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

  1. Nov 2, 2009 #1
    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, [tex] e^{-iV_ot/\hbar}[/tex].

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

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

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

    [tex]\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}[/tex]

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

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

    [tex]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}[/tex]

    [tex]\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}[/tex]

    [tex]\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[/tex]

    [tex]\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[/tex]

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

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

    [tex]\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[/tex]

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

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

    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. jcsd
  3. Nov 2, 2009 #2

    gabbagabbahey

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    Ermm.. isn't this kind of like saying [itex]2=2e^{-iV_{0}t/\hbar}[/itex]?...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 [itex]\Psi(x,t)[/itex] and [itex]\psi(x,t)[/itex] respectively)...

    What is the Schroedinger equation for [itex]\Psi(x,t)[/itex]? What is the Schroedinger equation for [itex]\psi(x,t)[/itex]?
     
  4. Nov 2, 2009 #3
    Ok, well, yeah, that's bad notation. But notational issues aside, though, I don't see how this changes my result.

    "Normal" Schrodinger equation:

    [tex]i\hbar\frac{d\psi}{dt} = \frac{-\hbar^2}{2m}\frac{d^2\psi}{dx^2} + V\psi[/tex]

    Schrodinger equation with the added potential factor:

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

    where [itex]\Psi=\psi e^{-iV_0t/\hbar}[/itex]

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

    [tex]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
    [/tex]

    What am I missing?
     
  5. Nov 2, 2009 #4
    Oh wait, I see it now.

    That second eqn should be:

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

    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?
     
  6. Nov 2, 2009 #5

    gabbagabbahey

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    Yup, exactly!:smile:
     
  7. Nov 2, 2009 #6
    Thanks for the help gabba!
     
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