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Time-Dependent Perturbation Theory

  1. Dec 6, 2016 #1
    Note this isn't actually a homework problem, I am working through my textbook making sure I understand the derivation of certain equations and have become stuck on one part of a derivation.

    1. The problem statement, all variables and given/known data


    I am working through my text (Quantum Mechanics 2nd Edition by B.H Bransden & C.J Joachain) section 9.1 on Time-Dependant Perturbation Theory. Part way through their derivation they give the equation

    [itex]i \hbar \sum_k \dot{c}_k (t)\psi_k^{(0)}e^{-\frac{iE_k^{(0)}t}{\hbar}}=\lambda \sum_kH'(t)c_k(t)\psi_k^{(0)}e^{-\frac{iE_k^{(0)}t}{\hbar}}[/itex]

    where ##c_k## denotes the time-dependent coefficients, any factor with a ##(0)## superscript denotes a factor relating to the "unperturbed" state, ##\lambda## is just a constant (usually equal to one for physical systems) and ##H'## denotes the "perturbed" Hamiltonian.

    From this equation the text states:

    "Taking the scalar product with a function ##\psi_b^{(0)}## belonging to the set ##y_k^{(0)}## of unperturbed energy eigenfunctions, and using the fact that ##\langle \psi_b^{(0)} | \psi_a^{(0)} \rangle = \delta_{bk}##, we find that

    [itex]\dot{c}_b (t) = i \hbar^{-1} \lambda \sum_k H'_{bk}(t)e^{i \omega_{bk} t} c_k(t)[/itex]

    where ##H'_{bk}=\langle \psi_{b}^{(0)} | H' (t) | \psi_k^{(0)} \rangle## and ##\omega_{bk}=\frac{E_b^{(0)}-E_k^{(0)}}{\hbar}## is the Bohr angular frequency.

    2. Relevant equations

    Not sure if there are any not given really, it's mostly just rearranging everything

    3. The attempt at a solution

    So first I multiply both sides by ##e^{\frac{iE_k^{(0)}t}{\hbar}}## cancel it out on the L.H.S and bring all the big terms to the R.H.S giving

    [itex]i \hbar \sum_k \dot{c}_k (t)\psi_k^{(0)}=\lambda \sum_k c_k(t)\psi_k^{(0)}e^{\frac{iE_k^{(0)}t}{\hbar}}H'(t)e^{-\frac{iE_k^{(0)}t}{\hbar}}[/itex]

    I then change all the ##\psi_k^{(0)}## eigenfunctions to their ket representation

    [itex]i \hbar \sum_k \dot{c}_k (t)| \psi_k^{(0)} \rangle=\lambda \sum_k c_k(t)e^{\frac{iE_k^{(0)}t}{\hbar}}H'(t)e^{-\frac{iE_k^{(0)}t}{\hbar}} | \psi_k^{(0)}\rangle[/itex]

    I then multiply (i.e take the inner product) of the L.H.S and the R.H.S with ##\langle\psi_b^{(0)}|## giving

    [itex]i \hbar \sum_k \dot{c}_k (t) \langle\psi_b^{(0)}| \psi_k^{(0)} \rangle=\lambda \sum_k c_k(t)\langle\psi_b^{(0)}|e^{\frac{iE_k^{(0)}t}{\hbar}}H'(t)e^{-\frac{iE_k^{(0)}t}{\hbar}} | \psi_k^{(0)}\rangle[/itex]

    The portion of this equation that is giving me issues is how the ##k## subscript in ##\langle \psi_b^{(0)} | e^{\frac{iE_k^{(0)}t}{\hbar}}## changes to ##b## giving me ##\langle\psi_b^{(0)}|e^{\frac{iE_k^{(0)}t}{\hbar}}##. If I can get this I belive all the exponential terms are not operators so I can pull them out of the braket term to give me ##H'_{bk}=\langle \psi_{b}^{(0)} | H' (t) | \psi_k^{(0)} \rangle## and and then I can just combine the exponentials to give me the Bohr angular frequency term.

    I'm sure I am doing something small (or something big) wrong somewhere but I just can't figure it out. Any help would be greatly appreciated!

    Thanks!
     
  2. jcsd
  3. Dec 6, 2016 #2
    This is certainly one of the big wrong things - the exponential factor is part of the sum and is not a global factor, so you can't just multiply both sides by it.
     
  4. Dec 6, 2016 #3
    Thanks, I see that now. Guess I'll be taking a look at it tomorrow and see if I can think of another way.
     
  5. Dec 6, 2016 #4

    vela

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    Start with what the text says and take the scalar product. Also, remember that ##H'(t)## is an operator, so you can't move ##\psi_k^{(0)}## past it like you did on the RHS.
     
  6. Dec 6, 2016 #5
    Thanks for the reply! I managed to figure it out. I was getting confused thinking ##e^{-iE_k^{(0)}t/\hbar}## was also an operator. When I get home I'll write out what I did to make sure it's correct.

    Thanks again!
     
  7. Dec 6, 2016 #6
    So I believe the solution is to just change the eigenfunctions to their ket representation and multiply through by ##\langle \psi_b^{(0)} |##

    [itex]i \hbar \sum_k \dot{c}_k (t)\langle \psi_b^{(0)} | \psi_k^{(0)} \rangle e^{-\frac{iE_k^{(0)}t}{\hbar}}=\lambda \sum_kc_k(t)\langle \psi_b^{(0)} | H'(t)| \psi_k^{(0)} \rangle e^{-\frac{iE_k^{(0)}t}{\hbar}}[/itex]

    Which reduces to

    [itex]i \hbar \dot{c}_k (t) e^{-\frac{iE_b^{(0)}t}{\hbar}}=\lambda \sum_kc_k(t)H'_{ba} e^{-\frac{iE_k^{(0)}t}{\hbar}}[/itex]

    then move the exponential over to the R.H.S and combine the terms for the Bohrn angular frequency to give what we want

    [itex] \dot{c}_k (t) =(i\hbar)^{-1}\lambda \sum_kH'_{bk} e^{-i\omega_{ba}t}c_k(t)[/itex]
     
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