2
$\begingroup$

The following formula has been given in 't Hooft's black holes notes ($|\Omega \rangle$ is the vacuum state of Minkowski space, O is a operator):

$$\langle \Omega| O|\Omega \rangle = \sum_{n \ge 0} \langle n | O | n \rangle e^{-2 \pi n \omega}(1-e^{-2 \pi \omega})=Tr(O \rho_{\Omega})$$

How does this mean that the radiation is thermal and follows Plancks black body law?

  1. Here, I read that $\langle O \rangle = \frac{1}{Z}\sum_n e^{-\beta E_n}\langle n |O|n\rangle$. How is this sum the same as that in the above expression?

  2. How is the density matrix related to the law of black body radiation? How can I derive Planck's law from the expectation value of O in the first expression?

$\endgroup$
  • $\begingroup$ $(1-e^{-2 \pi \omega})$ does not depends on $n$, so we don't care, it will be absorbed in the definition of $Z$ $\endgroup$ – Trimok Jul 8 '13 at 10:39
1
$\begingroup$

First, let's see the density matrix of thermal mixture (in some cases is the Planck's law): $$ \rho_{Thermal}=\frac{1}{Z}\sum_{n=0}^{N}e^{-\beta E_n}|n\rangle \langle n| $$

Now we note that the expectation value $\langle \Omega|O|\Omega \rangle$ is equal to $$ \frac{1}{Z}\sum_{n=0}^{N}e^{-\beta E_n}\langle n|O|n \rangle , $$

with $Z=(1-e^{-2 \pi \omega})^{-1}$ and $\beta E_n=2 \pi n \omega$ (this expression give us the $\beta$).

The operator $O$ in $|n\rangle 's$ basis is simple a matrix $O_{nm}=\langle n|O| m\rangle$. Now is easy to see that $\langle \Omega| O|\Omega \rangle = Tr(O \rho_{Thermal})=$ $$ \sum_{n=0}^{N}O_{n}\frac{e^{-\beta E_n}}{Z}. $$.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy