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Homework Help: Time-dependent Perturbation

  1. Sep 9, 2004 #1
    Given a system,
    [itex]H = H_0 + V[/itex]
    V is a small perturbation that does not depend on time.

    the system is in [itex]|E_0>[/itex] at time [itex]t_0[/itex]
    [itex]H_0 |E_n> = E_n |E_n> [/itex]
    [itex]H_0 |E_0> = E_0 |E_0> [/itex]

    Let [tex]|\Psi(t)>[/tex] be the solution of the system.
    Let [tex]|\Phi(t)>[/tex] be the solution of the system without perturbation.
    Let [tex]|u(t)> = |\Psi(t)> - |\Phi(t)>[/tex].

    Show that [tex]|<E_n|u(t)>|^2 = 4 |V_{n0}|^2 [{{\sin(t - t_0)(E_n - E_0)/\hbar}\over{E_n - E_0}}]^2[/tex]

    at lowest order
    No matter how many times I try, the answer I get is

    [tex]|<E_n|u(t)>|^2 = 2 |V_{n0}|^2 [{{1 - \cos(t - t_0)(E_n - E_0)/\hbar}\over{E_n - E_0}}]^2[/tex]

    Please help!!!
     
  2. jcsd
  3. Sep 9, 2004 #2
    I think you got the right answer, but u don't have the correct question, it should be show that ...sin(.../2), there is a divide by 2 missing in the sine function. I looked it up in my quantum book.
     
  4. Sep 9, 2004 #3
    Sorry,
    I typed wrongly,

    My answer was
    [tex]|<E_n|u(t)>|^2 = 2 |V_{n0}|^2 [{{1 - \cos(t - t_0)(E_n - E_0)/\hbar}\over{(E_n - E_0)^2}}][/tex]

    Ya, you are brilliantly right,

    Since,
    [tex]2\sin^2\theta/2 = 1 - \cos\theta[/tex]

    There should be a "divided by 2" inside the Sine
     
  5. Sep 10, 2004 #4
    How can i get certain solved papers in quantum mechanics?
    thanks
     
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