When is expectation value of momentum $0$?

Question

When is the expectation value of momentum for a quantum mechanical wavefunction $0$?

One of the possible cases, that I think of, is when the wavefunction is symmetric. In that case, the particle is equally likely to move in the positive and negative directions.

However, I've also come across the fact that the expectation value of momentum is "always" $0$, for a real wavefunction.

My question is, are the two cases above, independent of each other? For example, if the wavefunction is real, but not symmetric, is the momentum still $0$? Or if the wavefunction is complex, but symmetric, is the expectation value $0$?

Since I'm also led to believe that bound state wavefunctions are real, shouldn't one dimensional bound particles always have $0$ momentum, irrespective of symmetry, as they are 'real'? Are there bound states, whose wavefunctions are complex?

Any explanation would be highly appreciated.

Answer 1

Separate a wavefunction's real and imaginary parts into even and odd parts viz.$$\psi=e_r+o_r+ie_i+io_i$$with $e_r,\,e_i$ real and even and $o_r,\,o_i$ real and odd. The mean momentum is$$-i\hbar\int_{-\infty}^\infty\psi^\ast\psi^\prime dx=-i\hbar\int_{-\infty}^\infty(e_r+o_r-ie_i-io_i)(e_r^\prime+o_r^\prime+ie_i^\prime+io_i^\prime)dx.$$Expanding this integrand, there are sixteen terms of definite parity, of which eight are odd, so their contributions vanish. The mean is$$-i\hbar\int_{-\infty}^\infty(e_ro_r^\prime+o_re_r^\prime+e_io_i^\prime+o_ie_i^\prime)dx+\hbar\int_{-\infty}^\infty(e_ro_i^\prime+o_re_i^\prime-e_io_r^\prime-o_ie_r^\prime)dx.$$By unitarity, wavefunctions vanish at $\pm\infty$, as do their real and imaginary parts, and the even and odd parts thereof. So$$-i\hbar\int_{-\infty}^\infty(e_ro_r^\prime+o_re_r^\prime+e_io_i^\prime+o_ie_i^\prime)dx=-i\hbar[e_ro_r+e_io_i]_{-\infty}^\infty=0.$$Reassuringly, the mean is real, namely$$\hbar\int_{-\infty}^\infty(e_ro_i^\prime+o_re_i^\prime-e_io_r^\prime-o_ie_r^\prime)dx.$$This can vanish for several reasons. If $\psi$ is symmetric, $o_r=o_i=0$, so the integrand is $0$; if $\psi$ is real, $e_i=o_i=0$, so the integrand is $0$. As noted by which of the four DOFs vanish in each case, these are different if overlapping reasons.