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\title {Many Body Theory Problem Set 1}
\date{\today}

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\noindent

In the following problems, I use a unit system so that $\hbar =1 $ and $e =1$.

{\bf Problem One}\newline
a. The equation of motion is
\eq
\frac{d a (0)} {dt} =
-i [a, H]
\f
Substitute
\eq
a = \frac{1}{\sqrt{2}}
\left[\frac{1}{\sqrt{m\omega}} \p - i \sqrt{m\omega} \x \right]
\f
\eq
a^\dagger = \frac{1}{\sqrt{2}}
\left[\frac{1}{\sqrt{m\omega}} \p + i \sqrt{m\omega} \x \right]
\f
into
\eq
H = \omega ( a^\dagger a +\half )
\f
to get
\eq
H = \frac{1}{2m}\p^2 +\half m \omega^2 \x^2
\f
using $[\x_j,\p_k]=i\delta_{jk}$.  This shows that that $a$ and $a^\dagger$ can be considered as an operator factorization of H.  Note that these so called creation and annihilation operators satisfy the following commutation relationship,
\eq
[a,a^\dagger] =1
\f
Write out the commutator explicitly and commute the $a$ and $a^\dagger$ in the first term.
\eq
\frac{d a (0)} {dt} =
-i \omega (a a^\dagger a + \half a - a^\dagger a a -\half a) \\
= -i\omega (a^\dagger a a - a^\dagger a a + a)
= -i \omega a
\f
Upon integration
\eq
a(t) = a(0) e^{-i \omega t}
\f

b.
Given
\eq
a^\dagger(t) = a^\dagger(0) e^{i\omega t}
\f
And from part a
\eq
a^\dagger (0) = \frac{1}{\sqrt{2}}
\left[\frac{1}{\sqrt{m\omega}} \p(0) + i \sqrt{m\omega} \x(0) \right]
\f
so
\eq
a(t) = \frac{1}{\sqrt{2}}
\left[\frac{1}{\sqrt{m\omega}} \p(0) - i \sqrt{m\omega} \x(0) \right] e^{-i\omega t} \\
=
\left[\frac{1}{\sqrt{2m\omega}} \p(t) - i \sqrt{\half m\omega} \x(t) \right]
\f
\eq
a^\dagger(t) = \frac{1}{\sqrt{2}}
\left[\frac{1}{\sqrt{m\omega}} \p(0) + i \sqrt{m\omega} \x(0) \right] e^{i\omega t} \\
=
\left[\frac{1}{\sqrt{2 m\omega}} \p(t) + i \sqrt{\half m\omega} \x(t) \right]
\f
Add these
\eq
a(t)+a^\dagger(t)=
\sqrt{\frac{2}{m\omega}} \p(t)
=
\frac{1}{\sqrt{2 m \omega}} \p(0) \left[e^{-i\omega t}+e^{i \omega t}\right]
+ i\sqrt{\frac{m\omega}{2}} \x(0) \left[e^{i\omega t}-e^{-i\omega t}\right]
\f
Solve for $\p$
\eq
\p(t) =
\p(0) \cos \omega t
- m \omega \x(0) \sin \omega t
\f
To get $\x$, subtract
\eq
a^\dagger(t)-a(t)=
\sqrt{2 m\omega} \x(t)
=
\frac{1}{\sqrt{2 m \omega}} \p(0) \left[e^{i\omega t}-e^{-i \omega t}\right]
+ i\sqrt{\frac{m\omega}{2}} \x(0) \left[e^{i\omega t}+e^{-i\omega t}\right]
\f
Solve for $\x$
\eq
\x(t) =
\x(0) \cos \omega t  +
\frac{1}{m \omega} \p(0) \sin \omega t
\f

c. We know that $\la q \ra $ and  $\la p \ra $ obey the Erenfest
theorem and we can write:

\begin{eqnarray*}
\frac{ \la q \ra }{dt} &=&\frac{\la p \ra}{m} \\ \nonumber \frac{
\la p \ra }{dt}&=&-\la V' \ra . \\ \nonumber
\end{eqnarray*}
And they are equivalent to classical equations when $\la V' \ra $
= $\la V'_{cl} \ra $. If dispersion $(\Delta q)^2=\la q^2 \ra
-\la q \ra^2 $ of wave packet is small then we can write:
$$
V'(q)=V'_{cl}(\la q \ra)+(q-\la q \ra )V''_{cl}+(q-\la q \ra )^2
V'''_{cl}+ ...
$$
We see that $V'=V'_{cl}$ if all term higher that $V''$ are equal
to zero. Harmonic oscillator potential has quadratic form and
hence quantum and classical equation of motion look similar.

