# Find the eigenstates of a basis in terms of those of another basis?

1. Feb 19, 2014

### thecommexokid

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

This isn't exactly a homework question, but I figured this would be the best subforum for this sort of thing. For the sake of a concrete example, let's just say my question is:

Express the position operator's eigenstates in terms of the number operator's eigenstates.

2. Relevant equations

The number operator is given by
$$\hat{N} = \hat{a}^\dagger\hat{a}.$$
It has eigenvalues and eigenstates
$$\hat{N}|n\rangle = n|n\rangle.$$
The position operator is given by
$$\hat{x} = \sqrt{\frac{\hbar}{2m\omega}}\Big(\hat{a}^\dagger+\hat{a}\Big) \equiv \gamma\Big(\hat{a}^\dagger+\hat{a}\Big).$$
The action of the position operator on one of the number eigenstates is
$$\hat{x}|n\rangle = \gamma\Big(\hat{a}^\dagger|n\rangle + \hat{a}|n\rangle\Big) =\gamma\Big(\sqrt{n+1}|n+1\rangle + \sqrt{n}|n-1\rangle\Big).$$

3. The attempt at a solution

We'd like to find eigenstates of $\hat x$, that is, states $|x\rangle$ satisfying
$$\hat x|x\rangle = x|x\rangle.$$
The number basis is complete, so whatever these states we're looking for might be, they are representable as a linear combination of the number states:
$$|x\rangle = \sum_n |n\rangle\langle n|x\rangle;$$
I just need to know the coefficients $C_n \equiv \langle n|x\rangle$.

If I look again at the action of $\hat x$,
$$\hat x|x\rangle = \sum_n \hat x|n\rangle\langle n|x\rangle \\ \qquad = \sum_n \gamma\Big(\sqrt{n+1}|n+1\rangle + \sqrt{n}|n-1\rangle\Big) \langle n|x\rangle.$$
Suppose I attack this equation from the left with a particular $\langle n'|$. Then I get
$$x\langle n'|x\rangle = \sum_n \gamma\Big(\sqrt{n+1}\langle n'|n+1\rangle + \sqrt{n}\langle n'|n-1\rangle\Big) \langle n|x\rangle \\ \qquad = \gamma\Big(\sqrt{n'}\langle n'-1|x\rangle + \sqrt{n'+1}\langle n'+1|x\rangle\Big),$$
which allows me to develop a recurrence relationship
$$C_{n+1} = \frac{x}{\gamma\sqrt{n+1}}C_n-\frac{\sqrt{n}}{\sqrt{n+1}}C_{n-1}.$$

But I don't understand what this says. What is that $x$ doing in there? What does that even mean? And can I use this recurrence relation to get a closed-form answer? (At the very least I'd need to calculate $C_0$ and $C_1$ explicitly, which I don't know how to do.)

Also, I think at some point the position-space wavefunctions ought to come into it:
$$\langle x'|n\rangle = \sqrt[4]{\frac{m\omega}{\pi\hbar}}\frac{1}{\sqrt{2^nn!}}H_n\Big(\sqrt{\frac{1}{\sqrt{2}\gamma} x}\Big)e^{-x^2/2\gamma^2}$$
and
$$\langle x'|x\rangle = \delta(x-x').$$

2. Feb 20, 2014

### Goddar

Hi.
$\hat{x}$=$\gamma$($\hat{a}$†+ $\hat{a}$)
Shows that you have to find common eigenstates of creation and annihilation operators. Are you familiar with quantum coherent states?
If not, you can find out by starting for example with:
|$\alpha$>=$\sum$cn|n>,
$\hat{a}$|$\alpha$>= $\alpha$|$\alpha$>
Get a recurrence relation for the coefficients and see if you obtain a familiar function at the end (hint: in terms of |0>). It might take a while but that's the right path...