# QM Oscillator Question

1. Jan 10, 2008

### div curl F= 0

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

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

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

2. Relevant equations

3. The attempt at a solution

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

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

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

I'd be grateful if somebody could show me the way with this question, I've just hit a brick wall with it.

2. Jan 14, 2008

### µ³

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.

3. Jan 14, 2008

### StatusX

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.