%These equations of motion resemble the classical ones because the
%prescription used to solve the classical problem is a direct
%analogue.  Recall the classical poison bracket. \eq \{A,B\}_{PB} =
%\left[\frac{dA}{dx}\frac{dB}{dp}-\frac{dB}{dx}\frac{dA}{dp}\right]
%\f Sakurai notes the direct correspondence between the poison
%formalism and quantum mechanics; \eq \{A,B\}_{PB} \rightarrow -i
%[A,B] \f In the classical analogue,  we solve $\{p,H\}$ and
%$\{x,H\}$ and get the classical equations of motion. In the
%quantum regime, we solve $[\p,H]$ and $[\x,H]$ and get the
%quantum equations of motion.

%We chose
%\eq
%Q = p + i a x
%\f
%\eq
%P = p - i a q
%\f
%and solved the classical equations.  Mathematically, this procedure is analogous to what we did using operators.

% of motion to find
%\eq
%Q(t) = Q(0) e^{-i H t}
%\f
%\eq
%P(t) = P(0) e^{-i H t}
%\f

{\bf Problem Two}

\eq
\sx =\half \left( \Cup^\dagger\Cdn + \Cdn^\dagger\Cup\right)
\f
\eq
\sy =-\half i \left( \Cup^\dagger\Cdn - \Cdn^\dagger\Cup\right)
\f
\eq
\sz =\half \left( \Cup^\dagger\Cup - \Cdn^\dagger\Cdn\right)
\f
Check commutators
\eq
[\sx,\sy] =
-\frac{i}{4}
\left[
\Cup^\dagger\Cdn\Cup^\dagger\Cdn + \Cdn^\dagger\Cup\Cup^\dagger\Cdn
- \Cdn^\dagger\Cup\Cdn^\dagger\Cup -\Cup^\dagger\Cdn\Cdn^\dagger\Cup \right] \\
+\frac{i}{4}
\left[
\Cup^\dagger\Cdn\Cup^\dagger\Cdn + \Cup^\dagger\Cdn\Cdn^\dagger\Cup
- \Cdn^\dagger\Cup\Cdn^\dagger\Cup -\Cdn^\dagger\Cup\Cup^\dagger\Cdn \right]
\f
Use
\eq
\{c_\sigma,c^\dagger_{\sigma'}\} =
\delta_{\sigma\sigma'}
\f
Normal order.
Because $c_\sigma c_\sigma  = 0$ and $c^\dagger_\sigma c^\dagger_\sigma = 0$, several terms vanish.
\eq
[\sx,\sy] =
-\frac{i}{4}
\left[
-\Cdn^\dagger\Cup^\dagger\Cup\Cdn + \Cdn^\dagger\Cdn
+ \Cup^\dagger\Cdn^\dagger\Cdn\Cup -\Cup^\dagger\Cup
+\Cup^\dagger\Cdn^\dagger\Cdn\Cup + \Cdn^\dagger\Cdn
- \Cdn^\dagger\Cup^\dagger\Cup\Cdn -\Cup^\dagger\Cup
\right] \\
=
\frac{i}{2} \left[ \Cup^\dagger\Cup-\Cdn^\dagger\Cdn \right] = i \sz
\f

