$$\begin{eqnarray*} H_0 = \sum_\alpha \epsilon_\alpha a^\dagger_\alpha a_\alpha \end{eqnarray*}$$

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.

$$\begin{eqnarray*} \mbox{Trace}\left[e^{-\beta H_0} a_\gamma^\dagger a_\gamma \ri... ...beta H_0} a_\gamma^\dagger a_\gamma \right] \vert\{n_i\}\rangle \end{eqnarray*}$$

where $\vert\{n_i\}\rangle$ stands for a many particle state with a set, $\{n_i\}$, of occupation numbers. $a^\dagger_\gamma a_\gamma$ acting on $\vert\{n_i\}\rangle$ 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

$$\begin{eqnarray*} \mbox{Trace}\left[e^{-\beta H_0} a_\gamma^\dagger a_\gamma \ri... ...beta H_0} a_\gamma^\dagger a_\gamma \right] \vert\{n_i\}\rangle \end{eqnarray*}$$

Now, I let the density operators act on the various $\gamma$ states.

$$\begin{eqnarray*} \mbox{Trace}\left[e^{-\beta H_0} a_\gamma^\dagger a_\gamma \ri... ...a \prod_\alpha \left[e^{-\beta \epsilon_\alpha n_\alpha} \right] \end{eqnarray*}$$

Take the term which was generated by the state $\gamma$ out of the product.

$$\begin{eqnarray*} \mbox{Trace}\left[e^{-\beta H_0} a_\gamma^\dagger a_\gamma \ri... ...a\neq\gamma} \left[e^{-\beta \epsilon_\alpha n_{\alpha}} \right] \end{eqnarray*}$$

Do the sum over occupation numbers $\{n_i\}$ with $n_i=1$ or $0$.

$$\begin{eqnarray*} \mbox{Trace}\left[e^{-\beta H_0} a_\gamma^\dagger a_\gamma \ri... ...d_{\alpha\neq\gamma} \left[ 1+e^{-\beta \epsilon_\alpha} \right] \end{eqnarray*}$$

b. Solve

$$\begin{eqnarray*} \mbox{Trace}\left[e^{-\beta H_0} a_\gamma a_\gamma^\dagger \ri... ...gger a_\gamma \right] + \mbox{Trace}\left[e^{-\beta H_0} \right] \end{eqnarray*}$$

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.

$$\begin{eqnarray*} \mbox{Trace}\left[e^{-\beta H_0} \right] = \sum_{\{n_i\}} \pro... ...\rangle = \prod_\alpha \left(1+e^{-\beta \epsilon_\alpha}\right) \end{eqnarray*}$$

So

$$\begin{eqnarray*} \mbox{Trace}\left[e^{-\beta H_0} a_\gamma a_\gamma^\dagger \ri... ...}}\right] \prod_\alpha \left(1+e^{-\beta \epsilon_\alpha}\right) \end{eqnarray*}$$

Which can be rewritten

$$\begin{eqnarray*} \mbox{Trace}\left[e^{-\beta H_0} a_\gamma a_\gamma^\dagger \ri... ..._\gamma}} \prod_\alpha \left(1+e^{-\beta \epsilon_\alpha}\right) \end{eqnarray*}$$

c. Noting that

$$\begin{eqnarray*} e^{-\beta H_0} a^\dagger_\gamma a_\gamma = -\frac{1}{\beta}\frac{\partial }{\partial \epsilon_\gamma} e^{-\beta H_0} \end{eqnarray*}$$

We can find

$$\begin{eqnarray*} \mbox{Trace}\left[e^{-\beta H_0} a_\gamma^\dagger a_\gamma \ri... ...partial \epsilon_\gamma} \mbox{Trace}\left[e^{-\beta H_0}\right] \end{eqnarray*}$$

