Intro QM: Add a constant to the Potential, Effect on Wave Function?

1. Nov 2, 2009

dotman

Hi, I've got a problem with the following problem. This is 1.8 out of Griffiths QM text, and was previously covered on this forum for another user in https://www.physicsforums.com/showthread.php?t=152775", although that thread doesn't address my problem.

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, $$e^{-iV_ot/\hbar}$$.

My attempt at the solution was, as stated in the other thread, to sub in $$\psi(x,t) e^{-iV_0 t/\hbar}$$ and see if that solves the schrodinger eqn with the constant potential V0 added in:

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

$$\psi=\psi e^{-iV_{0}t/\hbar}$$

$$\Rightarrow\frac{\partial\psi}{\partial t}=\frac{\partial\psi}{\partial t}e^{-iV_{0}t/\hbar}-\frac{iV_{0}}{\hbar}\psi e^{-iV_{0}t/\hbar}$$

$$\frac{\partial\psi}{\partial x}=\frac{\partial\psi}{\partial x}e^{-iV_{0}t/\hbar}$$

$$\frac{\partial^{2}\psi}{\partial x^{2}}=\frac{\partial^{2}\psi}{\partial x^{2}}e^{-iV_{0}t/\hbar}$$

$$i\hbar\left[\frac{\partial\psi}{\partial t}e^{-iV_{0}t/\hbar}-\frac{iV_{0}}{\hbar}\psi e^{-iV_{0}t/\hbar}\right]=\frac{-\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}e^{-iV_{0}t/\hbar}+V\psi e^{-iV_{0}t/\hbar}$$

$$\Rightarrow i\hbar\left[-\frac{iV_{0}}{\hbar}\psi e^{-iV_{0}t/\hbar}\right]+i\hbar\frac{\partial\psi}{\partial t}e^{-iV_{0}t/\hbar}=\frac{-\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}e^{-iV_{0}t/\hbar}+V\psi e^{-iV_{0}t/\hbar}$$

$$\Rightarrow i\hbar\left[-\frac{iV_{0}}{\hbar}\psi\right]+i\hbar\frac{\partial\psi}{\partial t}=\frac{-\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}+V\psi$$

$$\Rightarrow-\frac{i^{2}V_{0}\hbar}{\hbar}\psi+i\hbar\frac{\partial\psi}{\partial t}=\frac{-\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}+V\psi$$

$$\Rightarrow V_{0}\psi+i\hbar\frac{\partial\psi}{\partial t}=\frac{-\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}+V\psi$$

$$\Rightarrow i\hbar\frac{\partial\psi}{\partial t}=\frac{-\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}+V\psi-V_{0}\psi$$

$$\Rightarrow i\hbar\frac{\partial\psi}{\partial t}=\frac{-\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}+\left(V-V_{0}\right)\psi$$

But we're supposed to have added the constant potential in, so it should look like:

$$i\hbar\frac{\partial\psi}{\partial t}=\frac{-\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}+\left(V+V_{0}\right)\psi$$

Have I made a mistake somewhere? Am I doing this wrong? Any help or advice would be appreciated.

Thanks!

Last edited by a moderator: Apr 24, 2017
2. Nov 2, 2009

gabbagabbahey

Ermm.. isn't this kind of like saying $2=2e^{-iV_{0}t/\hbar}$?...Use two different variables to differentiate the wave function with the added constant potential from the wave function without the added constant potential (I'd recommend using $\Psi(x,t)$ and $\psi(x,t)$ respectively)...

What is the Schroedinger equation for $\Psi(x,t)$? What is the Schroedinger equation for $\psi(x,t)$?

3. Nov 2, 2009

dotman

Ok, well, yeah, that's bad notation. But notational issues aside, though, I don't see how this changes my result.

"Normal" Schrodinger equation:

$$i\hbar\frac{d\psi}{dt} = \frac{-\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V\psi$$

Schrodinger equation with the added potential factor:

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

where $\Psi=\psi e^{-iV_0t/\hbar}$

And using the partial derivatives $\frac{\partial\Psi}{\partial t}$, etc, of this, the rest of my result is the same:

$$i\hbar\frac{d\Psi}{dt} = \frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2} + V(x)\Psi \Rightarrow i\hbar\frac{\partial\psi}{\partial t}=\frac{-\hbar^{2}}{2m}\frac{\partial^{2}\psi}{\partial x^{2}}+\left(V-V_{0}\right)\psi$$

What am I missing?

4. Nov 2, 2009

dotman

Oh wait, I see it now.

That second eqn should be:

$$i\hbar\frac{d\Psi}{dt} = \frac{-\hbar^2}{2m}\frac{d^2\Psi}{dx^2} + (V+V_0)\Psi$$

And so it reduces to the original equation, thus showing that it is, in fact a solution of the Schrodinger equation with the added potential.

Look right?

5. Nov 2, 2009

gabbagabbahey

Yup, exactly!

6. Nov 2, 2009

dotman

Thanks for the help gabba!