# Ladder operators and harmonic oscillator

1. Jun 11, 2009

### cleggy

1. Explain why any term (such as AA†A†A†)with unequal numbers of raising and lowering operators has zero expectation value in the ground state of a harmonic oscillator.

Explain why any term (such as AA†A†A) with a lowering operator on the extreme right has zero expectation value in the ground state of a harmonic oscillator.

2. Relevant equations

3. The attempt at a solution

Please help as I'm totally stuck with this. Does it have to do with orthonormality of different eigenfunctions?

2. Jun 11, 2009

### malawi_glenn

Yes it has, and also consider the ground state as the problem tells you to do.

what is the effect of A on |0> ? And what about it's dual? i.e. <0|A†

Now give it a try, star with your first operator AA†A†A† and tell us what you get and why

3. Jun 11, 2009

### cleggy

with AA†A†A† I get $$\psi$$$$_{}n+2$$. The first to terms convert $$\psi$$$$_{}n$$ into different eigenfunctions. Because $$\psi$$$$_{}n$$ is orthonormal to these eigenfunctions, these terms can be dropped from the integral.

Correct?

Also since the lowering operator is first to act on the ground-state harmonic iscillator, this takes it below the ground-state which cannot be allowed.

On the right track?

4. Jun 11, 2009

### malawi_glenn

I have a hard tome to understand your language..

with AA†A†A† acting on the state to the right (remember that this is sandwiched between $\psi^* _0$ and $\psi _0$, the two first will give you $\psi _2$ so what is left is

$\psi^* _0 AA^{\dagger} \psi_2$

do the rest of the operation:
$\psi^* _0 \psi_2$
thus integral gives zero

And yes, you can't go further down with the ground state, thus A|0> = 0