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  %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
  %% Indranil's LaTeX macros
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  \def\ek{\epsilon_{k}}
  \def\xk{\xi_k} 
  \def\la{{\langle\, }}
  \def\ra{{\,\rangle }}
  \def\p{\,|\,}
  \def\infintd3k{ \int_{-\infty}^\infty d^3r\,}
  \def\pm{\prime}
  % Equation environments
  \def\bea{\begin{eqnarray}}
  \def\eea{\end{eqnarray}}
  \def\ben{\begin{equation}}
  \def\een{\end{equation}}
  \def\benu{\begin{enumerate}}
  \def\enu{\end{enumerate}}

  
  
   \begin{document}

  
   
   \begin{center}
   Solution to Problem Set 3
   \end{center}
 

We will consider a system of fermions interacting via a short range
two body force with potential $V(r)$. The free part of the Hamiltonian
is given by,
\ben
\mathcal{H}_0 = \sum_k \ek a^{\dagger}_k a_k
\een
and the interacting part (which will be treated in perturbation
theory) is given by,
\[
\mathcal{H}_1 = \frac{1}{2} \int dr dr^{\pm} V(r-r^{\pm})\psi^{\dagger}(r)
\psi(r)\psi^{\dagger}(r^{\pm})\psi(r^{\pm})
\]
which in second quantized notation can be written as,
\ben
\mathcal{H}_1 = \frac{1}{2V}\sum_{k,k^{\pm},q}V(q) a^{\dagger}_{k+q} a_k
 a^{\dagger}_{k^{\pm}-q} a_{k^{\pm}}
\een
The Fourier transform of the potential is defined as,
\ben
V(r) = \frac{1}{V} \sum_q e^{iqr} V(q)
\een

\begin{enumerate}
\item To first order in perturbation, self energy corrections are given by
the Hartree and the Fock terms. 
\bea
\Sigma^1(p,\omega) &=& -i\frac{V(0)}{V}\sum^k \int \frac{d\Omega}{2\pi}
G^0(k,\Omega)e^{i\Omega\eta} + i\frac{1}{V}\sum_k V(p-q)\int
\frac{d\Omega}{2\pi}G^0(k,\Omega)e^{i\Omega\eta} \nonumber \\
&=& nV(0) - \frac{1}{V} \sum_{k \leq k_f} V(p-k)
\eea

From the definition of effective mass,
\[
\frac{p_f}{m^{\star}} = \frac{\partial \tilde{\epsilon_p}}{\partial p}
|_{p=p_f}
\]
we find that the Hartree term does not change the quasiparticle mass. If
the interaction is momentum dependent, the Fock term renormalizes the
mass to give,
\ben
\frac{p_f}{m^{\star}} = \frac{p_f}{m} - \frac{1}{V} \sum_{k \leq k_f}
\frac{\partial V(p-k)}{\partial p}|_{p=p_f}
\een



