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Hydrogen transition probability

  1. Jul 18, 2017 #1
    Hello! I have the following problem I'm trying to solve:

    1. The problem statement, all variables and given/known data
    An Hydrogen atom in the state |100> is found between the plates of a capacitor, where the electric field (weak and uniform) is: [itex]E(t) = \epsilon e^{-\alpha t / \tau}[/itex].

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

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

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

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

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

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

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

    For which I solved and took the limit [itex]T \rightarrow \infty[/itex] and got the result:

    [tex]\frac{\gamma^2 \epsilon^2}{\frac{9}{16} + \frac{\alpha^2}{\tau^2}}[/tex]

    Now, [itex]\gamma^2 = \frac{2^{15}}{3^{10}}[/itex] 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!
     
  2. jcsd
  3. Jul 18, 2017 #2
    Not that this is that much help but shouldn't the potential be proportional to [itex] r \cos \theta [/itex] not [itex] r[/itex]? With this you get zero for [itex]m=-1[/itex], but the same otherwise.
     
  4. Jul 18, 2017 #3
    Sorry, r would be a vector, for example r = xî + yĵ + zk, so you could have it in any of the three directions.
     
  5. Jul 18, 2017 #4
    Ah ok. It's just I would have thought you take the inner product of the potential, that is, the potential is your operator.
     
  6. Jul 18, 2017 #5
    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
     
  7. Jul 18, 2017 #6
    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.
     
  8. Jul 18, 2017 #7
    Well that's what I did, hence [itex] z = r \cos \theta [/itex], however to me that didn't give an obvious solution.
     
  9. Jul 18, 2017 #8
    Would the parameters [itex]\alpha = 3.67 \times 10^7[/itex], [itex]\epsilon = 1.93 \times 10^{11} \frac{V}{m}[/itex] and [itex]\tau = 1.109 \times 10^{-9} [/itex] 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 [itex]\tau[/itex] around the value of the half life for that transition, and calculating [itex]\alpha[/itex] with those two and the expression.
     
  10. Jul 18, 2017 #9
    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.
     
  11. Jul 18, 2017 #10
    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! :)
     
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