\eq
[\sz,\sy] =
-\frac{i}{4}
\left[
\Cup^\dagger\Cup\Cup^\dagger\Cdn
+ \Cdn^\dagger\Cdn\Cdn^\dagger\Cup
- \Cdn^\dagger\Cdn\Cup^\dagger\Cdn
-\Cup^\dagger\Cup\Cdn^\dagger\Cup \right] \\
+\frac{i}{4}
\left[
\Cup^\dagger\Cdn\Cup^\dagger\Cup + \Cdn^\dagger\Cup\Cdn^\dagger\Cdn
- \Cup^\dagger\Cdn\Cdn^\dagger\Cdn -\Cdn^\dagger\Cup\Cup^\dagger\Cup \right] \\
=
-\frac{i}{4}
\left[
-\Cup^\dagger\Cup^\dagger\Cup\Cdn +\Cup^\dagger\Cdn
- \Cdn^\dagger\Cdn^\dagger\Cdn\Cup +\Cdn^\dagger\Cup
+ \Cdn^\dagger\Cup^\dagger\Cdn\Cdn
+\Cup^\dagger\Cdn^\dagger\Cup\Cup\right] \\
-\frac{i}{4}
\left[
+\Cup^\dagger\Cup^\dagger\Cdn\Cup
+ \Cdn^\dagger\Cdn^\dagger\Cup\Cdn
- \Cup^\dagger\Cdn^\dagger\Cdn\Cdn +\Cup^\dagger\Cdn
-\Cdn^\dagger\Cup^\dagger\Cup\Cup +\Cdn^\dagger\Cup \right] \\
= -\half i \left( \Cup^\dagger\Cdn + \Cdn^\dagger\Cup\right)
= -i \sx
\f

\eq
[\sz,\sx] =
\frac{1}{4}
\left[
\Cup^\dagger\Cup\Cup^\dagger\Cdn
- \Cdn^\dagger\Cdn\Cup^\dagger\Cdn
- \Cdn^\dagger\Cdn\Cdn^\dagger\Cup
+\Cup^\dagger\Cup\Cdn^\dagger\Cup \right] \\
-\frac{1}{4}
\left[
\Cup^\dagger\Cdn\Cup^\dagger\Cup - \Cup^\dagger\Cdn\Cdn^\dagger\Cdn
- \Cdn^\dagger\Cup\Cdn^\dagger\Cdn +\Cdn^\dagger\Cup\Cup^\dagger\Cup \right] \\
=\frac{1}{4}
\left[
-\Cup^\dagger\Cup^\dagger\Cup\Cdn  +\Cup^\dagger\Cdn
+ \Cdn^\dagger\Cup^\dagger\Cdn\Cdn
+ \Cdn^\dagger\Cdn^\dagger\Cdn\Cup -\Cdn^\dagger\Cup
-\Cup^\dagger\Cdn^\dagger\Cup\Cup \right]\\
+\frac{1}{4}
\left[
+\Cup^\dagger\Cup^\dagger\Cdn\Cup
- \Cup^\dagger\Cdn^\dagger\Cdn\Cdn +\Cup^\dagger\Cdn
- \Cdn^\dagger\Cdn^\dagger\Cdn\Cdn
+\Cdn^\dagger\Cup^\dagger\Cup\Cup -\Cdn^\dagger\Cup\right] \\
= \half \left( \Cup^\dagger\Cdn - \Cdn^\dagger\Cup\right)
= i(-\frac{i}{2})\left( \Cup^\dagger\Cdn - \Cdn^\dagger\Cup\right)
= i \sy
\f

{\bf Problem Three}

We start with
\eq
H
=
\sum_{i,j} a^\dagger_i a_j \langle i| \frac{\p^2}{2 m} |j \rangle
+\half \sum_{i,j,k,n} a_i^\dagger a^\dagger_j a_k a_n
\langle i,j|
\frac{A}{4\pi} \oerR e^{-\lambda \R}
| k,n\rangle
\f
Insert the completeness relationship once in the first term, four times in the next term.  Rewrite the result equation in terms of wave-functions.
\eq
H =
-\sum_{i,j}\int d^3\r\; \frac{1}{2m} a_i^\dagger a_j \phi^\ast_i(\r) \nabla^2 \phi_j(\r)\\
+\frac{A}{8\pi}\sum_{i,j,k,n} a_i^\dagger a^\dagger_j a_k a_n \int d^3\r\;d^3\r'\; \oerR e^{-\lambda \R}
\phi_i^\ast(\r)\phi^\ast_j(\r') \phi_k(\r')\phi_n(\r)
\f
%Why are $\r$ and $\r'$ where they are?  The reason is that when I inserted the four completions relationships, I got a contraction, $\langle \r_a | \r'_b \rangle = \delta (\r_a-%\r'_b)$.  These match the inner terms first and then the outer ones.
Use the definitions of
\eq
\psi(\r) = \sum_j \phi_j(\r) a_j
\f
\eq
\psi^\dagger(\r) = \sum_i \phi^\ast_i(\r) a^\dagger_i
\f
to do the sums over the $i$,$j$, etc.  We get
\eq
H =
-\idr \frac{1}{2m} \psi^\dagger(\r) \nabla^2 \psi(\r)
+\frac{A}{8\pi}\idrr \psi^\dagger(\r) \psi^\dagger(\r') \oerR e^{-\lambda \R}
\psi(\r')\psi(\r)
\f
which is what we wanted.

