Q:

Potentials in QM vs. Classical Mechanics

I am getting a little confused by the notion of potentials.  In classical mechanics we are never worried about the value of a potential, we are always concerned mainly in the way a potential varies, or potential difference between points.  We can always set our potential to have any value anywhere we want (most of the times).  
On the other hand in quantum mechanics we explicitly use the value of the potential $V(\boldsymbol{r})$. Is there a difference in quantum mechanics between the potentials $V_1 (x) = V(x)$ and $V_2(x) = V(x) + C$? 
If we include both of these potentials into Schrodinger's equation there is a difference between the solutions we get. 

A:

You're correct that the solutions don't come out precisely the same.   However, if we look at Schrodinger's equation with the added constant,
$$ i\hbar \frac{\partial}{\partial t}|\psi\rangle = (\hat H+C)|\psi\rangle $$
we notice that if we let
$$|\psi\rangle = e^{-iCt/\hbar}|\psi'\rangle $$
then the Schrodinger equation becomes
$$i\hbar \frac{\partial}{\partial t}|\psi\rangle = e^{-iCt/\hbar}\cdot i\hbar \frac{\partial}{\partial t} |\psi'\rangle +Ce^{-iCt/\hbar}|\psi'\rangle= (\hat H+C)e^{-iCt/\hbar}|\psi'\rangle$$
Which simplifies to
$$i \hbar \frac{\partial}{\partial t}|\psi'\rangle = \hat H |\psi'\rangle$$
In other words, shifting the energy by some constant $C$ is equivalent to multiplying all of the states by the phase factor $e^{-iCt/\hbar}$.  Because this unitary factor is applied to every state, it cancels out in all physical calculations (e.g. expectation values) and so does not impact the predictions of the theory. 

Classical physics is insensitive to shifts in potential energy because measurable physical quantities depend only on derivatives of the potential.   Similarly, quantum mechanics is insensitive to shifts in potential energy because measurable physical quantities depend only on inner products between states, and the transformation induced by the potential shift ($|\psi\rangle \rightarrow e^{-iCt/\hbar}|\psi\rangle$) is unitary, meaning that it leaves the inner products unchanged.   Both of these are particular examples of the more general notion of gauge invariance.