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, $\langle n |\hat p| n\rangle = 0$.

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

$\hat q = q \hat x \implies\langle n |\hat p| 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

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

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, $\langle n |\hat p| n\rangle = 0$.

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

$\hat q = q \hat x \implies\langle n |\hat p| 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?

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, $\langle n |\hat p| n\rangle = 0$.

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

$\hat q = q \hat x \implies\langle n |\hat p| 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