} = e^{-i \ek t} c_k.
\]
Then,
\bea
\rq(t) &=& e^{i \ham_o t} \rq(0) e^{-i \ham_o t} \nonumber \\
&=& \sum_{k, \s} e^{i(\e_{k+q} - \ek )t} \cd_{k+q, \s} c_{k, \s}.
\eea
From equations (1) and (2) we get,
\bea
G_{\r}(t,q) &=& (-i) \sum_{k,\kp,\s,\sp} e^{i(\e_{k+q} - \ek )t} \left\{ 
\th(t) \la \cd_{k+q, \s} c_{k, \s} \cd_{\kp-q, \sp} c_{\kp, \sp} \ra \right. 
\nonumber \\
&& \left. + \th(-t)
\la \cd_{\kp-q, \sp} c_{\kp,\sp} \cd_{k+q, \s} c_{k, \s} \ra \right\}.
\eea
To evaluate the matrix elements in the above expression we have to remember 
two things. First, $\la \cd_{k,\s} c_{\kp,\sp} \ra = \d_{k,\kp} \d_{\s,\sp}
n_k$, since we can create a hole only from within the Fermi sea, and 
$\la c_{k,\s} \cd_{\kp,\sp} \ra = \d_{k,\kp} \d_{\s,\sp}(1-n_k) $, since we 
can create a particle only outside the Fermi sea. Here $n_k$ (zero temperature
Fermi function) is the particle occupation number of the state $k$. Second,
an expression  like $\la \cd_1 c_2 \cd_3 c_4 \ra$ is non-zero only when 
whatever particle is destroyed (or created) from the Fermi sea is created (or
destroyed) back. Thus the index in $c_4$ must match that of either $\cd_1$ or
$\cd_3$. In the jargon this is usually known as contarction. With these two
observations it is easy to see that,
\bea
\la \cd_{k+q, \s} c_{k, \s} \cd_{\kp-q, \sp} c_{\kp, \sp} \ra &=&
\la \cd_{k+q, \s} c_{k, \s} \ra \la \cd_{\kp-q, \sp} c_{\kp, \sp} \ra +
\la \cd_{k+q, \s}  c_{\kp, \sp} \ra \la c_{k, \s} \cd_{\kp-q, \sp} \ra 
\nonumber \\
&=& \d_{q,0} \ n_{k, \s} \ n_{\kp, \sp} + \d_{\s,\sp} \ \d_{\kp,k+q} \ 
n_{k+q,\s}(1- n_{k,\s}).
\nonumber
\eea
Similarly we get,
\[
\la \cd_{\kp-q, \sp} c_{\kp, \sp} \cd_{k+q, \s} c_{k, \s} \ra = \d_{q,0} 
\ n_{k, \s} \ n_{\kp, \sp} + \d_{\s,\sp} \ \d_{\kp,k+q} \ n_{k,\s} (1- 
n_{k+q,\s})
\]
Then, from equation (3) we get,
\ben
G_{\r}(t,q) = (-i)N^2 \d_{q,0} + (-2i) \sum_{k} e^{i(\e_{k+q} - \ek )t} 
\left\{ \th(t) n_{k+q} (1-n_k) + \th(-t) n_k(1-n_{k+q}) \right\},
\een
where the factor of two in the second term is from a sum over spin indices.
The first term is due to the constant density of the Fermi sea. For the 
purpose of studying density fluctuations the first term is irrelevant, and in
further analysis we drop it.

