# Does adding a constant to the potential of Schödinger's equation always make the solution pick up a factor $e^{-iV_0t/\hbar}$?

I've been going over problem 1.8 of [Griffiths, Introduction to Quantum Mechanics, 2nd edition], where it asks to prove the following: Say $$\Psi(x,t)$$ satisfies the time-dependent Schrödinger equation

$$\begin{equation} i\hbar \dfrac{\partial}{\partial t} \Psi = \Big[-\dfrac{\hbar^2}{2m}\nabla^2 + V(x)\Big] \Psi \end{equation}$$

Prove that adding a constant $$V_0$$ to the potential leads to the wavefunction picking up a factor $$e^{-iV_0t/\hbar}$$, i.e. the new solution is $$\tilde\Psi = \Psi e^{-iV_0t/\hbar}$$

What bothers me is that in every solution given to this question, it is always first assumed that the solution is separable, as in $$\Psi(x,t) = \phi(x)e^{ -iEt/\hbar}$$, and then show by substitution that it leads to the original equation but with potential $$V(x) + V_0$$. Obviously not every solution to this differential equation is separable, so is there a rigorous way to prove this fact?

I first thought it may be related to the Exponential shift theorem

$$P(D)(e^{at}y) = e^{at}P(D+a)y$$

where in this case, we have the polynomial $$P(u)=i\hbar u$$ and $$a=\dfrac{-iV_0t}{\hbar}$$, but using it only leads to triviality.

Another thing that seems very related to this is this property of Laplace transforms

$$e^{-as} \mathcal{L}(f(t)) = \mathcal{L}(f(t-a))$$

But again I still don't know what to do, so any guidance for a rigorous proof will be appreciated.

A lot of this comes down to the linearity of the time-dependent Schrödinger equation. The linearity means that it's usually straight-forward to prove well-posedness (in some Hilbert space), and uniqueness and existence of solutions. You already have existence of the original equation, $$i\hbar \dfrac{\partial}{\partial t} \Psi = \Big[-\dfrac{\hbar^2}{2m}\nabla^2 + V(x)\Big] \Psi,$$ by assumption. Because the operators are linear, you can prove uniqueness fairly straight-forwardly (just show that if $$\Psi_1$$ and $$\Psi_2$$ are two solutions, it must be that $$\Psi_2-\Psi_1$$ ends up being zero). Once you're armed with existence and uniqueness, you can use the deus ex machina information you're given, namely that $$\tilde\Psi=\Psi e^{-iV_0t/\hbar}$$. The logic is that if we find that $$\tilde\Psi$$ satisfies the new DE, we have proven existence. Uniqueness will be as easy as before. Therefore, $$\tilde\Psi$$ is the solution.

So suppose that we replace $$V$$ with $$V+V_0,$$ and let's assume the new solution is $$\tilde\Psi,$$ following your notation. Then, by definition, we must have \begin{align*} i\hbar \dfrac{\partial}{\partial t} \tilde\Psi &= \Big[-\dfrac{\hbar^2}{2m}\nabla^2 + V(x)+V_0\Big] \tilde\Psi \\ i\hbar \dfrac{\partial}{\partial t} \tilde\Psi &= \Big[-\dfrac{\hbar^2}{2m}\nabla^2 + V(x)\Big] \tilde\Psi+V_0\tilde\Psi \\ i\hbar \dfrac{\partial}{\partial t} \left(\Psi e^{-iV_0t/\hbar}\right) &= \Big[-\dfrac{\hbar^2}{2m}\nabla^2 + V(x)\Big] \Psi e^{-iV_0t/\hbar}+V_0\Psi e^{-iV_0t/\hbar} \\ i\hbar \left[e^{-iV_0t/\hbar}\dfrac{\partial}{\partial t} \Psi +\Psi\dfrac{\partial}{\partial t} \left( e^{-iV_0t/\hbar}\right)\right] &= \Big[-\dfrac{\hbar^2}{2m}\nabla^2 + V(x)\Big] \Psi e^{-iV_0t/\hbar}+V_0\Psi e^{-iV_0t/\hbar} \\ i\hbar \left[e^{-iV_0t/\hbar}\dfrac{\partial}{\partial t} \Psi -\Psi\,\frac{iV_0}{\hbar}e^{-iV_0t/\hbar}\right] &= \Big[-\dfrac{\hbar^2}{2m}\nabla^2 + V(x)\Big] \Psi e^{-iV_0t/\hbar}+V_0\Psi e^{-iV_0t/\hbar} \\ i\hbar \left[\dfrac{\partial}{\partial t} \Psi -\Psi\,\frac{iV_0}{\hbar}\right] &= \Big[-\dfrac{\hbar^2}{2m}\nabla^2 + V(x)\Big] \Psi+V_0\Psi \\ -i\hbar\Psi\,\frac{iV_0}{\hbar} &= V_0\Psi, \\ \end{align*} a true statement. Notice all these steps are reversible. So $$\tilde\Psi$$ satisfies the new DE, and by the logic above, we are done. Notice we didn't have to make any assumptions about whether any of the DE's are separable or not. We just needed linearity, which we have from the form of the equations.

Writing the general Schroedinger equation as

$$i\hbar \dot \phi = H\phi, \tag 0$$

given that $$\psi$$ is a solution:

$$i\hbar \dot \psi = H\psi, \tag 1$$

we compute, using the ordinary product rule for differentiation,

$$i\hbar \dot {(e^{-iV_0t / \hbar}\psi)} = i\hbar (-iV_0/\hbar e^{-iV_0t / \hbar} \psi + e^{-iV_0t / \hbar}\dot \psi)$$ $$= V_0 e^{-iV_0t / \hbar} \psi + e^{-iV_0t / \hbar} i\hbar \dot \psi$$ $$= V_0 e^{-iV_0t / \hbar} \psi + e^{-iV_0t / \hbar} H\psi = (H + V_0)(e^{-iV_0t / \hbar}\psi), \tag 2$$

and we see that $$e^{-iV_0t / \hbar}\psi$$ is a solution to (0) with the modified Hamiltonian $$H + V_0$$:

$$i\hbar \dot{(e^{-iV_0t / \hbar}\psi)} = (H + V_0)(e^{-iV_0t / \hbar}\psi) ; \tag 3$$

it is easily seen that this derivation may be reversed; that is, we may start with (3) and readily obtain the "secret sauce" of (2):

$$V_0 e^{-iV_0t / \hbar} \psi + e^{-iV_0t / \hbar} i\hbar \dot \psi$$ $$= V_0 e^{-iV_0t / \hbar} \psi + e^{-iV_0t / \hbar} H\psi, \tag4$$

equivalent to

$$e^{-iV_0t / \hbar} i\hbar \dot \psi = e^{-iV_0t / \hbar} H\psi, \tag 5$$

which after multiplication by $$e^{iV_0t / \hbar}$$ yields (1).

It will be observed that have had to make no "seperability" assumptions on the wave-function $$\psi$$, nor on the specific form of the Hamiltonian operator $$H$$; we have merely hypothesized that $$H$$ commutes with the scalar multiplication operator $$e^{-iV_0t/\hbar}$$, which appears to bind for a large family of Hamiltonians $$H$$.

We note in closing that (2) indicates that the exponential shift is essentially at play here, with

$$P(\psi) = i \hbar \dot \psi. \tag 6$$