\chapter{Transport and Thermodynamic Properties}
\label{chap:transport-eqs}
In this chapter we will use the results from chapter \ref{chap:HubbardModel}
to compute the transport properties of the Hubbard model with special
attention to thermoelectric properties.  As discussed in that chapter the NCA
results are not valid at very low temperatures so we will use the IPT method
in the second half of this chapter to extend the results to lower
temperatures.  However before we start computing the transport properties of
the model let us start with a brief look at thermoelectricity and some basic
transport theory.

\section{A brief overview of thermoelectricity}
\label{sec:thermoelectricity}
Thermoelectric effects which are responsible for the direct conversion of heat
energy into electrical energy and vice versa have in recent years received
renewed attention \cite{Mahan:1997}.  This is in large part due to progress in
material science and discoveries of new materials with promising
thermoelectric potential but theoretical methods have also come to the fore
and are now playing a key role in identifying likely candidates for
thermoelectric applications.

It is now almost 200 years since thermoelectric phenomena were discovered and
studied by Seebeck and although in hindsight he could in principle have
generated electricity with quite favorable efficiency compared to steam
engines there was little interest or progress in thermoelectricity during most
of the nineteenth century.  It was not until around 1910 that theory of
thermoelectric power generation and refrigeration was developed and it was
realized that the thermoelectric performance of materials could be quantified
with the so called dimensionless figure of merit given by
\begin{equation}
Z_{T}T = {S^{2}\sigma T \over \kappa} = {S\pi\over \kappa\rho}
\end{equation}
where $S$ is the thermopower or Seebeck coefficient, $\sigma = 1/\rho$ is the
electrical conductivity, $T$ is the temperature, $\kappa$ is the thermal
conductivity and $\pi = TS$ is the Peltier coefficient \cite{Kittel:ISSP}.  In
later chapters we will derive this equation more thoroughly but here we can
already understand the figure of merit in the following simple manner,
\begin{itemize}
\item $Z_{T}T \propto S$ so as to get a large voltage
drop for a given temperature difference. 
\item $Z_{T}T \propto \pi = \frac{J_Q}{J_C}$ so as to get a large heat
  current, $J_Q$ for a given electric current, $J_C$.
\item $Z_{T}T \propto {1 \over \rho}$ so as not to incur
Joule heating in the circuit. 
\item  $Z_{T}T \propto {1 \over \kappa}$ in order not to lose heat through
irreversible heat flow between the reservoirs. 
\end{itemize}
Early on the study of thermoelectricity focused mostly on metals and metallic
alloys but in the 1930 the attention shifted to semiconductors when it was
realized that they could possess large thermopower and give better efficiency
than metallic thermoelements.  The theory of thermopower in semiconductors was
developed around 1950 by Abraham Ioffe and subsequently the field enjoyed a
tremendous revival motivated by industrial and military applications.  The
launch of the space program and the need for reliable autonomous sources of
electricity was also an important factor in pushing forward the search for
good thermoelectric materials.

The best thermoelectric elements found during this period
\cite{CRC-Thermoelectrics} were doped bismuth telluride alloys (Bi$_2$Te$_3$)
and antimony telluride alloys (Sb$_2$Te$_3$) for use in refrigeration with
maximum operating temperatures of around 450 K.  For higher temperature
applications (1300K) such as power generation, the best materials have proven
to be lead telluride alloys (PbTe) and to a greater extent, silicon germanium
(SiGe) alloys.  All of these materials have dimensionless figure of merit
around 1.0 which is still today among the highest that can be found.

The search for new thermoelectric materials has in recent years been shifting
towards more structurally complex materials such as skutterudites like
CeFe$_{4}$Sb$_{12}$ or La(Fe,Co)$_{4}$Sb$_{12}$.  This shift can on one hand
be attributed to the fact that most simple semiconducting materials have
already be studied for this purpose but behind this is also the guiding
principle that a good thermoelectric material should conduct electricity like
a crystal but the thermal conductivity should be more characteristic of a
glass \cite{Sales:1997}.  So one of the reason for the favorable
thermoelectric properties of the materials above is the low value of thermal
conductivity which can be attributed to the thermal motion or rattling of the
rare earth ions in the crystal structure.

Another avenue that has recently been explored is to construct thermoelements
out of quantum well superlattices and both experimental \cite{Hicks:1996} and
theoretical work \cite{Sofo:1994,Broido:1995a,Broido:1995b} indicates that a
significant increase in the figure of merit is possible.  These effort are
mostly academic as yet and it is doubtful whether practical applications are
feasible.

Through the years some guidelines have been developed to aid in the search for
good thermoelectric semiconductors.  It is found that a good candidate for
thermoelectric use should have the following properties 
\cite[chapter 23]{CRC-Thermoelectrics}:
\begin{itemize}
\item A large value of the parameter, $A = \mu T (m^*)^{3/2}/\kappa_L$, where
  $\mu$ is the mobility, $m^*$ is the effective mass, T is the absolute
  temperature and $\kappa_L$ is the lattice thermal conductivity.
\item The band gap should be greater that $4k_BT$.
\item The material should be dopable to the extrinsic
regime.\footnote{This means that the conduction behavior is dominated
by the dopants and not by the intrinsic properties of the medium
\cite{Blakemore}.}
\end{itemize}
It is intuitively clear that $\kappa_L$ should be small and the mobility
should of course be large to minimize resistance.  The effective mass enters
here through the density of states at the Fermi level which also affects the
electric conductivity.  The restrictions on the band gap are made so as to
minimize the excitations of minority carriers across the gap which would
adversely affect the thermopower.  In addition the requirement on the doping
is so as to be able to have reasonable number of carriers while trying to
balance the deleterious effects that overdoping has on the thermopower.  

In a recent reevaluation of the analysis leading to the principles above Mahan
\cite{Mahan:1989} comes to essentially the same conclusion except that he
requires the band gap to be around $10k_BT$ but not larger to avoid the
increase of the lattice thermal conductivity that accompanies the increase of
the bandgap.  His conclusions seem to be consistent with the bismuth telluride
semiconductors which have band gaps in this range.

In a recent paper by Mahan and Sofo \cite{Mahan:1996} they consider
the general mathematical problem of finding the optimal conditions for
achieving good thermoelectric properties.  Their approach is to
separate the material characteristics from the expressions for the
transport parameters (see equations
(\ref{eq:transportParameters}),(\ref{eq:An-function}) and
(\ref{eq:MultiOrbitalAnFunction})) and define a function,
$\Sigma_{MS}$, that embodies these characteristics.  The function
$\Sigma_{MS}$, which corresponds to our transport function in equation
(\ref{eq:MultiOrbitalAnFunction}), is now taken as a free parameter
essentially and the figure of merit is optimized by varying this
function.  The resulting optimal $\Sigma_{MS}$ turns out to be a delta
function positioned slightly above the Fermi-surface.  In real materials
it is impossible for this function to be a real delta function but in
strongly correlated materials with narrow bands near the Fermi-surface
this function could be delta function like.  On the other hand, in
addition to the narrow correlated levels, there are usually also some
very broad, featureless bands that cross the Fermi-surface and
according to the analysis of Mahan and Sofo, those band have quite a
deleterious effect on the figure of merit.

The theoretical analysis and the computation of transport properties in the
semiconducting materials discussed above is all based on kinetic theory and
the Boltzmann equation.  However Boltzmann theory is not expected to be valid
for strongly correlated materials where correlation effects significantly
alter the electronic structure and the carriers in the system are in many
particle cooperative state rather than being simply described by single
electron physics. 

\section{Basic Transport Theory}
\label{sec:transport-theory}
In this section we will briefly review the basic notions of transport theory
as described in references
\cite{Callen:Thermodynamics,Heikes:Thermoelectricity,Barnard:Thermoelectricity,
  Mahan}.  Consider a thermodynamic system which is acted upon by external
forces $X_i$ which is turn induce currents, $J_i$ in the system.  Within the
linear response formalism we assume that the currents depend linearly on the
external forces according to the phenomenological equation:
\begin{equation}
  \label{eq:phenomenological}
  J_i = \sum_{j}Z^{ji}X_j.
\end{equation}
The kinetic coefficients, $Z^{ji}$ describe the response of the system and our
goal is to compute these coefficients and relate them to the transport
parameters of the system.  The choice of forces is to a certain extent
arbitrary, but they are to be chosen such that the Onsager relationships,
$Z^{ij} = Z^{ji}$ hold.  The criterion for this is that the internal entropy
generation be given by:
\begin{equation}
  \label{eq:OnsagerCriterion}
  \left.\frac{\partial s}{\partial t}\right|_{gen} = 
  \sum_{i}J_i\cdot X_i.
\end{equation}
We shall now deduce what these forces are in the case where there are only two
external driving forces: a temperature gradient and an electric potential
gradient.  Then we have the following equation for the total change in entropy
density:
\begin{equation}
  \label{eq:Entropy1}
  Tds = du-(\mu+qV)dn
\end{equation}
where $u$ is the internal energy\footnote{From our microscopic viewpoint this
  corresponds to the expectation value of the canonical Hamiltonian.}, $V$ is
the external potential and $q$ is the charge of the constituent particles.  It
is convenient to define here $\tilde{\mu} \equiv \mu+qV$.  From the equation
above we find that
\begin{equation}
  \label{eq:Entropy2}
  \left.\frac{\partial s}{\partial t}\right|_{total} = 
  \frac{1}{T}\left(\frac{\partial u}{\partial t}-
    \tilde{\mu}\frac{\partial n}{\partial t}\right).
\end{equation}
and using the continuity equation for the energy and particle currents:
\begin{equation}
  \label{eq:ContinuityEquations}
  \frac{\partial u}{\partial t} + \nabla\cdot J_{e}
  \qquad\mbox{and}\qquad
  \frac{\partial n}{\partial t} + \nabla\cdot J_{p}
\end{equation}
we get
\begin{eqnarray}
  \left.\frac{\partial s}{\partial t}\right|_{total} &=&
  \frac{-1}{T}\left( \nabla\cdot J_{e}-\tilde{\mu} \nabla\cdot J_{p}\right)
  \nonumber \\ &=&
  -\nabla\cdot\left(\frac{1}{T}J_{e}-\frac{\tilde{\mu}}{T}J_{p}\right)
  +\left\{J_e\cdot\nabla\left(\frac{1}{T}\right)-
    J_p\cdot\nabla\left(\frac{\tilde{\mu}}{T}\right)\right\}
\end{eqnarray}
The term $\frac{1}{T}\left(J_e-\tilde{\mu}J_p\right)$ is now recognized as the
entropy current and the heat current is $J_Q = TJ_s = J_e-\tilde{\mu}J_p$.
From this we get
\begin{eqnarray}
  \left.\frac{\partial s}{\partial t}\right|_{total}+\nabla\cdot J_s &=&
  \left.\frac{\partial s}{\partial t}\right|_{gen} 
  \nonumber \\ &=&
  J_e\cdot\nabla\left(\frac{1}{T}\right)-
  J_p\cdot\nabla\left(\frac{\tilde{\mu}}{T}\right)
  \nonumber \\ &=&
  J_Q\cdot\nabla\left(\frac{1}{T}\right)-
  \frac{1}{T}J_p\cdot\nabla\tilde{\mu}
\end{eqnarray}
Note that we have switched over from the energy current to the heat current
because it is really the heat current that is being measured in transport
measurements. It is worth remarking here that our definition of the heat
current does not agree with Mahan's definition \cite[Sec.~3.9B]{Mahan}, which
is just $J_Q = J_e-\mu J_p$.  On the other hand our definition does agree with
both \cite[Sec.~17.1]{Callen:Thermodynamics} and
\cite[Sec.~2.2]{Heikes:Thermoelectricity}.  We should note however that since
we are only considering linear response we will only need the currents in the
equilibrium state and there the external potential is constant and the shift
in the potential energy would simply be absorbed by the chemical potential
making our result agree with Mahan's.