b. Fourier transform everything.  The following will be useful:
\eq
\psi(\r) = \frac{1}{ (2\pi)^3 } \int d^3 \k \; c_{\k} \; e^{i \k\cdot\r }
\f
\eq
V(\r-\r')=\frac{1}{(2\pi)^3} \int d^3\q\; \frac{A}{q^2+\lambda^2} e^{i\q\cdot(\r-\r')}
\f
\eq
\delta( \k_2-\k_1) = \frac{1}{(2\pi)^3} \int d^3\r \; e^{i \r\cdot (\k_2-\k_1)}
\f
Plug these into the result from part a.
\eq
H =
\frac{1}{(2\pi)^6} \int d^3\r\; d^3 \k_1\; d^3 \k_2\;
\frac{1}{2m} k_2^2  c^\dagger_{\k_1} c_{\k_2} e^{i(\k_2-\k_1)\cdot\r}\\
+\frac{A}{16\pi^3} \frac{1}{(2\pi)^6}
\int d^3\r\;d^3\r'\;d^3 \k_1\; d^3 \k_2\; d^3 \k \; d^3 \k'\; d^3 \q\; \\
\frac{1}{q^2+\lambda^2}
c^\dagger_{\k_1}
c^\dagger_{\k_2}
c_{\k'}
c_{\k}
e^{i(\k_1+\q-\k)\cdot{\r}}
e^{i(\k_2-\q-\k')\cdot{\r'}}
\f
Pick out delta functions and do integrations over them.
\eq
H =
\frac{1}{(2\pi)^3} \int d^3 k\;
\frac{1}{2m} k^2  c^\dagger_{\k} c_{\k} \\
+\frac{A}{16\pi^3} \int d^3 k \; d^3 k'\; d^3 q\;
\frac{1}{q^2+\lambda^2}
c^\dagger_{\k-\q}
c^\dagger_{\k'+\q}
c_{\k'}
c_{\k}
\f

