# How to show that dipole moment of a system with definite parity is 0?

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

• 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? – Michael Seifert Jan 22 at 20:04
• Thanks for the notice, I have edited it. – Rico Jan 22 at 20:15
• 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 – Jakob Jan 22 at 20:18
• @Jakob your argument is exactly right. – ZeroTheHero Jan 22 at 20:24
• 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? – Rico Jan 22 at 20:30