\chapter{Fermi-liquid calculation of transport parameters}
\label{app:fermi-liq}
For small temperatures it is admissible in the calculation of transport
parameters to make the following expansion of the self-energy:
\begin{eqnarray}
\Sigma(\omega) &=& \Sigma'{\omega}+i\Sigma''{\omega} 
\nonumber\\
&=& \Sigma'(0)+\omega{\partial \Sigma'(0) \over \partial \omega}
+\Sigma'(0)+{1 \over 2}\omega^{2} {\partial^{2} \Sigma'(0) \over
\partial \omega^{2}}
\nonumber\\
&=& \Sigma'(0)+(1-{1 \over Z})\omega 
- \gamma_{0}((\kb T)^{2}+({\omega \over \pi})^{2})
\nonumber\\
&=& \Sigma'(0)+(1-{1 \over Z})\omega - \gamma(1+({\beta\omega \over
\pi})^{2})
\end{eqnarray}
Then we get (using $x = \beta\omega$),
\begin{eqnarray}
G(\omega,\epsilon) &=& {1 \over \omega+\mu-\epsilon-\Sigma(\omega)}
\nonumber \\
&=& {1 \over {x\kb T \over Z}+\tilde{\mu}-\epsilon
+ \gamma(1+({x \over \pi})^{2})}
\end{eqnarray}
The parameter $Z$ is the quasiparticle residue, $\gamma$ is the scattering
rate at the Fermi surface and $\tilde{\mu}$ is the renormalized Fermi energy.
The correlation functions $A_{n}$'s used in the calculation of the transport
parameters (see section \ref{sec:transport-eqs}) are given by
\begin{eqnarray}
A_{n} &=& {\pi N \over z}\int d\epsilon\Phi(\epsilon)\int d\omega
{\rho^{2}(\omega,\epsilon)(\beta\omega)^{n} \over
4\cosh^{2}({\beta\omega \over 2})}
\nonumber\\
&=& {\pi N \kb T \over z}\int d\epsilon\Phi(\epsilon)\int dx {x^{n} \over
4\cosh^{2}({x \over 2})}
\left\{{{\gamma \over \pi}[1+({x \over \pi})^{2}] \over 
[{xT \over Z}+\tilde{\mu}-\epsilon]^{2}+\gamma^{2} [1+({x \over
\pi})^{2}]^{2}}\right\}^{2}.
\end{eqnarray}
Now $\gamma$ is proportional to $T^{2}$ so at very low temperatures the
$\epsilon$ integral is concentrated around $\epsilon = \tilde{\mu}+x\kb T/Z$
but first note that
\begin{equation}
\int dx{\gamma/\pi \over x^{2}+\gamma^{2}} = 1 \quad \mbox{and}\quad
\int dx\left({\gamma/\pi \over x^{2}+\gamma^{2}}\right)^{2} 
= {1 \over 2\pi\gamma}
\end{equation}
and thus for small $\gamma$ we replace $\left({\gamma/\pi \over
    x^{2}+\gamma^{2}}\right)^{2}$ with ${\delta(x) \over 2\pi\gamma}$.  Then
we get
\begin{eqnarray}
A_{n} &=& {\pi N \kb T \over 2\pi z}\int dx {x^{n} \over 4\cosh^{2}({x
\over 2})} {\Phi(\tilde{\mu}+{x\kb T \over Z}) \over
\gamma [1+({x \over \pi})^{2}]}
\nonumber\\
&=& {N \kb T \over 2z\gamma}\int dx 
\frac{x^{n}}{4\cosh^{2}({x\over 2})}
\frac{
\left\{
\Phi(\tilde{\mu})+
{x\kb T \over Z} {d\Phi(\tilde{\mu}) \over d\tilde{\mu}}+ 
{1 \over 2}\left({x\kb T \over Z}\right)^{2} {d^{2}
\Phi(\tilde{\mu}) \over d\tilde{\mu}^{2}}
\right\}}
{[1+({x \over \pi})^{2}]}
\nonumber\\
&=& {N \kb T \over 2z\gamma}
\left\{
\Phi(\tilde{\mu})E_{n}+
{\kb T \over Z} {d\Phi(\tilde{\mu}) \over d\tilde{\mu}}E_{n+1}+ {1 \over
2}{(\kb T)^{2} \over Z^{2}} {d^{2}\Phi(\tilde{\mu}) \over
d\tilde{\mu}^{2}}E_{n+2}
\right\}
\end{eqnarray}
where
\begin{equation}
E_{m} = \int dx {x^{m} \over 4\cosh^{2}({x \over 2})[1+({x \over
\pi})^{2}]}.
\end{equation}
I believe that these numbers have to be calculated numerically, which
is easily done.  We note that $E_{m} = 0$ for $m$ odd and we get
\begin{equation}
E_{0} = 0.822467 
\qquad\mbox{and}\qquad
E_{2} = 1.752180
\end{equation}
The low temperature transport parameters are then:
\begin{eqnarray}
\sigma &=& E_0{Ne^{2} \over 2z\gamma}\Phi(\tilde{\mu})
\\
S &=& -\frac{\kb}{e}\frac{E_2}{E_0}\frac{\kb T}{Z}
{d \ln(\Phi(\tilde{\mu})) \over d\tilde{\mu}}
=  -\frac{\kb}{e}\frac{E_2}{E_0}\frac{\kb T}{Z}
{d \ln(\sigma)\over d\tilde{\mu}}
\\
\kappa &=& E_2{\kb N (\kb T) \over 2z\gamma}\Phi(\tilde{\mu})
\end{eqnarray}
The figure of merit is then given by
\begin{equation}
ZT = {S^{2}\sigma T \over \kappa} = \frac{E_2}{E_0}\left(
{\kb T \over Z}{d\ln\Phi(\tilde{\mu}) \over d\tilde{\mu}}
\right)^{2}.
\end{equation}
We note here that since the scattering rate goes as $T^2$ the conductivities
diverge at zero temperature, $\sigma \sim T^{-2}$ and $\kappa \sim T^{-1}$.
In this case we can of course neglect any contribution from the lattice
thermal conductivity when computing the figure of merit.