As an aside we will calculate the self energy using perturbation
theory and Wick's theorem (i.e. not use diagrams and Green's function),
and rederive eqn (4). To calculate energy upto first order in perturbation
we need only the zeroth order wavefunctions. Then, the energy of the 
interacting Fermi sea is given by, 
\[
E_0 = \la \phi_0 \p \mathcal{H}_0 \p \phi_0 \ra  +
\la \phi_0 \p \mathcal{H}_1 \p \phi_0 \ra 
\]
and the energy of the one quasiparticle state is given by,
\[
E_p = \la \phi_0 \p a_p \mathcal{H}_0 a^{\dagger}_p \p \phi_0 \ra +
\la \phi_0 \p a_p \mathcal{H}_1 a^{\dagger}_p \p \phi_0 \ra 
\]
Then, the quasiparticle energy is given by,
\bea
\tilde{\epsilon}_p &=& E_p - E_0 \nonumber \\
&=& \la \phi_0 \p a_p \mathcal{H}_0 a^{\dagger}_p \p \phi_0 \ra -
\la \phi_0 \p \mathcal{H}_0 \p \phi_0 \ra +
\la \phi_0 \p a_p \mathcal{H}_1 a^{\dagger}_p \p \phi_0 \ra -
\la \phi_0 \p \mathcal{H}_1 \p \phi_0 \ra \nonumber \\
&=& \epsilon_p + \Sigma^1_p \nonumber
\eea
Thus the self energy of the quasiparticle is given by,
\bea
\Sigma^1_p &=& \la \phi_0 \p a_p \mathcal{H}_1 a^{\dagger}_p \p \phi_0 \ra -
\la \phi_0 \p \mathcal{H}_1 \p \phi_0 \ra \nonumber \\
&=& \frac{1}{2V} \sum_{k_1,k_2,q} V(q) \la \phi_0 \p a_p 
a^{\dagger}_{k_1+q}a_{k_1}a^{\dagger}_{k_2-q}a_{k_2}
a^{\dagger}_p \p \phi_0 \ra - 
\frac{1}{2V} \sum_{k_1,k_2,q} V(q) \la \phi_0 \p
a^{\dagger}_{k_1+q}a_{k_1}a^{\dagger}_{k_2-q}a_{k_2}
\p \phi_0 \ra \nonumber
\eea
Now, we use Wick's theorem to calculate the above matrix elements. We
look at the first term above. There will be a contraction of the form
$  \la \phi_0 \p a_p a^{\dagger}_p \p \phi_0 \ra 
\la \phi_0 \p a^{\dagger}_{k_1+q}a_{k_1}a^{\dagger}_{k_2-q}a_{k_2}
\p \phi_0 \ra $ which will cancel exactly with the second term in the 
above expression. This corresponds to correction to the Fermi sea,
and in diagrammatic language, are the disconnected diagrams. The remaining
two contractions will be $ \la a^{\dagger}_{k_1+q}a_{k_1} \ra_0
\la a_p a^{\dagger}_{k_2-q} \ra_0 \la a_{k_2}a^{\dagger}_p \ra_0 $, which
gives the Hartree term, and $ \la a_p a^{\dagger}_{k_1+q} \ra_0
\la a_{k_1}a^{\dagger}_{k_2-q} \ra_0 \la a_{k_2} a^{\dagger}_p \ra_0 $,
which gives the Fock term. Interchanging $k_1$ and $k_2$ give a factor
of 2 which cancel with the 2 in the denominator. Finally we get,
\[
\Sigma^1_p  = n V(0) + \frac{1}{V} \sum_{k \geq k_f} V(p - k)
\]
which differ from equation (4) by a constant term. The origin of this
diffrence can be traced back to the fact that we have used 
$ a^{\dagger} a a^{\dagger} a $ in the form of the interaction, instead of 
$a^{\dagger} a^{\dagger} a a $ that is used in diagrammatic expansions.




\item The perturbed ground state $\p \psi_0 \ra $ is constructed from the 
unperturbed ground state $\p \phi_0 \ra $ by switching on
the interaction adiabatically. This is mathematically expressed as 
$\p \psi_0 \ra
= \hat{\Omega} \p \phi_0 \ra $, where $\hat{\Omega}$ is the evolution 
operator, and is given by,
\ben
\hat{\Omega} = T exp\{ \frac{1}{i}\int_{-\infty}^{\infty}d\tau
\mathcal{H}_1^{int} (\tau)e^{\eta \tau}\} \hspace{.1in} , 
\hspace{.1in} \eta \rightarrow 0^+
\een
The time evolution of the interaction term in the interaction picture
is given by,
\bea
\mathcal{H}_1^{int} (t) &=& e^{i\mathcal{H}_0 t}\mathcal{H}_1
e^{-i\mathcal{H}_0 t} \nonumber \\
&=& \frac{1}{2V} \sum_{k,k^{\prime},p} V(p) a^{\dagger}_{k+p} a_k
a^{\dagger}_{k^{\pm}-p} a_{k^{\pm}} e^{i \left(\epsilon_{k+p} 
+ \epsilon_{k^{\pm}-p} - \epsilon_k -  \epsilon_{k^{\pm}} \right) t}
\eea 
Expanding the evolution operator to first order in interaction we get,
\bea
\hat{\Omega} &=& 1 + \frac{1}{i} \int_{-\infty}^0 d\tau 
\mathcal{H}_1^{int}(\tau)e^{\eta \tau} \nonumber \\
&=& 1 + \frac{1}{2V}\sum_{k,k^{\prime},p} \frac{V(p)}
{\left( \epsilon_k + \epsilon_{k^{\pm}} - \epsilon_{k+p} -
\epsilon_{k^{\pm}-p} + i \eta \right) }
 a^{\dagger}_{k+p} a_k a^{\dagger}_{k^{\pm}-p} a_{k^{\pm}}
\eea
Then the many body state $ \p q \ra = \hat{\Omega} a^{\dagger}_q 
\p \phi_0 \ra $ , to lowest order 
in interaction is given by,
\ben
\p q \ra =  a^{\dagger}_q \p \phi_0 \ra + \frac{1}{2V}\sum_{k,k^{\prime},p}
\frac{V(p)}{\left( \epsilon_k + \epsilon_{k^{\pm}} - \epsilon_{k+p} -
\epsilon_{k^{\pm}-p} + i \eta \right) } a^{\dagger}_{k+p} a_k 
a^{\dagger}_{k^{\pm}-p} a_{k^{\pm}} a^{\dagger}_q \p \phi_0 \ra 
\een