We have now found the currents of interest and the corresponding forces which
are:
\begin{gather}
  J_1 = J_p, \qquad\mbox{and}\qquad X_1 = \frac{-1}{T}\nabla\tilde{\mu}
  \\
  J_2 = J_Q, \qquad\mbox{and}\qquad X_2 = \nabla\left(\frac{1}{T}\right)
\end{gather}
and the linear response equations are:
\begin{eqnarray}
  \label{eq:Phenomenological1}
  J_1 &=& -L^{11}\frac{1}{T}\nabla\tilde{\mu}
  +L^{21}\nabla\left(\frac{1}{T}\right)
  \\
  \label{eq:Phenomenological2}
  J_2 &=& -L^{12}\frac{1}{T}\nabla\tilde{\mu}
  +L^{22}\nabla\left(\frac{1}{T}\right)
\end{eqnarray}
where we now denote the kinetic coefficients corresponding to this choice of
currents by $L^{ij}$.  To see how these coefficients relate to the transport
parameters we need to consider how the measurements are made. Remember here
that the emf in the system is given by $\frac{-1}{q}\nabla\tilde{\mu}$ which
in a homogeneous system reduces to $-\nabla V$. 
\begin{enumerate}
\item Electrical conductivity:  Here we do not allow any temperature
  gradients. The electrical current can then be computed from
  equation (\ref{eq:Phenomenological1}),
  \begin{equation}
    J_C = qJ_p =  -L^{11}\frac{q}{T}\nabla\tilde{\mu}  
    =  L^{11}\frac{q^2}{T}E = \sigma E
  \end{equation}
  and hence the electrical conductivity is 
  \begin{equation}
    \sigma = \frac{q^2}{T} L^{11}
  \end{equation}
\item Thermal conductivity:  Here we do not allow any particle current.  From
  equation (\ref{eq:Phenomenological1}) we get,
  \begin{equation}
     \frac{1}{T}\nabla\tilde{\mu} 
     = \frac{L^{21}}{L^{11}}\nabla\left(\frac{1}{T}\right)
  \end{equation}
  and putting this into equation (\ref{eq:Phenomenological2}) we get,
  \begin{equation}
    J_Q =\frac{-1}{T^2} \left(L^{22}-\frac{[L^{12}]^2}{L^{11}}\right)
    \nabla T = -\kappa \nabla T
  \end{equation}
  and thus the thermal conductivity is 
  \begin{equation}
    \kappa = \frac{1}{T^2} \left(L^{22}-\frac{[L^{12}]^2}{L^{11}}\right).
  \end{equation}
\item Thermopower: Here no particle current is allowed so we obtain from
  equation (\ref{eq:Phenomenological1})
  \begin{equation}
    E = \frac{-1}{q}\nabla\tilde{\mu} = 
    \frac{1}{qT}\frac{L^{12}}{L^{11}}\nabla T 
    = S \nabla T 
  \end{equation}
  and hence the thermopower is given by:
  \begin{equation}
    S = \frac{1}{qT}\frac{L^{12}}{L^{11}}.
  \end{equation}
\end{enumerate}
It is instructive to rewrite the phenomenological equations in terms of the
transport parameters.  It is easy to solve for the kinetic coefficients in
terms of the transport parameters and plugging the solutions into the
equations we get:
\begin{eqnarray}
  J_p &=& -\frac{\sigma}{q^2}\nabla\tilde{\mu}
  +\frac{ST^2\sigma}{q}\nabla\left(\frac{1}{T}\right)
  \\
  J_Q &=& -\frac{ST\sigma}{q}\nabla\tilde{\mu}
  +\left(\kappa T^2 - S^2T^3\sigma\right)\nabla\left(\frac{1}{T}\right).
\end{eqnarray}
If we now solve for $\nabla\tilde{\mu}$ from the first equation and plug the
result into the second one we get:
\begin{equation}
  \label{eq:heat_particle_current}
  J_Q = STq J_p+\kappa T^2\nabla\left(\frac{1}{T}\right)
\end{equation}
and since the entropy current is now given by $TJ_S = J_Q$ we see that
\begin{equation}
  J_S = S J_C+\kappa T\nabla\left(\frac{1}{T}\right)
\end{equation}
where $J_C$ is the electric current.  From this equation we see that the
electric current carries entropy along with it which is directly proportional
to the thermopower.  Hence the thermopower can be regarded as the entropy
transported per charge carrier.

Finally let us see how the thermopower relates to the so called Peltier heat
and for that we imagine passing a current through a junction of two different
materials, A and B that are kept at the same temperature.  Due to the
isothermal conditions the heat current in these materials is given by 
\begin{equation}
  J_{Q,A} = S_AT J_{C,A} 
  \qquad\mbox{and}\qquad
  J_{Q,B} = S_BT J_{C,B}.
\end{equation}
The electric currents are now the same in both conductors so we see that if
the thermopowers in the two conductors are different, the heat carried in them
will be different.  Hence if the current passes from conductor A to B the heat
extracted or deposited at the junction will be:
\begin{equation}
  J_{Q,Junction} = J_{Q,B}-J_{Q,A} = \left(S_B-S_A\right)TJ_C.
\end{equation}
The Peltier coefficient for the junction is now defined as:
\begin{equation}
  \pi_{AB} = \frac{J_{Q,Junction}}{J_C} = \left(S_B-S_A\right)T.
\end{equation}

The kinetic coefficients are now evaluated for each model using the
Kubo formalism \cite{Mahan} and here we simple quote the results.  The
simplest expressions are obtained using Matsubara representation where
one must compute the time ordered current-current correlation
function\footnote{Note that our formula here includes a Boltzmann
constant that seems to have been missing in reference \cite{Mahan}.}:
\begin{equation}
  \label{eq:current-current}
  Z^{mn}(i\omega) = \frac{1}{\beta\kb}\int_{0}^{\beta}d\tau
  e^{i\omega\tau}\langle T_{\tau}J_{m}(\tau)J_{n}(0)\rangle
\end{equation}
and then we obtain our kinetic coefficients by analytically continuing to the
real axis and taking the limit $\omega \rightarrow 0$, i.e.
\begin{equation}
  L^{mn} = \lim_{\omega\rightarrow 0}\frac{1}{\omega}
  \imag Z^{mn}(i\omega\rightarrow \omega+i0).
\end{equation}

\section{Thermoelectric Performance and the Figure of Merit}
The thermoelectric power is of course what makes it possible to convert heat
energy directly into electrical energy and it is therefore natural to ask what
is the maximum conversion efficiency that can be achieved.  To see how this
can be related to material parameters consider the simple thermoelectric
refrigerator shown schematically in figure \ref{fig:Refrigerator}.
%-------------------
\begin{figure}[htb]
  \centering
  \epsfig{file=Figures-3/cooling.eps,width=0.5\linewidth}
\caption{A schematic diagram of thermoelectric refrigerator.}
\label{fig:Refrigerator}
\end{figure}
%-------------------
We imagine that the device is made of up both n-type and p-type conductors but
in general any two materials with different thermopowers will do.  The two
arms are connected to a metallic cooling element at the heat source.  The heat
currents that enter the two arms at the cooling element are given by
integrating the heat current densities in equation
(\ref{eq:heat_particle_current}) and subtracting the contribution from Joule
heating in the arms:
\begin{eqnarray}
  Q_n &=& -S_n I T_1-K_n(T_1-T_2)-\frac{1}{2}I^2 R_n
  \\
  Q_p &=& S_p I T_1-K_p(T_1-T_2)-\frac{1}{2}I^2 R_p.
\end{eqnarray}
Here $K_{n,p}$ and $R_{n,p}$ denote respectively the thermal conductance and
the electrical resistance of the two arms.  We should note that $S_n$ is
negative and thus the heat current induced by the particle current is
transferred from the source in both arms.  Some heat is also transferred from
the source due to thermal conduction and then we have used the fact that half
of the Joule heat generated in the arms will be returned to the source and
half goes the sink \cite{CRC-Thermoelectrics}.  The total heat current drawn
from the source is then
\begin{equation}
  Q_{total} = (S_p-S_n)IT_1-K(T_1-T_2)-\frac{1}{2}I^2 R
\end{equation}
and here we have denoted the total, parallel, thermal conductance by $K$ and
the total, series, resistance by $R$.  The current in the circuit can now be
tuned to obtain maximum cooling power, which is given by 
\begin{equation}
  \label{eq:MaximumCooling}
  Q_{max} = \frac{1}{2}\frac{S_{pn}^2}{R}T^2_1-K(T_1-T_2).
\end{equation}
We could now imagine running the refrigerator and thereby decreasing the
temperature in the heat source but we see from the equation above that at some
point the thermal conduction through the arms is going to balance the Peltier
heat transferred from the source.  Therefore the maximum temperature
difference between source and sink that can be attained is
\begin{equation}
  \label{eq:MaximumTemperatureDifference}
  (T_1-T_2)_{max} = \frac{1}{2}\frac{S_{pn}^2}{KR}T^2_1 
  = \frac{1}{2}Z^{pn}T^2_1. 
\end{equation}
Here we have defined the figure of merit, $Z^{pn}$ for the thermocouple and we
note that the efficiency of the refrigerator is controlled by this parameter.
In order to maximize the efficiency we have to maximize the difference between
the thermopowers of the arms and minimize the product $KR$.  The product
depends on the dimensions (length $L$ and cross sectional area $A$) of the
arms and if we write it in terms of the thermal conductivities and electrical
resistivities as
\begin{equation}
  KR = \left(\kappa_p\frac{A_p}{L_p}+\kappa_n\frac{A_n}{L_n}\right)
  \left(\rho_p\frac{L_p}{A_p}+\rho_n\frac{L_n}{A_n}\right)
  = \kappa_p\rho_p+\kappa_n\rho_n 
  + \frac{A_pL_n}{A_nL_p}\kappa_p\rho_n
  + \frac{A_nL_p}{A_pL_n}\kappa_n\rho_p
\end{equation}
and minimize with respect to one of the ratio, $\frac{A_pL_n}{A_nL_p}$ we find
that the minimum is given by:
\begin{equation}
  (KR)_{min} = \left(\sqrt{\kappa_p\rho_p}+\sqrt{\kappa_n\rho_n}\right)^2.
\end{equation}
The maximal figure of merit that depends only on the material parameters is
therefore:
\begin{equation}
  Z^{pn} = \frac{S_{pn}^2}
  {\left(\sqrt{\kappa_p\rho_p}+\sqrt{\kappa_n\rho_n}\right)^2}.
\end{equation}
It is evident that the figure of merit depends on both the materials that make
up the thermocouple and it would be useful to have some independent measure of
how good each material is for use in thermocouples.  For this purpose one
introduces the quantity
\begin{equation}
  Z_T = \frac{S^2}{\kappa\rho}
\end{equation}
and tries to maximize this figure of merit.  We see however that there is no
simple way of relating $Z_T$ for each of the materials to the figure of merit
for the thermocouple except maybe in some simple cases.  In practice however
this is a useful quantity to consider and we will compute it for our model and
consider how one would go about maximizing it with respect to material
parameters.

