- #1

KFC

- 488

- 4

[tex]H = \hbar\omega(n + 1/2)[/tex]

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

[tex]<H> = \frac{Tr(He^{\beta H})}{Tr(e^{\beta H})}[/tex]

But for multimode (assume no polarization), we have to consider all modes specified with [tex]k[/tex]

[tex]H = \sum_k \hbar\omega_k(n_k + 1/2)[/tex]

So the average energy would be

[tex]<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)]}

[/tex]

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

[tex]Z=Tr(e^{\beta H})[/tex]

so the energy average be

[tex]

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

[/tex]

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