Next we represent the theta function by the identity,
\ben
\th(t) = - \int_{-\infty}^{\infty} \frac{d \om^{\pm}}{2 \pi i} \ \frac{e^{-i 
\om^{\pm} t}}{\om^{\pm} + i \eta},
\een
because only for $t > 0$, we can close the contour in the lower half plane
( $Im \om^{\pm} < 0$), and pick the pole at $\om^{\pm} = -i \eta $. Similarly,
\ben
\th(-t) = \int_{-\infty}^{\infty} \frac{d \om^{\pm}}{2 \pi i} \ \frac{e^{-i 
\om^{\pm} t}}{\om^{\pm} - i \eta}.
\een
Using these identities we can write,
\[
G_{\r}(t,q)= (2i) \sum_{k} e^{i(\e_{k+q} - \ek )t} \ \int_{-\infty}^{\infty}
\frac{d \om^{\pm}}{2 \pi i} e^{-i \om^{\pm} t} \left\{ \frac{n_{k+q}(1-n_k)}
{\om^{\pm} + i \eta} - \frac{n_k (1- n_{k+q})}{\om^{\pm} - i \eta} \right\}.
\]
Now we can write the Fourier transform of the density-density correlator as,
\bea
G_{\r}(\om,q) &=& \int_{-\infty}^{\infty} dt e^{i \om t} G_{\r}(t,q) \nonumber
\\
&=& 2 \sum_k \left\{ \frac{n_{k+q}(1-n_k)}{\om + \e_{k+q} - \ek + i \eta}
- \frac{n_k (1- n_{k+q})}{\om + \e_{k+q} - \ek - i \eta} \right\}
\eea
The poles in the above expression are the energy for particle-hole excitations
of the Fermi sea. Hence, $G_{\r}(\om,q)$ is also called the particle-hole 
propagator. These exciations are possible modes by which the system can 
absorb energy and go to an excited state from its ground state.

The above expression can be simplified further if in the first term we 
replace $\vk$ by $- \vk - \vq$. Since $\e_{\vk}= \e_{- \vk}$, we get,
\ben
G_{\r}(\om,q) = 2 \sum_k n_k (1- n_{k+q}) \left\{ \frac{1}{\om + \e_k 
- \e_{k+q} + i \eta} - \frac{1}{\om + \e_{k+q} - \ek - i \eta} \right\}
\een

Next we have to evaluate the correlation function $G_{p}(t,q) = (-i)
\la T \left( \dq(t) \ddq(0) \right) \ra$, where $\ddq \equiv \sum_k \cd_{-k+q,
\ua}\cd_{k,\da}$. Proceeding as before we get,
\bea
G_{p}(t,q) &=& (-i) \left\{ \th(t) \la \dq(t) \ddq(0) \ra + \th(-t) \la
\ddq(0) \dq(t) \ra \right\} \nonumber \\
&=& (-i) \sum_{k,\kp} e^{-i (\ek + \e_{-k+q})t} \left\{ \th(t) \la 
c_{k,\da} c_{-k+q,\ua} \cd_{-\kp +q, \ua} \cd_{\kp,\da} \ra \right. \nonumber 
\\
&& \left. + \th(-t) \la
\cd_{-\kp +q, \ua} \cd_{\kp,\da} c_{k,\da} c_{-k+q,\ua} \ra \right\}.
\eea
Since,
\[
\la c_{k,\da} c_{-k+q,\ua} \cd_{-\kp +q, \ua} \cd_{\kp,\da} \ra = \d_{k,\kp}
(1- n_{k,\da})(1- n_{-k+q,\ua})
\]
and
\[
\la \cd_{-\kp +q, \ua} \cd_{\kp,\da} c_{k,\da} c_{-k+q,\ua} \ra = \d_{k,\kp}
n_{k,\da} n_{-k+q,\ua}, 
\]
we get,
\ben
G_{p}(t,q) = (-i) \sum_{k} e^{-i (\ek + \e_{-k+q})t} \left\{ \th(t)
(1- n_{k,\da})(1- n_{-k+q,\ua}) + \th(-t) n_{k,\da} n_{-k+q,\ua} \right\}
\een
We can now evaluate the Fourier transform of the correlator, as we did before.
The final result is, 
\ben
G_{p}(t,q) = \sum_{k} \left\{ \frac{(1- n_{k,\da})(1- n_{-k+q,\ua})}{\om -
\ek - \e_{-k+q} + i \eta} - \frac{n_{k,\da} n_{-k+q,\ua}}{\om - \ek - 
\e_{-k+q} - i \eta} \right\}
\een
The poles of this function are the energies required to create 
particle-particle or hole-hole excitations. The first term in the above 
expression is the retarded part which gives two particle excitations, and 
the second term is the advanced part that gives two hole excitations.