We can go through the same exercise assuming that the scattering can be
attributed to impurities in which case it would be finite at zero temperature
and essentially independent of frequency -- we end up with the same formulas
as before except that now the constants $E_m$ would be different.  Thus we get
\begin{eqnarray}
\sigma &=& {Ne^{2} \over 2z\gamma}\Phi(\tilde{\mu})
\\
S &=& -\frac{\kb}{e}\frac{E_2}{E_0}\frac{\kb T}{Z}
{d \ln(\Phi(\tilde{\mu})) \over d\tilde{\mu}}
=  -\frac{\kb}{e}\frac{E_2}{E_0}\frac{\kb T}{Z}
{d \ln(\sigma)\over d\tilde{\mu}}
\\
\kappa &=& {\pi^2\kb N (\kb T) \over 6z\gamma}\Phi(\tilde{\mu}).
\end{eqnarray}
Now of course the electrical conductivity is finite at zero temperature and
the thermal conductivity tends linearly to zero, which means that one would 
have to take the lattice thermal conductivity into account when the figure of
merit is computed.  

We should note here also that Wiedeman-Franz law holds for both cases but the
Lorenz ratio, $\kappa/\sigma T$, is different, namely
\begin{eqnarray}
L_{Fermi-Liquid} &=& \frac{E_2}{E_0}\left(\frac{\kb}{e}\right)^2
= 15.8 nW\Omega/K^2
\\ 
L_{Impurity} &=& 
\frac{\pi^2}{3}\left(\frac{\kb}{e}\right)^2 = 24.4 nW\Omega/K^2.
\end{eqnarray}
Hence one could use the Lorentz ratio as an indicator of the character of the
low temperature scattering.


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