{\bf Problem Four}

Show
\eq
e^{ -a (n_\uparrow -\half) (n_\downarrow -\half) }
=
\half e^{-\frac{a}{4}} \sum_{\sigma=\pm}
e^{a\sigma (n_\uparrow -n_\downarrow)}
\f
To prove, I'll manipulate each side into suggestive forms and then show that both sides are the same.  First, consider the left hand side of the equality.
\eq
\mbox{l.h.s.} =
e^{ -a (n_\uparrow -\half) (n_\downarrow -\half) }
= e^{ -\frac{a}{4}  +\half a \left( n_\uparrow + n_\downarrow \right)
-a n_\uparrow n_\downarrow }
\f
\eq
\mbox{r.h.s.} =
\half e^{-\frac{a}{4}} \sum_{\sigma=\pm}
e^{a\sigma (n_\uparrow -n_\downarrow)}\\
=
e^{-\frac{a}{4}}
\cosh \left[ a |n_\uparrow -n_\downarrow| \right]
\f
I put the absolute value sign to remind us that $\cosh(x) =\cosh(-x)$.\newline
For fermions,  $n_\sigma =0$ or $1$.\newline
If $n_\uparrow=1$ and $n_\downarrow=1$, then $|n_\uparrow -n_\downarrow|=0$\newline
If $n_\uparrow=0$ and $n_\downarrow=0$, then $|n_\uparrow -n_\downarrow|=0$\newline
If $n_\uparrow=1$ and $n_\downarrow=0$, then $|n_\uparrow -n_\downarrow|=1$\newline
If $n_\uparrow=0$ and $n_\downarrow=1$, then $|n_\uparrow -n_\downarrow|=1$\newline
In the first two cases, the hyperbolic cosine term, $\cosh (a |n_\uparrow -n_\downarrow|)$, gives one.  In the latter case, we can use the relation $\cosh \pm a = e^{\half a}$ and rewrite the $r.h.s.$
Equate the $r.h.s.$ to the $l.h.s$ and factor out common multiples of $e^{-\frac{a}{4}}$.  Take the log of both sides.  We'll be left with
\eq
-a n_\uparrow n_\downarrow
+\half a (n_\uparrow +n_\downarrow)
= \half a |n_\uparrow-n_\downarrow|
\f
Finally, I show that for all pairing of $n_\sigma=0,1$ this equality will be satisfied.
If $n_\uparrow=1$ and $n_\downarrow=1$, then $-a+a=0$\newline
If $n_\uparrow=0$ and $n_\downarrow=0$, then $0+0=0$\newline
If $n_\uparrow=1$ and $n_\downarrow=0$, then $0+\half a =\half a$\newline
If $n_\uparrow=0$ and $n_\downarrow=1$, then $0+\half a =\half a$\newline
so our relation is true for fermions.  All of this could have been rewritten in matrix notation.

{\bf Problem Five}

\eq
H_0 = \sum_\alpha \epsilon_\alpha a^\dagger_\alpha a_\alpha
\f
a.  I use $\gamma$ instead of what's on the homework assignment so that I can keep the labels straight.  $\beta$ has a negative sign so that the partition function is bounded.
\eq
\tr \left[e^{-\beta H_0} a_\gamma^\dagger a_\gamma \right] =
\sum_{\{n_i\}} \langle \{n_i\}|\left[e^{-\beta H_0}
 a_\gamma^\dagger a_\gamma \right] |\{n_i\}\rangle
\f
where $|\{n_i\}\rangle$ stands for a many particle state with
a set, $\{n_i\}$, of occupation numbers.
$a^\dagger_\gamma a_\gamma $ acting on $|\{ n_i\} \rangle$ returns 0 or 1
depending on whether there is a fermion in state $\gamma$ -
Alas, the notation is a bit degenerate -
We can likewise get the eigenvalues for the operators in the H-potential.  I will use
$e^{\sum\mbox{stuff}} = \prod e^{\mbox{stuff}}$, and write
\eq
\tr \left[e^{-\beta H_0} a_\gamma^\dagger a_\gamma \right] =
\sum_{\{n_i\}} \prod_{\alpha} \langle \{n_i\}|\left[e^{-\beta \epsilon_\alpha a^\dagger_\alpha a_\alpha}
 a_\gamma^\dagger a_\gamma \right] |\{ n_i\}\rangle
\f
Now, I let the density operators act on the various $\gamma$ states.
\eq
\tr \left[e^{-\beta H_0} a_\gamma^\dagger a_\gamma \right] =
\sum_{\{n_i\}} n_\gamma
\prod_\alpha \left[e^{-\beta \epsilon_\alpha n_\alpha} \right]
\f
Take the term which was generated by the state $\gamma$ out of the product.
\eq
\tr \left[e^{-\beta H_0} a_\gamma^\dagger a_\gamma \right] =
\sum_{\{n_i\}} n_\gamma e^{-\beta \epsilon_\gamma n_\gamma}
\prod_{\alpha\neq\gamma} \left[e^{-\beta \epsilon_\alpha n_{\alpha}} \right]
\f
Do the sum over occupation numbers $\{n_i\}$ with $n_i=1$ or $0$.
\eq
\tr \left[e^{-\beta H_0} a_\gamma^\dagger a_\gamma \right] =
e^{-\beta \epsilon_\gamma}
\prod_{\alpha\neq\gamma} \left[ 1+e^{-\beta \epsilon_\alpha} \right]
\f