\mbox{{\bf Finite Temperature Calculation}}

We will now evaluate the two correlation functions at finite temperature and
contrast the calculation with the zero temperature case. The density-density 
correlation function is defined as,
\bea
G_{\r}(\t,q) &=& - \frac{1}{Z} {\rm Tr} e^{-\b(\ham_o -\mu N)} T_{\t} \r_q(
\t) \r_{-q}(0) \nonumber \\
&=& - \la T_{\t} \r_q(\t) \r_{-q}(0) \ra, \nonumber
\eea
where now $\la \hat{O} \ra = \frac{1}{Z} {\rm Tr}( e^{-\b(\ham_o -\mu N)}
\hat{O} )$. $\t$ is now imaginary time defined as $\t = it$. The Heisenberg 
operators now evolve in imaginary time, and for a system of free particles
it is easy to show that 
\[
e^{(\ham_0 - \mu N) \t} \cd_k e^{-(\ham_0 - \mu N) \t}
= e^{(\ek - \mu) \t} \cd_k,
\]
and
\[
e^{(\ham_0 - \mu N) \t} c_k e^{-(\ham_0 - \mu N) \t}
= e^{-(\ek - \mu) \t} c_k.
\]
Then,
\[
\r_q(\t) = \sum_{k,\s} e^{(\e_{k+q} - \ek)\t} \cd_{k+q,\s} c_{k,\s}.
\]
Thus,
\ben
G_{\r}(\t,q) = - \sum_{k,\kp,\s,\sp} e^{(\e_{k+q} - \ek)\t} \left\{ \th(\t)
\la \cd_{k+q, \s} c_{k, \s} \cd_{\kp-q, \sp} c_{\kp, \sp} \ra +
\th(- \t) \la \cd_{\kp-q, \sp} c_{\kp, \sp} \cd_{k+q, \s} c_{k, \s} \ra \right
\}
\een
The evaluation of the matrix elements is similar to the zero temperature case.
The only difference is that now $\la \cd_{k,\s} c_{\kp,\sp} \ra = \d_{k,\kp}
\d_{\s,\sp} f_k $, where $f_k$ is the Fermi function that gives average 
particle occupation in the state $k$ at finite temperature. Similarly,
$\la c_{k,\s} \cd_{\kp,\sp} \ra = \d_{k,\kp} \d_{\s,\sp}(1-f_k) $. With this
observation we can write,
\[
\la \cd_{k+q, \s} c_{k, \s} \cd_{\kp-q, \sp} c_{\kp, \sp} \ra
= \d_{q,0} f_{k,\s} f_{\kp,\sp} + \d_{\kp,k+q} \d_{\s,\sp} f_{k+q,\s}(1-
f_{k,\s}),
\]
and
\[
\la \cd_{\kp-q, \sp} c_{\kp, \sp} \cd_{k+q, \s} c_{k, \s} \ra
= \d_{q,0} f_{k,\s} f_{\kp,\sp} + \d_{\kp,k+q} \d_{\s,\sp} f_{k,\s} (1-
f_{k+q,\s}).
\]
Ignoring the term from the constant average density, we get,
\ben
G_{\r}(\t,q) = - 2 \sum_{k} e^{(\e_{k+q} - \ek)\t} \left\{ \th(\t) f_{k+q}
(1- f_k) + \th(-\t) f_k (1- f_{k+q}) \right\}.
\een
The next thing to note is that $G_{\r}(\t)$ is periodic with period $\b$, as
we expect bosonic propagators to be. To verify this we will use the identity
\ben
\frac{f_k}{1- f_k} = e^{- \b \ek}.
\een
Then, for $\t <0$, we have,
\bea
G_{\r}(\t) &=& - 2 \sum_{k} e^{(\e_{k+q} - \ek)\t} f_k (1- f_{k+q}) \nonumber
 \\
&=& - 2 \sum_{k} e^{(\e_{k+q} - \ek)(\t + \b)} f_{k+q}(1- f_k) \nonumber \\
&=& G_{\r}(\t + \b), \nonumber
\eea
for $\t > - \b$. Thus, we have proved that over the range $-\b < \t < \b $,
$G_{\r}(\t,q)$ is periodic. The Fourier transform of $G_{\r}(\t)$ in imaginary
frequency is defined as,
\[
G_{\r}(\t) = \frac{1}{\b} \sum_n e^{-i \om_n \t} G_{\r}(i \om_n).
\]
To satisfy the periodic condition, one needs $e^{-i \om_n \b} = 1$, or
$\om_n = 2\pi n /\b$, i.e. bosonic Matsubara frequencies. The evaluation of 
$G_{\r}(i \om_n,q)$ is now straightforward, and we get,
\bea
G_{\r}(i \om_n,q) &=& \int_0^{\b} d \t e^{i \om_n \t} G_{\r}(\t,q) \nonumber 
\\
&=& (-2) \sum_k \int_0^{\b} d \t e^{(i \om_n + \e_{k+q} - \ek) \t}
f_{k+q}(1- f_k) \nonumber \\
&=& (-2) \sum_k \frac{f_{k+q}(1- f_k)}{i \om_n + \e_{k+q} - \ek} \left(
e^{\b(\e_{k+q} - \ek)} -1 \right) \nonumber \\
&=& 2 \sum_k \frac{f_{k+q} - f_k}{i \om_n + \e_{k+q} - \ek}
\eea

