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% 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\fds{f^{\dagger}_{\sigma}}
\def\fs{f_{\sigma}}
\def\fd{f^{\dagger}}
\def\la{{\langle\, }}
\def\ra{{\,\rangle }}
\def\ua{\uparrow}
\def\da{\downarrow}
\def\ptl{\partial}
\def\pm{\prime}
\def\d{\delta}
\def\t{\tau}
\def\m{\mu}
\def\s{\sigma}
\def\bs{\bar{\sigma}}
\def\b{\beta}
\def\om{\omega}
\def\ie{i\eta}
\def\th{\theta}

\begin{document}
\begin{center}
Solution to Problem Set 2
\end{center}

The Hubbard model in the atomic limit is given by,
\ben
\ham = -\m \sum_{\s} \fds \fs + U \fd_{\ua} f_{\ua} \fd_{\da} f_{\da}.
\een
The system has four possible states, namely $|0 \ra$ (no particle), $| \ua 
\ra$ (up electron), $| \da \ra$ (down electron), and $ | \ua \da \ra$ (doubly 
occopied). The grand canonical partition function for the system is given by,
\bea
Z &=& {\rm Tr} e^{- \b \ham} \nonumber \\
&=& 1 + 2 e^{\b \m} + e^{\b(2 \m - U)}.
\eea
\benu
\item
To calculate the retarded Green's function and the spectral function we will 
first calculate the temperature Green's function and then take appropriate 
analytic continuation. 

The temperature Green's function is defined as,
\bea
g_{\s}(\t) &=& - \la T_{\t} \fs (\t) \fds (0) \ra \nonumber \\
&=& - \frac{1}{Z} {\rm Tr} \left( e^{-\b \ham} T_{\t} \fs (\t) \fds (0) 
\right). \nonumber
\eea
The Fourier transform in terms of fermionic Matsubara frequency $\om_n = 
(2n+1) \pi/\b $, is 
\bea
g_{\s} (i \om_n) &=& \int_{0}^{\b} d \t g_{\s}(\t) e^{i \om_n \t} \nonumber \\
&=& - \frac{1}{Z} \int_{0}^{\b} d \t e^{i \om_n \t} \sum_n \la n \left| 
e^{-\b \ham} \fs (\t) \fds (0) \right| n \ra.
\eea
Since,
\bea
\sum_n \la n \left|e^{-\b \ham} \fs (\t) \fds (0) \right| n \ra
&=& \sum_n \la n \left|e^{-\b \ham} e^{\t \ham} \fs e^{- \t \ham} \fds 
\right| n \ra \nonumber \\
&=& \la 0 \left|e^{-\b \ham} e^{\t \ham} \fs e^{- \t \ham} \fds \right| 0 \ra
+ \la \bar{\s} \left|e^{-\b \ham} e^{\t \ham} \fs e^{- \t \ham} \fds \right| 
\bar{\s} \ra \nonumber \\
&=& e^{\t \m} + e^{\b \m + \t (\m - U)},
\eea
we get from equations (3) and (4),
\ben
g_{\s} (i \om_n) = \frac{1}{Z} \left\{ \frac{e^{\b \m} +1}{ i \om_n + \m}
+ \frac{e^{\b(2 \m - U)} + e^{\b \m}}{ i \om_n + \m - U} \right\}.
\een
Then, after the analytic continuation $i \om_n \rightarrow \om + \ie$, 
we have,
\ben
G^R_{\s} (\om) = \frac{1}{Z} \left\{ \frac{e^{\b \m} +1}{ \om + \m + \ie}
+ \frac{e^{\b(2 \m - U)} + e^{\b \m}}{ \om + \m - U + \ie} \right\}.
\een

The spectral function is given by,
\bea
A_{\s}(\om) &=& - \frac{1}{\pi} {\rm Im} G^R_{\s} (\om) \nonumber \\
&=& \left(\frac{e^{\b \m} +1}{Z} \right) \d ( \om + \m) + \left(\frac{
e^{\b(2 \m - U)} + e^{\b \m}}{Z} \right) \d(\om + \m -U).
\eea
In the non-interacting case the spectral function has a single pole with 
weight one, i.e, $A_{\s}(\om) = \d ( \om + \m)$. The effect of interaction 
is to produce two poles with finite energy gap $U$. The total spectral weight,
which is always one, is shared between the two poles.
\item
The Hubbard operators are defined as 
\bea
X_{\ua 0} &\equiv& \fd_{\ua} (1-n_{\da}) = | \ua \ra \la 0 | \nonumber \\
X_{d \da} &\equiv& \fd_{\ua} n_{\da} = | \ua \da \ra \la \da |.
\eea
Their conjugate operators are repectively,
\bea
X_{0 \ua} &\equiv& f_{\ua} (1-n_{\da}) = | 0 \ra \la \ua | \nonumber \\
X_{\da d} &\equiv& f_{\ua} n_{\da} = | \da \ra \la \ua \da |.
\eea

Now, it is easy to see that
\ben
\fd_{\ua} = X_{\ua 0} +X_{d \da} =| \ua \ra \la 0 |+| \ua \da \ra \la \da |.
\een

