# Homework Help: Perturbed Harmonic Oscillator

1. Nov 19, 2005

### Beer-monster

I've been given a question which asks to calculate the probability of finding an electron is an excited state for a Harmonic oscillator perturbed by an electric field pulse E(t) as t tends to infinity.

$$E= -exA\exp{(\frac{-t}{\tau})}$$

I knew I had to use the Time dependent perturbation theory and use the equations

$$c = \frac{1}{i\hbar} \int{V(nm)\exp(i \omega t).dt}$$
and $$P(t) = c^2$$

I was lucky enough to find the solution to a similar problem in a textbook: A.Levi's Applied Quantum Mechanics. However there are a few things about the presented solution I don't get (damn textbooks for skipping huge steps )

First they state that the only matrix element which contributes to the transition probability is the potential coupling the ground and first states i.e. V(10). Why is this?

In the calculation for this matrix element they go from
$$V(10) = -eAexp(\frac{-t}{\tau}) <m=1|x|n=0>$$
(Which I get and am fine with.)
To $$-eA exp(\frac{-t}{\tau}) (\frac{\hbar}{2m\omega})^1/2$$
Which I cannot replicate with the wavefunctions. Anyone have idea where this result comes from?
They then proceeded by calculating the Probability P using the squared of the integral times 1/hbar^2. Is there a reason they did not calculate the integral first and then square the result? And why when I try this to I not get a similar result?
I get $$P=\frac{e^2A^2\hbar}{2m\omega}(\tau^2+\frac{1}{\omega^2})$$
Which is nothing like there answer which preserves the exponent
Anyone familar with this problem can guess where I'm going wrong or can give me any tips?

Last edited: Nov 19, 2005
2. Nov 19, 2005

### Physics Monkey

Ok, first off you said you don't understand why $$\langle 1 | V | 0 \rangle$$ is the only important matrix element. Have you even thought about what the other matrix elements might be or tried calculating them? If you do this, you might find something interesting which easily explains the comment in the book. Remember your initial state is always $$| 0 \rangle$$.

Second, since everything in the potential is just a number except for the position operator, the matrix element reduces to
$$\langle 1 | V | 0 \rangle = -eAe^{-t/\tau} \langle 1 | x | 0 \rangle,$$
and you can calculate that position matrix element with ground and first excited state wavefunctions. It is a standard Gaussian integral. Perhaps you can tell me where exactly you're having trouble in the integral.

The answer they have given for the integral may be wrong. Assuming you start at $$t=0$$, you should have to do the following integral
$$\int^\infty_0 e^{-t/\tau + i \omega t} \,dt,$$
but this is a simple exponential integral and the result should be
$$-\left( -\frac{1}{\tau} + i \omega\right)^{-1},$$
which when squared gives
$$\frac{1}{\frac{1}{\tau^2} + \omega^2}.$$

Last edited: Nov 19, 2005
3. Nov 25, 2005

### Beer-monster

Right having found some time to come back to this problem. I tried the integral again and think I'm on the right track except when it comes to here.

$$2A_{1}A_{0} \alpha \int {x^2 e^{-\alpha^2 x^2}}.dx$$

Where $$A_{n} = [\frac{\alpha}{2^n n!\sqrt{\pi}}]^{1/2}$$ and $$\alpha = \sqrt{\frac{m \omega}{\hbar}}$$

Which I'm having some trouble integrating as a substitution for x^2 seems to leave extra x terms. By part doesn't seem to simplify anything, and any combination I've tried seems to fail. Any suggestions for how I would follow this.

I've started to notice my main problem with QM is that the calculations always lead to integrals much more complicated than any I had come across previously. However I'm hopeful that (with your help) if I can find out how the method to solve such integrals now, I should be familiar with them come exams time.

Physicsmonkey: I've not tried to calculate the other matrix elements. Mainly because looking at the higher order wavefunctions, and taking into account I can't do the above, i'm fairly certain they are beyond my(current) ability to calculate. What should I be looking for in the results?

4. Nov 25, 2005

### Physics Monkey

The integral you have run into is a standard Gaussian integral. You can look them up in table or calculate them yourself without too much trouble. Here is a website with details: http://www.chem.arizona.edu/~salzmanr/480a/480ants/integral/integral.html

Regarding the matrix elements $$\langle m | V | 0 \rangle$$, it turns out that they are all zero except when m=1. In other words, a system in the ground state can only transition to the first excited state under this perturbation (at least to first order in perturbation theory).

Do you know anything about the raising and lowering operators? They make such computations much easier.

5. Dec 8, 2008

### syang9

i feel really dumb asking this, but when i try to compute the other matrix elements, i get
$$\left\langle {n|V|m} \right\rangle = \delta _{mn} = \left\{ \begin{array}{c l} 1 & m = n \\ 0 & m \ne{n} \end{array} \right.$$

so i don't see how m = 1 would even contribute.. can someone set me straight?

Last edited: Dec 8, 2008