b.
Solve
\eq
\tr \left[e^{-\beta H_0} a_\gamma a_\gamma^\dagger \right] =
-\tr \left[e^{-\beta H_0} a_\gamma^\dagger a_\gamma \right]
+
\tr \left[e^{-\beta H_0} \right]
\f
using $\{a_\alpha,a_\alpha^\dagger\}=1$.  We know the first term on the left from part a.
The second term doesn't take too much effort to get.
\eq
\tr \left[e^{-\beta H_0} \right] = \sum_{\{n_i\}} \prod_\alpha
\langle \{n_i\}  | \left(e^{-\beta \epsilon_\alpha n_\alpha}\right) | \{n_i\}\rangle
= \prod_\alpha \left(1+e^{-\beta \epsilon_\alpha}\right)
\f
So
\eq
\tr \left[e^{-\beta H_0} a_\gamma a_\gamma^\dagger \right] =
-e^{-\beta \epsilon_\gamma}
\prod_{\alpha\neq\gamma} \left[1+e^{-\beta \epsilon_\alpha} \right]
+ \prod_\alpha \left(1+e^{-\beta \epsilon_\alpha}\right)\\
=  \left[
- \frac{e^{-\beta\epsilon_\gamma}}{1+e^{-\beta \epsilon_\alpha}}
+\frac{1+e^{-\beta \epsilon_\alpha}}{1+e^{-\beta \epsilon_\alpha}}\right]
\prod_\alpha \left(1+e^{-\beta \epsilon_\alpha}\right)
\f
Which can be rewritten
\eq
\tr \left[e^{-\beta H_0} a_\gamma a_\gamma^\dagger \right] =
\frac{1}{1+e^{-\beta\epsilon_\gamma}} \prod_\alpha \left(1+e^{-\beta \epsilon_\alpha}\right)
\f

c.
Noting that
\eq
e^{-\beta H_0} a^\dagger_\gamma a_\gamma =
-\frac{1}{\beta}\frac{\diff}{\diff\epsilon_\gamma} e^{-\beta H_0}
\f
We can find
\eq
\tr \left[e^{-\beta H_0} a_\gamma^\dagger a_\gamma \right] =
-\frac{1}{\beta}\frac{\diff}{\diff\epsilon_\gamma}
\tr \left[e^{-\beta H_0}\right]
\f
>From earlier,
\eq
\tr \left[e^{-\beta H_0}\right]
= \prod_\alpha \left[1+e^{-\beta \epsilon_\alpha}\right]
\f
whence it follows
\eq
\tr \left[e^{-\beta H_0} a_\gamma^\dagger a_\gamma \right] =
e^{-\beta \epsilon_\gamma}
\prod_{\alpha\neq\gamma} \left[1+e^{-\beta \epsilon_\alpha} \right]
\f
which is what I got in part a.

d.
\eq
Z =\tr \left[e^{-\beta H_0}\right]
= \prod_\alpha \left[1+e^{-\beta \epsilon_\alpha}\right]
\f
and
\eq
\tr \left[e^{-\beta H_0} a_\gamma^\dagger a_\gamma \right] =
e^{-\beta \epsilon_\gamma}
\prod_{\alpha\neq\gamma} \left[1+e^{-\beta \epsilon_\alpha }\right] \\
=
\frac{e^{-\beta \epsilon_\gamma}}{ 1+ e^{-\beta \epsilon_\gamma} }
\prod_{\alpha} \left[1+e^{-\beta \epsilon_\alpha }\right]
\f
so
\eq
\langle n_\gamma \rangle =
\tr \left[e^{-\beta H_0} a_\gamma^\dagger a_\gamma \right]
/  \tr \left[e^{-\beta H_0}\right] \\
= \frac{1}{1+e^{\beta\epsilon_\gamma}}
\f

