\documentclass[12pt]{article}

\textwidth=17cm
\textheight=22.5cm

\topmargin -1.5cm
\oddsidemargin -0.8cm
%\evensidemargin -0.8cm

\parskip 0.3cm

\pagestyle{plain}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% Indranil's LaTeX macros
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 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}}

% Special letters
\def\ham{{\mathcal H}}
\def\la{\langle}
\def\ra{\rangle}
\def\cd{c^{\dagger}}
\def\fd{f^{\dagger}}
\def\dl{\delta}
\def\ep{\epsilon}
\def\om{\omega}
\def\be{\beta}
\def\Ga{\Gamma}
\def\sg{\sigma}
\def\spec{\rho_{f \fd}(\omega)}
\def\mp{m^{\prime}}
\def\vk{{\bf k}}
\def\vS{{\bf S}}
\def\vs{{\bf s}}
\def\prl{\partial}
\begin{document}

\begin{center}
{\bf Solution to problem set 3}
\end{center}

\begin{enumerate}

\item
{\bf $SU(N)$ Anderson Impurity model}

We consider impurity electrons on a single site hybridizing with a bath of 
conduction electrons. The electrons carry $SU(N)$ spin. We consider the 
limiting case where the on-site Coulomb repulsion ($U$) at the impurity site
is infinitely large. Thus, at most a single electron can occupy the impurity 
site (multiple occupancy is not allowed). In this limit, the allowed states 
of the impurity site are $ | 0 \ra$ (i.e., no electron), and $| m \ra,$ where 
$m= (1, \cdots ,N)$ (i.e., a single electron with $SU(N)$ spin $m$). The 
Hubbard operators are defined as $X_{pq} = |p \ra \la q |$, where $p,q =
(0,1, \cdots, N)$. In this restricted subspace creation operator for an 
impurity electron $\fd_n \equiv |n \ra \la 0 | = X_{n0}$. Similarly, $f_n
\equiv X_{0n}$. The completeness relation in this subspace (of the impurity
level) is given by $X_{00} + \sum_n X_{nn} = 1$. Note, that the Hubbard 
operators satisfy the relation $[X_{pq} , X_{rs} ]_{\pm} = \dl_{qr} X_{ps} 
\pm \dl_{ps} X_{rq}$. This is unlike standard commutation or anticommutation 
of bosons and fermions. The Hamiltonian for the system is given by 
\[
\ham = \ham_b + \ham_f + \ham_I,
\]
where
$\ham_b = \sum_{\vk, n} \ep_k \cd_{\vk n} c_{\vk n}$ describes the bath of 
conduction electrons, $\ham_f = \ep_f \sum_m \fd_m f_m = \ep_f \sum_m X_{mm}$
describes the impurity electrons, and $\ham_I = \sum_{\vk m} ( V_{\vk} 
\cd_{\vk m} X_{0 m} + V^{\ast}_{\vk} X_{m 0} c_{\vk m})$ is the hybridization 
between the impurity electrons and the conduction electrons.

We consider the retarded Green's function $\la \la f_n \fd_n \ra \ra (\om)
= \la \la X_{0n} X_{n 0} \ra \ra (\om)$. The corresponding spectral function
is given by 
\[
\spec = - \frac{{\rm Im}}{ \pi} G_{f \fd} ( \om + i \dl).
\]

The first moment is given by $\la \{ X_{0n} , X_{n0} \} \ra = \la X_{00} 
+ X_{nn} \ra$. We can write $\la X_{00} \ra = 1 - \la n_f \ra$, where $\la n_f
\ra = \sum_n \la X_{nn} \ra$ is the expectation of electron occupancy at the 
impurity site. Also, $\la X_{nn} \ra = \la n_f \ra /N$. Thus,
\ben
\la \{ X_{0n} , X_{n0} \} \ra = 1 - \left( \frac{N-1}{N} \right) \la n_f \ra.
\een
Since,
\[
\int_{- \infty}^{\infty} d \om \spec  = \la \{ X_{0n} , X_{n0} \} \ra,
\]
the first moment gives the total spectral weight of the f-band.

For the calculation of the second moment we get
\[
[ \ham, X_{0n} ] = - \ep_f X_{0n} - \sum_{\vk} V^{\ast}_{\vk} c_{\vk n} X_{00}
- \sum_{\vk m } V^{\ast}_{\vk} c_{\vk m} X_{mn}.
\]
Then,
\ben
\la \{ [ \ham, X_{0n} ] , X_{n0} \} \ra = - \ep_f \la \{ X_{0n} , X_{n0} \} 
\ra - \sum_{\vk, m \neq n} V^{\ast}_{\vk} \la c_{\vk m} X_{m0} \ra.
\een
Since,
\ben
\int_{- \infty}^{\infty} \om \spec d \om = - \la \{ [ \ham, X_{0n} ] , X_{n0}
\} \ra,
\een
the centre of the f-band is given by
\[
\frac{\int_{- \infty}^{\infty} \om \spec d \om}{\int_{- \infty}^{\infty} 
d \om \spec} = \ep_f + \frac{1}{\la \{ X_{0n} , X_{n0} \} \ra}
\sum_{\vk, m \neq n} V^{\ast}_{\vk} \la c_{\vk m} X_{m0} \ra.
\]
Presumably (since we cannot conclusively say anything about the line shape)
the spectral function is peaked at this energy.
The second term is the shift in the energy of the f-level due to 
hybridization.

