
\documentstyle[12pt]{article}
\begin{document}
\begin{center}
{\bf  Problem Set 4, Due November 15th 2000}
\end{center}

$$H_{0} = \sum\limits_{k\sigma} \varepsilon_{k} c^{\dagger}_{k \sigma} c_{k \sigma}$$

$$N = \sum\limits_{k\sigma} c_{k \sigma}^{\dagger} c_{k \sigma}$$

$$\mid \psi_{0} \rangle  =  \prod\limits_{k < k _{F}}
c_{k \uparrow}^{\dagger} c_{k \downarrow}^{\dagger} \mid 0 \rangle$$

$$\rho_{q} \equiv \sum\limits_{k \sigma} \ \ c_{k+q \sigma}^{\dagger} c_{k \sigma}$$

$$\Delta_{q} \equiv \sum\limits_{k} c_{- k+q \sigma \uparrow}^{\dagger} c_{k \downarrow}^{\dagger}$$

a) zero temperature.\\
Calculate the Fourier transform of
$G_{\rho}(t,q)$, $G_{p}(t,q)$ in time.\\

$$G_{\rho}(t,q) = \langle T(\rho_{q}(t) \rho_{-q}(0)) \rangle$$

$$G_{p}(t,q) = \langle T(\Delta_{q}(t)\Delta_{-q}(0))\rangle$$
while a general expression involves integration for the case of $d$ dimensions
can you evaluate some of the integrals?


b)  Finite temperature - if $\tau$ is the imaginary time it is convenient to use Fourier
series
$$G_{\rho}(i \nu_{n},g)= \int_{o}^{\beta} e^{i \nu_{n}\tau} G_{\rho}(\tau,q)d \tau$$
$$G_{p}(i \nu_{n},q) = \int_{o}^{\beta} e^{i \nu_{n}} \tau G_{p}(\tau,q) d \tau.$$
Evaluate them in 3 and 2 dimensions.

Now $G_{g}$ and $G_{p}$ are defined by

$$G_{\rho}(\tau,q)= {1\over Z} tr e^{-\beta(H_{o} -\mu N)}\rho_{q}({\tau})\rho_{-q}(o) )$$
$$G_{\rho}(\tau,q)= {1\over Z} tr e^{-\beta(H_{o} -\mu N)}\Delta_{q}({\tau})\Delta_{-q}(o) )$$

%c)  What happens in 1 dimension?
c) Calculate
$$T\sum\limits_{n}\frac{e^{i\omega_n 0^+}}{i\omega_n -\varepsilon_k}  \mbox {   and   }
T\sum\limits_{n}\frac{e^{i\omega_n 0^-}}{i\omega_n -\varepsilon_k} .$$

Interpret the result.\\[0.5cm]


{\bf Hints to solve problem set 3}\\

Define
$$D_{\rho} (t, q) = -i   \langle T (\rho_q (t) \rho_{q}^{\dagger} (0)) \rangle$$
$$D_{\Delta} (t, q) = -i   \langle T (\Delta_q (t) \Delta_{q}^{\dagger} (0)) \rangle$$

$\rho_{q}{(t)} = \sum\limits_{k} c^{\dagger}_{k+q} c_{k}$; \ \
$\Delta_{q}{(t)} = \sum\limits_{k} c_{-k+q\uparrow } c_{k\downarrow}$; \ \
$ \langle   ...  \rangle $ is with
respect to a Fermi set.


a)  Use Wick's theorem to express $D_{\rho} (t,q)$ and $D_{\Delta}(t,q)$ in terms of

$-i \langle T (c_{k}(t) c_{k}^{\dagger} (0)) \rangle = G(k,t)$

b)  Compute $$G(k,\omega ) = \int_{-\infty}^{\infty} e^{i\omega t} G(k,t) dt$$
- show it is given by

$$G(k,\omega) = \frac{1}{i\omega - \varepsilon_{k} + i\delta sign (\mid k \mid-k_{F})},
\; \; \delta > 0, \;\;\; \delta \to 0.$$

c)  Combine (a) + (b) and use the fact that the Fourier transform
of the product is the solution of the Fourier transform to show that

$$D_{\rho}(\omega,q) = \int_{-\infty}^{\infty} dt e^{i\omega t}  D_{\rho}(t,q)$$
$$D_{\Delta}(\omega,q) = \int_{-\infty}^{\infty} dt e^{i\omega t}  D_{\Delta}(t,q)$$


are given by expressions like\\

$$D_{\rho}(\omega ,q)  \sim -i \int d\omega ' \sum_{k}  G(\omega '+\omega , k+q) G (\omega ',k)$$
$$D_{\Delta}(\omega ,q) \sim -i \int d\omega ' \sum_{k}  G(-\omega '+\omega , k+q) G (\omega ',k)$$

d)  Now carry out the frequency integral using the residue theorem.
Get expressions of the functions\\

$$D_{\rho}(\omega,q) \sim \sum_{k} \frac{[f(\varepsilon_{k+q}) - f(\varepsilon_{k})]}
{\varepsilon_{k+q} - \varepsilon_{k} -\omega - i \delta}, \ \ \omega > 0$$

$$D_{\Delta}(\omega,q) \sim \sum_{k} \frac{[1-(f(\varepsilon_{k+q}) + f(\varepsilon_{k}))]}
{\varepsilon_{k+q} + \varepsilon_{k} -\omega - i \delta}, \ \ \omega > 0$$



$$
f(\varepsilon_{k}) =\left\{
\begin{array}{ll}
1& \mbox{if } k<k_{F} \\
0& \mbox{if } k>k_{F}
\end{array}
\right .
$$

e)  Now everything is reduced to a $d $ dimensional integral over $k$.
One way to do them is say in the case of $D_{\rho}$

$$D_{\rho} \sim \sum\limits_{k} f(\varepsilon_{k})
\left[\frac{1}{\varepsilon_{k} - \varepsilon_{k+q} -\omega -i \delta} -
\frac{1}{\varepsilon_{k+q}-\varepsilon_{k} -\omega - i \sigma}\right]$$

Introduce polar coordinates and integrate over $d\theta $  and over
$dk$,\linebreak
 $0 \le  k \le k_{F}$. The 3d integral is done in  Fetter and Walecka page 158-162.

If you want, you can calculate the $D_{\rho}$ and  Im$D_{\rho}$ separately.

In $1d$ the integrals are even easier; $2d$ is more fun.\\

Repeat the same for $D_{\Delta}$.\\

Good Luck

\end{document}
