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Time dependant perturbation

  1. Mar 15, 2007 #1

    quasar987

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    Hi,

    I have a problem with the very nature of time dependant perturbation theory (TDPT). In TDPT, we consider a system of Hamiltonian H(t) = H_0 (for t<0), H(t)=H_0 + kW(t) (for t>0) [where k<<1], where H_0 is, for simplicity, discrete, non-degenerate and time-independent, and given that at t=0, the state of the system is |phi_i> (an eigenstate of H_0), we are interested in calculating P_if(t), the probability of finding the system in another eigenstate of H_0, |phi_f>, at time t.

    But this does not make sense because at t, the hamiltonian is no longer H_0, so it will, in general, not have |phi_f> as an eigenstates. But we know that the result of a measurement will project the wave-function into one of the eigenfunction. So as soon as |phi_f> is not an eigenstate of H(t), the probability of finding the system in |phi_f> will be 0.
     
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  3. Mar 16, 2007 #2

    Gza

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    "So as soon as |phi_f> is not an eigenstate of H(t), the probability of finding the system in |phi_f> will be 0"

    I'm not sure if i fully understand your question, but why does it worry you that |phi_f> will not in general be an eigenstate of H(t)? If i remember my last quarter of undergrad quantum correctly we are only worried about the probability of transition between |phi_i> and |phi_f> (both of which are eigenstates of H_0) under a time dependent pertubation that takes place for some delta t. As far as the results of measurements "forcing" the wave-function's projections onto one of the eigenstates, simply think about the TDP as having already measured the system before (hence our eigenstate |phi_i>) as well as after, (hence |phi_f>) and the TDP simply spits out the probablity of this transition within the abovementioned delta t.
     
  4. Mar 16, 2007 #3

    quasar987

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    Like you said, we have measured the system before the perturbation to be |phi_i>. Then the TDP is turned on and the wave function evolves according to the time dependant SE

    [tex]i\hbar\partial_t \Psi(t)=H(t)\Psi(t)[/tex]

    with initial condition Psi(0)=phi_i. Now say we measure the energy at a time t. The wave function Psi(t) will collapse into an eigenstate of H(t). It is most likely that phi_f will not be such an eigenstate for H(t), so let's suppose for simplicity that it isn't.

    Sure, mathematically, nothing stops us from expanding Psi(t) in a Fourier series in |phi_n> and we can even calculate the coefficient |<Psi(t)|phi_k>|², but it does not represent the probability of finding the system in state |phi_k>. According to the expansion postulate, this probability is 0, because phi_k is not an eigenstate of H(t).

    But the books say that the probability is |<Psi(t)|phi_k>|², so this is where I'm confused, and I'm asking "where in the above am I mistaken"?
     
    Last edited: Mar 16, 2007
  5. Mar 16, 2007 #4

    vanesch

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    You should actually consider the following:
    for t<0, we have H0, and for t>t1, we also have H0 as hamiltonian.
    We only consider the interaction "switched on" between t=0 and t=t1.
    We prepare the state at t<0 and we measure it after t=t1.

    The idea is that H0 is "good enough" as eigenstate generator, but not as time evolution generator. So we consider that, for short enough times, we can take the eigenstates of H0 as good approximations to the eigenstates of the full hamiltonian, but that the neglected parts do play a cumulative role in skewing this as time evolution.
     
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