\section{Transport Parameters of the Hubbard Model}
\label{sec:transport-eqs}
We turn now to the derivation of the operators for the particle and energy
currents of the Hubbard model.  Although we are mainly considering the case
when the interaction strength $U$ is infinite we will derive the equation for
the transport parameters assuming that $U$ is finite, but we will see that $U$
drops out of the equations and therefore they can be applied directly to the
infinite $U$ case.  The derivations will be done for the case of nearest
neighbor hopping on the hyper-cubic lattice.  The end results however will
only depend on the dispersion relation of the lattice and its derivatives and
is thus easily generalizable to other types of lattices or hoppings.

The current operators can be obtained from the corresponding density 
operators,
\begin{equation}
n_{i} = \sum_{\sigma}c_{i\sigma}^{+}c_{i\sigma}
\quad\mathrm{and}\quad
h_{i} = -t_{o}\sum_{r\sigma}c_{i+r\sigma}^{+}c_{i\sigma}
+ {U\over 2} \sum_{\sigma\neq\sigma'}n_{i\sigma}n_{i\sigma'} 
-\mu\sum_{\sigma}n_{i\sigma}
\end{equation}
by using the Heisenberg equations of motion and the continuity equations (here
in complex time since we want to work in the Matsubara formalism),
\begin{equation}
i{\partial n_{i} \over \partial \tau} + \nabla\cdot j_{i} = 0
\quad and \quad
i{\partial h_{i} \over \partial \tau} + \nabla\cdot Q_{i} = 0
\end{equation}
From the anti-commutation relations
\begin{equation}
\{c_{i\sigma}^{+},c_{j\sigma'}\} = \delta_{\sigma\sigma'}\delta_{ij},
\quad
\{c_{i\sigma}^{+},c_{j\sigma'}^{+}\} = 0,
\quad
\{c_{i,\sigma},c_{j,\sigma'}\} = 0,
\end{equation}
we find that
\begin{equation}
[n_{j\sigma'},c_{i\sigma}] = 
-\delta_{ij}\delta_{\sigma\sigma'}c_{i\sigma}
\quad\quad
[n_{j\sigma'},c^{+}_{i\sigma}] = 
\delta_{ij}\delta_{\sigma\sigma'}c^{+}_{i\sigma}
\end{equation}
and then we get
\begin{eqnarray}
{\partial c_{i\sigma}
\over 
\partial \tau} &=& [H,c_{i\sigma}] = -t_{o}\sum_{jr\sigma'}
[c_{j+r\sigma'}^{+}c_{j\sigma'},c_{i\sigma}]
+{U \over 2}\sum_{j\sigma'\neq\sigma''}
[n_{j\sigma'}n_{j\sigma''},c_{i\sigma}]
-\mu\sum_{j\sigma'}[n_{j\sigma'},c_{i\sigma}]
\nonumber \\ &=&
-t_{o}\sum_{jr\sigma}
[c_{j+r\sigma}^{+},c_{i\sigma}]c_{j\sigma}
+U\sum_{\sigma\neq\sigma'}
[n_{i\sigma},c_{i\sigma}]n_{\sigma'}
+\mu c_{i\sigma}
\nonumber \\ &=&
t_{o}\sum_{r}c_{i+r\sigma} + c_{i\sigma}
(\mu - U\sum_{\sigma'\neq\sigma}n_{i\sigma'}). 
\end{eqnarray}
Similarly we get
\begin{equation}
{\partial c_{i\sigma}^{+} \over \partial \tau} = 
-t_{o}\sum_{r} c_{i+r\sigma}^{+}
-c^{+}_{i\sigma}
(\mu - U\sum_{\sigma'\neq\sigma}n_{i\sigma'}). 
\end{equation}
Now we get for the number density
\begin{equation}
{\partial n_{i} \over \partial \tau}
= \sum_{\sigma}\left(
\dot{c}_{i\sigma}^{+}c_{i\sigma}
+c_{i\sigma}^{+}\dot{c}_{i\sigma}
\right)
=
-t\sum_{r\sigma}\left(
c_{i+r\sigma}^{+}c_{i\sigma} -
c_{i\sigma}^{+}c_{i-r\sigma}
\right).
\end{equation}
And similarly for the energy density
\begin{eqnarray}
{\partial h_{i} \over \partial \tau} &=&
-t_{o}\sum_{r\sigma}
(\dot{c}_{i+r\sigma}^{+}c_{i\sigma}+c_{i+r\sigma}^{+}\dot{c}_{i\sigma})
-\sum_{\sigma}\dot{n}_{i\sigma}(\mu - U\sum_{\sigma'\neq\sigma}n_{i\sigma'})
\nonumber \\ &=&
-t_{o}\sum_{r\sigma}
\left\{
\dot{c}_{i+r\sigma}^{+}c_{i\sigma}+c_{i+r\sigma}^{+}\dot{c}_{i\sigma}
-(c_{i+r\sigma}^{+}c_{i\sigma} - c_{i\sigma}^{+}c_{i-r\sigma})
(\mu - U\sum_{\sigma'\neq\sigma}n_{i\sigma'})
\right\}
\nonumber \\ &=&
-t_{o}\sum_{r\sigma}
\left(
\dot{c}_{i+r\sigma}^{+}c_{i\sigma}-
\dot{c}_{i\sigma}^{+}c_{i-r\sigma}
\right)
\end{eqnarray}
To differentiate on the lattice we use the formulas
\begin{equation}
f(x+a) = f(x)+af'(x) 
\quad\mathrm{and}\quad
f(x-a) = f(x)-af'(x)
\end{equation}
where $a$ is the lattice constant.  Using these formulas we can 
easily identify the currents and if we consider specifically the
$x$-component we have  
\begin{eqnarray}
\label{eq:currents}
j_{i}^{x} &=& it_{o}a\sum_{\sigma}\left(
c_{i+x\sigma}^{+}
-c_{i-x\sigma}^{+}
\right)c_{i\sigma} 
\\
Q_{i}^{x} &=& it_{o}a\sum_{\sigma}\left(
\dot{c}_{i+x\sigma}^{+}
-\dot{c}_{i-x\sigma}^{+}
\right)c_{i\sigma} 
\end{eqnarray}
Now we Fourier transform this by writing 
\begin{equation}
c_{i\sigma} = {1 \over V}\sum_{k}e^{iR_{j}k}c_{k\sigma}
\quad\mathrm{and}\quad
c_{i\sigma}^{+} = {1 \over V}\sum_{k}e^{-iR_{j}k}c_{k\sigma}^{+}
\end{equation}
and thus
\begin{eqnarray}
j_{j}^{x} &=& {iat_{o} \over V^{2}}\sum_{kq\sigma}
c_{k\sigma}^{+}c_{q\sigma}e^{iR_{j}(q-k)}
\left(e^{-iak_{x}}-e^{iak_{x}}
\right) = {2at_{o} \over V^{2}}\sum_{kq\sigma}
sin(ak_{x})
c_{k\sigma}^{+}c_{k+q\sigma}e^{iR_{j}q}
\nonumber
\end{eqnarray}
so
\begin{eqnarray}
j_{x}(q) &=& {2at_{o} \over V}\sum_{k\sigma}
sin(ak_{x})c_{k\sigma}^{+}c_{k+q\sigma}
\\
Q_{x}(q) &=& {2at_{o} \over V}\sum_{k\sigma}
sin(ak_{x})\dot{c}_{k\sigma}^{+}c_{k+q\sigma}
\end{eqnarray}
Remember now that the dispersion relation for the cubic lattice is
\begin{equation}
\label{eq:dispersion}
\varepsilon_{k} = -2t_{o}\sum_{i}cos(ak_{i})
\end{equation}
and thus the equations above can be written on the form
\begin{eqnarray}
j_{x}(q) &=& {1 \over V}\sum_{k\sigma}
\left({\partial \varepsilon_{k} \over \partial k_{x}}\right)
c_{k\sigma}^{+}c_{k+q\sigma}
\\
Q_{x}(q) &=& {1 \over V}\sum_{k\sigma}
\left({\partial \varepsilon_{k} \over \partial k_{x}}\right)
\dot{c}_{k\sigma}^{+}c_{k+q\sigma}
\end{eqnarray}
Now that we have obtained expressions for the currents we can turn to the task
of deriving the transport coefficient of the model.  As noted before, the
transport coefficients are most easily expressed in the Matsubara formalism as
follows:
\begin{equation}
L^{mn} = \lim_{\omega \rightarrow 0} {1 \over \omega}\imag 
Z^{mn}(\omega + i0)
\end{equation}
where 
\begin{equation}
Z^{mn}(i\omega) = \frac{1}{\beta\kb}\int_{0}^{\beta}d\tau e^{i\omega\tau}
\langle
T_{\tau}j_{m}(\tau)j_{n}(0)
\rangle,
\end{equation}
with $j_{1} = j_{x}$ and $j_{2} = Q_{x}$ and the currents are
evaluated at $q = 0$. Putting in the expressions above and summing
over spin we get,
\begin{eqnarray}
\langle
T_{\tau}j_{x}(\tau)j_{x}(0)
\rangle
&=& 
{1 \over V^{2}}\sum_{k\sigma,k'\sigma'}
\left({\partial \varepsilon_{k} \over \partial k_{x}}\right)
\left({\partial \varepsilon_{k'} \over \partial k_{x}'}\right)
\langle
T_{\tau}c_{k\sigma}^{+}(\tau)c_{k\sigma}(\tau)
c_{k'\sigma'}^{+}c_{k'\sigma'}
\rangle
\nonumber \\ &=&
-{N \over V}\sum_{k}
\left({\partial \varepsilon_{k} \over \partial k_{x}}\right)^{2}
G_{k}(\tau)G_{k}(-\tau)
\\
\langle
T_{\tau}j_{x}(\tau)Q_{x}(0)
\rangle
&=& 
{1 \over V^{2}}\sum_{k\sigma,k'\sigma'}
\left({\partial \varepsilon_{k} \over \partial k_{x}}\right)
\left({\partial \varepsilon_{k'} \over \partial k_{x}'}\right)
\langle
T_{\tau}c_{k\sigma}^{+}(\tau)c_{k\sigma}(\tau)
\dot{c}_{k'\sigma'}^{+}c_{k'\sigma'}
\rangle
\nonumber \\
&=&
{N \over V}\sum_{k}
\left({\partial \varepsilon_{k} \over \partial k_{x}}\right)^{2}
\dot{G}_{k}(\tau)G_{k}(-\tau)
\\
\langle
T_{\tau}Q_{x}(\tau)Q_{x}(0)
\rangle
&=& 
{1 \over V^{2}}\sum_{k\sigma,k'\sigma'}
\left({\partial \varepsilon_{k} \over \partial k_{x}}\right)
\left({\partial \varepsilon_{k'} \over \partial k_{x}'}\right)
\langle
T_{\tau}\dot{c}_{k\sigma}^{+}(\tau)c_{k\sigma}(\tau)
\dot{c}_{k'\sigma'}^{+}c_{k'\sigma'}
\rangle
\nonumber \\
&=&
{N \over V}\sum_{k}
\left({\partial \varepsilon_{k} \over \partial k_{x}}\right)^{2}
\dot{G}_{k}(\tau)\dot{G}_{k}(-\tau)
\end{eqnarray}
In calculating the correlation functions here above we have used the fact that
in the mean field limit $z \rightarrow \infty$ all vertex corrections to the
current-current correlator drop out and we are left with a single bubble
diagram \cite{Georges:1996}.  We have also used equations from Appendix
\ref{app:spectral} to relate the expectation value of the time derivative of
the creation and annihilation operators to the time derivative of the Green's
function.  To proceed further we make use of the spectral representation for
the Green's function.
\begin{eqnarray}
G_{k}(\tau) &=& -\langle
T_{\tau}c_{k\sigma}(\tau)c_{k\sigma}^{+}(0)
\rangle
= T\sum_{i\omega_{n}}e^{-i\omega\tau}G_{k}(i\omega).
\\
G_{k}(i\omega) &=& \int_{-\infty}^{\infty}d\omega'{\rho_{k}(\omega') \over 
i\omega - \omega'}
\end{eqnarray}
where
\begin{equation}
\rho_{k}(\omega) = {-1 \over \pi}\imag G_{k}(\omega+i0)
\end{equation}
is the spectral function.  Defining
\begin{eqnarray}
Z_{0}(i\omega,k) &=& \frac{1}{\beta\kb}\int_{0}^{\beta}d\tau e^{i\omega \tau}
G_{k}(\tau)G_{k}(-\tau)
\\
Z_{1}(i\omega,k) &=& \frac{1}{\beta\kb}\int_{0}^{\beta}d\tau e^{i\omega \tau}