>From earlier,

$$\begin{eqnarray*} \mbox{Trace}\left[e^{-\beta H_0}\right] = \prod_\alpha \left[1+e^{-\beta \epsilon_\alpha}\right] \end{eqnarray*}$$

whence it follows

$$\begin{eqnarray*} \mbox{Trace}\left[e^{-\beta H_0} a_\gamma^\dagger a_\gamma \ri... ...d_{\alpha\neq\gamma} \left[ 1+e^{-\beta \epsilon_\alpha} \right] \end{eqnarray*}$$

which is what I got in part a.

d.

$$\begin{eqnarray*} Z =\mbox{Trace}\left[e^{-\beta H_0}\right] = \prod_\alpha \left[1+e^{-\beta \epsilon_\alpha}\right] \end{eqnarray*}$$

and

$$\begin{eqnarray*} \mbox{Trace}\left[e^{-\beta H_0} a_\gamma^\dagger a_\gamma \ri... ...mma} } \prod_{\alpha} \left[1+e^{-\beta \epsilon_\alpha }\right] \end{eqnarray*}$$

so

$$\begin{eqnarray*} \langle n_\gamma \rangle = \mbox{Trace}\left[e^{-\beta H_0} a_... ...e^{-\beta H_0}\right] \\ = \frac{1}{1+e^{\beta\epsilon_\gamma}} \end{eqnarray*}$$

e. For bosons,

$$\begin{eqnarray*}[a,a^\dagger]=1 \end{eqnarray*}$$

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.

$$\begin{eqnarray*} \mbox{Trace}\left[ e^{-\beta H_0}\right] = Z = \sum_{\{n_i\}} ... ...ha}}\\ = \prod_{\alpha} \frac{1}{1-e^{-\beta \epsilon_\alpha} } \end{eqnarray*}$$

Then, we use a nice little trick.

$$\begin{eqnarray*} \mbox{Trace}\left[e^{-\beta H_0} a_\gamma^\dagger a_\gamma \ri... ...\alpha} } \prod_{\alpha} \frac{1}{1-e^{-\beta \epsilon_\alpha} } \end{eqnarray*}$$

$$\begin{eqnarray*} \langle n_\gamma \rangle = \mbox{Trace}\left[e^{-\beta H_0} a_... ...{-\beta H_0} \right]\\ = \frac{-1}{1-e^{\beta \epsilon_\gamma}} \end{eqnarray*}$$

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 $\mbox{Trace}\left[e^{-\beta H_0} a_\gamma a_\gamma^\dagger \right]$. Use the commutator to shuffle terms.

$$\begin{eqnarray*} \mbox{Trace}\left[e^{-\beta H_0} a_\gamma a_\gamma^\dagger \ri... ...ce}\left[e^{-\beta H_0} \right] = \langle n_\gamma \rangle Z + Z \end{eqnarray*}$$

f. We work with bosons.

$$\begin{eqnarray*}[a_\gamma, H_0]= \sum_\alpha \epsilon_\alpha [a_\gamma, a^\dagg... ...\gamma,a_\gamma a^\dagger_\gamma] \\ = \epsilon_\gamma a_\gamma \end{eqnarray*}$$

$$\begin{eqnarray*}[a^\dagger_\gamma, H_0]= \sum_\alpha \epsilon_\alpha [a^\dagger... ...\dagger_\gamma a_\gamma] \\ = -\epsilon_\gamma a^\dagger_\gamma \end{eqnarray*}$$

So $[H_0,a_\gamma]=-\epsilon_\gamma a_\gamma$ and $[H_0,a^\dagger_\gamma]=\epsilon_\gamma a^\dagger_\gamma$.

$$\begin{eqnarray*} \left( \begin{array}{c} a_\gamma^\dagger (\tau) \\ a_\gamma (... ...ilon_\gamma \tau}\\ \end{array} \right) e^{\tau H_0 -\tau H_0 } \end{eqnarray*}$$

So commuting the $a$ and $a^\dagger$s past the $H$, we find

$$\begin{eqnarray*} \left( \begin{array}{c} a_\gamma^\dagger (\tau)\\ a_\gamma (\... ...\\ a_\gamma \; e^{-\epsilon_\gamma \tau}\\ \end{array} \right) \end{eqnarray*}$$