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QM simple harmonic oscillator

by ehrenfest
Tags: harmonic, oscillator, simple, solved
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Oct16-07, 07:25 PM
P: 1,996
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
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Oct16-07, 11:08 PM
P: 1,996
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?
Oct17-07, 12:07 AM
Sci Advisor
PF Gold
Gokul43201's Avatar
P: 11,155
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.

Oct17-07, 02:31 AM
P: 1,996
QM simple harmonic oscillator

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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