QM Oscillator: Find Eigenvalues & Eigenvectors of \hat{a}^2

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The discussion focuses on finding the eigenvalues and eigenvectors of the operator \hat{a}^2 in the energy state basis |n>. The operator is expressed as \hat{a}^2 = \sum_{n=0}^{\infty} \sqrt{n(n-1)} |n-2><n|. The challenge arises in determining the eigenvalues, with suggestions to use coherent states and the relationship between eigenvectors of \hat{a} and \hat{a}^2. It is noted that while eigenvectors of \hat{a} remain eigenvectors of \hat{a}^2, new eigenvectors can emerge from linear combinations of eigenvectors with opposite eigenvalues. The discussion concludes that with further analysis, these new eigenvectors can be identified.
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Homework Statement



"Write down the operator \hat{a}^2 in the basis of the energy states |n&gt;. Determine the eigenvalues and eigenvectors of the operator \hat{a}^2 working in the same basis.

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


Homework Equations





The Attempt at a Solution




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

\hat{a}^2 = \sum_{n=0}^{\infty} \sqrt{n(n-1)} |n-2&gt;&lt;n|

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: \hat{a} |n&gt; = \lambda |n&gt;, and I've tried a ket composed on the basis n: |\psi&gt; = \sum_{n=0}^{\infty} C_n |n&gt;.


I'd be grateful if somebody could show me the way with this question, I've just hit a brick wall with it.
 
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Well, \hat{a}^2 commutes with \hat{a} so any eigenvector of the latter is an eigenvector of the former. So to compute the eigenvalues, just operate \hat{a}^2 on a coherent state.
 
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 each other. With a little work you can show these are the only new eigenvectors.
 
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