I am currently reading David Tong's notes on statistical physics (Page 33) where, just before finishing the derivation of the classical partition function, he obtains the following equation $$ Z=\int{dqdpdp' e^{-\beta V(q)} \langle{p|q}\rangle \langle{p|e^{-\beta\hat{p}^2/2m}|p'}\rangle \langle{p'|q}\rangle }. $$ I am stuck at this point, mainly because I am not that used to work with quantum physics, so I don't know how to compute $\langle{p|e^{-\beta\hat{p}^2/2m}|p'}\rangle$

Just by looking at the solution and trying to guess, the only thing I can come up with is that$$ \langle{p|e^{-\beta\hat{p}^2/2m}|p'}\rangle = e^{-\beta p^2/2m} \langle{p|p'}\rangle = e^{-\beta p^2/2m} \delta (p'-p) $$ But even if that was correct and not a stupid guess, I still wouldn't know why it holds.


1 Answer 1


The exponential of an operator is defined as

$$e^\hat{A}=\sum_{n=0}^\infty \frac{(\hat{A})^n}{n!}.$$

In your case, since $|p'\rangle$ is an eigenstate of $\hat p$ with eigenvalue $p'$, the eigenvalue of the operator $\exp(-\beta\hat{p}^2/2m)$ is, indeed, $\exp(-\beta (p')^2/2m)$. So $$e^{-\beta\hat{p}^2/2m}|p'\rangle=e^{-\beta (p')^2/2m}|p'\rangle.$$

Note that you got the $p$ without the $'$, so it is as if you had applied the exponential operator to the bra $\langle p|$, which in this case is also correct because the operator in the exponent is hermitian. Anyway, since there is a delta later, the result is the same in both cases.


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