1. Apr 27, 2009

### 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

2. Apr 27, 2009

### clem

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

3. Apr 29, 2009

### ice109

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> ?

4. Apr 29, 2009

### dx

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

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

Last edited: Apr 29, 2009
5. Apr 29, 2009

### KFC

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

6. Apr 29, 2009

### 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 $\hbar \omega ( 0 + \frac{1}{2})$.

7. Apr 29, 2009

### KFC

Oh, got it :)

8. Apr 29, 2009

### Pengwuino

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

9. Apr 29, 2009

### Ben Niehoff

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

<0|0> = 1

10. May 2, 2009

### 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) ?

11. May 2, 2009

### Pengwuino

Technically speaking, it is the null vector.