- #1

genneth

- 980

- 2

## Homework Statement

We wish, for masochistic purposes, to calculate the occupation number of the electromagnetic field in thermal equilibrium using density matrices. It's all pretty routine, except for one little factorisation that, whilst I can convince myself that it's correct, might not convince anyone else.

## Homework Equations

In thermal equilibrium, the state is given by:

[tex]\hat{\rho} = \prod_{\mathrm{k}} \frac{1}{Z_{\mathrm{k}}} \exp\left(-\beta\hbar\omega_{\mathrm{k}}\hat{n}_{\mathrm{k}}}\right)[/tex]

Where [tex]\mathrm{k}[/tex] ranges over all the modes of the field, [tex]\beta=k_B T[/tex], [tex]\hbar\omega_{\mathrm{k}}[/tex] is the energy of the mode, and [tex]\hat{n}_{\mathrm{k}} = \hat{a}^\dagger_{\mathrm{k}}\hat{a}_{\mathrm{k}}[/tex] is the number operator (in terms of the creation/annihilation operators) for that mode. [tex]Z_\mathrm{k}[/tex] is the partition function for the mode:

[tex]Z_\mathrm{k} = \sum_{n=0}^\infty \exp\left(-\beta\hbar\omega_\mathrm{k} n \right) = \left[1 - \exp\left(-\beta\hbar\omega_\mathrm{k} \right) \right]^{-1}[/tex]

We wish to calculate [tex]\langle\hat{n}_\mathrm{k}\rangle=\mathrm{Tr}\left[\hat{\rho}\hat{n}_\mathrm{k} \right][/tex]

## The Attempt at a Solution

Let's label the Fock states as [tex]\left|n_{\mathrm{k}_1}, n_{\mathrm{k}_2}, n_{\mathrm{k}_3}, \ldots \right>[/tex], ignoring issues of countability aside (we can always regularise and take limits). These are a complete set of orthonormal basis for the space, so we can compute the trace by summing over the appropriate matrix elements:

[tex]\begin{align*}

\mathrm{Tr}\left[\hat{\rho}\hat{n}_\mathrm{k} \right]

&= \sum_{n_{\mathrm{k}_1}, n_{\mathrm{k}_2}, n_{\mathrm{k}_3}, \ldots} \left< n_{\mathrm{k}_1}, n_{\mathrm{k}_2}, n_{\mathrm{k}_3}, \ldots | \hat{\rho}\hat{n}_\mathrm{k} | n_{\mathrm{k}_1}, n_{\mathrm{k}_2}, n_{\mathrm{k}_3}, \ldots \right> \\

&= \sum_{n_{\mathrm{k}_1}, n_{\mathrm{k}_2}, n_{\mathrm{k}_3}, \ldots} n_\mathrm{k} \prod_{\mathrm{k}_m} \frac{1}{Z_{\mathrm{k}_m}} \exp\left(-\beta\hbar\omega_{\mathrm{k}_m}n_{\mathrm{k}_m}\right) \\

\end{align*}[/tex]

Note that the summation is over all possible values for the numbers, and the [tex]n_\mathrm{k}[/tex] is necessary one of the [tex]k_m[/tex] being product'ed over. After staring at if for a bit, I convinced myself of the following:

[tex]\mathrm{Tr}\left[\hat{\rho}\hat{n}_\mathrm{k} \right] = \sum_{n_\mathrm{k}} \frac{n_\mathrm{k}}{Z_\mathrm{k}} \exp\left( -\beta\hbar\omega_\mathrm{k}n_\mathrm{k}\right) \prod_{m, \mathrm{s.t.}\,\mathrm{k}_m \neq \mathrm{k}} \sum_{n_{\mathrm{k}_m}} \frac{1}{Z_{\mathrm{k}_m}} \exp\left(-\beta\hbar\omega_{\mathrm{k}_m}n_{\mathrm{k}_m}\right)[/tex]

**Can anyone think of a better way to show the manipulation that gives this?**

Now, in case anyone else is interested, the rest is trivial:

[tex]\begin{align*}

\mathrm{Tr}\left[\hat{\rho}\hat{n}_\mathrm{k} \right]

&= \sum_{n_\mathrm{k}} \frac{n_\mathrm{k}}{Z_\mathrm{k}} \exp\left( -\beta\hbar\omega_\mathrm{k}n_\mathrm{k}\right) \\

&= \left[1-\exp\left(-\beta\hbar\omega_\mathrm{k}\right)\right]^{-1} - 1 \\

&= \frac{1}{\exp\left(\beta\hbar\omega_\mathrm{k}\right) - 1}

\end{align*}[/tex]

Which is what we would expect for a Bose system with zero chemical potential.