For the third moment we get
\bea
\la \{ [ \ham, X_{0n} ], [ \ham, X_{n0} ] \} \ra
&=& \la \{ - \ep_f X_{0n} - \sum_{vk} V_{\vk}^{\ast} c_{\vk n} X_{00} - 
\sum_{\vk m} V_{\vk}^{\ast} c_{\vk m} X_{mn} \ , \ \ep_f X_{n0} 
\nonumber \\
&& + \sum_{\vk} V_{\vk} \cd_{\vk n} X_{00} + \sum_{\vk m} V_{\vk} \cd_{\vk m} 
X_{nm} \} \ra
\nonumber \\
&=& - \ep_f^2 \la \{ X_{0n} , X_{n0} \} \ra - \ep_f \sum_{\vk, 
m \neq n} V_{\vk}^{\ast} \la c_{\vk m} X_{m0} \ra + \ep_f \sum_{\vk, m \neq n}
V_{\vk} \la \cd_{\vk m} X_{0m} \ra
\nonumber \\
&& - \sum_{\vk} |V_{\vk} |^2 + \sum_{\vk_1 \vk_2, m_1 m_2} V_{\vk_1}^{\ast}
V_{\vk_2} \la \cd_{\vk_2 m_2} c_{\vk_1 m_1} X_{m_1 m_2} \ra
\nonumber \\
&& - \sum_{\vk_1 \vk_2, m} V_{\vk_1}^{\ast} V_{\vk_2} \la \cd_{\vk_2 m} 
c_{\vk_1 m} X_{nn} \ra.
\eea
We can show that 
\bea
\la \{ [ \ham, X_{0n} ] , X_{n0} \} \ra 
&=& \frac{1}{Z} \sum_{nm} \left( e^{- \be E_n} + e^{- \be E_m} \right) 
\left( E_n - E_m \right) \la n | X_{0n} | m \ra \la m | X_{n0} | n \ra
\nonumber \\
&=& - \la \{ X_{0n} , [ \ham , X_{n0} ] \} \ra.
\eea
This is due to time translation invariance of the correlation function in 
equilibrium. Using this result we get,
\ben
- \sum_{\vk, m \neq n} V_{\vk}^{\ast} \la c_{\vk m} X_{m0} \ra
= \sum_{\vk, m \neq n} V_{\vk} \la \cd_{\vk m} X_{0m} \ra.
\een
The $SU(N)$ spin at the impurity site can be written as $\vS = \sum_{m_1 m_2}
\Ga^M_{m_1 m_2} X_{m_1 m_2}$, where $\Ga^M$ are the generators of $SU(N)$
group, with $M = (1, \cdots ,(N^2-1))$. The spin of the conduction electrons
in the momentum basis is similarly given by $\vs_{\vk_2 \vk_1} = \sum_{m_1 
m_2} \cd_{\vk_2 m_1} \Ga^M_{m_1 m_2} c_{\vk_1 m_2} $. The generators of the 
group satisfy the completeness relation $\sum_M \Ga^M_{m_1 m_2} 
\Ga^M_{m_3 m_4} = \dl_{m_1
m_4} \dl_{m_2 m_3} - \frac{1}{N} \dl_{m_1 m_2} \dl_{m_3 m_4}$. Using this 
we can show that
\ben
\sum_{\vk_1 \vk_2} V_{\vk_1}^{\ast} V_{\vk_2} \la \vs_{\vk_2 \vk_1} \cdot 
\vS \ra = \sum_{\vk_1 \vk_2, m_1 m_2} V_{\vk_1}^{\ast}
V_{\vk_2} \la \cd_{\vk_2 m_1} c_{\vk_1 m_2} X_{m_2 m_1} \ra
- \frac{1}{N} \sum_{\vk_1 \vk_2, m \mp} V_{\vk_1}^{\ast} V_{\vk_2} \la 
\cd_{\vk_2 m} c_{\vk_1 m} X_{\mp \mp} \ra.
\een
Thus, we can write
\bea
\la \{ [ \ham, X_{0n} ] , [\ham, X_{n0} ] \} \ra &=& 
- \ep_f^2 \la \{ X_{0n} , X_{n0} \} \ra - 2 \ep_f \sum_{\vk, m \neq n} 
V_{\vk}^{\ast} \la c_{\vk m} X_{m0} \ra - \sum_{\vk} |V_{\vk} |^2
\nonumber \\
&& + \sum_{\vk_1 \vk_2} V_{\vk_1}^{\ast} V_{\vk_2} \la \vs_{\vk_2 \vk_1} 
\cdot \vS \ra.
\eea
We identify the last term in the above equation as a spin-spin correlation
function. The sum rule for the third moment is given by 
\ben
\int_{- \infty}^{\infty} \om^2 \spec d \om = \la \{ [ \ham, [\ham, X_{0n} ] ]
, X_{n0} \} \ra = - \la \{ [ \ham, X_{0n} ] , [ \ham, X_{n0} ] \} \ra.
\een
The width of the impurity band (standard deviation of a distribution, in
general) is given by
\bea
\lefteqn{
\frac{\int_{- \infty}^{\infty} \om^2 \spec d \om}{{\int_{- \infty}^{\infty} 
d \om \spec}} - \left( \frac{\int_{- \infty}^{\infty} \om \spec d \om}
{\int_{- \infty}^{\infty} d \om \spec} \right)^2 =}
\nonumber \\
&& - \frac{1}{\la \{ X_{0n} , X_{n0} \} \ra}\sum_{\vk_1 \vk_2} 
V_{\vk_1}^{\ast} V_{\vk_2} \la \vs_{\vk_2 \vk_1} \cdot \vS \ra 
- \left( \frac{1}{\la \{ X_{0n} 
, X_{n0} \} \ra} \sum_{\vk, m \neq n} V^{\ast}_{\vk} \la c_{\vk m} X_{m0} 
\ra \right)^2
\eea
Originally the f-level was a delta function peak at $\ep_f$. Due to 
hybridization there is a shift in the energy level, and also a broadening of 
the band. 

