- #1

- 296

- 0

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.

[tex] E= -exA\exp{(\frac{-t}{\tau})}[/tex]

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

[tex] c = \frac{1}{i\hbar} \int{V(nm)\exp(i \omega t).dt}[/tex]

and [tex]P(t) = c^2 [/tex]

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

[tex] V(10) = -eAexp(\frac{-t}{\tau}) <m=1|x|n=0>[/tex]

(Which I get and am fine with.)

To [tex]-eA exp(\frac{-t}{\tau}) (\frac{\hbar}{2m\omega})^1/2 [/tex]

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 [tex]P=\frac{e^2A^2\hbar}{2m\omega}(\tau^2+\frac{1}{\omega^2})[/tex]

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?

[tex] E= -exA\exp{(\frac{-t}{\tau})}[/tex]

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

[tex] c = \frac{1}{i\hbar} \int{V(nm)\exp(i \omega t).dt}[/tex]

and [tex]P(t) = c^2 [/tex]

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

[tex] V(10) = -eAexp(\frac{-t}{\tau}) <m=1|x|n=0>[/tex]

(Which I get and am fine with.)

To [tex]-eA exp(\frac{-t}{\tau}) (\frac{\hbar}{2m\omega})^1/2 [/tex]

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 [tex]P=\frac{e^2A^2\hbar}{2m\omega}(\tau^2+\frac{1}{\omega^2})[/tex]

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: