# Creation operator

1. Feb 6, 2009

### KFC

For creation operator of hamonic oscillator, we have

$$a^\dagger |n> = \sqrt{n+1}|n+1>$$

if I consider the creation operator operate on the bar vector, should I also get the same thing? namely

$$<n|a^\dagger = \sqrt{n+1}<n+1|$$

2. Feb 6, 2009

### Avodyne

No. Take the hermitian conjugate of your first equation; what do you get?

3. Feb 6, 2009

### malawi_glenn

You have to consider the dual correspondence:

$$X|\alpha >$$ corresponds to $$< \alpha | X^{\dagger}$$

Also you can consider the number operator: $$N = a^{\dagger}a$$, with $$N|n> = n |n>$$

Sandwhich it between: $$<n|N|n>$$, try it!