\dot{G}_{k}(\tau)G_{k}(-\tau)
\\
Z_{2}(i\omega,k) &=& \frac{1}{\beta\kb}\int_{0}^{\beta}d\tau e^{i\omega \tau}
\dot{G}_{k}(\tau)\dot{G}_{k}(-\tau)
\end{eqnarray}
and using the representation above for the Green's function we 
find that
\begin{eqnarray}
\lim_{\omega\rightarrow 0}{1 \over \omega}\imag Z_{0}(\omega+i0,k) 
= -\frac{\pi}{\kb}\int_{-\infty}^{\infty}d\omega \rho_{k}^{2}(\omega)
\left(-\kb T{df(\omega) \over d\omega}\right)
\\
\lim_{\omega\rightarrow 0}{1 \over \omega}\imag Z_{1}(\omega+i0,k) 
= \frac{\pi}{\kb}\int_{-\infty}^{\infty}d\omega \rho_{k}^{2}(\omega)\omega
\left(-\kb T{df(\omega) \over d\omega}\right)
\\
\lim_{\omega\rightarrow 0}{1 \over \omega}\imag Z_{2}(\omega+i0,k) 
= \frac{\pi}{\kb}\int_{-\infty}^{\infty}d\omega \rho_{k}^{2}(\omega)\omega^{2}
\left(-\kb T{df(\omega) \over d\omega}\right)
\end{eqnarray}
From this we get
\begin{equation}
L^{11} = \frac{A_{0}}{\kb}, \quad L^{12} = {1 \over \beta\kb}A_{1} 
\quad\mbox{and}\quad 
L^{22} = {1 \over \beta^{2}\kb}A_{2}
\end{equation}
where
\begin{equation}
A_{n} = 
N\pi{1 \over V}\sum_{k}
\left({\partial \varepsilon_{k} \over \partial k_{x}}\right)^{2}
\int_{-\infty}^{\infty}d\omega {\rho_{k}^{2}(\omega)(\omega\beta)^{n}
\over 4\cosh^{2}({\beta\omega \over 2})}
\end{equation}
The spectral density depends on momentum only through the lattice
dispersion relation $\varepsilon_k$ and hence we can write
\begin{equation}
\label{eq:An-function}
A_{n} = {N\pi \over z}
\int_{-\infty}^{\infty}d\omega d\varepsilon{\rho^{2}(\varepsilon,\omega)
(\omega\beta)^{n}\over 4\cosh^{2}({\beta\omega \over 2})}\Phi(\varepsilon)
\end{equation}
where 
\begin{equation}
\label{eq:phifunction}
\Phi(\varepsilon) = {z \over V}\sum_{k}
\left({\partial \varepsilon_{k} \over k_{x}}\right)^{2}
\delta(\varepsilon-\varepsilon(k)) = 
(2at)^{2}\int_{-{\pi \over a}}^{\pi \over a}
{d^{d}k \over (2\pi)^{d}}\sin^{2}(ak_{x})
\delta(\varepsilon-\varepsilon(k))
\end{equation}
Notice that this function is zero for 
$|\varepsilon| \geq zt_{o} = \sqrt{z}t$.
The Fourier transform of this quantity can easily be calculated
analytically as follows
\begin{eqnarray}
\mytilde{\Phi}(s) &=& \int_{-\infty}^{\infty}
d\varepsilon e^{-is\varepsilon}\Phi(\varepsilon) 
\quad = \quad 
(2at)^{2}\int_{-{\pi \over a}}^{\pi \over a} {d^{d}k \over (2\pi)^{d}}
e^{-is\varepsilon_{k}}\sin^{2}(ak_{x})
\nonumber \\
&=& {(2at)^{2} \over a^{d}}\left(
\int_{-\pi}^{\pi} {dk \over 2\pi}
e^{i2t_{o}s\cos(k)}
\right)^{d-1}
\left(
\int_{-\pi}^{\pi} {dk \over 2\pi}
e^{i2t_{o}s\cos(k)}\sin^{2}(k)
\right)
\nonumber \\
&=& 
{(2at)^{2} \over a^{d}}
J_{0}^{d-1}(2t_{o}s){J_{1}(2t_{o}s) \over 2t_{o}s}. 
\end{eqnarray}
Here $J_{0}$ and $J_{1}$ are Bessel functions and we have used the formula
\cite{Gradshteyn}
\begin{equation}
  \label{eq:BesselFunction}
  J_{\nu}(\zeta) = 
  {\left({\zeta \over 2}\right)^{\nu} \over \Gamma(\nu + {1 \over 2})
    \Gamma({1 \over 2})}\int_{0}^{\pi}d\phi e^{\pm i\zeta\cos(\phi)}
  \sin^{2\nu}(\phi).
\end{equation}
The Fourier transform can now be used to calculate $\Phi(\varepsilon)$
numerically and in figure (\ref{fig:phifunc}) we have plotted the result for
$d = 3$ and $d = \infty$.  In $z = 2d = \infty$, $\Phi$ can be calculated
exactly and is given by
\begin{equation}
\Phi(\varepsilon) = {(2at)^{2} \over 2a^{d}\sqrt{2\pi t^{2}}}\exp\left(-
{\varepsilon^{2} \over 2t^{2}}
\right)
 = {(2at)^{2} \over 2}D_{\infty}(\varepsilon).
\end{equation}
%-------------------
\begin{center}
\begin{figure}[htb]
  \epsfig{file=Figures-3/sinsities.eps,width=0.9\linewidth}
\caption{The function $\Phi(\epsilon)$ for (a) $d = 3$ and (b) $d = \infty$.} 
\label{fig:phifunc}
\end{figure}
\end{center}
%-------------------
The thermal conductivity, the electrical conductivity and the thermopower are
now given in terms of the correlation functions $L^{ij}$ and the $A_{n}$'s as
follows,
\begin{eqnarray}
\label{eq:transportParameters}
\sigma &=& {e^{2} \over T} L^{11} = e^{2}\beta A_{0} 
\\
\kappa &=& {1 \over T^{2}}
\left(L^{22}-{(L^{12})^{2} \over L^{11}}\right)
= \kb\left(A_{2}-{A_{1}^{2} \over A_{0}}\right)
\\
S &=& -{1\over eT}{L^{12} \over L^{11}} = -{\kb \over e}{A_{1} \over A_{0}}.
\end{eqnarray}
Note that as $z = 2d$ goes to infinity the functions $A_{n}$ defined in
(\ref{eq:An-function}) tend to zero and thus the conductivities also go to
zero.  This is not surprising if we look at the formulas for the currents in
(\ref{eq:currents}) and then apply the scaling to the hopping integral
$t_{o}$; the currents simply go to zero in the limit $z \rightarrow \infty$.
What is finite however is the sum of the conductivities in all directions,
i.e. $z\sigma$ and $z\kappa$.  At this point it is in order to clarify the
philosophy behind the infinite $d$ calculations.  In the calculations of the
local Green's functions the infinite $d$ limit has already been taken and
although we profess to be doing calculations for the cubic lattice we use the
self consistency conditions for the infinite Bethe lattice.  The reason for
doing that is that apart from some Van Hove singularities the density of states
for the 3D cubic lattice and the Bethe lattice are very similar.  Now when
calculating correlation functions we saw earlier that upon taking the infinite
$d$ limit we got rid of all vertex corrections which was a good thing.
However, in the remaining expressions we do not necessarily have to take this
limit and in fact we will henceforth take $z = 1$ in our calculations and take
the $\Phi$ corresponding to the infinite dimensional Bethe lattice.  One might
ask here, how we can calculate the $\Phi$ for the Bethe lattice since the
expression (\ref{eq:phifunction}) obviously cannot be used.  Well, we simply
assume that on the Bethe lattice we obtain the same expression as on the
infinite hyper-cubic lattice, i.e.
\begin{equation}
\Phi_{Bethe}(\varepsilon) = {(2at)^{2} \over 2}D_{Bethe}(\varepsilon)
= {(2at)^{2} \over 2}{1 \over \pi t}
\sqrt{1-\left({\varepsilon \over 2t}\right)^{2}}
\end{equation}
This enables us to do the $\varepsilon$ integral in the expression for the
$A_{n}$'s analytically (see appendix \ref{app:bethe-lattice}) but we will also
later compare our result to the results obtained when the proper cubic 3D
version of $\Phi$ is used.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Results: Transport parameters}
\label{sec:res-transport}
In figure \ref{fig:transfig1} the transport parameters for density $n = 0.90$
are displayed versus temperature.  At very low temperature the thermopower is
seen to diverge and actually the same goes for the thermal and electrical
conductivities due to the pathology of the solution mentioned in
\article\ref{sec:res-green-self}.
%----------------------------
\begin{figure}[htb]
\centering
\begin{tabular}{c@{\quad}c}
\mbox{\epsfig{file=Figures-3/res-n0.90.eps,width=0.45\linewidth}} &
\mbox{\epsfig{file=Figures-3/thcon-n0.90.eps,width=0.45\linewidth}} \\
\mbox{\epsfig{file=Figures-3/thpow-n0.90.eps,width=0.45\linewidth}} &
\mbox{\epsfig{file=Figures-3/fom-n0.90.eps,width=0.45\linewidth}}
\end{tabular}
\caption{Results for transport parameters from NCA+DMFT calculations
  at density $n = 0.90$.  $\rho$ denotes resistance and is measured in units
  of ${z\hbar a \over e^{2}}$, $\kappa$ denotes thermal conductivity and is
  measured in units of ${\kb D \over za \hbar}$, $S$ denotes thermopower and
  is measured in units of ${\kb \over e}$ and $ZT$ denotes the dimensionless
  figure of merit.  Here $a$ is the lattice constant, $D = 2t$ is the bare
  half-width of the band and $z$ is the coordination number of the lattice.  }
\label{fig:transfig1}
\end{figure}
%----------------------------
Our results here are quite consistent with the results of Oguri and Maekawa
\cite{Oguri:1990} who computed the transport parameters for the infinite $U$
Hubbard model close to half filling using the retraceable path approximation
for the motion of holes in a paramagnetic spin background.  We see that the
electrical resistivity is linear almost all the way down to the lowest
temperatures displayed in the graph.  The thermal conductivity displays a
powerlaw behavior in the high temperature regime going as $1/T^2$ but at
lower temperatures it reaches a maximum at temperatures around quarter of the
bandwidth.  Similar maxima were seen in the data by Oguri and Maekawa but at
somewhat larger temperature.  The thermopower on the other hand tends to a
finite constant at large temperatures but starts decreasing at temperatures
slightly below the bandwidth, becomes electron-like and seems to be diverging.