The technology of evaluating the particle-particle (or hole-hole) correlation
function is along similar lines, and we get,
\bea
G_p(\t,q) &=& - \frac{1}{Z} {\rm Tr} e^{-\b(\ham_o -\mu N)} T_{\t} \left(
\dq(\t) \ddq(0) \right) \nonumber \\
&=& - \sum_{k,\kp} e^{-(\ek + \e_{-k+q} -2\mu)\t} \left\{ \th(\t) 
\la c_{k,\da} c_{-k+q,\ua} \cd_{-\kp +q, \ua} \cd_{\kp,\da} \ra + \th(-\t)
\la \cd_{-\kp +q, \ua} \cd_{\kp,\da} c_{k,\da} c_{-k+q,\ua} \ra \right\} 
\nonumber \\
&=& - \sum_k e^{-(\ek + \e_{-k+q} -2\mu)\t} \left\{ \th(\t) (1- f_{k,\da})
(1- f_{-k+q,\ua}) + \th(-\t) f_{k,\da} f_{-k+q,\ua} \right\}
\eea
As in the previous case, we can now show that $G_p(\t,q)$ is periodic over
the range $-\b < \t < \b$, so that the Fourier modes in imaginary frequencies
are only the bosonic Matsubara frequencies. Finally, the result of the Fourier
transform is,
\bea
G_p(i \om_n,q) &=& \int_0^{\b} d \t e^{i \om_n \t} G_p(\t,q) \nonumber \\
&=& -\sum_k (1- f_{k,\da})(1- f_{-k+q,\ua}) \int_0^{\b} d \t 
e^{(i \om_n -\ek - \e_{-k+q} +2\mu)\t} \nonumber \\
&=& -\sum_k \frac{(1- f_{k,\da})(1- f_{-k+q,\ua})}{i \om_n -\ek - \e_{-k+q} 
+2\mu} \left( e^{-\b(\ek + \e_{-k+q}- 2\mu)} - 1 \right) \nonumber \\
&=& \sum_k \frac{1- f_{k,\da} - f_{-k+q,\ua}}{i \om_n -\ek - \e_{-k+q} +2\mu}
\eea

