# Ladder operators for 2-D Isotropic Harmonic Oscillator

I am confuse how to work with raising and lowering operators for 2-D quantum harmonic oscillator. What I'm trying to calculate is:

$$\langle01|\hat{a}_1^\dagger\hat{a}_2|10\rangle$$

What I don't understand is, What happens when lowering operator hits $|0\rangle$? e.g.

$$\hat{a}_2|10\rangle$$

Can ladder operators act on bra? e.g.

$$\langle01|\hat{a}_1^\dagger$$

Also can I split it such as?

$$\langle01|\hat{a}_1^\dagger\hat{a}_2|10\rangle = \langle0|\hat{a}_1^\dagger|1\rangle\langle1|\hat{a}_2|0\rangle$$

Any helpful comment will be appreciated.

## 1 Answer

What happens when lowering operator hits $|0\rangle$?

There's no difference to the 1D case, do you know what $a|0\rangle$ is?`

Can ladder operators act on bra?

As with any operator, $$\langle \psi | X^\dagger = \left( X | \psi \rangle \right)^\dagger .$$

Also can I split it such as [...]

Yes you can, that is basically the definition.

• Thanks a lot for your answer. I understand that a|0> = 0. Mar 9, 2017 at 13:26