By definition, $\hat q = q \hat r$ is the dipole moment. Given that the system has definite parity, show that the dipole moment is $0$ for a stationary state, that is, $\langle n |\hat q| n\rangle = 0$.

I wanted to try this problem in one dimension first, so what I did was:

$\hat q = q \hat x \implies\langle n |\hat q| n\rangle = \langle n |q\hat x| n\rangle =q \langle n| \langle\hat x |n\rangle. $ and I was stuck here.

Definite parity given from the problem means the stationary states function must be either even or odd, is this something I can incorporate into solving this problem?

If I were to use this fact, then suppose $n$ is even, then $\langle\hat x| n\rangle$ is even$\implies\langle n |\hat x| n\rangle $ is even, and if $n$ is odd, still $\\\langle n |\hat x| n\rangle $ is even

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    $\begingroup$ You seem to have mixed up your $p$'s and $q$'s in a few places; can you check them and edit to correct as necessary? $\endgroup$ – Michael Seifert Jan 22 at 20:04
  • $\begingroup$ Thanks for the notice, I have edited it. $\endgroup$ – Rico Jan 22 at 20:15
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    $\begingroup$ I don't know if this is very rigorous, but I guess one can calculate $\langle n|\hat{x}|n\rangle = \int \mathrm{d}x\, |\varphi_n(x)|^2 x$, where $\varphi_n(x) \equiv \langle x|\varphi\rangle$. Then you integrate over an odd function. For a finite intervall, i.e. from -a to a, this will vanish. However I am not sure if this is the case where you consider the whole real line. Edit: See also this post physics.stackexchange.com/questions/426037/… and the answers therein $\endgroup$ – Jakob Jan 22 at 20:18
  • $\begingroup$ @Jakob your argument is exactly right. $\endgroup$ – ZeroTheHero Jan 22 at 20:24
  • $\begingroup$ Thanks for the response, I believe I can prove it in the 1-D case now. How would I extend it into the 3-D case? $\endgroup$ – Rico Jan 22 at 20:30

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