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Homework Help: QM Oscillator Question

  1. Jan 10, 2008 #1
    1. The problem statement, all variables and given/known data

    "Write down the operator [tex] \hat{a}^2 [/tex] in the basis of the energy states [tex] |n> [/tex]. Determine the eigenvalues and eigenvectors of the operator [tex] \hat{a}^2 [/tex] working in the same basis.

    You may use the relation: [tex] \sum_{k = 0}^{\infty} \frac{|x|^{2k}}{(2k)!} = cosh(|x|) [/tex]"

    2. Relevant equations

    3. The attempt at a solution

    For the first part, I've got the abstract version of the operator to be:

    [tex] \hat{a}^2 = \sum_{n=0}^{\infty} \sqrt{n(n-1)} |n-2><n| [/tex]

    but the second part is giving me some trouble. I'm not too sure how to set about it, I've tried a few different approaches but nothing ends up using the above relation. I've tried a coherent state: [tex] \hat{a} |n> = \lambda |n> [/tex], and I've tried a ket composed on the basis n: [tex] |\psi> = \sum_{n=0}^{\infty} C_n |n> [/tex].

    I'd be grateful if somebody could show me the way with this question, I've just hit a brick wall with it.
  2. jcsd
  3. Jan 14, 2008 #2
    Well, [itex]\hat{a}^2[/itex] commutes with [itex]\hat{a}[/itex] so any eigenvector of the latter is an eigenvector of the former. So to compute the eigenvalues, just operate [itex]\hat{a}^2[/itex] on a coherent state.
  4. Jan 14, 2008 #3


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    Homework Helper

    To continue along the lines of the previous post, assume |A> is an eigenvector of a^2 with eigenvalue A^2. Does this mean it is an eigenvector of a with eigenvalue A? Well, first of all, we see it could just as well have eigenvalue -A. And because of this, we could in general expect to find:

    |A> = |B> + |C>

    where a|B>=A|B> and a|C>=-A|C>, giving:

    a|A> = A |B> - A |C>

    so that |A> is not an eigenvector of a unless either |B> or |C> is zero, and yet:

    a^2 |A> = A^2 |B> + A^2 |C> = A^2 |A>

    In other words, while all the eigenvectors of a will still be eigenvectors of a^2, there will in general be some new ones as well, given by linear combinations of a-eigenvectors with eigenvalues that are negatives of eachother. With a little work you can show these are the only new eigenvectors.
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