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QM simple harmonic oscillator

  1. Oct 16, 2007 #1
    [SOLVED] QM simple harmonic oscillator

    1. The problem statement, all variables and given/known data
    If I have a particle in an SHO potential and an electric field, I can represent its potential as:

    [tex] V(x) = 0.5 * m \omega^2 (x - \frac{qE}{mw^2})^2 - \frac{1}{2m}(\frac{qE}{\omega})^2 [/tex]

    I know the solutions to the TISE:

    [tex] -\hbar^2 /2m \frac{d^2 \psi}{ dx^2} + 0.5 m\omege^2 x^2\psi(x) = E\psi(x) [/tex] (*)

    (Those are different Es)


    So, I plug V(x) into the TISE and get:

    [tex] -\hbar^2 /2m \frac{d^2 \psi}{ dx^2} + (0.5 * m \omega^2 (x - \frac{qE}{mw^2})^2 - \frac{1}{2m}(\frac{qE}{\omega})^2) \psi(x) = E\psi(x) [/tex]


    Now, since we only shift and translated the potential, I should be able to find a substitution for x that yields the equation (*) in a new variable y = f(x), right?



    The problem is, after I move the constant term to the RHS, I cannot find the right substitution. What am I doing wrong?

    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Oct 16, 2007
  2. jcsd
  3. Oct 16, 2007 #2
    I think that I can even prove that there is no constant that you can add to x to find a suitable substitution. Something must be wrong here?
     
  4. Oct 17, 2007 #3

    Gokul43201

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    Have you tried the obvious substitution: [itex]\xi = x-qE/m\omega ^2 ~[/itex] ?

    You can ignore the additive constant and refer all energies relative to that value.
     
  5. Oct 17, 2007 #4
    Yes, I figured it out. The problem was that I was under the false impression that I had to substitute epsilon for every x in that equation, which does not work.

    I realized, however, that you can substitute for psi(x) separately since it is a factor on both sides of the equation.
     
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