0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Thanks a lot for your answer. I understand that a|0> = 0. $\endgroup$ – ad1v7 Mar 9 '17 at 13:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.