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!