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 Physics and Atronomy, Louisiana State University, LA 70803).  This 
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\title{Chapter 6: The Fermi Liquid}
\author{L.D.\ Landau}
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\pagebreak
\Large

\section{introduction: The Electronic Fermi Liquid}

As we have seen, the electronic and lattice degrees of freedom
decouple, to a good approximation, in solids.  This is due
to the different time scales involved in these systems.
\beq
\tau_{ion}\sim 1/\om_D\;\gg\;\tau_{electron}\sim\frac{\hbar}{E_F}
\eeq
where $E_F$ is the electronic Fermi energy.  The electrons may be thought of 
as instantly reacting to the (slow) motion of the lattice, while remaining 
essentially in the electronic ground state.  Thus, to a good approximation 
the electronic and lattice degrees of freedom separate, and the small 
electron-lattice (phonon) interaction (responsible for resistivity, 
superconductivity etc) may be treated as a perturbation (with $\om_D/E_F$
as an expansion parameter); that is if we are 
capable of solving the problem of the remaining purely electronic system.

     At first glance the remaining electronic problem would also appear to be 
hopeless since the (non-perturbative) electron-electron interactions are 
as large as the combined electronic kinetic energy and the potential energy
due to interactions with the static ions (the latter energy, or rather
the corresponding part of the Hamiltonian, composes the solvable portion 
of the problem).  However, the Pauli principle keeps low-lying orbitals
from being multiply occupied, so
is often justified to ignore the electron-electron interactions, or treat 
them as a renormalization of the non-interacting problem (effective mass) etc.
This will be the initial assumption of this chapter, in which we will cover
\begin{itemize}
\item the non-interacting Fermi liquid, and
\item the renormalized Landau Fermi liquid (Pines \& Nozieres).
\end{itemize}

	These relatively simple theories resolve some of the most 
important puzzles involving metals at the turn of the century.  Perhaps 
the most intriguing of these is the metallic specific heat.  Except in
certain ``heavy fermion'' metals, the electronic contribution to
the specific heat is always orders of magnitude smaller than the
phonon contribution.  However, from the classical theorem of equipartition,
if each lattice site contributes just one electron to the conduction
band, one would expect the contributions from these sources to be
similar ($C_{electron}\approx C_{phonon}\approx 3Nrk_B$).  This puzzle 
is resolved, at the simplest level:  that of the non-interacting 
Fermi gas.

\section{The Non-Interacting Fermi Gas}

\subsection{Infinite-Square-Well Potential }

     We will proceed to treat the electronic degrees of freedom,
ignoring the electron-electron interaction, and even the electron-lattice
interaction.  In general, the electronic degrees of freedom are split into 
electrons which are bound to their atomic cores with wavefunctions which are 
essentially atomic, unaffected by the lattice, and those valence (or near 
valence) electrons which react and adapt to their environment.  For the 
most part, we are only interested in the valence electrons.  Their
environment described by the potential due to the ions and the core 
electrons-- the core potential.  Thus, ignoring the electron-electron
interactions, the electronic Hamiltonian is
\beq
H=\frac{P^2}{2m} + V(\r)\,.
\eeq
As shown in Fig.~\ref{fig:corepot}, the core potential $V(\r)$, like the lattice, 
is periodic
\begin{figure}[htb]
\centerline{\includegraphics[width=0.8\textwidth,clip=true]{1d.pdf}}
\caption[]{\em{Schematic core potential (solid line) for a one-dimensional
lattice with lattice constant $a$.}}
\label{fig:corepot}
\end{figure}


For the moment, ignore the core potential, then the electronic wave 
functions are plane waves $\psi\sim e^{i\k\cdot\r}$.  Now consider the
core potential as a perturbation.  The electrons will be strongly effected 
by the periodicity of the potential when $\lambda=2\pi/k\sim a$ 
\footnote{Interestingly, when $\lambda\sim a$, the Bragg condition 
$2d\sth\approx a\approx \lambda$ may easily be satisfied, so the electrons, 
which may be though of as DeBroglie waves, scatter off of the lattice.
Consequently states for which $\lambda=2\pi/k\sim a$ are often
forbidden.  This is the source of gaps in the band structure, to be 
discussed in the next chapter.}.  However, when $\k$ is small so that
$\lambda\gg a$ (or when $\k$ is large, so $\lambda\ll a$) the structure of 
the potential may be neglected, or we can assume $V(\r)=V_0 $ anywhere 
within the material.  The potential still acts to confine the electrons (and 
so maintain charge neutrality), so $V(\r)=\infty $ anywhere outside the 
material.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{squarewell.pdf}}
\caption[]{\em{Infinite square-well potential. $V(\r)=V_0$ within
the well, and $V(\r)=\infty$ outside to confine the electrons and
maintain charge neutrality.}}
\label{fig:squarewell}
\end{figure}

Thus we will approximate the potential of a cubic solid with linear
dimension $L$ as an infinite square-well potential.
\beq
V(\r)=\left\{
\begin{array}{ll}
V_0	& 0<r_i<L\\
\infty	& \mbox{otherwise}
\end{array}\right.
\eeq
The electronic wavefunctions in this potential satisfy
\beq
-\frac{\hbar^2}{2m}\lap\psi(\r)=\lep E'-V_0\rip\psi(\r)=E\psi(\r)
\eeq
The normalize plane wave solution to this model is
\beq
\psi(\r)=\prod_{i=1}^3\lep\frac{2}{L}\rip^{1/2}\sin k_ix_i\;\;
\mbox{where $i=x,y,\;{\rm{or}}\;z$}
\eeq
and $k_iL=n_i\pi$ in order to satisfy the boundary condition that
$\psi=0$ on the surface of the cube.  Furthermore, solutions with $n_i<0$ 
are not independent of solutions with $n_i>0$ and may be excluded.
Solutions with $n_i=0$ cannot be normalized and are excluded (they correspond 
to no electron in the state).  
\begin{figure}[htb]
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{1stquad.pdf}}
\caption[]{\em{Allowed k-states for an electron confined by a infinite-square
potential.  Each state has a volume of $\lep\pi/L\rip^3$ in k-space.}}
\label{fig:1stquad}
\end{figure}
The eigenenergies of the wavefunctions
are
\beq
-\frac{\hbar^2\lap}{2m}\psi =\frac{\hbar^2}{2m}\sum_ik_i^2
=\frac{\hbar^2\pi^2}{2mL^2} \lep n_x^2+n_y^2+n_z^2\rip
\eeq
and as a result of these restrictions, states in k-space are confined to the 
first quadrant (c.f.\ Fig.~\ref{fig:1stquad}). 
Each state has a volume $\lep\pi/L\rip^3$ of k-space.  Thus as $L\to\infty$, 
the number of states with energies $E(\k)<E<E(\k)+dE$ is 
\beq
dZ'=\frac{(4\pi k^2dk)/8}{(\pi/L)^3}\,.
\eeq
Then, since $E=\frac{\hbar^2 k^2}{2m}$, so 
$k^2dk=\frac{m}{\hbar^2}\sqrt{2mE/\hbar^2} dE$
\beq
dZ=dZ'/L^3=\frac{1}{4\pi^2} \lep\frac{2m}{\hbar^2}\rip^{3/2} E^{1/2} dE\,.
\eeq
or, the density of state per unit volume is
\beq
D(E)=\der{Z}{E}=\frac{1}{4\pi^2} \lep\frac{2m}{\hbar^2}\rip^{3/2} E^{1/2}\,.
\eeq
Up until now, we have ignored the properties of electrons. 
However, for the DOS, it is useful to recall that the electrons
are spin-1/2 thus $2S+1=2$ electrons can fill each orbital or k-state, 
one of spin up the other spin down.  If we account for this spin
degeneracy in $D$, then
\beq
D(E)=\frac{1}{2\pi^2} \lep\frac{2m}{\hbar^2}\rip^{3/2} E^{1/2}\,.
\eeq


\subsection{The Fermi Gas}

\subsubsection{$T=0$, The Pauli Principle}

Electrons, as are all half-integer spin particles, are Fermions.  Thus, by 
the Pauli Principle, no two of them may occupy the same state.
For example, if we calculate the density of electrons per unit volume
\beq
n=\int_0^\infty D(E) f(E,T) dE\,,
\eeq
where $f(E,T)$ is the probability that a state of energy $E$ is occupied,
the factor $f(E,T)$ must enforce this restriction.  However, $f$ is
just the statistical factor; c.f.\ for classical particles
\beq
f(E,T)=e^{-E/k_BT}\;\;\mbox{for classical particles}\,,
\eeq
which for $T=0$ would require all the electrons to go into the
ground state $f(0,0)=1$.   Clearly, this violates the Pauli
principle.

