Does the average momentum vanish for an eigenstate of the simple harmonic oscillator? Suppose we have a simple harmonic oscillator, let's consider the ground state, $|0\rangle$ and the first excited state $|1\rangle$.
$\langle 0|\hat p|0 \rangle$ is zero right? Since the particle can either be travelling to the left or right, where $\hat p$ is the momentum operator.
Similarly, I think $\langle 1|\hat p|1 \rangle = 0$
But, $\langle 0 | \hat p | 1 \rangle$ is non-zero, right? Since they are different states. Also, since $\hat p$ is Hermitian, $\langle 0 | \hat p | 1 \rangle = \langle 1 | \hat p | 0 \rangle $, right?
 A: Your correct intuition stems from a symmetry of the Hamiltonian called parity invariance, which means that it is the same under reflection (so there is nothing to distinguish left and right). Try looking at the parity operator, i.e. the unitary operation $\Pi$ that takes $x\to -x$ and $p\to - p$. Specifically, you can write $\Pi x\Pi = -x$ and $\Pi p \Pi = -p$. You should also be able to show that:
1) The Hamiltonian commutes with $\Pi$ (since $\Pi H \Pi = H$) and therefore the energy eigenstates $\lvert n \rangle$ are eigenstates of $\Pi$.
2) $\Pi^2 = 1$. What does this tell you about the eigenvalues $\pi_n$ of $\Pi$?
3) $\langle n \rvert p \lvert n \rangle = \langle n \rvert \Pi^2 p \Pi^2 \lvert n \rangle = - \pi_n^2 \langle n \rvert p \lvert n \rangle$. Why does this imply that $\langle n \rvert p \lvert n \rangle = 0$?
Finally, your last point does not follow from the Hermiticity of $p$. Actually you have only the weaker condition $\langle 0\rvert p \lvert 1 \rangle = \left(\langle 1\rvert p^{\dagger} \lvert 0 \rangle\right)^{\ast} = \langle 1\rvert p \lvert 0 \rangle^{\ast}.$