The behavior of the transport parameters in the high temperature region can
be deduced from our results on spectral function in previous chapter.  At high
temperatures we have already seen that the spectral function behaves as a
rigid band centered on $\omega = -\mu$ and $\mu$ is a linear function of
temperature, i.e. $\mu/T = \alpha = const.$ Thus we write
\begin{equation}
\mytilde{\rho}(\omega,\epsilon) = \rho(\omega-\mu,\epsilon).
\end{equation}
Hence the coefficients $A_{n}$ from \article\ref{sec:transport-eqs}
become\footnote{Remember that we take $z = 1$.}:
\begin{equation}
A_{n} = {\pi N \over 4}\int d\omega{(\beta\omega-\alpha)^{n} \over 
\cosh^{2}({\beta\omega-\alpha \over 2})}
\int d\epsilon \mytilde{\rho}^{2}(\omega,\epsilon)\Phi(\epsilon).
\end{equation}
We now expand the hyperbolic cosine in Taylor series around $\beta = 0$.
\begin{equation}
{1 \over \cosh^{2}({\beta\omega-\alpha \over 2})} = 
{1 \over \cosh^{2}({\alpha \over 2})}
\left(
1+\beta\omega\tanh({\alpha \over 2})+
{\omega^{2}\beta^{2} \over 4}\left[
3\tanh({\alpha \over 2})-1\right]
\right).
\end{equation}
Before we go any further we also define 
\begin{equation}
\gamma_{n} = \int d\epsilon d\omega \omega^{n}
\mytilde{\rho}^{2}(\omega,\epsilon)\Phi(\epsilon)
\quad\mbox{and let}\quad \tau = \tanh({\alpha \over 2})
\quad\mbox{and}\quad \zeta = {1 \over 4\cosh^{2}({\alpha \over 2})}.
\end{equation}
Then calculating to second order in $\beta$ we get
\begin{eqnarray}
A_{0} &=& \pi N \zeta
\left(\gamma_{0} + \gamma_{1}\beta\tau+{1 \over 4}
\gamma_{2}\beta^{2}[3\tau-1]\right)
\\
A_{1} &=& \pi N \zeta
\left(-\alpha\gamma_{0} + \gamma_{1}\beta\left[
1-\alpha\tau\right]+\gamma_{2}\beta^{2}\left[\tau
-{\alpha \over 4}(3\tau-1)\right]\right)
\\
A_{2} &=& \pi N \zeta
\left(\alpha^{2}\gamma_{0} + \gamma_{1}\beta\left[
\alpha^{2}\tau-2\alpha\right]+\gamma_{2}\beta^{2}\left[1
-2\alpha\tau+{\alpha^{2} \over 4}(3\tau-1)\right]\right)
\end{eqnarray}
Now we put these expressions into the formulas for the transport coefficient
and calculate again to second order in $\beta$.
\begin{eqnarray}
\sigma &=& e^{2}\pi N \zeta \beta 
\left(\gamma_{0} + \gamma_{1}\beta\tau+{1 \over 4}
\gamma_{2}\beta^{2}[3\tau-1]\right)
\\
S &=& {\kb \over e}\left(\alpha-{\gamma_{1} \over \gamma_{0}}\beta
-{\tau(\gamma_{2}\gamma_{0}-\gamma_{1}^{2}) \over \gamma_{0}^{2}}
\beta^{2}\right)
\\
\kappa &=& {\kb\pi N\zeta
(\gamma_{2}\gamma_{0}-\gamma_{1}^{2}) \over \gamma_{0}}\beta^{2}.
\end{eqnarray}
The figure of merit is then to lowest order in $\beta$ given by
\begin{equation}
ZT = {S^{2}\sigma T \over \kappa} = 
\alpha^{2}(kT)^{2}
{\gamma_{0}^{2} \over \gamma_{2}\gamma_{0}-\gamma_{1}^{2}}.
\end{equation}
It is interesting at this point to look at the spectral moments, $\gamma_{n}$,
and see how they depend on temperature and density.
%---------------------------------------------------------------------------
\begin{figure}[htb]
\centering
\begin{tabular}{c@{\quad}c}
\mbox{\epsfig{file=Figures-3/mom-n0.90.eps,width=0.45\linewidth}} &
\mbox{\epsfig{file=Figures-3/mom-t5.0.eps,width=0.45\linewidth}} 
\end{tabular}
\caption{The spectral moments, $\gamma_{n}$, for $N = 2$.
  Versus doping, $\delta = 1-n$, at $T = 5.0$ and versus temperature at
  density, $n = 0.90$.}
\label{fig:transfig3}
\end{figure}
%---------------------------------------------------------------------------
As we can see from figure \ref{fig:transfig3} both $\gamma_{0}$ and
$\gamma_{2}$ tend to a constant at high temperature as could be expected from
our previous discussion in section \ref{sec:res-green-self}.  We see also that
the asymmetric part of the spectral density which is measured by $\gamma_{1}$
goes as $1/T$.  If we look at the doping dependence we see essentially the
same behavior as on the temperature graph, i.e. $\gamma_{0}$ and $\gamma_{2}$
seem to be tending to a constant while $\gamma_{1}$ goes as $\delta$.  In some
sense low doping corresponds to high temperature as we will see more clearly
in section \ref{ssec:thermodynamic-results}.  We should also note the apparent
increase of $\gamma_{0}$ and $\gamma_{2}$ around 30\% doping or so.  The
reason is that the spectral moments scale approximately as $[n/N+(1-n)]^2$,
i.e.
\begin{equation}
\gamma_{0} \simeq \tilde{\gamma}_{0}[n/N+(1-n)]^2
\qquad\mbox{and}\qquad
\gamma_{2} \simeq \tilde{\gamma}_{2}[n/N+(1-n)]^2
\end{equation}
where $\tilde{\gamma}_{0}$ and $\tilde{\gamma}_{2}$ are constants, independent
of doping.  The factor $[n/N+(1-n)]^2$ is nothing but the square of the total
integrated spectral weight of $\mytilde{\rho}$ and looking at the definition
of the spectral moments this is indeed what we would have expected.

Now $\alpha$ is given by 
\begin{equation}
\alpha = \ln\left({n \over N(1-n)}\right)
\quad\mbox{so}\quad
\zeta = {nN(1-n) \over (n+N(1-n))^{2}}.
\end{equation}
so finally we have to lowest order in $\beta$
\begin{eqnarray}
  \label{eq:ncaHighTemp}
\sigma &=& e^{2}\pi {nN^{2}(1-n) \over (n+N(1-n))^{2}}\gamma_{0}\beta
\simeq e^{2}\pi n(1-n)\tilde{\gamma}_{0}\beta
\\
S &=& {\kb \over e}\ln\left({n \over N(1-n)}\right)
\\
\kappa &=& \kb\pi{nN^{2}(1-n) \over
(n+N(1-n))^{2}}\gamma_{2}\beta^{2}
\simeq \kb\pi n(1-n)\tilde{\gamma}_{2}\beta^{2}
\\
ZT &=& \ln^{2}\left({n \over N(1-n)}\right)(\kb T)^{2}
\frac{\gamma_{0}}{\gamma_{2}}
\end{eqnarray}
Of the equations above the one for the thermopower is well known, at least for
the case $N = 2$ and goes by the name of Heikes formula \cite{Chaikin:1976}.
%---------------------------------------------------------------------------
\begin{figure}[htb]
\centering
\begin{tabular}{c@{\quad}c}
\mbox{\epsfig{file=Figures-3/thpow-inflim.eps,width=0.45\linewidth}} &
\mbox{\epsfig{file=Figures-3/thcon-n0.90log.eps,width=0.45\linewidth}} 
\end{tabular}
\caption{Results for transport parameters from NCA+DMFT calculations
  at density $n = 0.90$.  I: The thermopower, $S_{\infty}-S(T)$ where
  $S_{\infty}$ is the infinite temperature limit given by Heikes formula.  II:
  The thermal conductivity, $\kappa$.}
\label{fig:transfig2}
\end{figure}
%---------------------------------------------------------------------------

From these formulas we see that indeed the resistivity should be linear in
temperature at high temperatures and from figure \ref{fig:transfig2} we see
that the thermal conductivity and the next to leading order term in the
thermopower go as $T^{-2}$ in agreement with the formulas above.  We see also
that the figure of merit increases quadratically with temperature and thus
tends to infinity as $T$ goes to infinity and furthermore it goes to infinity
at finite temperature if the density goes to one or zero.  This behavior seems
striking given that all known materials have figures of merit which are in the
range of 0.0-2.0. The reason is of course that here we have neglected to take
into account the lattice contribution to the thermal conductivity.  If we
assume a finite lattice contribution to the thermal conductivity then we can
of course neglect the electronic contribution since it tends to zero at high
temperatures and then we have:
\begin{equation}
  \label{eq:ncaMerFiniteLattice}
  ZT \simeq \frac{\kb\pi}{\kappa_{L}}\ln^{2}\left({n \over N(1-n)}\right)
  \tilde{\gamma}_{0}n(1-n).
\end{equation}
Hence the figure of merit will actually go to zero at $n = 0$ and $n = 1$ and
if lattice contribution is taken independent of temperature the FOM will tend
to a constant at high temperatures.