     At $T=0$ we need to put just one particle in each state, starting
from the lowest energy state, until we are out of particles.  Since 
$E\propto k^2$ in our simple square-well model, will fill up all k-states 
until we reach some Fermi radius $k_F$, corresponding to some Fermi Energy 
$E_F$
\begin{figure}[htb]
\centerline{\includegraphics[width=0.75\textwidth,clip=true]{1stqdfermi.pdf}}
\caption[]{\em{Due to the Pauli principle, all k-states up to $k_F$,
and all states with energies up to $E_f$ are filled at zero temperature.}}
\label{fig:1stqdfermi}
\end{figure}
\beq
E_F=\frac{\hbar^2 k_F^2}{2m}\,,
\eeq
thus,
\beq
f(E,T=0)=\theta(E_F-E)
\eeq
and
\beqa
n&=&\int_0^\infty D(E) f(E,T) dE=\int_0^{E_F} D(E) DE\nonumber\\
 &=&\lep\frac{2m}{\hbar^2}\rip^{3/2}\frac{1}{2\pi^2} \int_0^{E_F} E^{1/2} DE\nonumber\\
 &=&\lep\frac{2m}{\hbar^2}\rip^{3/2}\frac{1}{2\pi^2}\frac23 E_F^{3/2}\,,
\eeqa
or
\beq
E_F=\frac{\hbar^2}{2m} \lep 3\pi^2 n\rip^{2/3}=k_B T_F
\eeq
which also defines the Fermi temperature $T_F$.  Thus for metals, in which  
$n\approx 10^{23}/{\rm{cm}}^3$, 
$E_F\approx 10^{-11}{\rm{erg}} \approx 10 {\rm{eV}}\approx k_B 10^5K$.

Notice that due to the Pauli principle, the average energy of the
electrons will be finite, even at $T=0$!
\beq
\CE=\int_0^{E_F} D(E) E dE =\frac35 nE_F\,.
\eeq
However, it is the electrons near $E_F$  in energy which may be
excited and are therefore important.  These have a DeBroglie wavelength
of roughly
\beq
\lambda_e=\frac{12.3\AA}{\lep E({\rm{eV}})\rip^{1/2}}\approx 4\AA
\eeq
thus our original approximation of a square well potential,
ignoring the lattice structure, is questionable for electrons near the
Fermi surface, and should be regarded as yielding only qualitative results.

\subsubsection{$T\neq 0$, Fermi Statistics}

At finite temperatures some of the states will be thermally excited.
The energy available for these excitations is roughly $k_BT$, and the
only possible excitations are from filled to unfilled electronic states.
Therefore, only the states within $k_BT$ ($E_F-k_BT<E< E_F+k_BT$)
of the Fermi surface may be excited.  $f(E,T)$ must be modified accordingly.

     What we need is then $f(E,T)$ at finite T which also satisfies
the Pauli principle.  Lets return to our model of a periodic solid 
which is constructed by bringing individual atoms together from an 
infinite separation.  First, just consider a solid constructed from only 
two atoms, each with a single orbital (Fig.~\ref{fig:chempot}).
\begin{figure}[htb]
\centerline{\includegraphics[width=0.49\textwidth,clip=true]{chempot.pdf}}
\caption[]{\em{Exchange of electrons in a solid composed of two orbitals.}}
\label{fig:chempot}
\end{figure}
For this system, in equilibrium,
\beq
0=\delta F=\sum_i\pde{F}{n_i} \delta n_i
\eeq
electrons are conserved so $\sum_i \delta n_i=0$.  Thus, for our two orbital 
system
\beq
\pde{F}{n_1} \delta n_1 + \pde{F}{n_2} \delta n_2 =0
\;\;\;\mbox{and}\;\;\;
\delta n_1 + \delta n_2=0
\eeq
or 
\beq
\pde{F}{n_1}=\pde{F}{n_2}
\eeq
A similar relation holds for an arbitrary number of particles.
Apparently this quantity, the increased free energy needed to add
a particle to the system, is a constant
\beq
\pde{F}{n_i}=\mu
\eeq
for all $i$.  $\mu$ is called the {\em{chemical potential}}.

Now consider an ensemble of orbitals.  We will treat the thermodynamics of 
this system within the canonical ensemble (i.e., the system is in contact
with a thermal bath, and the particle number is conserved) for which 
${\cal{F}}={\cal{E}}-T{\cal{S}}$
is the appropriate potential.  The system energy ${\cal{E}}$ and Entropy 
${\cal{S}}$ may be written as functions of the orbital energies $E_i$ and
occupancies $n_i$ and the degeneracy $g_i$ of the state of energy $E_i$.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.55\textwidth,clip=true]{entropy.pdf}}
\caption[]{\em{states from an ensemble of orbitals.}}
\label{fig:entropy}
\end{figure}
For example, 
\beq
{\cal{E}}=\sum_i n_iE_i\,.
\eeq
The entropy ${\cal{S}}$ requires a bit more thought.  If $P$ is the number of
ways of distributing the electrons among the states, then
\beq
{\cal{S}}=k_B \ln P\,.
\eeq
Consider a set of $g_i$ states with energy $E_i$.  The number of ways
of distributing the first electron in these states is $g_i$.  For a
second electron we then have $g_i-1$ ways...   etc.  So for $n_i$ electrons 
there are
\beq
\frac{g_i!}{n_i!(g_i-n_i)!}
\eeq
possible ways of accommodating the $n_i$ (indistinguishable) electrons in 
$g_i$ states.  

The number of ways of making the whole system (ie, filling energy levels
with $E_i\neq E_j$) is then
\beq
P=\prod_i \frac{g_i!}{n_i!(g_i-n_i)!}\,,
\eeq
and so, the entropy
\beq
{\cal{S}}=k_B\sum_i \ln g_i!-\ln n_i!-\ln(g_i-n_i)!\,.
\eeq
For large $n$, $\ln n!\approx n\ln n-n$, so
\beq
{\cal{S}}=k_B\sum_i g_i\ln g_i -n_i\ln n_i -(g_i-n_i)\ln(g_i-n_i)
\eeq
and 
\beq
{\cal{F}}=\sum_i n_iE_i -k_B T\sum_i g_i\ln g_i -n_i\ln n_i -(g_i-n_i)\ln(g_i-n_i)
\eeq

We will want to use the chemical potential $\mu$ in our thermodynamic 
calculations
\beq
\mu=\pde{{\cal{F}}}{n_k}=E_k+k_BT\lep \ln n_k+1-\ln(g_k-n_k)-1\rip\,,
\eeq
where $\beta=1/k_BT$.  Solving for $n_k$
\beq
n_k=\frac{g_k}{1+e^{\beta(E_k-\mu)}}\,.
\eeq
Thus the probability that a quantum state with energy $E$ is
occupied, is (the Fermi function)
\beq
f(E,T)=\frac{1}{1+e^{\beta(E_k-\mu)}}\,.
\eeq
At $T=0$, $\beta=\infty$, and  $f(E,0)=\theta(\mu-E)$.  Thus $\mu(T=0)=E_F$.  
However in general $\mu$ is temperature dependent, since it must be adjusted 
to keep the particle number fixed.  
\begin{figure}[htb]
%%BoundingBox: 28 75 524 333
\centerline{\includegraphics[width=0.65\textwidth,clip=true]{fermi.pdf}}
\caption[]{\em{Plot of the Fermi function $1/\lep e^{-\beta(\om-\mu)}+1\rip$ 
when $\beta=1/k_BT=20$ and $\mu=1$.  Not that at energies $\om\approx\mu$
the Fermi function displays a smooth step of width $\approx k_BT=0.05$.
This allows thermal excitations of particles near the Fermi surface.}}
\label{fig:fermi}
\end{figure}
In addition, when $T\neq 0$, $f$ becomes 
less sharp at energies $E\approx \mu$.  This reflects the fact that particles  
with energies $E-\mu\approx k_BT$ may be excited to higher energy states.

\paragraph{Specific Heat}
The form of $f(E,T)$ also clarifies why the electronic specific heat of metals 
is so small compared to the classical result $C_{classical}=\frac32 nk_BT$.  
The reason is simple:  only the electrons with energies within about $k_BT$  
of the Fermi surface may be excited (about $\frac{k_BT}{E_F}$ of the electron
density)  each with excitation energy of about $k_BT$.  Therefore,
\beq
U_{excitation}\approx k_BT n \frac{k_BT}{E_F}=nk_BT\frac{T}{T_F}
\eeq
so
\beq
C\approx nk_B\frac{T}{T_F}
\eeq
Then as $T\ll T_F$ ($T_F$ is typically about $10^5$K in most metals\footnote{
Heavy Fermion systems are the exception to this rule.  There $T_F$ can
be as small as a fraction of a degree Kelvin.  As a result, they may have
very large electronic specific heats.}) $C\approx nk_B\frac{T}{T_F}\ll
C_{classical}\approx nk_B$.  Thus at temperatures where the phonons
contribute essentially a classical result  to the specific heat,
the electronic contribution is vanishingly small.  In general
this holds except at very low $T$ where the phonon contribution
$C_{phonon}\sim T^3$ goes to zero faster than the electronic contribution
to the specific heat.