\item
{\bf f-sum rule}

The retarded current-current correlation function is defined by
\ben
\Pi_{jj}^R (\om) = \la \la j,j \ra \ra (\om) = \frac{1}{Z} \sum_{nm}
\left( e^{- \be E_n} - e^{- \be E_m} \right) \left| \la n | j | m \ra \right|
^2 \frac{1}{\om + E_n - E_m + i \eta }.
\een
The corresponding spectral function is
\bea 
\rho_{jj} (\om) &=& - \frac{1}{\pi} {\rm Im} \Pi_{jj}^R (\om)
\nonumber \\
&=& \frac{1}{Z} \sum_{nm} \left( e^{- \be E_n} - e^{- \be E_m} \right) \left| 
\la n | j | m \ra \right|^2 \dl(\om + E_n - E_m)
\nonumber \\
&=& \int_{- \infty}^{\infty} \frac{dt}{2 \pi} e^{i \om t} \la [ j(t) , j] \ra.
\eea
In terms of the spectral function, the real part of the conductivity (which
is what enters the f-sum rule) is given by,
\ben
\sg (\om) = - \frac{1}{V} \frac{1}{\om} {\rm Im} \Pi_{jj}^R (\om) 
= \frac{\pi}{V} \rho_{jj} (\om).
\een
We define the polarization operator $P$ such that $\frac{\prl P}{\prl t} = j$.
Then,
\bea
\rho_{jj} (\om) &=& \int_{- \infty}^{\infty} \frac{dt}{2 \pi} e^{i \om t}
\la [ \frac{\prl P}{\prl t}, j ] \ra 
\nonumber \\
&=& - i \om \int_{- \infty}^{\infty} \frac{dt}{2 \pi} e^{i \om t} \la [ P(t),
j ] \ra
\nonumber \\
&=& -i \om \rho_{Pj} (\om),
\eea
where
\ben
\rho_{Pj} (\om) = \int_{- \infty}^{\infty} \frac{dt}{2 \pi} e^{i \om t} \la 
[ P(t), j ] \ra.
\een
Then,
\bea
\int_{- \infty}^{\infty} d \om \sg (\om) &=& - \frac{\pi i}{V} \int_{- \infty}
^{\infty} d \om \rho_{Pj} (\om)
\nonumber \\
&=& - \frac{\pi i}{V} \la [P(0), j ] \ra.
\eea
The current operator is given by $j = \sum_{i = 1}^N (e p_i)/m$, where $p_i$ 
is the momentum of the i-th particle. The polarization operator is given by
$P = \sum_{i = 1}^N  e r_i$, where $r_i$ is the position of the i-th particle.
Since,
\[
[P, j] = \frac{e^2}{m} \sum_{ij} [ r_i , p_j] = \frac{N e^2 i}{m},
\]
(in units where $\hbar = 1$), we finally get the f-sum rule as
\ben
\int_{- \infty}^{\infty} \frac{d \om} {\pi} \sg (\om) = \frac{n e^2}{m},
\een
where $n = N/V$ is the density of particles.
\end{enumerate}
\end{document} 
