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{\bf  Problem Set 3}
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$H_{o} = \sum\limits_{k} \varepsilon\limits_{k} c^{\dagger}_{k \sigma} c_{k \sigma}$
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$N = \sum\limits_{k} c_{k \sigma}^{\dagger} c_{k \sigma}$
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$\mid \psi_{o} \rangle  =  \prod\limits_{k < k _{F}} c_{k \uparrow}^{\dagger} c_{k \downarrow}^{\dagger} \mid o \rangle$
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$\rho_{q} \equiv \sum\limits_{k \sigma} \ \ c_{t g \sigma}^{\dagger} c_{k \sigma}$
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$\Delta_{q} \equiv \sum\limits_{k} c_{-ktq \sigma \uparrow}^{\dagger} c_{k \downarrow}^{\dagger}$
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a) zero temperature.\\
Calculate the Fourrier transform of 
$G_{\rho}(t,q) G_{p}(t,q)$ in time.\\
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$G_{\rho}(t,q) = \langle T(\rho_{q}(t) \rho_{-q}(0)) \rangle$
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$G_{p}(t,q) = \langle T(\Delta_{q}(t)\Delta_{+q}(o))\rangle$
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in 3 dimensions and in 2 dimensions.
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b)  Finite temperature - if $\tau$ is the imaginary time it is convenient to use Fourrier series.  $G_{\rho}(i \nu_{n},g)= \int_{o}^{\beta} e^{i \nu_{n}\tau} G_{\rho}(\tau,_{t}^{o})d \tau$ 
$G_{p}(i \nu_{n},q) = \int_{o}^{\beta} e^{i \nu_{n}} \tau G_{p}(\tau,_{t}^{o}) d \tau$  
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\underline{Evaluate them} in 3 and 2 dimensions.  

Now G$_{g}$ and G$_{p}$ are defined by
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$G_{\rho}(\tau,q)$ $= tr e^{-\beta(H_{o} -\mu N)}$ $\rho_{t}
^{o}(\rho_{q}(o))$
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$G_{p}(\tau, q) = t_{r}$ $e^{-\beta(H_{o} -\mu N)}$ $\Delta_{q}(\rho) \Delta_{t q}(o)$
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c)  What happens in 1 dimension?

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Hints to solve problem set 3\\

Define  $\rho_{g} (t, t^{o}) = -i  \ \ \langle T (\rho q (t) \rho t q^{t} (o)) \rangle$\\

$\rho_{\Delta}(t, q) = -i \ \ \langle (\Delta_{q} (t)  \ \Delta_{t}^{t} (o) \rangle)$
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$\rho_{q}^{(t)} = \sum\limit_{k} c_{-k q \uparrow} c_{k \downarrow}$
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a)  Use Wick's theorem to express $\rho_{g} (t,q)$ and $\rho_{\Delta}(t,q)$ in terms of\\

$-i \langle T (c_{k}(t) c_{k}^{t} (o)) \rangle = G(kt)$
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b)  Compute G(k,w) = $\int_{\infty}^{\infty} e^{iwt} G(k,t) dt$ - show it as given by\\ 

$G(k,w) = \frac{1}{iw - \varepsilon_{k} ti \rho sign (\mid k \mid-k_{F})}$ $\rho \rangle 0$
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c)  Combine (a) + (b) and use the fact that the Fourrier transfer of the product is the solution of the Fourrier transfer to show that

$\rho_{g} (w,q) = \int_{-\infty}^{\infty} dt e^{iwt}  \rho_{g} (t,q)$\\
$\rho_{\Delta} (w,q) = \int_{-\infty}^{\infty}dt e^{iwt} \rho{\Delta} (t,q)$\\


are given Bg expressions line\\

$\rho_{g} (w,q) ~ -i \int dw' \sum_{k}  G(w'tw, ke) G (w',k)$\\

$\rho_{\Delta} (w,q) ~ -i \int dw' \sum_{k}  G(-w'tw, ktq) G(w',k)$\\
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d)  Now carry out the frequency integral using the residue theorem.  Get expressions of the fermi\\
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$\rho_{g}(w,q) ~ \sum_{k} \frac{[f(\varepsilon_{ktq}) - f(\varepsilon_{k})]}{\varepsilon_{ktq} - \varepsilon_{k} -U - i \sigma} \ \ w > o$\\

$\rho_{\Delta}(w,q) ~ \sum_{k} \frac{[f(\varepsilon_{ktq}) - f(\varepsilon_{k})]}{\varepsilon_{ktq} + \varepsilon_{k} - w -i \sigma} w > o



$f(\eplison_{k}) = {1 lf k<k_{F}}$ \atop
                   $ 1 o  k<k_{F}$

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e)  Now everything is reduced to a d dimensional integral over k.  One way to do them is stay in the case of $\rho_{g}$
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$\rho_{g} ~ \sum_{k} f(\varepsilon_{k} [\frac{1}{\varepsilon_{k} - \varepsilon_{ktq} -w -i \sigma} - \frac{1}{\varepsilon_{ktq}-\varepsilon_{k} -w - i \sigma}]$

Introduce polar ? and intergate over d$\sigma$ and over dk $o \Delta k ,k_{F}$; The 3d integral is done in  Fetter and Walecva page 158-162.

If you want you can calculate the $\rho_{g}$ and $\rho_{g}$ separately.

In 1d the integrals are even easier; 2d is more fun.
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Repeat the same for $\rho_{\Delta}$.
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Good Luck
  
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