\paragraph{Specific Heat Calculation}  Of course, since we know the
free energy of the non-interacting Fermi gas, we can calculate the
form of the specific heat.  Here we will follow Ibach and L\"uth and 
Kittel;  however, since the chemical potential does depend upon 
the temperature, I would like to make the approximations we make
a bit more explicit.

     Upon heating from $T=0$ to finite $T$, the Fermi gas will gain energy
\beq
U(T)=\int_0^\infty dE\, E D(E) f(E,T)-\int_0^{E_F}dE\, E D(E)
\eeq
so
\beq
C_V=\der{U}{T}=\int_0^\infty dE\, E D(E) \der{f(E,T)}{T}\,.
\eeq
Then since at constant volume the electronic density is constant, so
$\der{n}{T}=0$, and $n=\int_0^\infty dE D(E) f(E,T)$, 
\beq
0=E_F\der{n}{T}=\int_0^\infty dE\, E_F D(E) \der{f(E,T)}{T}
\eeq
so we may write
\beq
C_V=\int_0^\infty dE\lep E-E_F\rip D(E) \der{f}{T}\,.
\eeq
In $f$, the temperature $T$ enters through both $\beta=1/k_BT$ and $\mu$
\beqa
\der{f}{T}&=&\pde{f}{\beta}\pde{\beta}{T} +  \pde{f}{\mu}\pde{\mu}{T}\nonumber\\
 &=& \frac{\beta e^{\beta(E-\mu)}}{\lep e^{\beta(E-\mu)} +1\rip^2}
\leb\frac{E-\mu}{T}-\pde{\mu}{T}\rib
\eeqa

However, $\pde{\mu}{T}$ depends upon the details of the density of states
near the Fermi surface, which can differ greatly from material to material.
Furthermore, $\left|\pde{\mu}{T}\right|<1$ especially in common metals at temperatures
$T\ll T_F$, and the first term $\frac{E-\mu}{T}$ is of order one (c.f.\
Fig.~\ref{fig:fermi}). Thus, for now we will neglect $\pde{\mu}{T}$ relative
to $\frac{E-\mu}{T}$ (you will explore the validity of this approximation
in your homework), and, consistent with this approximation, replace
$\mu$ by $E_F$, so
\beq
\der{f}{T}\approx \frac{\beta e^{\beta(E-E_F)}}{\lep e^{\beta(E-E_F)} +1\rip^2}
\frac{E-E_F}{T}\,,
\eeq
and
\beqa
C_V &\approx& \frac{1}{k_BT^2} \int_0^\infty dE 
\frac{\lep E-E_F\rip^2 D(E)\beta e^{\beta(E-E_F)}}{\lep e^{\beta(E-E_F)} +1\rip^2} \;\;\; 
\mbox{let $x=\beta(E-E_F)$}\nonumber\\
&\approx & k_BT\int_{-\beta E_F}^\infty dx\, D\lep\frac{x}{\beta} + E_F\rip
x^2 \frac{e^x}{\lep e^x+1\rip^2} 
\eeqa
As shown in Fig.~\ref{fig:dfermi}, the function $x^2 \frac{e^x}{\lep e^x+1\rip^2}$
is only large in the region $-10<x<10$.  
\begin{figure}[htb]
%%BoundingBox: 28 75 524 333
\centerline{\includegraphics[width=0.75\textwidth,clip=true]{dfermi.pdf}}
\caption[]{\em{Plot of $x^2 \frac{e^x}{\lep e^x+1\rip^2}$ vs.\ $x$.
Note that this function is only finite for roughly $-10<x<10$.  Thus,
at temperatures $T\ll T_F \sim 10^5$K, we can approximate 
$D\lep\frac{x}{\beta} + E_F\rip\approx D\lep E_F\rip$ in Eq.~\ref{eq:cv}.}}
\label{fig:dfermi}
\end{figure}
In this region, and for temperatures 
$T\ll T_F$, $D\lep\frac{x}{\beta} + E_F\rip\approx D\lep E_F\rip$, since
the density of states usually does not have features which are sharp on
the energy scale of $10 k_BT$.  Thus
\beqa
C_V &\approx&  k_BT D\lep E_F\rip\int_{-\beta E_F}^\infty 
dx\, x^2 \frac{e^x}{\lep e^x+1\rip^2}\nonumber\\
& \approx & \frac{\pi^2}{3} k_B^2T D(E_F)\,.
\label{eq:cv}
\eeqa
Note that no assumption about the form of D(E) was made other than the
assumption that it is smooth within $k_BT$ of the Fermi surface.
Thus, experimental measurements of the specific heat at constant volume
of the electrons, gives us information about the density of electronic
states at the Fermi surface.

     Now let's reconsider the DOS for the 3-D box potential.
\beq
D(E)=\frac1{2\pi^2} \lep\frac{2m}{\hbar}\rip^{3/2} E^{1/2}=
D(E_F)\lep\frac{E}{E_F}\rip^{1/2}
\eeq
For which $n=\int_0^{E_F} D(E) dE = D(E_F) \frac23 E_F$, so
\beq
C_V=\frac{\pi^2}{2} nk_B \frac{T}{T_F} \ll \frac32 nk_B
\eeq
where the last term on the right is the classical result.
For room temperatures $T\sim 300$K, which is also of the same order of
magnitude as the  Debye temperatures $\theta_D$,
\beq
C_{V\, phonon}\sim \frac32 nk_B \gg C_{V\, electron}
\eeq
So, the only way to measure the electronic specific heat in most materials
is to go to very low temperatures $T\ll \theta_D$, for which 
$C_{V\, phonon}\sim T^3$.  Here the total specific heat
\beq
C_V\approx \gamma T + \beta T^3
\eeq
We will see that  gives us some measurement of the electronic effective mass 
for our Fermi liquid theory.  I.e.\ it tell us something about electron-
electron interactions.

\section{The Weakly Correlated Electronic Liquid}

\subsection{Thomas-Fermi Screening}

     As an introduction to the effect of electronic correlations, consider 
the effect of a charged oxygen defect in one of the copper-oxygen planes of
a cuprate superconductor shown in Fig.~\ref{fig:cupdef}.  Assume that the
oxygen defect captures two electrons from the metallic band, going
from a 2s$^2$2p$^4$ to a 2s$^2$2p$^6$ configuration.  
\begin{figure}[htb]
\centerline{\includegraphics[width=0.6\textwidth,clip=true]{cupdef.pdf}}
\caption[]{\em{A charged oxygen defect is introduced into one of the 
copper-oxygen planes of a cuprate superconductor.  The oxygen defect captures 
two electrons from the metallic band, going from a 2s$^2$2p$^4$ to a 
2s$^2$2p$^6$ configuration.}}
\label{fig:cupdef}
\end{figure}
The defect will then become a cation, and have a net charge of two electrons.
In the vicinity of this oxygen defect, the electrostatic potential
and the electronic charge density will be reduced.  

