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I Fermi's golden rule derivation

  1. Jul 19, 2017 #1
    I'm reading Modern Particle Physics by Mark Thomson and watching Susskind's lecture on QM. In Thompson's book, equation (2.41) the wavefunction is expressed in terms of complete set of states of the unperturbed Hamiltonian as
    [tex] \Psi(\textbf{x}, t) = \sum_{k} c_k(t)\phi_k(\textbf{x})e^{-iE_kt} [/tex]

    Susskind explains the same thing and the result is
    [tex] |\Psi(\textbf{x}, t)\rangle = \sum_{j} \alpha_j(0)e^{-iE_jt}|j\rangle [/tex]

    Is it correct to change
    [tex] \alpha_j(0) \rightarrow \alpha_j(t) [/tex]
    and use
    [tex] |\Psi(\textbf{x}, t)\rangle = \sum_{j} \alpha_j(t)e^{-iE_jt}|j\rangle [/tex]
    in the presence of an interaction Hamiltonian?

    After introducing time-dependent coefficients αj, Susskind's and Thompson's expressions should be "equal", right?

    Many thanks :)
  2. jcsd
  3. Jul 21, 2017 #2


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    I don't really get what you are trying to ask. The first and second equations are clearly inequivalent since the second only applies to time-independent potentials whereas the first is for more general potential. I don't know why you want to change from ##\alpha_j(0)## to ##\alpha_j(t)## if your potential is not a function of time.
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