\item
The charge density operator is given by,
\[
\rho(r) = \frac{1}{V} \sum_q e^{iqr} \rho(q)
\]
where,
\[
\rho(q) = \sum_p  a^{\dagger}_{p+q}a_p
\]
The expectation of the charge density operator in the state $ \p k \ra $
is given by,
\[
\la \rho(q) \ra = \sum_p \la k \p a^{\dagger}_{p+q}a_p \p k \ra
\]
Looking at the momentum of the states, it is easy to conclude that the
above expectation values are zero except for $ q = 0 $. $ \rho(q = 0) $
gives  the total particle number, which is a conserved quantity. For a
N particle ground state $ \p \phi_0 \ra $, we have $ \la \rho (q = 0)
\ra = N+1 $. Then,
\ben
\la \rho(r) \ra = \frac{1}{V}\la \rho (q = 0) \ra = \frac{N+1}{V}
\een
Thus, the charge density is uniform and is delocalized over the entire
system.

The current density operator is given by,
\[
\bar{j}(r) =  \frac{1}{V} \sum_q e^{iqr} \bar{j}(q)
\]
where,
\[
\bar{j}(q) =  \sum_p \frac{1}{m} (\bar{p} - \frac{\bar{q}}{2}) 
a^{\dagger}_{p+q}a_p
\]
By the same argument as above the average quantity $ \la k \p \bar{j}(q) 
\p k \ra $ is zero except for $ q = 0$. For a translationally invariant
system, total momentum is a conserved quatity. Particularly, $ \left[
\hat{\Omega}, \sum_p \bar{p} a^{\dagger}_{p}a_p \right] = 0 $, so that
$ \la k \p \bar{j}(q = 0) \p k \ra = \frac{\bar{k}}{m} $. It is to be 
noted that what enters in the expression for current density is the
bare particle mass and not the renormalized quasiparticle mass, and
that this is a consequence of translational invariance. Thus,
\ben
\la \bar{j}(r) \ra =  \frac{1}{V} \left( \frac{\bar{k}}{m} \right)
\een
The current density due to the quasiparticle excitation is uniformly
spread.


\item 
We consider the wave-packet state $ \p x_0 \ra = \sum_q A_q \p q \ra $,
where $ A_q = e^{- \alpha (q - q_0)^2}e^{-iqx_0} $, constructed out of 
single particle excitations. We expect this state to behave as an excitation
that is localized around $x_0$. To find whether indeed that is the case,
we look at the average charge density $\la \rho_q \ra $ in this
state. We have,
\ben
\la x_0 \p \rho_q \p x_0 \ra = \sum_{k,p} A_k A^{\star}_{k+q}
\la k+q \p a^{\dagger}_{p+q} a_p \p k \ra
\een
As before the $ q=0 $ component measures the total particle number,
which is $ N+1 $ for a N particle ground state. Thus,
\ben
\la x_0 \p \rho_0 \p x_0 \ra = (N+1) \sum_k A_k A^{\star}_k
\een
For $ q \neq 0 $ components we expand the many particle wavefunction 
$ \p k \ra $ in perturbation theory. Then the zeroth order contribution
to the matrix element on the right hand side of eqn(12) is 
$ \la \phi_0 \p a_{k+q} a^{\dagger}
_{p+q} a_p a^{\dagger}_k \p \phi_0 \ra = \delta_{p,k} $. From the
result of part (2) we get the first order contribution as,
\[
\frac{1}{V} \sum_{k_1,k_2,k_3} \frac{V(k_3)}{\epsilon_{k_1} + \epsilon_{k_2}
- \epsilon_{k_1 + k_3} - \epsilon_{k_2 - k_3} + i \eta} \la \phi_0 \p a_{k+q}
\left[ a^{\dagger}_{p+q} a_p , a^{\dagger}_{k_1+k_3} a_{k_1} 
 a^{\dagger}_{k_2-k_3} a_{k_2} \right] a^{\dagger}_k \p \phi_0 \ra
\]
Using Wick's theorem to simplify the above expression, we find that 
there are only two non-zero terms, and we get,
\ben
\la \rho_q \ra = \sum_k A_k A^{\star}_{k+q} \left\{ 1 - \frac{1}{V} 
\sum_p \frac{V(q) - V(k-p)}{q\cdot(v_p - v_k)}\left( f^0_p -f^0_{p+q}\right)
\right\}
\een
where $f^0$ is equilibrium Fermi distribution.
The most important observation is that the second term in the right
hand side above has a nonzero limit as $ q \rightarrow 0 $. Thus $
\rho_q $ is discontinuous at $ q=0 $. The total charge that is localized
is given by,
\ben
\lim_{q \rightarrow 0} \rho_q = \sum_k A_k A^{\star}_{k}
\left\{ 1 - \frac{1}{V} \sum_p \frac{V(0) - V(k-p)}{q\cdot(v_p - v_k)}
\delta(\epsilon_p)(v_p \cdot q) \right\}
\een
Since $\lim_{q \rightarrow 0} \rho_q \neq \delta \rho_{q=0} = 
\sum_k A_k A^{\star}_{k} $, the entire charge of the excitation is not
localized. The delocalized part is given by the second term on the
right hand side of eqn(15). Thus we find that it is not possible to 
construct entirely localized objects from single particle excitations. And
the reason for this is that in systems with Fermi surfaces there are
large number of excitations with arbirary small energies. These low
energy processes are responsible for the delocalization of the charge.

