Stuck on Intro Quantum Mechanics Question

[G01]

1. Suppose a constant potential energy,Vo, independent of x and t is added to a particle's potential energy. Show that this adds a time-dependent phase factor, $$e^{-iV_ot/\hbar}$$Right now I'm completely lost. Here's what I think so far:

2.
$$i\hbar\frac{d\Psi}{dt} = \frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2} + V(x)\Psi$$


As I said, I'm lost(Maybe I'm just to tired.) I added in the constant Potential Energy Term and that changes the Shrodinger Eq. like so:



3.$$i\hbar\frac{d\Psi}{dt} = \frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2} + V(x)\Psi + V_o\Psi$$

Now I'm stuck. As I write this I'm thinking about moving the constant term to the time side, separating ans solving the time side, since I know the answer is Time-dependent. Is this the right way to proceed? I'm not asking for a solution to the problem, just a hint about where to go. For reference this is problem 1.8 out of Griffith. Thank-you for any hints you can give me. I really appreciate it. If I need to show more work tell me and If I have accomplished anything else I will post it.

 

[Physics Monkey]

You know $$\psi(x,t)$$ satisfies the Schrödinger equation without $$V_0$$, I would see what equation $$\psi(x,t) e^{-iV_0 t/\hbar}$$ satisfies. Hint: take a time derivative and use the equation satisfied by $$\psi$$.

 

[G01]

Ok I think I may have got it. I did what I said and solved the time side with the new Potential Energy component. and The Timedependent part of $$\Psi$$ becomes:

$$e^{Ct}e^{-iV_ot/\hbar}$$ Does this look correct?

 

[G01]

I'll try what you said monkey...

 

[Physics Monkey]

What I suggested is basically equivalent to what you did where your $$C = - i E t /\hbar$$ is the energy without $$V_0$$. The difference is that with my method you don't have to assume that $$\psi$$ is originally an energy eigenstate.

 

[G01]

I see I see... Thanks for your help. I went about your method PM, and I put the new psi into the SHRO. EQ. and after symplifying the SHRO EQ becomes the equation for the original Wavefunction but with the extra potential energy term added. This seems like what was supposed to happen, so thank you for your help.

 

[Physics Monkey]

Great, you're welcome.