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

[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.

 

[olgranpappy]

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.

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

 

[KFC]

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

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

 

[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?

 

[olgranpappy]

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?

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.