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Superposition of eigenstates in QM

  1. Oct 11, 2008 #1
    I am having trouble understanding the subtleties of this topic, which we've just covered in my QM course.

    I guess, if I understand correctly, that the superposition of eigenstates is not necessary an eigenstate itself, unless the states are degenerate. I'm not sure if I really understand why this is (unless this is one of those things that "just is" in quantum and I'm not supposed to get it).

    Yet it seems (also from my course discussions) that the superposition of solutions to the Schrodinger equation is also a solution, regardless to things being degenerate, or what not.

    And I am not sure if I understand how this relates to the commutator relations - if two operators commute, they share some eigenstate, right? How does it relate to the superposition and degeneracy?

    If anyone can elucidate this subject I'd be all sorts of grateful!
  2. jcsd
  3. Oct 11, 2008 #2


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    If you prepare a quantum system in such a way that you know it has some energy [tex] E_n [/tex], that means that it is in an energy eigenstate [tex] \psi_n [/tex]. In general, you don't know for certain what energy the system has; you know only the probabilities for various states. But you always know that its energy must be one of the eigenvalues of the energy operator. In this general case, we say that the system is in a superposition of the eigenstates, and it is always possible to write its wave function [tex] \psi [/tex] as a superposition of the eigenfunctions:

    [tex] \psi = \sum c_i \psi_{i} [/tex],

    where [tex] |c_i|^2 [/tex] is the probability that it has energy [tex] E_i [/tex].
  4. Oct 11, 2008 #3


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    (Incidentally, your first statement assumes that you are considering eigenstates of your operator corresponding to different eigenvalues)

    I suspect you're just overthinking it, and so you haven't looked for easy answers. Both of these happen to be fairly trivial arithmetic facts. The first one should be clear if you tried any example at all, except for the most utterly trivial ones. (I suggest working through it for the eigenvectors of 2x2 diagonal matrix with distinct diagonal entries) The second one is nothing more than the distributive rule.

    Or... maybe it's not really these arithmetic facts that's giving you trouble, but instead that this appears to conflict with your understanding of other things? Of course, if that's the case, we can't really help if you don't tell us what's bothering you about it.
  5. Oct 12, 2008 #4
    yea you shouldn't be concerned with whether the superposition of two eigenstates is an eigenstate. it's whether any state is a superposition of eigenstates that is important.
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