About ladder operator

It is the number zero. If a were the identity operator, you would get |0> back.

 

you sure about that? i know that the books say zero but those operators aren't hermitian anyway hence unphysical so what does it matter what it does to |0> ?

 

Neither. [tex] a|0\rangle = [/tex] the zero ket.

EDIT: The zero ket and [tex] |0\rangle [/tex] are different things.

 

No. The zero ket is the zero vector in the state space. [tex] |0\rangle [/tex] is not the zero vector, it's just the vector with eigenvalue [itex] \hbar \omega ( 0 + \frac{1}{2}) [/itex].

 

Or better yet, a|0> = null vector.

 

Yes. The thing to remember is that all physical states must be normalizable. So

<0|0> = 1

 

Technically speaking, it is the null vector.