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Simple time-independent non-degenerate quantum perturbation

  1. Feb 12, 2013 #1
    I'm reading through this pdf (http://www.pa.msu.edu/~mmoore/TIPT.pdf) on simple quantum perturbation theory and I'm quite confused with equations 32 through 34.

    They have [tex]E_{n}^{(2)} = <n^{(0)}|V|n^{(1)}> = - \sum_{m \neq 0}{\frac{|V_{mn}|^{2}}{E_{mn}}}[/tex] but I would have done [tex]E_{n}^{(2)} = <n^{(0)}|V|n^{(1)}> - <n^{(0)}|E_{n}^{(1)}|n^{(1)}>[/tex] and then plugged in [itex]E_{n}^{(1)} = V_{nn}[/itex] from their earlier solution for first order terms. I don't know where I would have gone form there and I certainly couldn't even take a gander at how they end up with a summation either in this equation or in equations 33 and 34. Are there steps being omitted and/or can this be explained conceptually?

    I have similar complaints about equations 33 and 34, though in equation 34 I have the first right hand side they end up with, but then again I have no idea about the summation that suddenly appears in the final answer. What am I overlooking/not thinking about?
     
  2. jcsd
  3. Feb 12, 2013 #2
    Looking at Equation (24),
    [tex]
    \left\langle n^{(0)} \right.\left|n^{(1)}\right\rangle = -\frac{1}{2} \sum_{k=1}^{1-1}\left\langle n^{(1-k)}\right.\left| n^{(k)}\right\rangle = 0
    [/tex]
    so the [itex]\left\langle n^{(0)} \right.\left|n^{(1)}\right\rangle[/itex] term in what you wrote is zero.
     
  4. Feb 12, 2013 #3

    TSny

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    The pdf document has a small typo in that the summation index should have m ≠ n rather than m ≠ 0

    See if you can show the last term on the right is zero using equation (31).

    Note that any vector can be expanded in the basis set ##\{|m^{(0)}>\}##. So, in particular the vector ##|n^{(1)}>## can be expanded as ##|n^{(1)}> =\displaystyle \sum\limits_{m \neq n}{c_m|m^{(0)}>}##. Equation (31) allows the sum to be restricted to m≠n.

    Now use equation (30) to identify the constants ##c_m##. See what you get if you substitute this expansion of ##|n^{(1)}>## into ##E_{n}^{(2)} = <n^{(0)}|V|n^{(1)}>##
     
  5. Feb 12, 2013 #4

    vela

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    He skipped a couple of steps. Equations 30 and 31 are
    \begin{align*}
    \langle m^{(0)} | n^{(1)} \rangle &= -\frac{V_{mn}}{E_{mn}} \\
    \langle n^{(0)} | n^{(1)} \rangle &= 0
    \end{align*} If you expand ##\lvert n^{(1)} \rangle## in terms of the eigenstates of the unperturbed Hamiltonian, you get
    $$\lvert n^{(1)} \rangle = \sum_{m} \lvert m^{(0)} \rangle\langle m^{(0)} \lvert n^{(1)} \rangle.$$ Using equations 30 and 31, you end up with
    $$\lvert n^{(1)} \rangle = \sum_{m \ne n} -\frac{V_{mn}}{E_{mn}}\lvert m^{(0)} \rangle.$$ When you plug this into the first line of equation 32, you get the second line.
     
  6. Feb 13, 2013 #5
    Ok, that definitely cleared everything up. Thank you for everything, I especially wouldn't have guessed that they were expanding corrections to the eigenstates in terms of unperturbed eigenstates. Wow, you'd think that would have been a part of the derivations they would have spent more than nothing on...

    The nuances are beginning to make sense..
     
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