How to Determine Parameters for Hydrogen Transition Probability?

[svletana]

Hello! I have the following problem I'm trying to solve:

Homework Statement

An Hydrogen atom in the state |100> is found between the plates of a capacitor, where the electric field (weak and uniform) is: E(t) = \epsilon e^{-\alpha t / \tau}.

Calculate the parameters of the potential (\epsilon, \alpha, \tau) so that for a time t \gg \tau the transition probability to any of the n=2 states is equal to 0.1.

Homework Equations

The field is asumed to be in an arbitrary r = (x,y,z) direction, so that W = \epsilon e^{-\alpha t / \tau} r.

The formula for transition probability is (using atomic units):
P = \left| \int_0^{T \gg \tau} e^{i \omega t} <100 | r | 21m> \epsilon e^{-\alpha t / \tau}dt \right|^2

where \omega = \frac{E_{21m} - E_{100}}{\hbar} = \frac{-3}{4}.

For the <100|r|21m> elements we have the results, for each m:
<100|r|200> = 0
<100|r|210> = \frac{2^7 \sqrt{2} \hat{z}}{3^5}
<100|r|21\pm1> = \frac{2^7}{3^5}(\mp\hat{x} - i\hat{y})

The Attempt at a Solution

I solved the integral for an arbitrary m and l=1, calling the result of the <100|r|21m>=\gamma. Since it does not depend on t we can take it out of the integral along with all the constants. The integral is then:

P = \gamma^2 \epsilon^2 \left| \int_0^{T \gg \tau} e^{-\left( \frac{3i}{4} + \frac{\alpha}{\tau}\right)t} dt \right|^2

For which I solved and took the limit T \rightarrow \infty and got the result:

\frac{\gamma^2 \epsilon^2}{\frac{9}{16} + \frac{\alpha^2}{\tau^2}}

Now, \gamma^2 = \frac{2^{15}}{3^{10}} for any value of m. And that result must be equal to 0.1 according to the guidelines.

I don't see how I could possibly calculate 3 parameters from this equation, what am I missing? Thanks in advance!

 

[Dazed&Confused]

Not that this is that much help but shouldn't the potential be proportional to r \cos \theta not r? With this you get zero for m=-1, but the same otherwise.

 

[svletana]

Not that this is that much help but shouldn't the potential be proportional to r \cos \theta not r? With this you get zero for m=-1, but the same otherwise.

Sorry, r would be a vector, for example r = xî + yĵ + zk, so you could have it in any of the three directions.

 

[Dazed&Confused]

Ah ok. It's just I would have thought you take the inner product of the potential, that is, the potential is your operator.

 

[svletana]

I just talked to my teacher and he said that i should choose the field going in the z direction so the only transition possible is to the |210> state and then choose parameters that would fit the problem... Guess that settles it :P

 

[Dazed&Confused]

But I'm also not sure how you can get 3 parameters from one equation. It is strange that it says for any n=2 state though, when at least if l=1, m=0, you got 0 as the probability. Also perhaps that the electric field is weak needs to be employed.

 

[Dazed&Confused]

Well that's what I did, hence z = r \cos \theta, however to me that didn't give an obvious solution.

 

[svletana]

Would the parameters \alpha = 3.67 \times 10^7, \epsilon = 1.93 \times 10^{11} \frac{V}{m} and \tau = 1.109 \times 10^{-9} seconds work?

I had in mind that it would have to be an electric field strong enough to make the transition but not so big so hydrogen is ionized, a value of \tau around the value of the half life for that transition, and calculating \alpha with those two and the expression.

 

[Dazed&Confused]

That sounds reasonable. The wording of question doesn't imply you have some freedom in choosing your parameters, but since the number of equations is under determined I would say you are likely right.

 

[svletana]

Yeah I only really found out because I emailed my teacher asking how would I get 3 parameters from one equation, but that's exactly what he wrote...

Thanks for your help! :)