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Time Dependent Perturbation Theory

  1. Dec 28, 2008 #1
    1. The problem statement, all variables and given/known data

    Question: If a particle is in the ground state at time t<0, use the 1st order time dependent perturbation theory to calculate the probability that the particle will still be in the ground state at time t.

    Suppose we turn on the perturbation at time t=0 H(x) = ax

    Suppose this particle is in an usual one dimensional infinite square well with width going from x=0 to x=a.

    Then the question asks what happens to the probability of finding the particle will still be in the ground state if t tends to infinity and justify the answer? This is the bit I don't understand because somehow I have to refer this to the sum of the probabilities = 1???

    2. Relevant equations

    3. The attempt at a solution

    I get the matrix element conjugate state Fi(0) * H(x) * state Fi(0) with respect to x
    and I get ax*1/2 for my matrix element.

    Then I integrate this matrix element with respect to time from 0 to t. Then I take the mod square of the whole expression.
    I get (ax*1/2*1/h bar * t)^2.

    Is this correct or not?

    And how do I answer the next part of the question?

    Thank you!
  2. jcsd
  3. Dec 28, 2008 #2
    Hello wam mi :smile:

    This problem is not actually a time dependent perturbation theory problem; it is, in fact, a time independent perturbation theory problem, because the potential doesn't change with time.

    If the particle is in the ground state for t < 0 and there is NO perturbation, what would the probability be? This is just elementary quantum mechanics; the answer is simpler than you think! :smile:

    Now, I think what you mean by the second part is that a potential, [tex]V(x)=\alpha x[/tex], is suddenly added (not adiabatically, but if you don't know what that means, never mind). In this case, the wavefunction of the particle stays the same, but the eigenstates (stationary states) change. How do we find these?
  4. Dec 28, 2008 #3

    Hello Domnu,

    First thank you very much for your reply!

    The probability of finding the particle in ground state just =1 for t<0,

    And then I move on and compute the probability of finding the particle at ground state
    Do I do this by computing first the matrix element and then integrate matrix element times exp(iwt) with respect to time going from t=0 to t=t? And since exp(iwt)=0 for w=0 if the particle remains at its ground stsate, then I got the answer of (ax*1/2*1/h bar * t)^2 after taking the mod square of the integral. Is this correct?

    Thank you very much for your help by the way.
  5. Dec 28, 2008 #4
    You're very welcome! Well, you got the first part, so nice job! :smile: Aaaah! You're trying too much for the second part.. you could try working it out a bit easier.. if you're familiar with bra-ket notation, you could try working something out where you have [tex]\psi[/tex], our initial ground state ket before any perturbation (the standard ground state for the infinite well), and try expanding that. In other words, we'd have something like

    [tex]|\psi\rangle = \sum_{k} |e_i\rangle \langle e_i| \psi \rangle e^{-i E_i t/\hbar}[/tex]​

    Now, this shows us how it evolves in time, right? Can you show where to go from here? :smile:

    Now, I noticed that you had "exp(iwt) = 0," which, remember, can never happen.. try to work out why this is so on your own.. it's a good exercise! Good luck with the rest of the problem, and if you have problems still, ask away! :smile:
  6. Dec 30, 2008 #5

    Hey Domnu,

    Thanks for your reply.

    (i) So I think the probability of finding the particle at ground state at t<0 is just 1.

    (ii) Hm, since E(final) = E(initial) for we want to find the probability of finding the particle at ground state at t>0, so exp i[E(final)-E(initial)]t/h) = 1. Is that right?
    Then we just work out the matrix element, and then integrate this matrix element with respect to time, with the limits 0 to t. May I ask if I am in the right direction?

    (iii) Then I take the mod square of everything is that correct?

    (iv) But then I will have the answer directly proportional to t^2.

    Thank you.

    Yours sincerely
  7. Dec 30, 2008 #6
    Well, the first part is correct. Now, the second part is correct, but when you say you have to work out the matrix element, could you say exactly what you mean? I'm pretty sure you're on the right track. I could give you a hint: remember that [tex]\langle \alpha | \beta \rangle[/tex] is the probability density of getting a measurement of state [tex]\alpha[/tex] when you're in state [tex]\beta[/tex]. If I understand you correctly, part III is correct, but I don't think the answer will be proportional to t^2.
  8. Dec 31, 2008 #7

    Hi Domnu,

    I think the matrix element would be <alpha l perturbation l beta>.
    What I don't understand is since the matrix element is going to be time independent, so when we integrate this matrix element from 0 to t, we get the coefficient to be matrix element*t. Then when we take the mod.square of this coefficient, we get the probability. So that's why I was wondering if the answer is just lmatrix element * tl^2 which then we 'll have t^2 for our answer.

    Thank you,

    Happy New Year's Eve!
  9. Dec 31, 2008 #8
    Happy New Year's Eve! :smile:

    Ah, okay. I think I see what you're doing now. To be honest, I haven't seen that formula before. Could you explain exactly what the formula is? I know that to find the first order energies, you would do [tex]\langle \psi | \text{perturbation} | \psi \rangle[/tex] where [tex]\psi[/tex] is the eigenket without the perturbation; now I see that this is similar to what you were doing, excepting that you have two different [tex]\psi[/tex]s. Now, I can tell you that this isn't correct, because your matrix element is going to have units of Energy (and if you mod square this, you get Energy^2 (!)... this is because each of the eigenkets *always* has dimensions 1/length ). And we can't have that :smile:

    Now, look at this idea :smile:

    We know that after the perturbation has been added, we're currently in the state [tex]\psi_0[/tex], which is the ground state of the infinite well, without any perturbations (since the change is immediate). Now, after the perturbation has been added, we have new ground states! It's now up to us to figure out what these ground state eigenkets are going to be.

    Here's where your part comes in :smile: : we can calculate the first order energy differences and add them to the infinite square well energies; we get that

    [tex]\langle \psi_0 | \alpha x | \psi_0 \rangle = \alpha[/tex]​

    after some algebra, which isn't too difficult. Also here, I'm using [tex]\alpha x[/tex] as the perturbation instead of [tex]a x[/tex] since the width of the well is also [tex] a [/tex]... that may cause confusion, so I'm sticking to alpha for now :smile: Anyways, so we have our first order ground energy to now just be [tex]E_0 + \alpha[/tex].

    Okay, so this doesn't look like we're going anywhere, right? But wait! We can use the WKB approximation to get an approximate wavefunction for the ground state (see if you can do this). Let this wavefunction which we got from the WKB approximation be [tex]\psi_1[/tex]. Now, we are currently in [tex]\psi_0[/tex], the wavestate of the infinite potential well, and we're trying to find the probability of us being in [tex]\psi_1[/tex].

    So here are two questions: i) What is the probability amplitude, and ii) What is the probability?

    If I don't get back to you until a bit later, Happy New Year! :smile:
  10. Jan 5, 2009 #9
    Happy New Year Domnu,

    Thank you for your reply! I hope you are well.

    Many thanks for your help. Btw I'll try to get back to you ASAP. I'm skiing in France at the moment so yeah things 're good here! Très Bien!
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