We should also note that associated with the divergency of the figure of merit
around density $n = 0,1$ is the divergence of the chemical potential.  Now of
course this would never happen in a real physical system, eventually other
bands would come into play and participate in the dynamics and thus the use of
our model would no longer be justified.  In this respect our model is somewhat
unphysical and one needs to be aware of that when looking at results close to
the two extremes in density.




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Thermodynamics}
\label{sec:thermodynamics}
As we saw in section \ref{sec:transport-theory} there is a close connection
between the thermopower of the system and the entropy -- the thermopower being
essentially the entropy per carrier.  It is therefore instructive to take a
brief look at the energy and specific heat of the model.

\subsection{Thermodynamic Equations}
\label{ssec:thermodynamic-equations}
We now turn to the calculation of the total energy of the Hubbard model.  The
total energy corresponds to the expectation value of the canonical
Hamiltonian:
\begin{equation}
H_{c} = -t_{o}\sum_{jr\sigma}c_{j+r\sigma}^{+}c_{j\sigma}
+ {U \over 2}\sum_{j\sigma\neq\sigma'}n_{j\sigma}n_{j\sigma'}. 
\end{equation}
Note however that all observables evolve according to the grand canonical
Hamiltonian which we have been using so far.  The total energy is thus:
\begin{eqnarray}
E &=& \langle H_{c} \rangle =
{1 \over Z}Tr(e^{-\beta H}H_{c}) = 
{1 \over Z}\sum_{n}
\langle \Psi_{n}|e^{-\beta H}H_{c}|\Psi_{n}\rangle
\nonumber \\
&=&
\sum_{k\sigma}\epsilon_{k}
\langle c_{k\sigma}^{+}c_{k\sigma}\rangle
+{U \over 2}\sum_{i\sigma\neq\sigma'}
\langle n_{i\sigma}n_{i\sigma'}\rangle
\end{eqnarray}
To calculate the expectation value involving the number operators we use the
equation of motion for the creation operator which we have derived earlier and
we find that
\begin{equation}
{U \over 2}\sum_{i\sigma\neq\sigma'}n_{i\sigma}n_{i\sigma'} = 
{1 \over 2}\left(
\sum_{k\sigma}\dot{c}_{k\sigma}^{+}c_{k\sigma}-
\sum_{k\sigma}(\epsilon_{k}-\mu)c_{k\sigma}^{+}c_{k\sigma}
\right)
\end{equation}
and hence
\begin{eqnarray}
\langle H_{c} \rangle &=&
{1 \over 2}\sum_{k\sigma}(\epsilon_{k}+\mu)
\langle c_{k\sigma}^{+}c_{k\sigma}\rangle +
{1 \over 2}\sum_{k\sigma}
\langle \dot{c}_{k\sigma}^{+}c_{k\sigma}\rangle 
\nonumber \\
&=&
{N \over 2}\sum_{k}(\epsilon_{k}+\mu)
\lim_{\tau\rightarrow 0^{+}}G_{k}(-\tau)+
{N \over 2}\sum_{k}
\lim_{\tau\rightarrow 0^{+}}{d \over d\tau}G_{k}(-\tau)
\nonumber \\
&=&
{N \over 2}\sum_{k}T\sum_{i\omega_{n}}(\epsilon_{k}+\mu+i\omega_{n})
G_{k}(i\omega_{n})e^{i\omega_{n}0^{+}}
\end{eqnarray}
Here we have used 
\begin{equation}
G_{k}(\tau) = -\langle T_{\tau}c_{k}(\tau)c_{k}^{+} \rangle =
T\sum_{i\omega_{n}}e^{-i\omega_{n}\tau}G_{k}(i\omega_{n})
\end{equation}
so
\begin{equation}
\langle c_{k}^{+}c_{k} \rangle = \lim_{\tau\rightarrow 0^{+}}G_{k}(-\tau)
\quad\mbox{and similarly} \quad
\langle \dot{c}_{k}^{+}c_{k} \rangle = 
\lim_{\tau\rightarrow 0^{+}}{d \over d\tau}G_{k}(-\tau)
\end{equation}
Using the spectral representation for the Green's function,
\begin{equation}
G_{k}(i\omega_{n}) = \int{\rho_{k}(\varepsilon)d\varepsilon \over 
i\omega_{n}-\varepsilon}
\end{equation}
we get 
\begin{equation}
\langle H_{c} \rangle = {N \over 2}\sum_{k}
 \int d\varepsilon\rho_{k}(\varepsilon)
T\sum_{i\omega_{n}}{\epsilon_{k}+\mu+i\omega_{n} \over
i\omega_{n}-\varepsilon}e^{i\omega_{n}0^{+}}
\end{equation}
The sum over $i\omega_{n}$ can be done by contour integration if
properly regulated.  The convergence factor is $e^{i\omega_{n}0^{+}}$
which tends to zero in the half-plane $\imag (i\omega) <0$ so the
proper function to use to pick out the poles in the sum is the Fermi
function which tends to zero in the other half-plane.  Thus we get
\begin{eqnarray}
\langle H_{c} \rangle &=& {N \over 2}\sum_{k}
\int d\omega\rho_{k}(\omega)(\epsilon_{k}+\mu+\omega)f(\omega)
\nonumber \\ &=&
{N \over 2}\int d\omega d\varepsilon
\rho(\varepsilon,\omega)(\varepsilon+\mu+\omega)f(\omega)D(\varepsilon)
\nonumber \\ &=&
{N \over 2}\int d\omega d\varepsilon
\rho(\varepsilon,\omega)\varepsilon f(\omega)D(\varepsilon)
+{N \over 2}\int d\omega 
\rho(\omega)_{loc}(\omega+\mu)f(\omega)
\nonumber \\ &=&
{N \over 2}\int d\omega d\varepsilon
\rho(\varepsilon,\omega)\varepsilon f(\omega)D(\varepsilon)
+{N \over 2}\int d\omega 
\rho_{loc}(\omega)\omega f(\omega)+{1 \over 2}\mu n.
\end{eqnarray}
The first line in the expression above is the general formula.  In the
next three lines we have used the fact that in the mean field theory the sum
over momenta can be replaced by an integral weighted with the lattice spectral
density $D(\varepsilon)$, and then we use the fact that the local spectral
density $\rho_{loc}(\omega)$ is given by the $\varepsilon$ integral over
$\rho(\varepsilon,\omega)D(\varepsilon)$.

A further simplification arises in the infinite $U$ Hubbard model.  In that
case the potential energy is zero so the total energy equals the kinetic
energy which corresponds to the first integral in the final expression above,
i.e.
\begin{equation}
\langle H_{c} \rangle_{U = \infty} = N \int d\omega d\varepsilon
\rho(\varepsilon,\omega)\varepsilon f(\omega)D(\varepsilon)
\end{equation}
On the Bethe lattice the integral over the density of states can be done
exactly (see appendix \ref{app:bethe-lattice}) and combining that with the
equations above we get the following two expressions for the energy.
\begin{equation}
\langle H_{c} \rangle_{U = \infty} = N\int d\omega 
\rho_{loc}(\omega)\omega f(\omega) + \mu n = 
2Nt^{2}\int d\omega f(\omega) 
\rho_{loc}(\omega)Re G_{loc}(\omega).
\end{equation}
Note that this expression includes only the local spectral density so the
breakdown of the self energy which was mentioned earlier would not be noticed
in these computations.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\subsection{Thermodynamic Results}
\label{ssec:thermodynamic-results}
The results for the energy and specific heat of the model are displayed in
figure \ref{fig:thermofig1}.  Over the whole accessible temperature range the
energy turned out to be independent of degeneracy, and goes as $n(1-n)$ with
density, i.e. it is directly proportional to the product of the densities of
electrons and holes.  This seems reasonable since the energy is obviously zero
in the limits $n = 0$ and $n = 1$.  We note that at very low temperatures
something interesting seems to be happening in the system at density $n =
0.90$ where a little more time was invested in solving the system at low
temperatures.  It is however unfortunate that with the method that we used to
solve the NCA equations we are not able to go to very low temperatures and the
lowest temperature that was obtained with $n = 0.9$ was around 0.007.
%----------------------------
\begin{figure}[htb]
\centering
\begin{tabular}{c@{\quad}c}
\mbox{\epsfig{file=Figures-3/energy.eps,width=0.45\linewidth}} &
\mbox{\epsfig{file=Figures-3/specheat.eps,width=0.45\linewidth}} \\
\mbox{\epsfig{file=Figures-3/energylog.eps,width=0.45\linewidth}} &
\mbox{\epsfig{file=Figures-3/encollap.eps,width=0.45\linewidth}}
\end{tabular}
\caption{Thermodynamics for $N = 2$.  I: Energy.  II: Specific heat.
  III: Same data as in figure (I), graph shows clearly the $1/T$ behavior of
  energy in the high temperature regime.  IV: The energy divided by $n(1-n)$
  collapses onto a single curve for densities 0.1 - 0.9.  The energy is
  measured in units of the bare band half-width, $D$, and the specific heat is
  measured in units of $\kb$.}
\label{fig:thermofig1}
\end{figure}
%----------------------------
Let us now use the density scaling of the energy and the data from $n = 0.90$
to calculate the entropy of the model using the formula:
\begin{equation}
S(T) = S_{\infty}+\int_{\infty}^{T}{c_{V}(T') \over T'}
\end{equation}
where $S_{\infty}$ is the entropy at infinite temperature given by equation
(\ref{eq:entropy}) and we take $T = 5$ to correspond to infinite temperature.
The results from doing this are displayed in figure \ref{fig:thermofig2}.
Notice that the entropy becomes negative for all the densities except $n =
0.90$.  This indicates that the density scaling cannot possibly persist down
to zero temperature if we are to satisfy the requirement that the entropy be
positive and go to zero when $T$ goes to zero.  Therefore the temperature at
which the entropy crosses zero can be regarded as some sort of breakdown
temperature for the density scaling (it actually has to break down at a little
higher temperature).