\mbox{{\bf Evaluation of Integrals at Zero temperature}}

\begin{enumerate}

\item
{\bf $G_{\r}(\om,q)$ in 3 dimensions}

From equation (8) we note that $G_{\r}(\om,q)$ is even in $\om$. So we can
restrict our discussion to $\om > 0$. Then,
\[
G_{\r}^R(\om,q) = 2 \sum_k n_k(1- n_{k+q}) \left\{ P \frac{1}{ \om + \ek
- \e_{k+q}} - P \frac{1}{ \om + \e_{k+q} - \ek} \right\}.
\]
Since under the transformation $\vk \rightarrow -\vk -\vq $, the term $n_k 
n_{k+q}\left(1/(\om + \ek- \e_{k+q}) - 1/(\om + \e_{k+q} - \ek) \right)$ 
changes sign, therefore it must be zero. Then,
\ben
G_{\r}^R(\om,q) = 2 \sum_k n_k \left\{ P \frac{1}{ \om + \ek
- \e_{k+q}} - P \frac{1}{ \om + \e_{k+q} - \ek} \right\}.
\een
In the continuum limit this can be written as,
\[
G_{\r}^R(\om,q) = \frac{4 \pi}{(2 \pi)^3} \int_0^{k_F} dk k^2 \int_{-1}^1
dz \left\{ \frac{1}{\om -\e_q - \frac{kqz}{m}} - \frac{1}{\om + \e_q +
\frac{kqz}{m}} \right\}.
\]
We will introduce the dimensionless parameters $q^{\pm} = q/k_F , k^{\pm}
= k/k_F$ and $\om^{\pm} = \om m/k_F^2 $.  Dropping the primes we can rewrite 
the above expression in terms of dimensionless parameters as,
\ben
G_{\r}^R(\om,q) = \frac{k_F m}{2 \pi^2} \int_0^1 dk k^2 \int_{-1}^1dz 
\left\{ \frac{1}{\om - q^2/2 - kqz} - \frac{1}{\om + q^2/2 + kqz} \right\}.
\een
Performing the $z$ integral we get,
\[
G_{\r}^R(\om,q) = \frac{k_F m}{2 \pi^2 q} \int_0^1 dk k \left\{
\ln \left| \frac{\om/q - q/2 +k}{\om/q - q/2 -k} \right| - 
\ln \left| \frac{\om/q + q/2 +k}{\om/q + q/2 -k} \right| \right\}.
\]
Now, using the identity
\[
\int_0^1 dx x \ln(a +x) = \frac{1}{2}\left(1 - a^2 \right) \ln \left| a +1
\right| + \frac{a^2}{2} \ln \left| a\right| - \frac{1}{4} + \frac{a}{2},
\]
we get the final expression,
\bea
G_{\r}^R(\om,q) &=& \frac{k_F m}{2 \pi^2} \left[ -1 + \frac{1}{2q} \left\{
1- \left( \om/q - q/2 \right)^2 \right \} \ln \left| \frac{1+ \left( \om/q
- q/2 \right)}{1- \left( \om/q - q/2 \right)} \right| \right. \nonumber \\
&-& \left.
\frac{1}{2q} \left\{1- \left( \om/q + q/2 \right)^2 \right \}\ln \left|
\frac{1+ \left( \om/q + q/2 \right)}{1- \left( \om/q + q/2 \right)} \right|
\right],
\eea
where all the variables are dimensionless.

