Momentum conservation in Quantum Mechanics Most quantum mechanical potentials are not translationally invariant and therefore the expectation value of momentum varies. 
The question is then where has this momentum been transferred to? Because as a whole the system must conserve momentum.   
In electrodynamics of course there is a similar question with for example a particle in a constant electric field. This is resolved by saying the momentum has been transferred to the field. 
So in the quantum mechanical case does the momentum transfer to the source of the potential?
 A: 
Most quantum mechanical potentials are not translationally invariant...

... because those potentials are externally imposed from some overarching system whose dynamics you're not explicitly considering. The momentum goes to that overarching system.
As an example, consider the hamiltonian of the electron in an $\rm H_2^+$ molecular ion, which reads
$$
\hat H_\mathrm{electronic} = \frac{\hat{\mathbf p}^2}{2m} - \frac{e^2}{4\pi\epsilon_0}\left[\frac{1}{|\hat{\mathbf r}-\mathbf R_1|}+\frac{1}{|\hat{\mathbf r}-\mathbf R_2|}\right],
$$
in the Born-Oppenheimer approximation, with the two protons clamped at positions $\mathbf R_1$ and $\mathbf R_2$. This is not translation-invariant in $\mathbf r$, so its conjugate $\mathbf p$ is not conserved. To see where that momentum goes, simply extend your description of the dynamics so that the overarching system that imposes the potential (in this case, the protons) is fully included within the dynamics, i.e.
$$
\hat H_\mathrm{full} = \frac{\hat{\mathbf p}_1^2}{2m} + \frac{\hat{\mathbf p}_2^2}{2m} + \frac{\hat{\mathbf p}^2}{2m} - \frac{e^2}{4\pi\epsilon_0}\left[\frac{1}{|\hat{\mathbf r}-\hat{\mathbf R}_1|}+\frac{1}{|\hat{\mathbf r}-\hat{\mathbf R}_2|}\right].
$$
Then the hamiltonian does become translationally invariant (where the translation must move all of $\mathbf R_1$, $\mathbf R_2$ and $\mathbf r$ by the same amount), and the conjugate variable (the center-of-mass momentum, $\mathbf P = \mathbf p_1 + \mathbf p_2 + \mathbf p$) is conserved.
Identical considerations hold in every other such situation.