There is of course a more fundamental reason why the density scaling and the
apparent independence of degeneracy cannot persist down to zero temperature.
The fact is that the high temperature entropy density:
\begin{equation}
  S = -n\ln(n/N)-(1-n)\ln(1-n)
\end{equation}
does not scale with $n(1-n)$ and does depend on the degeneracy $N$.  Hence if
we want to get rid of all this entropy and no more as we turn the temperature
down from infinity to zero the specific heat must have a more complicated
density and degeneracy dependence at low temperatures.
%----------------------------
\begin{figure}[htb]
\centering
\begin{tabular}{c@{\quad}c}
\mbox{\epsfig{file=Figures-3/entropy.eps,width=0.45\linewidth}} &
\mbox{\epsfig{file=Figures-3/transtemp.eps,width=0.45\linewidth}} 
\end{tabular}
\caption{I: Entropy for N = 2 calculated from the $n = 0.9$ data using
  the density scaling $n(1-n)$.  The inset shows that the scaling breaks down
  at temperatures below 0.05.  II.  Carrier crossover temperature.  In the
  temperature-density plane above the curve the carriers are localized, below
  it they are extended (see text).}
\label{fig:thermofig2}
\end{figure}
%----------------------------
At high temperatures the electrons in the system can be regarded as
essentially localized and this is reflected in the high value of entropy at $T
= \infty$.  As temperature decreases the electrons delocalize and at the same
time the entropy decreases.  This suggests that the entropy can be used as a
measure of the localization of the carriers in the system and although we do
not have any rigorous way of characterizing the crossover from localized to
extended carriers we can define operationally a crossover temperature by the
following criterion:
\begin{equation}
{S(T_{xo}) \over S_{\infty}} = {1 \over 2}.
\end{equation}
That is, we say that the carriers in the system are being localized when the
entropy of the system has reached half its maximum value.  This definition
gives us the curves shown in figure \ref{fig:thermofig2}.  We see that the
crossover temperature is very low for $n = 0.9$ and therefore we were
required to spend more time on calculation for this particular density.  It is
not surprising that the crossover temperature is so low for $n = 0.9$, since
we are then close to half filling and the carriers do not have much room to
delocalize.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{Iterated perturbation theory}
\label{sec:ipt}
To complement the result from the NCA which are mainly reliable in the high
temperature regime, at least as far as the transport calculations are
concerned, we did a few calculations using the iterated perturbation theory
(IPT) modified for finite doping as described in \cite{Kajueter:1996b}.  This
method is exact in various limits at zero temperature and has been extended to
finite temperature in \cite{Kajueter:thesis}.  The IPT is a perturbation
theory in the interaction strength, $U$, so it does not correspond exactly to
the model we have considered so far.  $U$ will nevertheless be the largest
scale in the problem so one can expect that the results obtained will be
somewhat indicative of what happens in models with large interaction.  We will
not delve further into the specifics of this method, it provides us with the
local spectral density as the NCA and the calculation of response functions
proceeds the same way as before.
%----------------------------
\begin{figure}[htb]
\centering
\begin{tabular}{c@{\quad}c}
\mbox{\epsfig{file=Figures-3/iptres.eps,width=0.45\linewidth}} &
\mbox{\epsfig{file=Figures-3/iptthcon.eps,width=0.45\linewidth}} \\
\mbox{\epsfig{file=Figures-3/iptthpow.eps,width=0.45\linewidth}} &
\mbox{\epsfig{file=Figures-3/iptfom.eps,width=0.45\linewidth}} 
\end{tabular}
\caption{Results for transport parameters from IPT+DMFT calculations
  for $U = 2.82$. Here $\rho$ denotes resistance and is measured in units of
  ${z\hbar a \over e^{2}}$, $\kappa$ denotes thermal conductivity and is
  measured in units of ${\kb D \over za \hbar}$, $S$ denotes thermopower and
  is measured in units of ${\kb \over e}$ and $ZT$ denotes the dimensionless
  figure of merit.  Here $a$ is the lattice constant, $D = 2t$ is the bare
  half-width of the band and $z$ is the coordination number of the lattice.  }
\label{fig:iptfig1}
\end{figure}
%----------------------------
The results for the transport parameters are displayed in figure
\ref{fig:iptfig1}.  These results should be discussed in the context of the
low temperature Fermi liquid \cite{Nozierez,Abrikosov:QFT} estimate for the
transport parameters given by the following equations which are derived in
appendix \ref{app:fermi-liq}:
\begin{alignat}{2}
\sigma &= E_0{Ne^{2} \over 2\gamma}\Phi(\tilde{\mu}) & \qquad
\kappa &=  E_2{\kb N(\kb T) \over 2\gamma}\Phi(\tilde{\mu})
\nonumber \\
S &= -{\kb \over e}\frac{E_2}{E_0}{\kb T \over Z}
{d \ln(\Phi(\tilde{\mu})) \over d\tilde{\mu}} & \qquad
ZT &= \frac{E_2}{E_0}\left(
{\kb T \over Z}{d\ln\Phi(\tilde{\mu}) \over d\tilde{\mu}}
\right)^{2}.
\end{alignat}
Here $\gamma = \imag\Sigma(0)$ is the scattering rate at the Fermi-level, $Z =
(1-{\partial Re\Sigma \over \partial \omega}|_{\omega = 0})^{-1}$ is the
quasiparticle residue at the Fermi-level, $\tilde{\mu} = \mu-Re\Sigma(0)$ is
the effective chemical potential and the function $\Phi$ is defined in section
\ref{sec:transport-eqs}.  The quantities $E_n$ are constants with $E_0 = 0.82$
and $E_2 = 1.75$.  These equations are derived assuming that the imaginary
part of the self energy can be approximated by:
\begin{equation}
  \label{eq:FLscattering}
  \imag \Sigma(\omega,T) = 
  \gamma_0\left[(\kb T)^2+\left(\frac{\omega}{\pi}\right)^2\right]
  \qquad\mbox{where}\qquad \gamma_0 = \mbox{constant}
\end{equation}
to a good accuracy with several $\kb T$ of the Fermi surface while the other
parameters are assumed to be nearly temperature independent.  Notice that the
Wiedeman-Franz law holds here just as it does for transport dominated by
impurity scattering.  However since the scattering is frequency dependent
around the Fermi surface the ratio of $\sigma T$ and $\kappa$ is not equal to
the classical Lorenz number.

The scattering at the Fermi surface is now $\gamma = \gamma_0 (kT)^2$ and
thus we see from the equations above that the temperature behavior of the
transport parameters should be:
\begin{equation}
\rho \sim T^{2} \qquad \kappa \sim T^{-1} \qquad S \sim T
\qquad ZT \sim T^{2}
\end{equation}
and indeed it looks like the transport parameters are behaving this way,
except for the thermal conductivity which reaches a maximum at a finite
temperature and then seems to be going linearly to zero.  This discrepancy can
be attributed to an artificial numerical broadening of the spectral density,
i.e. the scattering rate has a constant term which is put by hand into the
computer code to ensure convergence.

%----------------------------
\begin{figure}[htb]
\centering
\begin{tabular}{c@{\quad}c}
\mbox{\epsfig{file=Figures-3/iptFLres.ps,angle=-90,width=0.45\linewidth}} &
\mbox{\epsfig{file=Figures-3/iptScatt.ps,angle=-90,width=0.45\linewidth}} \\
\end{tabular}
\caption{The electrical resistivity for $U=2.82$ and $n=0.80$ along with its
  Fermi liquid estimate.  Also shown is the scattering rate at the Fermi
  surface as a function of $T^2$.}
\label{fig:iptfig2.1}
\end{figure}
%---------------------------------------------------------------------------
\begin{figure}[htb]
\centering
\begin{tabular}{c@{\quad}c}
\mbox{\epsfig{file=Figures-3/iptFLthc.ps,angle=-90,width=0.45\linewidth}} &
\mbox{\epsfig{file=Figures-3/iptQuasi.ps,angle=-90,width=0.45\linewidth}} \\
\end{tabular}
\caption{The electrical resistivity for $U=2.82$ and $n=0.80$ along with its
  Fermi liquid estimate.  Also shown is the quasiparticle residue at the Fermi
  surface.}
\label{fig:iptfig2.2}
\end{figure}
%---------------------------------------------------------------------------
\begin{figure}[htb]
\centering
\begin{tabular}{c@{\quad}c}
\mbox{\epsfig{file=Figures-3/iptFLseb.ps,angle=-90,width=0.45\linewidth}} &
\mbox{\epsfig{file=Figures-3/iptEchem.ps,angle=-90,width=0.45\linewidth}} 
\end{tabular}
\caption{The thermopower for $U=2.82$ and $n=0.80$ along with its Fermi liquid
  estimate.  Also shown is the effective chemical potential.}
\label{fig:iptfig2.3}
\end{figure}
%---------------------------------------------------------------------------
\begin{figure}[htb]
\centering
\begin{tabular}{c@{\quad}c}
\mbox{\epsfig{file=Figures-3/iptReSigma.ps,angle=-90,width=0.45\linewidth}} &
\mbox{\epsfig{file=Figures-3/iptImSigma.ps,angle=-90,width=0.45\linewidth}} \\
\end{tabular}
\caption{The real and the imaginary part of the self energy for $U=2.82$ and
  $n=0.80$ along with the Fermi liquid approximation of these quantities.  The
  curves labeled with an integer value of $k$ correspond to the weight
  functions $f(\omega)f(-\omega)(\beta\omega)^k$ that are used to compute the
  transport functions.}
\label{fig:iptfig4}
\end{figure}
%---------------------------------------------------------------------------
In figures \ref{fig:iptfig2.1}, \ref{fig:iptfig2.2} and \ref{fig:iptfig2.3} we
have plotted the scattering rate, the effective chemical potential and the
quasiparticle residue for $U = 2.82$ and $n = 0.80$ along with a comparison of
the numerical and Fermi liquid calculation of the transport parameters
according to the equations above.  We see indeed that the scattering rate goes
as $T^{2}$ apart from a very small constant part and the other coefficients
are nearly temperature independent at temperatures below $T = 0.05$ so as we
indicated the main temperature dependence of the transport parameters should
come from the scattering rate.

For the resistivity the two methods compare quite favorably with deviations
occurring only for temperatures above 0.1.  For the thermal conductivity and
the thermopower however the Fermi liquid estimate is not as good although it
does give the correct qualitative behavior.  This indicates that the data
does not obey the assumptions that we set out with although the resistivity
seems to be less sensitive to this than the other quantities.

The reason for the discrepancy can be inferred from figure \ref{fig:iptfig4}
where we have displayed the self energy close to the Fermi level along with
the Fermi liquid approximations to the to self energy and the weights that are
used in the calculation of the transport parameters.  In order for our simple
Fermi liquid calculation to work the imaginary part of the self energy must
obey equation (\ref{eq:FLscattering}) within an interval of width at least
$8\kb T$ around the Fermi level.  We also see that for the resistivity this
bound can be relaxed somewhat since the weight function associated with the
computation of the resistivity is narrower than the one for the other
parameters.

