Is $\langle\psi_1|p\psi_1\rangle$ necessarily 0 for eigenstates? Is $\langle\psi_1|p\psi_1\rangle$ necessarily 0 for harmonic oscillator eigenstates?
If $\Psi(x,t)= c_0\psi_0(x)e^{-iE_0t/\hbar}+c_1\psi_1(x)e^{-iE_1t/\hbar}$,
is the following true? Where $p$ is the momentum operator,
$$\langle\psi_1|p\psi_1\rangle = \langle\psi_0|p\psi_0\rangle=0.$$
Any hint would be appreciated.
 A: Which system are you talking about?
I believe this is not true in general...
However, eigenstates of all the simplest quantum mechanical systems (free particle, harmonic oscillator, hydrogen atom etc) have definite parity.
Suppose the $\lvert\psi_1\rangle$ you are talking about has $$P\lvert\psi_1\rangle=+1\lvert\psi_1\rangle$$
Also, in the coordinate representation, the momentum operator is $$\hat{p}_x=-i\partial_x$$ which being a derivative changes the parity of the state to which it is applied.
$$P\lvert p\psi_1\rangle=-1\lvert p\psi_1\rangle$$
Therefore you are projecting a state of definite parity onto a state of opposite definite parity.
That is why you have a $0$, maybe...
A: Recall that $p = i\sqrt{\frac{m \omega\hbar}{2}}(a^{\dagger}-a)$. The result follows immediately. Although @FredericoCarta 's method is perhaps more insightful.
A: As @FredericoCarta mentionned, you need to be more precise with your system, especially with the momentum operator $P$ which is very tricky as it is hermitian, but not necessarily self adjoint depending on its domain of definition $\mathcal D(P)$.
Hence, you may find a state $|\Psi\rangle$, that is indeed eingenstate of the momentum operator, (for example $e^{i p x}$) but which doesn't belong to your initial Hilbert space, hence you have $p=0$.
Note that this state then usually belongs to $\mathcal D(P^\dagger)$), as $\mathcal D(P)\subset \mathcal D(P^\dagger)$) 
Hope this helps.
A: For the harmonic oscillator, $\langle p\rangle$ is always going to be $0$ for a single energy eigenstate. An energy eigenstate is a "stationary state", which means that only the phase of the wavefunction changes with time. Consider what would happen if $\langle p\rangle$ were positive or negative: the wavefunction would be more likely to be moving to the right (positive) or left (negative). This would result in movement of the average value of $x$, but $\langle x \rangle$ can't change because you're in a stationary state.
Another way to look at it is that the hamiltonian is symmetric in both $x$ and $p$. If $p$ were nonzero and pointed in one direction or another, what broke the symmetry?
A: In general, the expectation value of momentum for any non-relativistic, eigen-, bound-state is zero, which includes the harmonic oscillator eigenstates. Since  at one hand
$$ \langle n | [x, H] | n \rangle
= \langle n | [x, \frac{p^2}{2m}] | n \rangle = \frac{1}{m} \langle n | [x, p] p | n \rangle = \frac{i}{m} \langle n| p |n \rangle \tag{1} $$
on the other hand,
$$ \langle n | [x, H] | n \rangle =  \langle n | x H - Hx| n \rangle  = \langle n | x H |n \rangle- \langle n| Hx| n \rangle = E_n \langle n|x| n \rangle -  E_n \langle n|x| n \rangle \tag{2} $$
Since for bound state, $\langle n | x |n \rangle$ is a well-defined number, the right-hand-side of Eq (2) is zero.
The requirement of bound state is necessary. Since 
(i) the expectation value of momentum for a momentum eigenstate is in-general, non-zero (we are answering what is the momentum for a momentum eigenstate);
(ii) for momentum eigenstate, $\langle x \rangle $ is not a well-defined number, which corresponds to the average position for a plane wave...