Next we evaluate the imaginary part, which for $\om > 0$ is,
\[
G_{\r}^I(\om,q) = - 2 \pi \sum_k n_k \left(1 - n_{k+q} \right) \d
( \om - \e_{k+q} + \ek ).
\]
In the continuum limit, the integral, in terms of dimensionless variables,
can be written as,
\ben
G_{\r}^I(\om,q) = - \frac{k_F m}{2 \pi} \int_0^1 dk k^2 \int_{-1}^1 dz
\th( \left| \vk + \vq \right| -1 ) \d( \om - q^2/2 - kqz ).
\een
The above integral is tricky because we have to satisfy both the 
delta-function and the theta-function restrictions. Since, $|kz| < 1$, 
we get from
the delta-function, $-q + q^2/2 \leq \om \leq q + q^2/2$. This is a necessary 
condition. Next we note that whenever the delta-function is satisfied, the 
z-integral gives a factor of $1/kq$, so that we are left with 
\ben
G_{\r}^I(\om,q) = - \frac{k_F m}{2 \pi q} \int_{\km}^{\kma} dk k,
\een
where $\kma$ and $\km$ are determined by the delta-function or the theta-
function. It is easy to see that $\kma =1$. To determine $\km$ we note that 
the theta-function gives $(k^2 + q^2 + 2kqz) >1$, while from the 
delta-function we have $qkz = \om - q^2/2$. From these we get, $\km = (1-
2 \om)^{1/2}$. This is the $\km$ from the theta-function. Now in the delta-
function, since $|z| \leq 1$, we have $-qk + q^2/2 \leq \om \leq qk + q^2/2$.
This gives $\km = |\om/q - q/2 |$. This is $\km$ from the delta-function. We 
have to choose the larger of the two $\km$. It is easy to see that the 
condition when the second $\km$ dominates is when $\om > q - q^2/2$. Thus, 
finally we have, for $0 < \om < q - q^2/2$, and $q<2$ (since $\om >0$),
\ben
G_{\r}^I(\om,q) = - \frac{k_F m}{2 \pi q} \int_{(1- 2\om)^{1/2}}^{1} dk k
= - \frac{k_F m \om}{2 \pi q},
\een
and for $q - q^2/2 < \om < q+ q^2/2$,
\ben
G_{\r}^I(\om,q) = - \frac{k_F m}{2 \pi q} \int_{(|\om/q - q/2|}^{1} dk k
= - \frac{k_F m}{2 \pi q} \left\{ 1 - \left( \om/q - q/2 \right)^2 \right\}.
\een

\item
{\bf $G_{\r}(\om,q)$ in 2 dimensions}

 In dimensionless parameters the real part is,
\[
G_{\r}^R(\om,q) = \frac{m}{2 \pi^2} \int_0^1 dk k \int_0 ^{2 \pi} d \f 
\left\{ \frac{1}{\om - q^2/2 - kq \cos \f } - \frac{1}{\om + q^2/2 +
kq \cos \f } \right\}.
\]
The $\f$ integral can be done using the identity
\ben
\int_0 ^{2 \pi} \frac{d \f}{a + b \cos \f} = \frac{2 \pi}{\sqrt{a^2 -b^2}}.
\een
Then we have,
\bea
G_{\r}^R(\om,q) &=& \frac{m}{\pi} \int_0^1 dk k \left\{ \frac{1}{\sqrt{
(\om - q^2/2)^2 - k^2q^2}} - \frac{1}{\sqrt{(\om + q^2/2)^2 - k^2q^2}}
\right\} \nonumber \\
&=& - \frac{m}{\pi q} \left\{ \sqrt{ ( \om/q -q/2)^2 -1} - \sqrt{ ( \om/q 
+q/2)^2 -1} \right. \nonumber \\
&+& \left. \left| \om/q +q/2 \right| -  \left| \om/q -q/2 \right| \right\}
\eea

