Ladder Operator for Harmonic Oscillator: a|0> = |0>

[KFC]

For harmonic oscillator, let |0> be the ground state, so which statement is correct?

a|0> =|0>

or

a|0> = 0 (number)

where a is the destroy operator

 

[Meir Achuz]

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

 

[ice109]

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> ?

 

[dx]

Neither. a|0\rangle = the zero ket.

EDIT: The zero ket and |0\rangle are different things.

 

[KFC]

Neither. a|0\rangle = the zero ket.

zero ket? Do you mean a|0\rangle = |0\rangle ?

 

[dx]

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

 

[KFC]

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

Oh, got it :)

 

[Pengwuino]

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

 

[Ben Niehoff]

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

<0|0> = 1

 

[KFC]

and what about \hat{N}|0\rangle, where \hat{N}=\hat{a}^\dagger\hat{a} is the number operator? Should it be

\hat{N}|0\rangle = 0|0\rangle = 0 (number) ?

 

[Pengwuino]

and what about \hat{N}|0\rangle, where \hat{N}=\hat{a}^\dagger\hat{a} is the number operator? Should it be

\hat{N}|0\rangle = 0|0\rangle = 0 (number) ?

Technically speaking, it is the null vector.