e.
For bosons,
\eq
[a,a^\dagger]=1
\f
First, we get the partition function.  Notice that all for each $n_i$, all positive integer values are allowed.  The sum is a geometric series and can be done explicitly.
\eq
\tr \left[ e^{-\beta H_0}\right] = Z
= \sum_{\{n_i\}} \prod_\alpha e^{-\beta \epsilon_\alpha n_{\alpha}}\\
= \prod_{\alpha} \frac{1}{1-e^{-\beta \epsilon_\alpha} }
\f
Then, we use a nice little trick.
\eq
\tr \left[e^{-\beta H_0} a_\gamma^\dagger a_\gamma \right] =
-\frac{1}{\beta} \frac{\diff}{\diff \epsilon_\gamma}
\tr \left[ e^{-\beta H_0} \right] \\
= \frac{ e^{-\beta \epsilon_\alpha }}{ 1-e^{-\beta \epsilon_\alpha} }
\prod_{\alpha} \frac{1}{1-e^{-\beta \epsilon_\alpha} }
\f
\eq
\langle n_\gamma \rangle =  \tr \left[e^{-\beta H_0} a_\gamma^\dagger a_\gamma \right] / \tr \left[ e^{-\beta H_0} \right]\\
=
\frac{-1}{1-e^{\beta \epsilon_\gamma}}
\f
When one of the energies is zero or $\beta\rightarrow 0$,  the bose function diverges, and we get a condensate.

Finally, I need to find $\tr \left[e^{-\beta H_0} a_\gamma a_\gamma^\dagger \right]$. Use the commutator to shuffle terms.
\eq
\tr \left[e^{-\beta H_0} a_\gamma a_\gamma^\dagger \right] =
\tr \left[e^{-\beta H_0} a_\gamma^\dagger a_\gamma \right]
+
\tr \left[e^{-\beta H_0} \right] = \langle n_\gamma \rangle Z + Z
\f
%so
%\eq
%\tr \left[e^{-\beta H_0} a_\gamma a_\gamma^\dagger \right] =
%-\frac{1}{1-e^{-\beta\epsilon_\gamma}}  \prod_\alpha \frac{1}{1-e^{-\beta \epsilon_\alpha}}
%\f


f.
We work with bosons.
\eq
[a_\gamma, H_0] =
\sum_\alpha \epsilon_\alpha [a_\gamma, a^\dagger_\alpha a_\alpha] \\
= \sum_{\alpha\neq \gamma} \epsilon_\alpha [ a_\gamma a_\alpha^\dagger a_\alpha - a_\alpha^\dagger a_\alpha a_\gamma ] +\epsilon_\gamma [a_\gamma,a_\gamma a^\dagger_\gamma] \\
= \epsilon_\gamma a_\gamma
\f
\eq
[a^\dagger_\gamma, H_0] =
\sum_\alpha \epsilon_\alpha [a^\dagger_\gamma, a^\dagger_\alpha a_\alpha] \\
= \sum_{\alpha\neq \gamma} \epsilon_\alpha [ a^\dagger_\gamma a_\alpha^\dagger a_\alpha - a_\alpha^\dagger a_\alpha a^\dagger_\gamma ] +\epsilon_\gamma [a^\dagger_\gamma,a^\dagger_\gamma a_\gamma] \\
= -\epsilon_\gamma a^\dagger_\gamma
\f
So $[H_0,a_\gamma]=-\epsilon_\gamma a_\gamma$ and $[H_0,a^\dagger_\gamma]=\epsilon_\gamma a^\dagger_\gamma$.
\eq
\left( \begin{array}{c}
a_\gamma^\dagger (\tau) \\
a_\gamma (\tau)\\
\end{array} \right)
=
e^{\tau H_0}
\left( \begin{array}{c}
a_\gamma^\dagger \\
a_\gamma \\
\end{array} \right)
e^{-\tau H_0} \\
=
\left( \begin{array}{c}
a_\gamma^\dagger \; e^{\epsilon_\gamma \tau}\\
a_\gamma \; e^{-\epsilon_\gamma \tau}\\
\end{array} \right)
e^{\tau H_0 -\tau H_0 }
\f
So commuting the $a$ and $a^\dagger$s past the $H$, we find
\eq
\left( \begin{array}{c}
a_\gamma^\dagger (\tau)\\
a_\gamma (\tau)\\
\end{array} \right)
=
\left( \begin{array}{c}
a_\gamma^\dagger \; e^{\epsilon_\gamma \tau}\\
a_\gamma \; e^{-\epsilon_\gamma \tau}\\
\end{array} \right)
\f




\end{document}