\item
Similarly, we have,
\ben
\fd_{\da}=X_{\da 0} +X_{d \ua}=| \da \ra \la 0 |+| \ua \da \ra \la \ua |.
\een
\item
The anticommutator
\bea
\left\{ X_{\ua 0}, X_{0 \ua} \right\} &=& \left\{ \fd_{\ua}(1- n_{\da}), 
(1- n_{\da})f_{\ua} \right\} \nonumber \\
&=& (1- n_{\da})^2 \left\{ \fd_{\ua},f_{\ua} \right\} \nonumber \\
&=& (1- n_{\da}),
\eea
since $(1- n_{\da})$ is either zero or one. 
Thus, 
\ben
\la\{ X_{\ua 0}, X_{0 \ua}\}\ra = 1 - \la n_{\da} \ra
=\frac{e^{\b \m} +1}{Z},
\een
is the average occupation of a $\da$-hole. We also note, from equation (7), 
that this is also the spectral weight for the pole corresponding to single
occupancy.

Similarly, the anticommutator $\left\{ X_{d \da}, X_{\da d} \right\} 
= n_{\da}$. Thus, 
\ben
\la \{X_{d \da}, X_{\da d} \} \ra = \la n_{\da} \ra
=\frac{e^{\b(2 \m - U)} + e^{\b \m}}{Z},
\een
is the average occupation of a $\da$-electron. Again from equation (7) we 
note that this is the spectral weight for the pole corresponding to double 
occupancy.

\item
To evaluate the commutator $[\ham, X_{0\ua}]$, it is enough to consider the
state $| \ua \ra$, since all other states give zero when acted on by 
$X_{0\ua}$. We have $[\ham, X_{0\ua}] | \ua \ra = \m | 0 \ra = \m X_{0\ua}
| \ua \ra$. Thus, $[\ham, X_{0\ua}] = \m X_{0\ua}$, and 
\ben
\la \left\{ X_{\ua 0}, \left[ \ham, X_{0\ua} \right] \right\} \ra 
= \m \la\left\{ X_{\ua 0}, X_{0 \ua}\right\}\ra
= \m \left(1 - \la n_{\da} \ra \right).
\een
Comparing with equation (7) we find that the energy coefficient in the above
expression gives the pole for the singly occupied state. 

Similarly, we can show that $[\ham, X_{\da d}] = (\m - U) X_{\da d}$, and
\ben
\la \left\{ X_{d \da}, \left[ \ham, X_{\da d} \right] \right\} \ra
= \left( \m - U \right) \la n_{\da} \ra.
\een
Here the energy coefficient gives the pole for the doubly occupied state.
 
\item{Comment}

To put parts (1), (4) and (5) into proper perspective we will re-evaluate the 
retarded Green's function using the Hubbard operators.  From definition,
\bea
G^R_{\s} (t) &=& -i \th(t) \la \left\{ \fs (t), \fds (0) \right\} \ra
\nonumber \\
&=& -i \th(t) \la \left\{ X_{0 \s}(t) + X_{\bs d}(t), X_{\s 0}(0) + X_{d \bs}
(0) \right\} \ra \nonumber \\
&=& -i \th(t) \left[ \la \left\{X_{0 \s}(t), X_{\s 0}(0) \right\} \ra +
\la \left\{ X_{\bs d}(t), X_{d \bs}(0) \right\} \ra \right]. \nonumber
\eea
Here $\bs$ is the spin opposite to $\s$. In the last line above, we have 
used the fact that cross-terms are zero. Now, from our discussion in part (5),
we know that $[\ham, X_{0\s}] = \m X_{0\s}$ and that $[\ham, X_{\bs d}] = 
(\m - U) X_{\bs d}$. Then, $X_{0 \s}(t) = e^{i \m t}X_{0 \s}(0)$, and
$X_{\bs d}(t)= e^{i(\m - U)t}X_{\bs d}(0)$. We can write,
\[
G^R_{\s} (t)= -i \th(t) \left[ e^{i \m t} \la \left\{X_{0 \s}, X_{\s 0}
\right\} \ra + e^{i(\m - U)t} \la \left\{ X_{\bs d}, X_{d \bs}\right\} \ra 
\right].
\]
From our discussion in part (4) we know that $\la\{ X_{\s 0}, X_{0 \s}\}\ra 
= 1 - \la n_{\bs} \ra$ and that $\la \{ X_{\bs d}, X_{d \bs} \} \ra
=\la n_{\bs} \ra$. So, finally we have,
\ben
G^R_{\s} (t) = -i \th(t) \left\{ (1 - \la n_{\bs} \ra)e^{i \m t} +
\la n_{\bs} \ra e^{i(\m - U)t} \right\}.
\een
Taking the Fourier transform of the above equation we get,
\ben
G^R_{\s} (\om) = \frac{1 - \la n_{\bs} \ra}{\om + \m + \ie} + \frac{\la 
n_{\bs} \ra}{\om + \m -U + \ie}.
\een
This is nothing but equation (6) in a more compact form.


\enu
\end{document}
