# How to deal with index in the problem of thermal field?

1. Jan 25, 2009

### KFC

In the problem of harmonic oscillator, for single mode, that is, the energy

$$H = \hbar\omega(n + 1/2)$$

It is easy to find the average of energy by considering the density operator

$$<H> = \frac{Tr(He^{\beta H})}{Tr(e^{\beta H})}$$

But for multimode (assume no polarization), we have to consider all modes specified with $$k$$

$$H = \sum_k \hbar\omega_k(n_k + 1/2)$$

So the average energy would be

$$<H> = \prod_k\frac{Tr(He^{\beta H})}{Tr(e^{\beta H})} = \prod_k\frac{\sum_{k_1} \hbar\omega_{k_1}(n_{k_1} + 1/2) \exp[\beta\sum_{k_2}\hbar\omega_{k_2}(n_{k_2}+1/2)] }{\exp[\beta\sum_{k_3}\hbar\omega_{k_3}(n_{k_3}+1/2)]}$$

Since each sum and prod should have independent index so I use k, k1, k2, k3. I am very confusing how to deal with the index to get the simplifed expression.

In some textbook, it first let the partition function be
$$Z=Tr(e^{\beta H})$$

so the energy average be

$$<H> = -\frac{\partial \ln Z}{\partial \beta} = \sum_{k}\dfrac{\hbar\omega_{k}}{\exp\{\beta\hbar\omega_{k}\}-1}$$

My question is how to deal with index if I don't use the partion function.

2. Jan 26, 2009

### olgranpappy

same way as with the partition function. it's just easier to calculate the partition function and take a derivative.

3. Jan 26, 2009

### KFC

But there is three different index, how can I make it into one index?

4. Jan 26, 2009

### KFC

Well, I found one problem. For single mode, which way is the correct one to write the partition function?

$$Z= \prod_k\exp[\beta\sum_{k}\hbar\omega_{k}(n_{k}+1/2)]$$

or

$$Z= \prod_k\exp[\beta\hbar\omega_{k}(n_{k}+1/2)]$$

?

And what about the hamiltonian in the numerator?

5. Jan 26, 2009

### olgranpappy

Neither is the correct expression for the partition function. This is apparent since the occupation numbers still appear on the LHS...

But, before going on, note that
$$\prod_j e^{f_j} =e^{f_1}e^{f_2}\ldots e^{f_N}\ldots=e^{f_1+f_2+\ldots} =e^{\sum_j f_j}\;.$$

The correct partition function is
$$\sum_{n_1}\sum_{n_2}\ldots\sum_{n_N}\ldots e^{-\beta(\sum_k(n_k+1/2)\hbar\omega_k)} =\sum_{n_1}\sum_{n_2}\ldots\sum_{n_N}\ldots e^{-\beta((n_1+1/2)\hbar\omega_k)} e^{-\beta((n_2+1/2)\hbar\omega_k)}\ldots = \sum_{n_1}e^{-\beta((n_1+1/2)\hbar\omega_k)}\sum_{n_2}e^{-\beta((n_2+1/2)\hbar\omega_k)}\ldots$$
and this equals
$$\prod_k \sum_{n_k}e^{-\beta(n_k+1/2)\hbar\omega_k}\;,$$
where the sum, for bosons, runs from 0 to infinity.