If we model the electronic density of states
in this material with our box-potential DOS, we can think of this reduction
in the local charge density in terms of raising the DOS parabola near the 
defect (cf.\ Fig.~\ref{fig:cupdefpara}).  
\begin{figure}[htb]
\centerline{\includegraphics[width=0.6\textwidth,clip=true]{cupdefpara.pdf}}
\caption[]{\em{The shift in the DOS parabola near a charged defect.}}
\label{fig:cupdefpara}
\end{figure}
This will cause the free electronic charge to flow away from the defect.
Near the defect (since $e<0$ and hence $e\delta U(\r_{near}) < 0$)
\beq
n(\r_{near}) \approx \int_0^{E_F +e\delta U(\r_{near})} D(E) DE
\eeq
While away from the defect, $\delta U(\r_{away})=0$, so
\beq
n(\r_{away}) \approx \int_0^{E_F} D(E) DE
\eeq
or
\beq
\delta n(\r) \approx \int_0^{E_F +e\delta U(\r)} D(E) DE-\int_0^{E_F} D(E) DE
\eeq
If $|e\delta U| \ll E_F$, then
\beq
%\delta n(\r) \approx D(E_F)\leb E_F +e\delta U -E_F\rib = e\delta U D(E_F)\,.
\delta n(\r) \approx  e\delta U D(E_F)\,.
\eeq
We can solve for the change in the electrostatic potential by
solving Poisson equation.
\beq
\lap \delta U = 4\pi \delta\rho=4\pi e\delta n=4\pi e^2 D(E_F) \delta U\,.
\eeq
Let $\lambda^2=4\pi e^2 D(E_F)$, then $\lap \delta U = \lambda^2 \delta U$
has the solution\footnote{The solution is actually $Ce^{-\lambda r}/r$,
where $C$ is a constant.  $C$ may be deterined by letting $D(E_F)=0$, so
the medium in which the charge is embedded becomes vacuum.  Then the
potential of the charge is $q/r$, so $C=q$.}
\beq
\delta U(\r)=\frac{q e^{-\lambda r}}{r}
\eeq
The length $1/\lambda=r_{TF}$ is known as the Thomas-Fermi screening length.
\beq
r_{TF}=\lep 4\pi e^2 D(E_F) \rip^{-1/2}
\eeq

Lets estimate this distance for our square-well model, 
\beqa
r_{TF}^2 &=& \frac{a_0\pi}{3(3\pi^2n)^{1/3}}\approx 
\frac{a_0}{4n^{1/3}}\nonumber\\
r_{TF}\approx \frac12 \lep\frac{n}{a_0^3}\rip^{-1/6}
\eeqa
In Cu, for which $n\approx 10^{23}$ cm$^{-3}$ (and since $a_0=0.53\AA$)
\beq
r_{TFCu}\approx 
\frac12 \frac{\lep 10^{23}\rip^{-1/6}}{\lep 0.5\times 10^{-8}\rip^{-1/2}}
\approx 0.5\times 10^{-8} \mbox{ cm }=0.5 \AA
\eeq
Thus, if we add a charge defect to Cu metal, the effect of
the defect's ionic potential is screened away for distances $r>\frac12\AA$!  

\subsection{The Mott Transition}

\begin{figure}[htb]
\centerline{\includegraphics[width=0.7\textwidth,clip=true]{screenpot.pdf}}
\caption[]{\em{Screened defect potentials.  As the screening length increases,
states that were free, become bound.}}
\label{fig:screenpot}
\end{figure}
Now consider an electron bound to an ion in Cu or some other metal.  As shown
in Fig.~\ref{fig:screenpot} the screening length decreases, and bound states 
rise up in energy.  
In a weak metal (i.e., something like YBCO), in which
the valence state is barely free, a reduction in the number of carriers 
(electrons) will increase the screening length, since
\beq
r_{TF}\sim n^{-1/6}\,.
\eeq
This will extend the range of the potential, causing it to trap or bind
more states--making the one free valance state bound.  

Now imagine that instead of a single defect, we have a concentrated system
of such ions, and suppose that we decrease the density of carriers 
(i.e., in Si-based semiconductors, this is done by doping certain 
compensating dopants, or even by modulating the pressure).  This will in 
turn, increase the screening length, causing some states that were free 
to become bound, causing an abrupt transition from a metal to an insulator, 
and is believed to explain the MI transition in some transition-metal oxides, 
glasses, amorphous semiconductors, etc.This metal-insulator transition was 
first proposed by N.\ Mott, and is called the Mott transition, more significantly
Mott proposed a criterion based on the relevant electronic density for 
when this transition should occur.  

In Mott's criterion, a metal-insulator transition occurs when this potential, 
in this case coming from the addition of an ionic impurity, can just bind an 
electronic state.  If the state is bound, the impurity band is localized.  
If the state is not bound, then the impurity band is extended.  The critical 
value of $\lambda=\lambda_c$ may be determined numerically\cite{Yukawa} with 
$\lambda_c/a_0 = 1.19$, which yields the Mott criterion of
\beq
2.8 a_0 \approx n_c^{-1/3} \, ,
\eeq 
where $a_0$ is the Bohr radius.  Despite the fact 
that electronic interactions are only incorporated in the extremely weak coupling 
limit, Thomas-Fermi Screening, Mott's criterion even works for moderately and 
strongly interacting systems\cite{mottworkswell}.

\pagebreak
\subsection{Fermi liquids}

The purpose of these next several lectures is to introduce you to
the theory of the Fermi liquid, which is, in its simplest form, a 
collection of Fermions in a box plus interactions.

In reality , the only physical analog is a gas of $^3$He, which due
its nuclear spin (the nucleus has two protons, one neutron), obeys 
Fermi statistics for sufficiently low energies or temperatures. In addition, 
simple metals, from the first or second column of the periodic table,
for which we may approximate the ionic potential
\beq
V(\R)=V_0
\eeq
are a close approximant to Fermi liquids.

Moreover, Fermi Liquid theory only describes the "gaseous" phase of these 
quantum fermion systems.  For example, $^3$He also has a superfluid 
(triplet), and at least in $^4$He-$^3$He mixtures,
a solid phase exists which is not described by Fermi Liquid Theory.  One
should note; however, that the Fermi liquid theory state does serve as the starting
point for the theories of superconductivity and super fluidity.

One may construct Fermi liquid theory either starting from a many-body
diagrammatic or phenomenological viewpoint.  We, as Landau, will
choose the latter.  Fermi liquid theory has 3 basic tenants:
\begin{enumerate}
\item momentum and spin remain good quantum numbers to describe
     the (quasi) particles.

\item the interacting system may be obtained by adiabatically turning 
on a particle-particle interaction over some time $t$.

\item the resulting excitations may be described as quasi-particles 
with lifetimes $\gg t$.
\end{enumerate}

\subsection{Quasi-particles}
     The last assumption involves a new concept, that of the
quasiparticles which requires some explanation.

\subsubsection{Particles and Holes}

Particles and Holes are excitations of the non-interacting system at 
zero temperature.  Consider a system of $N$ free Fermions each of mass $m$ in 
a volume $V$.  The eigenstates are the anti-symmetrized combinations 
(Slater determinants) of $N$ different single particle states.
\beq
\psi_\p(\r)=\frac{1}{\sqrt{V}}e^{i\p\cdot\r/\hbar}
\eeq
The occupation of each of these states is given by $n_\p=\theta(p-p_F)$
where $p_F$ is the radius of the Fermi sphere.  The energy of the system
is
\beq
E=\sum_\p n_\p \frac{p^2}{2m}
\eeq
and $p_F$ is given by
\beq
\frac{N}{V}=\frac1{3\pi^2}\lep\frac{p_F}{\hbar}\rip^3
\eeq

     Now lets add a particle to the lowest available state $p=p_F$ then,
for $T=0$,
\beq
\mu=E_0(N+1)-E_0(N)=\pde{E_0}{N}=\frac{p_F^2}{2m}\,.
\eeq
If we now excite the system, we will promote a certain number of
particles across the Fermi surface $S_F$ yielding particles above and an 
equal number of vacancies or holes below the Fermi surface.  These are our 
elementary excitations, and they are quantified by $\delta n_p=n_p-n_p^0$
\beq
\delta n_p=
\left\{
\begin{array}{ll} 
\delta_{p,p'} \mbox{ for a particle $p'>p_F$}\\
-\delta_{p,p'} \mbox{ for a hole $p'< p_F$}
\end{array}
\right.\,.
\eeq
\begin{figure}[htb]
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{ph_excitation.pdf}}
\caption[]{\em{Particle and hole excitations of the Fermi gas.}}
\label{fig:ph_excitation}
\end{figure}
If we consider excitations created by thermal fluctuations, then 
$\delta n_p\sim 1$ only for excitations of energy within $k_BT$ of
$E_F$.  The energy of the non-interacting system is
completely characterized as a functional of the occupation
\beq
E-E_0=\sum_\p\frac{p^2}{2m} (n_p-n_p^0)=\sum_\p\frac{p^2}{2m}\delta n_p\,.
\eeq

\begin{figure}[htb]
\centerline{\includegraphics[width=0.49\textwidth,clip=true]{ph_fe.pdf}}
\caption[]{\em{Since $\mu=p_F^2/2m$, the free energy of a particle or a hole 
is $\delta F = \left| p^2/2m-\mu\right| >0$, so the system is stable to 
these excitations.}}
\label{fig:ph_fe}
\end{figure}
Now lets take our system and place it in contact with a particle
bath.  Then the appropriate potential is the free energy, which
for $T=0$, is $F=E- \mu N$, and 
\beq
F-F_0=\sum_p\lep\frac{p^2}{2m}-\mu\rip\delta n_p\,.
\eeq
The free energy of a particle, with momentum $\p$ and 
$\delta n_{\p'}=\delta_{\p,\p'}$ is $\frac{p^2}{2m}-\mu$ and it corresponds 
to an excitation outside $S_F$.  The free energy of a hole 
$\delta n_{\p'}=-\delta_{\p,\p'}$ is $\mu-\frac{p^2}{2m}$, which corresponds 
to an excitation within $S_F$.  However, since $\mu=p_F^2/2m$, the free energy 
of either at $p=p_F$ is zero, hence the free energy of an excitation is
\beq
\left| p^2/2m-\mu\right|\,,
\eeq
which is always positive;  ie., the system is stable to excitations.