The Fermi liquid assumption holds quite well for the real part of the self
energy but the imaginary part does show significant asymmetry within this
interval.  The states below the Fermi surface are more strongly scattered than
the ones above the Fermi surface with the result that the electron
contribution to the thermopower is enhanced over the hole contribution and
therefore the thermopower becomes more negative than the Fermi liquid estimate
would indicate.  According to the Fermi liquid estimate the thermopower should
not depend on the scattering at all and be wholly dependent on the band
structure through the transport function.  This is a direct consequence of the
assumption of symmetric scattering rate.

At higher temperature the transport parameters seem to be tending to some
asymptotic high temperature values, and the thermopower is actually quite
close to the high temperature value given by Heikes formula.  This is
consistent with the equation of Chaikin and Beni \cite{Chaikin:1976} for the
thermopower at intermediate temperatures.  In the high temperature regime
which for this model is well above the interaction strength $U$ the asymptotic
value for the thermopower is not given by Heikes formula but rather by:
\begin{equation}
  \label{eq:ChaikinBeni}
  S = \frac{\kb}{e}\ln\left(\frac{n}{2-n}\right).
\end{equation}
This equation gives always lower thermopower than Heikes formula, although
they do converge at low densities.  Moreover the asymptotic value of the
thermopower is always negative, tending to zero at zero doping.  Thus if we
could access higher temperatures within this model we would expect that the
thermopower would decrease very slowly down from the Heikes limit towards the
limit given by the equation above as we reach temperatures higher than the
interaction strength.

For the other transport parameters however we have no good way of estimating
the exact intermediate and high temperature behavior within this model.

\section{Application to La$_{1-x}$Sr$_{x}$TiO$_{3}$}
\label{sec:titanate}
Let us now turn to quantitative calculations of the thermopower in the
intermediate temperature range, which is the range most pertinent to
experiment and possible applications.  In the class of materials that we want
to consider, i.e. a system that is doped away from the Mott insulating state,
the one that is best characterized is the perovskite-like
La$_{1-x}$Sr$_{x}$TiO$_{3}$ \cite{Tokura:1993a}.  The end compound LaTiO$_3$
is a Mott insulator with the valence on the titanium being +3.0 and is
antiferromagnetically ordered below around 150K.  By doping with strontium the
insulating magnetic state is destroyed and the system becomes a Pauli
paramagnetic metal.  The doping percentage is found to give an accurate
measure of the carrier density in the titanium 3d-bands and experiments
indicate that system is a Fermi-liquid at low temperatures almost all the way
up to zero doping.

We model this system with a Hubbard model on a cubic lattice, with
half-bandwidth, $D = 0.5eV$ \cite{Crandles:1992} and interaction strength $U =
2.0eV$ in the Mott insulating end at $x=0$.  The crystalline lattice of
LaTiO$_3$ displays orthorhombic distortions that reduce the Ti-O-Ti bond angle
from the ideal 180$^{\circ}$ to 154$^{\circ}$ \cite{Tokura:1993b}. As the
system is doped with strontium the bond angle straightens out and reaches
180$^{\circ}$ around $x = 0.77$.  The variation of the bond angle with doping
is close to being linear as we can see from figure \ref{fig:bondAngle}.
%---------------------------------------------------------------------------
\begin{figure}[htb]
\centering
\begin{tabular}{c@{\quad}c}
\mbox{\epsfig{file=Figures-3/bondAngle.ps,angle=-90,width=0.45\linewidth}} &
\mbox{\epsfig{file=Figures-3/bandWidth.ps,angle=-90,width=0.45\linewidth}}
\end{tabular}
\caption{The ab plane Ti-O-Ti bond angle vs. doping for the system 
  La$_{1-x}$Sr$_{x}$TiO$_{3}$ based on data from \cite{Sunstrom:1992}.  The
  figure on the right shows the corresponding change in the titanium half
  bandwidth of the titanium 3d-band.}
\label{fig:bondAngle}
\end{figure}
%---------------------------------------------------------------------------
The hopping interaction between the titanium atoms is governed by
supertransfer interaction over the oxygen sites and thus the band width of the
system depends on the Ti-O-Ti bond angle with the exact dependence given by
\cite[page 1141]{Imada:1998}:
\begin{equation}
  \label{eq:BandwidthVsBondangle}
  D(\theta) = D(180^{\circ})\cos^2\theta
\end{equation}
We should note here that the on-site interaction is assumed not to change with
doping and therefore must be scaled appropriately with the bandwidth.  Since
we take the bandwidth in our programs to be equal to one the interaction in
the Mott insulating end corresponds to $U=4.0$ and then decreases down to
around $3.2$ in the other end of the doping scale.

The results of the computation of the thermopower for this system
using IPT is displayed in figure \ref{fig:latio3-1}.
%---------------------------------------------------------------------------
\begin{figure}[htb]
\centering
\begin{tabular}{c@{\quad}c}
\mbox{\epsfig{file=Figures-3/sebLoDop.ps,angle=-90,width=0.45\linewidth}} &
\mbox{\epsfig{file=Figures-3/sebHiDop.ps,angle=-90,width=0.45\linewidth}}
\end{tabular}
\caption{The thermopower in La$_{1-x}$Sr$_{x}$TiO$_{3}$ for various different
  values of doping.}
\label{fig:latio3-1}
\end{figure}
%---------------------------------------------------------------------------
\begin{figure}[htb]
\centering
\epsfig{file=Figures-3/goodenough.ps,angle=-90,width=0.90\linewidth}
\caption{Experimental values of thermopower in
La$_{1-x}$Sr$_{x}$TiO$_{3}$ for various different values of doping.
The figure is reproduced from \cite{Hays:1999}.} 
\label{fig:goodenough}
\end{figure}
%---------------------------------------------------------------------------
We see that for all computed values of doping the thermopower displays
electron-like Fermi-liquid behavior at low temperature although at the
smallest doping values the thermopower actually becomes positive again
but that can be attributed to breakdown of the numerics in the IPT
code.  The transition temperature at which the Fermi liquid picture
breaks down is clearly dependent on the proximity to the Mott
transition and we see that for the lowest doping values the Fermi
liquid range is quite small.  Similar results have been observed in
experiments by Onoda et.al.~\cite{Onoda:1998} and Hays et.al.~ (see
figure \ref{fig:goodenough}) but in our case the transition to a
hole-like thermopower occurs at lower temperature than indicated by
experiments.  In the higher doping range however the Fermi liquid
behavior extends at least up to 300K and both this range and the value
of the thermopower agrees quite well with experiments by Moos
et.al.~\cite{Moos:1995} on La$_{1-x}$Sr$_{x}$TiO$_{3}$ ceramics.
Between 400 and 1000K the experimental data shows logarithmic,
semiconducting behavior with the thermopower changing by around
$-300\mu V/K$ per decade.  In our data we see similar logarithmic
behavior in this temperature range for the largest doping values but
the change per decade depends on the doping and is only in the range
of $-70$ to $-90\mu V/K$.  This discrepancy could be due the fact that
we assume here that we are dealing with a single crystal system but we
should also note that IPT has its range of validity and is most
dependable at lower temperatures.

\section{Optimization of the figure of merit}
\label{sec:optfom} 
To conclude this chapter let us try to assess how efficient these
strongly correlated materials would be for thermoelectric purposes.  We
continue with the La$_{1-x}$Sr$_{x}$TiO$_{3}$ as in the previous section using
the same parameters as before for the IPT.  For the NCA with use the same
doping dependent bandwidth and take the degeneracy equal to 2.  To get a
reasonable estimate for the figure of merit we have to assume that the lattice
contribution to the thermal conductivity is finite.  We are not aware any
experiments on the thermal conductivity in these titanate compounds, so for
the purpose of getting an order of magnitude estimate we use a value of 2.0
W/mK for $\kappa_L$ based on measurements made on manganite oxides
\cite{Visser:1997}.  We will moreover assume that this value does not depend on
either temperature or doping.
%---------------------------------------------------------------------------
\begin{figure}[htb]
\centering
\begin{tabular}{c@{\quad}c}
\mbox{\epsfig{file=Figures-3/iptMerMax1.ps,angle=-90,width=0.45\linewidth}} &
\mbox{\epsfig{file=Figures-3/iptMerMax2.ps,angle=-90,width=0.45\linewidth}}
\end{tabular}
\caption{The dimensionless figure of merit for the La$_{1-x}$Sr$_{x}$TiO$_{3}$
  system, computed using the IPT.}
\label{fig:iptMerMax}
\end{figure}
%---------------------------------------------------------------------------
\begin{figure}[htb]
\centering
\begin{tabular}{c@{\quad}c}
\mbox{\epsfig{file=Figures-3/ncaMerMax1.ps,angle=-90,width=0.45\linewidth}} &
\mbox{\epsfig{file=Figures-3/ncaMerMax2.ps,angle=-90,width=0.45\linewidth}}
\end{tabular}
\caption{The dimensionless figure of merit for the La$_{1-x}$Sr$_{x}$TiO$_{3}$
  system, computed using the NCA.  The figure on the left shows the expected
  density dependence of the dimensionless figure of merit within the NCA (see
  equation (\ref{eq:ncaMerFiniteLattice}).}
\label{fig:ncaMerMax}
\end{figure}
%---------------------------------------------------------------------------
In figures \ref{fig:iptMerMax} and \ref{fig:ncaMerMax} we have plotted the
figures of merit computed with both the IPT and the NCA.  Admittedly the
temperature scale for the NCA figure is rather large, but as we have seen the
NCA does brake down at low temperature preventing us from reaching lower
temperatures. We can clearly see this breakdown in the divergence of the
figure of merit for the two higher doping values.  We have also displayed the
expected doping dependence of the high temperature figure of merit within the
NCA assuming a finite thermal lattice conduction.  As we can see here the
figure of merit does not diverge at all as we approach the Mott transition,
instead it reaches a low maximum just below half filling and then tends to
zero in a non-analytic fashion.  It is clear that in this context the analysis
of the NCA is not expected to be at all valid except in system with smaller
bandwidth where it is easier to access the high temperature regime.  We could
also try to access the high temperature region by increasing the doping and in
fact judging by the figure on the right in \ref{fig:ncaMerMax} we would do a
much better job of maximizing the figure of merit that way.  

This conclusion is borne out in the IPT data, which is much more pertinent to
real systems.  There we see that the figure of merit dramatically increases as
we increase the doping.  In the low temperature region we see typical metallic
Fermi liquid behavior, characterized by small figure of merit but at higher
temperatures the system shows more of a semiconducting behavior as discussed
earlier in section \ref{sec:titanate} with quite a sizeable figure of merit.

In light of our results we have to conclude that it seems quite unlikely that
a significant figure merit can be attained in this class of systems.  In fact
for the case of La$_{1-x}$Sr$_{x}$TiO$_{3}$ our results indicate that the
semiconducting regime close to the strontium filled end of this system is more
likely to give rise to large figures or merit.  This of course is consistent
with the general rule that large figure of merit are mainly to be obtained in
semiconductors.

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