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

[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!

 

[gabbagabbahey]

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

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)?

 

[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<br />

What am I missing?

 

[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<br />

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?

 

[gabbagabbahey]

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<br />

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?

Yup, exactly!

 

[dotman]

Thanks for the help gabba!