\subsubsection{Quasiparticles and Quasiholes at $T=0$}

\begin{figure}[htb]
\centerline{\includegraphics[width=0.49\textwidth,clip=true]{fl_sp.pdf}}
\caption[]{\em{Model for a fermi liquid: a set of interacting
particles an average distance $a$ apart bound within an infinite
square-well potential.}}
\label{fig:fl_sp}
\end{figure}
     Now let's consider a system with interacting particles an average 
distance $a$ apart, so that the characteristic energy of interaction is 
$\frac{e^2}{a}e^{-a/r_{TF}}$.  We will imagine that this system evolves slowly 
from an ideal or noninteracting system in time $t$ (i.e., the interaction 
$U\approx \frac{e^2}{a} e^{-a/r_{TF}}$ is turned on slowly, so that the 
non-interacting system evolves while remaining in the ground state into 
an interacting system in time $t$).

          If the eigenstate of the ideal system is characterized
by $n_p^0$, then the interacting system eigenstate will evolve 
quasistatistically from $n_p^0$ to $n_p$.  In fact if the system is 
isotropic and remains in its ground state, then $n_p^0 = n_p$.   However,
clearly in some situations (superconductivity, magnetism)  we
will neglect some eigenstates of the interacting system in this way.

     Now let's add a particle of momentum $\p$ to the non-interacting ideal 
system, and slowly turn on the interaction.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.7\textwidth,clip=true]{qpevolve.pdf}}
\caption[]{\em{We add a particle with momentum $\p$ to our noninteracting
($U=0$) Fermi liquid at time $t=0$, and slowly increase the interaction
to its full value $U$ at time $t$.  As the particle and system evolve, the
particle becomes dressed by interactions with the system (shown as a shaded
ellipse) which changes the effective mass but not the momentum of this 
single-particle excitation (now called a quasi-particle).}}
\label{fig:qpevolve}
\end{figure}
As $U$ is switched on, we slowly begin to perturb the particles
close to the additional particle, so the particle becomes
dressed by these interactions.  However since momentum is conserved, we
have created an excitation (particle and its cloud) of momentum $\p$.
We call this particle and cloud a quasiparticle.  In the
same way, if we had introduced a hole of momentum $\p$ below the
Fermi surface, and slowly turned on the interaction, we would have produced 
a quasihole.

     Note that this adiabatic switching on procedure will have
difficulties if the lifetime of the quasi-particle $\tau<t$.  If 
so, then the the quasiparticle could decay into a number of other 
quasiparticles and quasiholes.  If we shorten $t$ so that again 
$\tau\gg t$, then switching on U could excite the system out of 
its ground state).  We will show that such difficulties do not 
arise so long as the energy of the particle is close to the Fermi 
energy.  Here there are few states accessible for creating 
particle-hole excitations so collisions are rare.

\begin{figure}[htb]
\centerline{\includegraphics[width=0.45\textwidth,clip=true]{billiards.pdf}}
\caption[]{\em{A particle with momentum $\p_1$ above the Fermi
surface ($p_1 > p_F$) interacts with one of the particles 
below the Fermi surface with momentum $\p_2$.  As a result, two new
particles appear above the Fermi surface (all other states are full)
with momenta $\p_3$ and $\p_4$..}}
\label{fig:billiards}
\end{figure}
To estimate this lifetime consider the following argument
from AGD:  A particle with momentum $\p_1$ above the Fermi
surface ($p_1 > p_F$) interacts with one of the particles 
below the Fermi surface with momentum $\p_2$.  As a result, two new
particles appear above the Fermi surface (all other states are full)
with momenta $\p_3$ and $\p_4$.  This may also be interpreted as
a particle of momentum $\p_1$ decaying into particles with momenta
$\p_3$ and $\p_4$ and a hole with momentum $\p_2$.  By Fermi's
golden rule, the total probability of such a process if proportional to
\beq
\frac{1}{\tau}\propto \int\delta\lep\vep_1+\vep_2-\vep_3-\vep_4\rip d^3p_2d^3p_3
\eeq
where $\vep_1=\frac{p_1^2}{2m}-E_F$, and the integral is subject to the 
constraints of energy and momentum conservation and that
\beq
p_2 <p_F\,,\;\;\;p_3>p_F\,,\;\;\;p_4=\left|\p_1+\p_2-\p_3\right|>p_F
\eeq
It must be that $\vep_1+\vep_2=\vep_3+\vep_4>0$ since both particles
3 and 4 must be above the Fermi surface.  However, since
$\vep_2<0$, if $\vep_1$ is small, then $|\vep_2|\alt\vep_1$ is also small,
so only of order $\vep_1/E_F$ states may scatter with the state
$\k_1$, conserve energy, and obey the Pauli principle.  Thus, restricting
$\vep_2$ to a narrow shell of width $\vep_1/E_F$ near the Fermi surface,
and reducing the scattering probability $1/\tau$ by the same factor.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.6\textwidth,clip=true]{quasilife.pdf}}
\caption[]{\em{A quasiparticle of momentum $\p_1$ decays via a particle-hole
excitation into a quasiparticle of momentum $\p_4$.  This may also be 
interpreted as a particle of momentum $\p_1$ decaying into particles with 
momenta $\p_3$ and $\p_4$ and a hole with momentum $\p_2$.  Energy 
conservation requires $|\vep_2|\alt\vep_1$.  Thus, restricting $\vep_2$ to a 
narrow shell of width $\vep_1/E_F$ near the Fermi surface.  Momentum 
conservation $\k_1-\k_3=\k_4-\k_2$ further restricts the available
states by a factor of about $\vep_1/E_F$.  Thus the lifetime of a
quasiparticle is proportional to $\lep\frac{\vep_1}{E_F}\rip^{-2}$.}}
\label{fig:quasilife}
\end{figure}

	Now consider the constraints placed on states $\k_3$ and
$\k_4$ by momentum conservation
\beq
\k_1-\k_3=\k_4-\k_2\,.
\eeq
Since $\vep_1$ and $\vep_2$ are confined to a narrow shell around the
Fermi surface, so too are $\vep_3$ and $\vep_4$.  This can be seen
in Fig.~\ref{fig:quasilife}, where the requirement that $\k_1-\k_3=\k_4-\k_2$
limits the allowed states for particles 3 and 4.  If we take $\k_1$
fixed, then the allowed states for 2 and 3 are obtained by rotating
the vectors $\k_1-\k_3=\k_4-\k_2$; however, this rotation is severely limited
by the fact that particle 3 must remain above, and particle 2 below, the
Fermi surface. This restriction on the final states further reduces the 
scattering probability by a factor of $\vep_1/E_F$.  

Thus, the scattering  rate $1/\tau$ is proportional to 
$\lep\frac{\vep_1}{E_F}\rip^2$ so that excitations of sufficiently small 
energy will always be sufficiently long lived to satisfy the constraints of 
reversibility.  Finally, the fact that the quasiparticle only interacts with 
a small number of other particles due to Thomas-Fermi screening (i.e., those 
within a distance $\approx r_{TF}$), also significantly reduces the scattering
rate.  

\subsection{Energy of Quasiparticles.}

     As in the non-interacting system, excitations will be
quantified by the deviation of the occupation from the ground
state occupation $n_\p^0$
\beq
\delta n_\p=n_\p-n_\p^0\,.
\eeq
At low temperatures $\delta n_\p\sim 1$ only for $p\approx p_F$ where the
particles are sufficiently long lived that $\tau \gg t$.
It is important to emphasize that only $\delta  n_\p$ not $n_\p^0$
or $n_\p$, will be physically relevant.  This is important since it does 
not make much sense to talk about quasiparticle states, described by $n_\p$, 
far from the Fermi surface since they are not stable and cannot be excited
by the perturbations we are considering (i.e., thermal, or in a transport
experiment).

