2
$\begingroup$

I was wondering if anyone could point me towards the analytical solution for the probability distribution for the momentum of a quantum harmonic oscillator in the canonical ensemble. I've come across the position distribution which one can obtain by taking the diagonal of the thermal density matrix in the position representation $$\langle x'|\hat{\rho}|x\rangle \propto \exp(- \frac{m \omega}{4 \hbar}[ (q+q')^2 \tanh(\beta \omega/2) + (q-q')^2 \coth (\beta \omega/2)])\quad .$$ I tried to obtain the momentum distribution by Fourier transforming this but it didn't seem to work out. Alternatively, I tried to work out $$\int dx' \int dx \langle p| x' \rangle \langle x' |\hat{\rho}|x\rangle \langle x | p \rangle \quad ,$$ change variable to $u=q+q'$, $v=q-q'$ and perform the integrals. I obtained $$\langle p | \hat{\rho} | p\rangle \propto \exp(-p^2/\coth(\beta \omega/2))$$ but I'm also unconvinced by this result.

$\endgroup$
4
  • 3
    $\begingroup$ Well, the Hamiltonian is of the form $H\sim x^2 +p^2$, so I guess the result will more or less the same as in the position representation. In other words, check if you can compute step by step with the eigenfunctions in the momentum representation a similar result for the representation of $\rho$. $\endgroup$ Commented Mar 2 at 18:20
  • $\begingroup$ Hi, thanks for your comment. I had tried something similar at the start (seemed like the easiest way to go about the transformation) but I guess I made a silly mistake. Making the substitution $m \to 1/(m \omega^2)$ in the expression for $\langle x|\hat{\rho}|x\rangle$ indeed gives me the correct expression for $\langle p|\hat{\rho}|p\rangle$ I believe. Thanks again! $\endgroup$ Commented Mar 2 at 19:00
  • $\begingroup$ Good. Consider to write an answer yourself. It may help future readers with a similar problem. $\endgroup$ Commented Mar 2 at 19:07
  • 1
    $\begingroup$ @aQuarkyName You have basically the right expression; it's what you obtained in your final equation by Fourier transforming. It should (and does) look very similar to the diagonal part of the real-space expression that you came across. Just set $q=q'$ in your real space expression and use $\coth = 1/\tanh$ to see how similar the two expressions are... $\endgroup$
    – hft
    Commented Mar 2 at 21:24

1 Answer 1

1
$\begingroup$

...point me towards the... probability distribution for the momentum of a quantum harmonic oscillator in the canonical ensemble.

To establish some notation, I note that the distribution you are seeking is $$ f(p) = \sum_n \tilde P_n(p)e^{-\beta E_n} = \sum_n \tilde \psi_n(p) \tilde \psi_n^*(p)e^{-\beta E_n}\tag{1} $$ $$ =\sum_n \langle p|\psi_n\rangle\langle\psi_n| e^{-\beta \hat H}|p\rangle $$ $$ =\langle p|e^{-\beta H}|p\rangle \equiv \langle p|\hat \rho |p\rangle $$

I've come across the position distribution which one can obtain by taking the diagonal of the thermal density matrix in the position representation $$\langle x'|\hat{\rho}|x\rangle \propto \exp(- \frac{m \omega}{4 \hbar}[ (q+q')^2 \tanh(\beta \omega/2) + (q-q')^2 \coth (\beta \omega/2)])\quad .$$ I tried to obtain the momentum distribution by Fourier transforming this but it didn't seem to work out.

First of all, the LHS of the above-quoted equation uses symbols $x$ and $x'$, but the RHS uses symbols $q$ and $q'$. So, that's confusing... It would probably be best to not switch the symbol you are using for position within a single equation...

To find the appropriate transformation, you can perform the following manipulations starting from Eq. (1): $$ f(p) = \sum_n \tilde \psi_n(p) \tilde \psi_n^*(p)e^{-\beta E_n} $$ $$ =\sum_n \int dx \int dy \psi_n(x) e^{-ipx} \psi_n^*(x') e^{ipx'} e^{\beta E_n} $$ $$ =\int dx \int dy e^{-ip(x-x')}\langle x|\hat \rho|x'\rangle\;. $$ So, indeed, you need to perform a specific Fourier transform of the real-space density matrix in order to get the momentum-space distribution. In particular, you are lucky enough to have only the different between the two real-space locations to worry about, which makes life a lot easier.

Alternatively, I tried to work out $$\int dx' \int dx \langle p| x' \rangle \langle x' |\hat{\rho}|x\rangle \langle x | p \rangle \quad ,$$

This isn't really an "alternative," it is just another way to write the Fourier transform, since for example $\langle x | p \rangle = e^{ipx}$...

I obtained $$\langle p | \hat{\rho} | p\rangle \propto \exp(-p^2/\coth(\beta \omega/2))$$ but I'm also unconvinced by this result.

OK, If you performed the Fourier transforms correctly, then there is no reason to be unconvinced... I'm not going to check all your work, but this looks right to me (at least with $m=\hbar=\omega=1$). Probably it should look something more like: $$ f(p) \propto e^{-\frac{p^2}{\hbar m\omega}\tanh(\beta\omega/2)} $$

To understand why it "looks right" please note that if you were to set $q=q'$ in your above-quoted expression for the density matrix (i.e., if you are only looking at the diagonal elements) then you will see something that looks very much like the expression you arrived at for the diagonal elements in position space. (Please also note that $1/\coth = \tanh$.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.