The average current density in the state $ \p x_0 \ra $ is given by,
\ben
\la x_0 \p \bar{j}_q \p x_0 \ra = \sum_k A_k A^{\star}_{k+q}
\sum_p \frac{1}{m} \left( \bar{p} - \frac{\bar{q}}{2} \right)
\la k+q \p a^{\dagger}_{p+q} a_p \p k \ra
\een
The $ q=0$ component gives the total current in the system. Since total
momentum is conserved, we find,
\ben
\la \bar{j}_0 \ra = \sum_k A_k A^{\star}_k \left( \frac{\bar{k}}{m} \right)
= \frac{\bar{q}_0}{m} \sum_k A_k A^{\star}_k
\een
while for $q \neq 0$ components we have
\ben
\la \bar{j}_q \ra = \sum_k A_k A^{\star}_{k+q} 
\sum_p \frac{1}{m} \left( \bar{p} - \frac{\bar{q}}{2} \right)
\left\{ \delta_{p,k} - \frac{1}{V} \frac{V(q) - V(k-p)}{q\cdot(v_p - v_k)}
\left( f^0_p -f^0_{p+q}\right) \right\}
\een
As before, $\lim_{q \rightarrow 0}\la \bar{j}_q \ra \neq \la \bar{j}_0 \ra $
, and we find that part of the current is delocalized.

\item
To find an expression for conductivity, we look at the Boltzman equation,
\[
\frac{\partial n(k,r,t)}{\partial t} +\frac{\partial n}{\partial r} 
 \frac{\partial \epsilon}{\partial k} - \frac{\partial n}{\partial k}
 \frac{\partial \epsilon}{\partial r} = 0
\]
Here we are disregarding the effect of collision that appears in the 
right hand side of the Boltzman equation. The collision frquency
$ \nu $ goes as $T^2$, and so at low enough temperature the above 
approximation is valid. Let $n(k,r,t) = n_0(k) + \delta n(k,r,t)$,
where $n_0$ is the equilibrium Fermi distribution function. Linearizing
the Boltzman equation we get,
\[
\frac{\partial \delta n}{\partial t} + \frac{\partial \delta n}{\partial r} 
\frac{\partial \epsilon^0_k}{\partial k} - \frac{\partial n_0}{\partial k}
\frac{\partial \epsilon}{\partial r} = 0
\]
Let there be an external electric field $ \bar{E} = \bar{E}_0 e^{iqr -
i\omega t} $ where $\omega \gg \nu $. Then,
\[
\epsilon_k = \epsilon^0_k + \sum_{k^{\pm}} f(k, k^{\pm}) \delta n(k^{\pm})
- eEr
\]
Let the shift from the equilibrium be $ \delta n(k,r,t) = \delta n(k)
e^{iqr - i\omega t} $. Then, from the Boltzman equation we get,
\[
\left( q \cdot v_k - \omega \right)  \delta n(k) + q \cdot v_k \delta 
(\epsilon_k
- \mu) \sum_{k^{\pm}} f(k, k^{\pm}) \delta n(k^{\pm}) + ie E_0 \cdot v_k
\delta (\epsilon_k - \mu) = 0
\]
For $q=0$, i.e. a spatially uniform external field we get,
\ben
\delta n(k) = \frac{ie E_0 \cdot v_k}{\omega} \delta (\epsilon_k - \mu)
\een
The current density is given by $\bar{j} = e \sum_k \delta n(k)
\bar{j}_k $, where for a translationally invariant system we have already 
found that $ \bar{j}_k = \frac{\bar{k}}{m} $. Then,
\bea
\bar{j} &=& \frac{ie^2}{\omega} \sum_k E_0 \cdot v_k  \frac{\bar{k}}{m}
\delta (\epsilon_k - \mu) \nonumber \\
&=& \frac{iNe^2}{m \omega} \bar{E}_0 
\eea
so that $ \sigma = \frac{iNe^2}{m \omega} $. Thus the constant C has
the value,
\ben
C = \frac{- Ne^2}{m}
\een
Due to Galilean invariance the quasiparticle current density is 
$\frac{\bar{k}}{m}$, and so the expression for conductivity has a
factor of $m$ and not $m^{\star}$. Thus there is no correction from $V(r)$.

\end{enumerate}
\end{document}