For the ideal system
\beq
E-E_0=\sum_\p\frac{p^2}{2m} \delta n_\p\,.
\eeq
For the interacting system $E[n_\p]$ becomes much more complicated.  If however 
$\delta n_\p$ is small (so that the system is close to its ground state) 
then we may expand:
\beq
E[n_\p]=E_o +\sum_\p \ep_\p \delta n_\p +{\cal{O}}(\delta n_\p^2)\,,
\eeq
where $\ep_\p=\delta E/\delta n_\p$.  Note that $\ep_\p$ is intensive
(ie.\ it is independent of the system volume).  If 
$\delta n_\p=\delta_{\p,\p'}$, then $E\approx E_0+\ep_{\p'}$; i.e., the 
energy of the quasiparticle of momentum $\p'$ is $\ep_{\p'}$.

     In practice we will only need $\ep_p$ near the Fermi surface 
where $\delta n_\p$ is finite.  So we may approximate
\beq
\ep_\p\approx \mu+(\p-\p_F)\cdot\grad_p\left.\ep_\p\right|_{p_F}
\eeq
where $\grad_p\ep_\p=v_\p$, the group velocity of the quasiparticle.  
The ground state of the $N+1$ particle system is obtained by adding 
a particle with $\ep_\p=\ep_F=\mu=\pde{E_0}{N}$ (at zero temperature); 
which defines the chemical potential $\mu$.  We make learn more
about $\ep_\p$ by employing the symmetries of our system.
If we explicitly display the spin-dependence,
\beqa
\ep_{\p,\sigma}&=&\ep_{-\p,-\sigma}\;\;\mbox{under time-reversal}\\
\ep_{\p,\sigma}&=&\ep_{-\p,\sigma}\;\;\mbox{under BZ reflection}
\eeqa
So $\ep_{\p,\sigma}=\ep_{-\p,\sigma}=\ep_{\p,-\sigma}$; i.e., in the
absence of an external magnetic field, $\ep_{\p,\sigma}$  does not depend 
upon $\sigma$ if.  Furthermore, for an isotropic system $\ep_\p$ 
depends only upon the magnitude of $\p$, $|\p|$, so $\p$ and
$ v_\p=\grad \ep_\p(|\p|)=\frac{\p}{|p|} \der{\ep_\p(|\p|)}{|\p|}$
are parallel.  Let us define $m^*$ as the constant of proportionality
at the fermi surface
\beq
v_{p_F}=p_F/m^*
\eeq

Using $m^*$ it is useful to define the density of states at the
fermi surface.  Recall, that in the non-interacting system,
\beq
D(E_F)=\frac1{2\pi^2}\lep\frac{2m}{\hbar^2}\rip^{3/2} E_F^{1/2}
=
\frac{mp_F}{\pi\hbar^3}
\eeq
where $p=\hbar k$, and $E=p^2/2m$.
Thus, for the interacting system at the Fermi surface
\beq
D_{interacting}(E_F)=\frac{m^*p_F}{\pi\hbar^3}\,,
\eeq
where the $m^*$ (generally $>m$, but not always)  accounts for the
fact that the quasiparticle may be viewed as a dressed particle, and must
``drag'' this dressing along with it.  I.e., the effective mass 
to some extent accounts for the interaction between the particles.

\section{Interactions between Particles:  Landau Fermi Liquid}

\subsection{The free energy, and interparticle interactions}

The thermodynamics of the system depends upon the free energy $F$,
which at zero temperature is
\beq
F-F_0=E-E_0-\mu(N-N_0)\,.
\eeq
Since our quasiparticles are formed by
adiabatically switching on the interaction in the $N+1$  particle
ideal system, adding one quasiparticle to the system adds one
real particle.  Thus,
\beq
N-N_0=\sum_\p\delta n_\p\,,
\eeq
and since
\beq
E-E_0\approx \sum_\p \ep_\p \delta n_\p\,,
\eeq
we get
\beq
F-F_0\approx\sum_\p \lep\ep_\p-\mu\rip\delta n_\p\,.
\eeq
As shown in Fig.~\ref{fig:fs_distort}, we will be interested in excitations of the 
system which distort the Fermi surface by an amount proportional to $\delta$.
\begin{figure}[htb]
\centerline{\includegraphics[width=0.45\textwidth,clip=true]{fs_distort.pdf}}
\caption[]{\em{We consider small distortions of the fermi surface, 
proportional to $\delta$, so that $\frac1{N}\sum_\p|\delta n_\p |\ll 1$.}}
\label{fig:fs_distort}
\end{figure}
For our theory/expansion to remain valid, we must have
\beq
\frac1{N}\sum_\p|\delta n_\p |\ll 1\,.
\eeq
Where $\delta n_\p\neq 0$,  $\ep_\p-\mu$ will also be of order $\delta$.  
Thus,
\beq
\sum_\p \lep\ep_\p-\mu\rip\delta n_\p\sim \CO(\delta^2)\,,
\eeq
so, to be consistent we must add the next term in the Taylor series
expansion of the energy to the expression for the free energy.
\beq
F-F_0=\sum_\p \lep\ep_\p-\mu\rip\delta n_\p
+\frac12\sum_{\p,\p'} f_{\p,\p'} \delta n_\p\delta n_{\p'}
+\CO(\delta^3)
\label{eq:deltafree}
\eeq
where
\beq
f_{\p,\p'} =\frac{\delta E}{\delta n_\p\delta n_{\p'}}
\eeq
The term, proportional to $f_{\p,\p'}$, was added (to the Sommerfeld theory)  
by L.D.\ Landau.  Since each sum over $\p$ is proportional to the
volume $V$, as is $F$, it must be that $f_{\p,\p'}\sim 1/V$.
However, it is also clear that $f_{\p,\p'}$ is an interaction between
quasiparticles, each of which is spread out over the whole volume $V$, so
the probability that they will interact is $\sim r_{TF}^3/V$, thus
\beq
f_{\p,\p'}\sim r_{TF}^3/V^2
\eeq

     In general, since $\delta n_\p$ is only of order one near the 
Fermi surface, we will only care about $f_{\p,\p'}$ on the Fermi surface 
(assuming that it is continuous and changes slowly as we cross the Fermi 
surface.
\beq
\mbox{Interested only in $\left. f_{\p,\p'}\right|_{\ep_\p=\ep_{\p'}=\mu}$ !}
\eeq
Thus, $f_{\p,\p'}$ only depends upon the angle between $\p$ and $\p'$.

We can also reduce the spin dependence of $f_{\p,\p'}$ to a
symmetric and anti symmetric part.  First in the absence of an external
field, the system should be invariant under time-reversal, so
\beq
f_{\p\sigma,\p'\sigma'}=f_{-\p-\sigma,-\p'-\sigma'}\,,
\eeq
and, in a system with reflection symmetry 
\beq
f_{\p\sigma,\p'\sigma'}=f_{-\p\sigma,-\p'\sigma'}\,.
\eeq
Then
\beq
f_{\p\sigma,\p'\sigma'}=f_{\p-\sigma,\p'-\sigma'}\,.
\eeq
It must be then that $f$ depends only upon the relative orientations of the
spins $\sigma$ and $\sigma'$, so there are only two independent components 
$f_{\p\uparrow,\p'\uparrow}$ and $f_{\p\uparrow,\p'\downarrow}$. We can 
split these into symmetric and antisymmetric parts.
\beq
f_{\p,\p'}^a=\frac12 \lep f_{\p\uparrow,\p'\uparrow}- f_{\p\uparrow,\p'\downarrow}\rip 
\;\;\;\;\;\;
f_{\p,\p'}^s=\frac12 \lep f_{\p\uparrow,\p'\uparrow}+ f_{\p\uparrow,\p'\downarrow}\rip \,.
\eeq
$f_{\p,\p'}^a$ may be interpreted as an exchange interaction, or
\beq
f_{\p\sigma,\p'\sigma'}=
f_{\p,\p'}^s + \bsi\cdot\bsi' f_{\p,\p'}^a
\eeq
where $\bsi$ and $\bsi'$  are the Pauli matrices for the spins.

