0
$\begingroup$

In 29:22 here, the professor says that the expectation value calculated using the partition function when finding the specific heat of an Einstein solid is both a quantum mechanical and a statistical mechanical expectation value. It isn't clear to me why this should be the case. The probability assumed here was the statistical one which maximizes entropy $p_i \propto e^{- \beta E_i}$. The quantum mechanical probability here depends on the wavefunction and what sort of linear combination of the eigenfunctions of energy it is. Why should the two be the same here?

Empirically I suppose, I could say that if I measure the number of atoms with energy $E_n$, and make a bar chart, then the distribution should match both the quantum mechanical and the statistical mechanics distribution since they are both describing the same thing. However, theoretically I don't see what the connection is. They come from completely different ideas. One is about maximizing entropy while the other is related to the Schrodinger equation and the wavefunction.

$\endgroup$

1 Answer 1

0
$\begingroup$

You have to remember that in quantum mechanics we can only calculate the expectation value of the energy $\langle H \rangle = E$. This step is the first, quantum mechanical averaging. Then after that you take the statistical average. So you really need to average twice if you want to find the expected energy of an ensemble.

The average of some operator $O_S$, that only acts on the system and not the environment, is defined as $\langle O_S \rangle = \mathrm{tr}(\rho O_s)$ in statistical qm. Here $\rho$ is the density matrix which can be written as $\rho = \sum_i |\psi_i\rangle \langle \psi_i| w_i$ such that $|\psi_i\rangle$ are the eigenfunctions of the systems Hamiltonian and $w_i$ is the probability to be in the state $i$. Expanding this definition \begin{align*} \mathrm{tr}(\rho O_S) &= \sum_n \langle \psi_n| \rho O_S|\psi_n\rangle = \sum_n \langle \psi_n| \sum_i |\psi_i \rangle \langle \psi_i| w_i O_S |\psi_n\rangle = \sum_{i, n} w_i \langle \psi_n | \psi_i \rangle \langle \psi_i | O_S | \psi_n\rangle \\ &= \sum_n w_n \langle \psi_n | O_S |\psi_n\rangle \end{align*} which shows the double averaging.

$\endgroup$
2
  • $\begingroup$ Isn't $\langle H \rangle$ the ensemble average, with all the ensemble elements being in the same state? $\endgroup$ Sep 13, 2021 at 14:14
  • $\begingroup$ I edited my answer to include some more detail. $\endgroup$
    – Wihtedeka
    Sep 13, 2021 at 14:27

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.