The imaginary part can be written as,
\bea
G_{\r}^I(\om,q) &=& - \frac{m}{2\pi} \int_0^1 dk k \int_0 ^{2 \pi}d \f 
\th( \left| \vk +\vq \right| -1 ) \d( \om -q^2/2 - kq \cos \f) \nonumber \\
&=& - \frac{m}{2\pi} \int_0^1 dk k \int_{-1}^1 dz \frac{1}{\sqrt{1-z^2}}
\th( k^2 + q^2 + 2kqz -1 ) \d (\om -q^2/2 - kqz). 
\eea
The arguement now is similar to that in 3 dimensions. The integral over z
gives a factor of $1/kq$ and the k integral is restricted to $\km$ and 
$\kma =1$. Also since, $z= (\om - q^2/2)/kq $, we have,
\bea
G_{\r}^I(\om,q) &=& - \frac{m}{2\pi q} \int_{\km}^1 dk k \frac{1}{\sqrt{k^2-
(\om/q - q/2)^2}} \nonumber \\
&=& - \frac{m}{2\pi q} \left\{ \sqrt{1- (\om/q - q/2)^2} - \sqrt{\km^2- 
(\om/q - q/2)^2} \right\}.
\eea
Now,  for $0 \leq \om \leq q- q^2/2$, and $q <2$, $\km = (1-2 \om)^{1/2}$.
Then,
\ben
G_{\r}^I(\om,q) = - \frac{m}{2\pi q} \left\{ \sqrt{1- (\om/q - q/2)^2} -
\sqrt{1- (\om/q + q/2)^2} \right\},
\een
and for $q- q^2/2 \leq \om \leq q+ q^2/2$,
\ben
G_{\r}^I(\om,q) = - \frac{m}{2\pi q} \sqrt{1- (\om/q - q/2)^2}.
\een

\item
{\bf $G_{\r}(\om,q)$ in 1 dimension}
\bea
G_{\r}^R(\om,q) &=& \frac{m}{k_F \pi} \int_{-1}^1 dk \left\{ \frac{1}
{\om -q^2/2 - kq} - \frac{1}{\om +q^2/2 + kq} \right\} \nonumber \\
&=& - \frac{m}{k_F \pi q} \left\{ \ln \left| \frac{1+ (\om/q + q/2)}
{1- (\om/q + q/2)} \right| + \ln \left| \frac{1- (\om/q - q/2)}
{1+ (\om/q - q/2)} \right| \right\},
\eea
and
\bea
G_{\r}^I(\om,q)&=& - \frac{m}{k_F} \int_{1-q}^1 dk \d (\om -q^2/2 - kq)
\nonumber \\
&=& - \frac{m}{k_F q} \ \ \ \mbox{for $q- q^2/2 \leq \om \leq q + q^2/2$}, 
\nonumber
\\
&=& 0 \ \ \ \mbox{otherwise}.
\eea






\end{enumerate}

\mbox{{\bf Evaluation of Matsubara Sums (Problem 3) }}

The result of the two Matsubara sums $\frac{1}{\b} \sum_n \frac{e^{i \om_n
0^+}}{i \om_n - \ek}$ and $\frac{1}{\b} \sum_n \frac{e^{i \om_n
0^-}}{i \om_n - \ek}$ can be easily guessed if we remember the definition of
Fourier transform of temperature Green's function. Since,
\bea
g(k,\t) &=& - \la T_{\t} c_k(\t) \cd_k (0) \ra \nonumber \\
   &=& \frac{1}{\b} \sum_n e^{-i \om_n \t} g(k,i \om_n),
\eea
and 
\ben
g(k,i \om_n) = \frac{1}{i \om_n -\ek},
\een
we can easily write,
\ben
\frac{1}{\b} \sum_n \frac{e^{i \om_n 0^+}}{i \om_n - \ek}
= \la \cd_k (0) c_k(0^-) \ra =  f_k,
\een
and
\ben
\frac{1}{\b} \sum_n \frac{e^{i \om_n 0^-}}{i \om_n - \ek}
= - \la c_k(0^+) \cd_k (0) \ra = - (1-f_k)
\een
From this discussion it is obvious that depending on whether $\t > 0 $ or 
$\t < 0$, the Green's function represents particle excitation or hole 
excitation respectively. This is the reason why the two sums differ.