Our ideal system is isotropic in momentum.  Thus, $f_{\p,\p'}^a$ and 
$f_{\p,\p'}^s$ will only depend upon the angle $\theta$ between $\p$
and $\p'$, and so we may expand either $f_{\p,\p'}^a$ and $f_{\p,\p'}^s$
\beq
f_{\p,\p'}^\alpha =\sum_{l=0}^\infty f_l^\alpha P_l(\cth)\,.
\eeq
Conventionally these f parameters are expressed in terms of reduced units.
\beq
D(E_F) f_l^\alpha=\frac{Vm^*p_F}{\pi^2\hbar^3} f_l^\alpha =F_l^\alpha\,.
\eeq

\subsection{Local Energy of a Quasiparticle}
\label{sec:local_energy}

 \begin{figure}[htb]
\centerline{\includegraphics[width=0.4\textwidth,clip=true]{qp_inhom.pdf}}
\caption[]{\em{The addition of another particle to a homogeneous system 
will yeilds in forces on the quasiparticle which tend to restore equilibrium.}}
\label{fig:qp_inhom}
\end{figure}
    Now consider an interacting system with a certain distribution of excited 
quasiparticles $\delta n_{\p'}$.  To this, add another quasiparticle of 
momentum $\p$ ($\delta n_\p'\to \delta n_\p'+ \delta_{\p,\p'}$).  From 
Eq.~\ref{eq:deltafree} 
the free energy of the additional quasiparticle is
\beq
\epti_\p-\mu=\ep_\p-\mu +\sum_{\p'}f_{\p',\p}\delta n_{\p'}\,,
\label{eq:addfree}
\eeq
(recall that $f_{\p,\p'}=f_{\p',\p}$).  Both terms here are  $\CO(\delta)$.  
The second term describes the free energy of a quasiparticle due to the other
quasiparticles in the system (some sort of Hartree-like term).

The term $\epti_\p$ plays the part of the {\em{local energy of a 
quasiparticle}}.  For example, the gradient of $\epti_\p$ is the force the  system exerts on the additional quasiparticle.  When the quasiparticle is
added to the system, the system is inhomogeneous so 
that $\delta n_{\p'}=\delta n_{\p'}(\r)$. The system will react to this 
inhomogeneity by minimizing its free energy so that $\grad_rF=0$. 
However, only the additional free energy due the added particle 
(Eq.~\ref{eq:addfree}) is inhomogeneous, and has a non-zero gradient.  Thus, 
the system will exert a force
\beq
-\grad_r\epti=-\grad_r\sum_{\p'} f_{\p',\p} \delta n_{\p'}(\r)
\eeq
on the added quasiparticle resulting from interactions with other
quasiparticles.

	

\subsubsection{Equilibrium Distribution of Quasiparticles at Finite $T$}

$\epti_\p$ also plays an important role in the finite-temperature
properties of the system.  If we write
\beq
E-E_0=\sum_\p\ep_\p \delta n_\p + 
\frac12\sum_{\p,\p'}f_{\p',\p} \delta n_{\p'}\delta n_{\p}
\eeq
Now suppose that $\sum_\p\left|\langle \delta n_{\p}\rangle\right| \ll N$, 
as needed for the expansion above to be valid, so that
\beq
\delta n_{\p}= \langle \delta n_{\p}\rangle
+\lep \delta n_{\p}-\langle \delta n_{\p}\rangle \rip
\eeq
where the first term is $\CO(\delta)$, and the second $\CO(\delta^2)$.
Thus,
\beq
\delta n_{\p}\delta n_{\p'}\approx
-\langle \delta n_{\p}\rangle\langle \delta n_{\p'}\rangle
+\langle \delta n_{\p}\rangle\delta n_{\p'}
+ \langle \delta n_{\p'}\rangle\delta n_{\p}
\eeq
We may use this to rewrite the energy of our interacting system
\beqa
E-E_0&\approx&
\sum_\p\ep_\p\delta n_\p 
-\frac12\sum_{\p,\p'}f_{\p',\p}
	\langle \delta n_{\p}\rangle\langle \delta n_{\p'}\rangle
+ \sum_{\p,\p'}f_{\p',\p}\langle \delta n_{\p}\rangle\delta n_{\p'}\nonumber\\
&\approx&
\sum_\p \lep \ep_\p +\sum_{\p'}f_{\p',\p}\langle \delta n_{\p'}\rangle
\rip\delta n_\p
-\frac12\sum_{\p,\p'}f_{\p',\p}
	\langle \delta n_{\p}\rangle\langle \delta n_{\p'}\rangle\nonumber\\
&\approx&
\sum_\p \langle\epti_\p\rangle \delta n_\p
-\frac12\sum_{\p,\p'}f_{\p',\p}
	\langle \delta n_{\p}\rangle\langle \delta n_{\p'}\rangle+\CO(\delta^4)
\eeqa
At this point, we may repeat the arguments made earlier to
determine the fermion occupation probability for non-interacting
Fermions (the constant factor on the right hand-side has no effect).  
We will obtain
\beq
n_\p(T,\mu)=\frac1{1+\exp{\beta(\langle\epti_\p\rangle-\mu)}}\,,
\eeq
or
\beq
\delta n_\p(T,\mu)=\frac1{1+\exp{\beta(\langle\epti_\p\rangle-\mu)}}
-\theta(p_f-p)\,.
\eeq
However, at least for an isotropic system, this expression bears
closer investigation.  Here, the molecular field (evaluated
within $k_BT$ of the Fermi surface)
\beq
\langle\epti_\p-\ep_\p\rangle=\sum_{\p'}f_{\p',\p}\langle \delta n_{\p'}\rangle
\eeq
must be independent of the location of $\p$ on the Fermi surface (and of 
course, spin), and is thus constant.  To see this, reconsider the
Legendre polynomial expansion discussed earlier
\beqa
\langle\epti_\p-\ep_\p\rangle
&=&\sum_{\p'}f_{\p',\p}\langle \delta n_{\p'}\rangle\nonumber\\
&\propto& \sum_l\int d^3p f_lP_l(\cth)\langle \delta n_{\p'}\rangle\nonumber\\
&\propto& f_0\int d^3p \langle \delta n_{\p'}\rangle=0\nonumber\\
\eeqa
In going from the second to the third line above, we made use of the
isotropy of the system, so that  $\langle \delta n_{\p'}\rangle$ is 
independent of the angle $\th$.  The evaluation in the third line,
follows from particle number conservation.  Thus, to lowest order in
$\delta$
\beq
n_\p(T,\mu)=\frac1{1+\exp{\beta(\ep_\p-\mu)}} + \CO(\delta^4)
\eeq

\subsubsection{Local Equilibrium Distribution}

	Now suppose we introduce a local weak perturbation such as the 
type discussed in Sec.~\ref{sec:local_energy} to an isotropic system at
zero temperature. Such a perturbation could be caused by, eg. a sound 
wave or a weak magnetic field (which you will explore in your homework, 
to calculate the sound velocity and susceptibility of a Landau Fermi 
liquid).  This pertubation will cause a small deviation of the 
equilibrium distribution function, leading to a new "local equilibrium" 
distribution
\beq
\bar n_\p= n_\p(\epti_\p-\mu)
\eeq
where the argument of the RHS indicates that this is the distribution
corresponding to the local energy discussed above. The gradient of
the local energy yields a force which tries to restore the equilibrium 
distribution $n_\p(\ep_\p-\mu)$ derived above).  The deviation from
true equilibrium is
\beqa
\delta n_\p &=& n_\p -\bar n_\p \\
	    &=& \delta \bar n_\p + \frac{\partial n_\p (\ep_\p-\mu)}
	                                {\partial \ep_\p}
				    \lep \epti_\p - \ep_\p \rip
\label{eq:deltanp}
\eeqa
Using Eq.~\ref{eq:addfree}, we find 
\beq
\delta n_\p = \delta \bar n_\p - \frac{\partial n_\p }{\partial \ep_\p}
				 \sum_{\p'} f_{\p,\p'} \delta n_{\p'}
\eeq
At zero temperature, the factor $\frac{\partial n_\p }{\partial \ep_\p}=
-\delta (\ep_\p-\mu)$, so both $\delta n_\p$ and  $ \delta \bar n_\p$
are restricted to the fermi surface.  Since the pertubation of interest
is small, we may expand both $\delta n_\p$ and  $ \delta \bar n_\p$
in a series of Legendre polynomials, and we will also split
them into symmetric and antisymmetric parts (as we did with $f_{\p,\p'}$
previously).  For example,
\beq
\delta n_\p^s = \sum_l \delta (\ep_\p-\mu) \delta n_l^s P_l
\eeq
If we make a similar expansion for the antisymmetric and symmetric
parts of $\delta \bar n_\p$, and substitute this back into
Eq.~\ref{eq:deltanp}, then we find
\beqa
\delta \bar n_l^a &=& \lep 1+ \frac{F_l^a}{2l + 1}\rip \delta n_l^a\\
\delta \bar n_l^s &=& \lep 1+ \frac{F_l^s}{2l + 1}\rip \delta n_l^s
\eeqa


\subsection{Effective Mass $m^*$ of Quasiparticles}

This argument most closely follows that of AGD, and we will follow their
notation as closely as possible (without introducing any new symbols).
In particular, since an integration by parts is necessary, we will
use a momentum integral (as opposed to a momentum sum) notation
\beq
\sum_\p\to V\int \frac{d^3p}{(2\pi\hbar)^3}\,.
\eeq
The net momentum of the volume $V$ of quasiparticles is
\beq
\P_{qp}=2V \int \frac{d^3p}{(2\pi\hbar)^3} \p n_\p\mbox{ net quasiparticle momentum}
\eeq
which is also the momentum of the Fermi liquid.  On the other
hand since the number of particles equals the number of quasiparticles,
the quasiparticle and particle currents must also be equal
\beq
\J_{qp}=\J_{p}=2V \int \frac{d^3p}{(2\pi\hbar)^3} \v_\p n_\p\mbox{ net 
quasiparticle and particle current}
\eeq
or, since the momentum is just the particle mass times this current
\beq
\P_{p}=2Vm \int \frac{d^3p}{(2\pi\hbar)^3} \v_\p n_\p\mbox{   net quasiparticle 
and particle current}
\eeq
where $\v_\p=\grad_p\epti_\p$, is the velocity of the quasiparticle.  So
\beq
\int \frac{d^3p}{(2\pi\hbar)^3} \p n_\p = 
m \int \frac{d^3p}{(2\pi\hbar)^3} \grad_p\epti_\p n_\p
\label{eq:massequal}
\eeq
Now make an arbitrary change of $n_\p$ and recall that $\epti_\p$ depends 
upon $n_\p$, so that
\beq
\delta \epti_\p=V \sum_{\sigma'} \int \frac{d^3p}{(2\pi\hbar)^3} f_{\p,\p'} \delta n_{\p'}\,.
\eeq
For Eq.~\ref{eq:massequal}, this means that 
\beqa
\int \frac{d^3p}{(2\pi\hbar)^3} \p \delta n_\p&=& 
m \int \frac{d^3p}{(2\pi\hbar)^3} \grad_p\epti_\p \delta n_\p \\
&& +
m V \int \frac{d^3p}{(2\pi\hbar)^3} \sum_{\sigma'} 
\int \frac{d^3p'}{(2\pi\hbar)^3}
\grad_p\lep  f_{\p,\p'} \delta n_{\p'}\rip n_\p\,,\nonumber
\eeqa
or integrating by parts (and renaming $\p\to\p'$ in the last part), we get
\beqa
\int \frac{d^3p}{(2\pi\hbar)^3} \frac{\p}{m} \delta n_\p &=& 
\int \frac{d^3p}{(2\pi\hbar)^3} \grad_p\epti_\p \delta n_\p  \\
&&-
V \sum_{\sigma'}\int \frac{d^3p'}{(2\pi\hbar)^3} \int \frac{d^3p}{(2\pi\hbar)^3}
\delta n_\p  f_{\p,\p'} \grad_{p'} n_{\p'} \,,\nonumber
\eeqa
Then, since $\delta n_\p$ is arbitrary, it must be that the
integrands themselves are equal
\beq
\frac{\p}{m}=\grad_p\epti_\p
-\sum_{\sigma'}V \int \frac{d^3p'}{(2\pi\hbar)^3}
f_{\p,\p'} \grad_{p'} n_{\p'}
\eeq
The factor $\grad_{p'} n_{\p'}=-\frac{\p'}{p'}\delta(p'-\p_F)$.  The 
integral may be evaluated by taking advantage of the system isotropy,
and setting $\p$ parallel to the z-axis, since we mostly interested
in the properties of the system on the Fermi surface we take $p=p_F$,
let $\th$ be the angle between $\p$ (or the z-axis) and $\p'$, and finally
note that on the Fermi surface 
$\left|\left.\grad_p\epti_p\right|_{p=p_F}\right|=v_F=p_F/m^*$.  Thus,
\beq
\frac{p_F}{m}=\frac{p_F}{m^*}
+
\sum_{\sigma'}\int \frac{p'^2 dpd\Omega}{(2\pi\hbar)^3}
f_{\p\sigma,\p'\sigma'} \frac{\p'}{p'} \delta(p'-p_F)
\eeq
However, since both $p$ and $p'$ are restricted to the Fermi surface
$\frac{\p'}{p'}=\cth$, and evaluating the integral over $p$, we get
\beq
\frac{1}{m}=\frac{1}{m^*}+
\frac{Vp_F}{2}\sum_{\sigma,\sigma'} 
\int \frac{d\Omega}{(2\pi\hbar)^3}
f_{\p\sigma,\p'\sigma'} \cth \,,
\eeq
where the additional factor of $\frac12$ compensates for the additional
spin sum.  If we now sum over both spins, $\sigma$ and $\sigma'$,
only the symmetric part of $f$ survives (the sum yields $4f^s$), so
\beq
\frac{1}{m}=\frac{1}{m^*}+
\frac{4\pi Vp_F}{(2\pi\hbar)^3} 
\int d\lep\cth\rip f^s(\th) \cth \,,
\eeq
We now expand $f$ in a Legendre polynomial series
\beq
f^\alpha(\th)=\sum_l f_l^\alpha P_l(\cth)\,,
\eeq
and recall that $P_0(x)=1$, $P_1(x)=x$, .... that
\beq
\int_{-1}^1 dx P_n(x) P_m(x) dx = \frac{2}{2n+1}\delta_{nm}
\eeq
and finally that 
\beq
D(0) f_l^\alpha=\frac{Vm^*p_F}{\pi^2\hbar^3} f_l^\alpha =F_l^\alpha\,,
\eeq
we find that
\beq
\frac{1}{m}=\frac{1}{m^*} + \frac{F_1^s}{3m^*}\,,
\eeq
or $m^*/m= 1+F_1^s/3$.

\begin{table}[htb]
\begin{center}
\begin{tabular}{|l|l|l|}\hline
Quantity		& Fermi Liquid & Fermi Liquid/Fermi Gas\\\hline
Specific Heat		& $C_v=\frac{m^*p_F}{3\hbar^3}k_B^2 T$		&$\frac{C_V}{C_{V0}}=\frac{m^*}{m}=1+F_1^s/3$\\
Compressibility		& 						&$\frac{\kappa}{\kappa_0}=\frac{1+F_0^s}{1+F_0^s/3}$\\
Sound Velocity		& $c^2=\frac{p_F^2}{3mm^*} \lep 1+F_0^s\rip$	&$\lep\frac{c}{c_0}\rip^2=\frac{1+F_0^s}{1+F_1^s/3}$\\
Spin Susceptibility	& $\chi=\frac{m^*p_F}{\pi^2\hbar^3}\frac{\beta^2}{1+F_0^a}$	&$\frac{\chi}{\chi_0}=\frac{1+F_1^s/3}{1+F_0^a}$\\
\hline
\end{tabular}
\caption[]{\em{Fermi Liquid relations between the Landau parameters $F^\alpha_n$
and some experimentally measurable quantities.  For the latter, a zero
subscript indicates the value for the non-interacting Fermi gas.}}
\label{fig:Lanparam}
\end{center}
\end{table}
	The effective mass cannot be experimentally measured directly;
however, it appears in many physically relevant measurable quantities,
including the specific heat
\beq
C_V=\lep\pde{E/V}{T}\rip_{VN}=\frac1{V}\pde{}{T}\sum_\p \epti_\p n_\p.
\eeq
To lowest order in $\delta$, we may neglect $f_{\p,\p'}$ in both $\epti_\p$
and $n_\p$, so
\beq
C_V=\frac1{V}\sum_\p \ep_\p \pde{n_\p}{T}\,.
\eeq
Recall that the density of states $D(E)=\sum_\p\delta(E-\ep_p)$, and
making the same assumption that we made for the non-interacting system,
that $\pde{\mu}{T}$ is negligible, we get,
\beq
C_V=\frac{1}{V} \int d\ep D(\ep)\ep\pde{}{T}\frac{1}{\exp{\beta(\ep-\mu)}+1}\,.
\eeq
This integral is identical to the one we had to evaluate for the 
non-interacting system, and yields the result
\beqa
C_V&=& \frac{\pi^2}{3V} k_B^2TD(E_F)\nonumber\\
   &=& \frac{k_B^2Tm^*p_F}{3\hbar^3}\,.
\eeqa
Thus, measuring the electronic contribution to the specific heat $C_V$ 
yields information about the effective mass $m^*$, and hence $F_1^s$.
Other measurements are related to some of the remaining Landau parameters,
as summarized in table~\ref{fig:Lanparam}.


\begin{thebibliography}{00}

%% \bibitem must have the following form:
%%   \bibitem{key}...
%%

\bibitem{Yukawa} https://arxiv.org/pdf/hep-ph/0407258v2.pdf

\bibitem{mottworkswell} https://arxiv.org/pdf/1411.4372.pdf

\end{thebibliography}

\edo
