# Perturbed harmonic oscillator

1. Mar 7, 2012

### Demon117

1. The problem statement, all variables and given/known data
I showed earlier this semester that in the presence of a "constant force", $F_{o}$, i.e. $V=-Fx$, that the eigenvalues for the Harmonic oscillator are shifted by

$\frac{F^{2}}{2m\omega^{2}}$

from the "unperturbed" case. It was also discussed that $x\rightarrow x-\frac{F}{m\omega^{2}}$.

An oscillator is initially in its ground state (n=0). At t=0, a perturbation V is suddenly applied. What is the probability of finding the system in its (new) ground state for t>0, i.e. find $|a_{o}|^{2}$.

2. Relevant equations
For this $|a_{n}|^{2}=|\int \Phi^{*}_{n}(x)\Psi_{o}(x)dx|^{2}$ over all space.

3. The attempt at a solution

For t>0, the state of the system is $\Psi(x,t)=\sum a_{n}exp(-i(\frac{E_{n}}{\hbar})t)\Phi_{n}(x)$. Here $\Phi_{n}(x)$ is an eigenvector of H. And the coefficients $a_{n}$ are obtained by expanding $\Psi_{o}(x)$, the ground state of $H_{o}$, in terms of $\Phi_{n}(x)$.

I also know that the basis states $\Phi_{n}(x)$ as well as $\Psi_{o}(x)$ are Hermite polynomials.

With that in mind my assumption would simply be to integrate the following:

$|a_{o}|^{2}=|\int \Phi_{o}(x) \Psi_{o}(x) dx|^{2} =|\int 1*1 dx|^{2}$

If I integrate this over all space I end up with a probability that goes to infinity. . . .Maybe I am missing something as far as Hermite polynomials go. . . or maybe I have the wrong idea about this problem. Any suggestions would be helpful.

Last edited: Mar 7, 2012
2. Mar 10, 2012

### dracobook

Hi matumich, sorry I'm not sure if I can help but if you could so kindly explain to me where you got the relavant equation for |a_n|^2 and for psi(x,t) I would greatly appreciate it. Also, could you explain to me why the last line is equal to |integral 1*1 dx|^2?

3. Mar 12, 2012

### Demon117

.

Well I made several mistakes. The equation $|a_{n}|^{2}=|∫Φ^{∗}_{n}(x)Ψ_{o}(x)dx|^{2}$ was given in class. The coefficients $a_{n}$ are obtained by expanding $Ψ_{o}(x)$, the ground state of $H_{o}$, in terms of $Φ_{n}(x)$. From this it follows that the probability of finding the system in some state (t>0) is given by that integral.

This was just purely as mistake and since the eigenfunctions $Φ_{n}(x)$ correspond to the state once the perturbation is applied they can be expanded in terms of their basis elements, which in this case are the Hermite polynomials. Similarly, the unperturbed wave function $Ψ_{o}(x)$ can be expressed in terms of the Hermite polynomials. Therefore we have:

$Φ_{n}(x) = C exp(-\frac{1}{2}\alpha^{2}(x-\frac{F}{m \omega^{2})^{2})H_{n}(x)$

&

$Ψ_{o}(x) = Cexp(-\frac{1}{2}\alpha^{2} x^{2})H_{o}(x) = exp(-\frac{1}{2}\alpha^{2} x^{2})$

The probability would be along the lines of

$|a_{o}|^{2}=|∫C exp(-\frac{1}{2}\alpha^{2}(x-\frac{F}{m \omega^{2})^{2})exp(-\frac{1}{2}\alpha^{2} x^{2})dx|^{2}$

After normalizing both wave functions you can integrate this and find the appropriate probability of finding the system in the ground state for t>0.