Creation Operator of Harmonic Oscillator

[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

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

 

[Avodyne]

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

 

[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!