However, now we will evaluate the sums by the usual method as an exercise 
in doing Matsubara sums. In particular, we will evaluate the expression,
\[
g((k,\t) = \frac{1}{\b} \sum_n \frac{e^{-i \om_n \t}}{i \om_n - \ek},
\]
for $\t > 0 $ and $\t < 0$. The important observation is that the Fermi 
function $f(z) = 1/(e^{\b z} +1)$, has poles at fermionic Matsubara 
frequencies $\om_n = (2n +1)\pi/\b$, with residue $-1/\b$. Then, the integral
(refer to figure 1)
\ben
- \oint_C \frac{dz}{2 \pi i} \left( \frac{1}{e^{\b z} +1} \right) \left( 
\frac{1}{z- \ek} \right)e^{-z \t} = \frac{1}{\b} \sum_n \frac{e^{-i \om_n \t}}
{i \om_n - \ek}.
\een  
The integrand in the above expression has another pole at $z= \ek$. The 
contour $C$ can now be deformed into the contour $C^{\pm}$ (refer to figure
2). Thus we can write,
\ben
\frac{1}{\b} \sum_n \frac{e^{-i \om_n \t}}{i \om_n - \ek} =- \oint_{C^{\pm}}
 \frac{dz} {2 \pi i} \left( \frac{1}{e^{\b z} +1} \right) \left( \frac{1}
{z- \ek} \right) e^{-z \t}.
\een
Now, if $\t < 0$, as $\left| z \right| \rightarrow \infty $, the integrand
$\sim \frac{1}{z} e^{-(\b + \t)z}$, for $Re(z) > 0$. This is exponentially
supressed because $\t > - \b$, always. And for $Re(z) < 0$, the integrand
$\sim \frac{1}{z} e^{-z \t}$, which is also exponentially supressed. Thus,
in the contour $C^{\pm}$ the contribution from the large arcs is zero, and the
integral in the above expression only picks up the pole at $z=\ek$. Thus, 
for $\t < 0$,
\ben
\frac{1}{\b} \sum_n \frac{e^{-i \om_n \t}}{i \om_n - \ek} = e^{-\ek \t} f_k,
\een
which confirms the result of equation (35). Now for $\t > 0$, we cannot use
the same arguement as above because for $Re(z) < 0$, the integrand 
$\sim \frac{1}{z} e^{-z \t}$ will diverge exponentially. So, we now introduce 
the function $e^{\b z}/(e^{\b z} +1)$, which also has poles at fermionic 
Matsubara frequencies, with residue $+1/\b$, and which is able to supress the 
above mentioned divergence. Then we can write, for $\t > 0$,
\bea
\frac{1}{\b} \sum_n \frac{e^{-i \om_n \t}}{i \om_n - \ek}
&=& \oint_C \frac{dz}{2 \pi i} \left( \frac{e^{\b z}}{e^{\b z} +1} \right) 
\left( \frac{1}{z- \ek} \right)e^{-z \t} \nonumber \\
&=& \oint_{C^{\pm}} \frac{dz}{2 \pi i} \left( \frac{e^{\b z}}{e^{\b z} +1} 
\right) \left( \frac{1}{z- \ek} \right)e^{-z \t}, \nonumber \\
&=& -(1- f_k)e^{-\ek \t}
\eea
where in the last step we have used the fact that with this choice of function
the integrand vanishes over the large arcs of the contour $C^{\pm}$, and so
picks up only the pole at $z=\ek$. This confirms the result in equation (36).

 
\newpage

\begin{figure}[h]
\unitlength1cm
\begin{picture}(10.0,7.0)
\special{psfile=contour1.eps angle=0 hoffset=100 voffset=-10 hscale=80 
vscale=80}
\end{picture}
\caption{Initial contour C. It picks up the poles at fermionic Matsubara
frequencies.}
\end{figure}
\vspace{2.0cm}

\begin{figure}[h]
\unitlength1cm
\begin{picture}(10.0,7.0)
\special{psfile=contour2.eps angle=0 hoffset=100 voffset=-15 hscale=80 
vscale=80}
\end{picture}
\caption{Deformed contour. The contribution from the arcs is zero.}
\end{figure}


\end{document}