Is $0 | \psi \rangle=0$? For example, the spin operator for spin 1 particle is
$$
\hat{S}_z\doteq\hbar\begin{pmatrix} 1&&\\&0&\\&&-1\end{pmatrix}
{\qquad}
\text{for state}
{\qquad}
\left|\psi\right> \doteq\begin{pmatrix}0\\1\\0\end{pmatrix}
\,.$$ 
Questions:


*

*We have $\hat{S}_z \left|\psi\right>=0 \left| \psi \right>$, but we won't write $0 \left|\psi \right>=0$, right?

*For the annihilation operator, we have $\hat{a} \left| 0 \right> = 0 \left|0\right> = 0$, according to this question.  So, what's the difference?
 A: (1) We could very well write $0|\psi\rangle=0$ but we must keep in mind that the first $0$ is a scalar, the one that belongs to the field over which the vector space (where $|\psi\rangle$ lives) is defined, while the second $0$ is the zero vector of that vector space. (We always use the same symbol to denote both, so we ought to be careful when writing these sorts of things.)
Now, why couldn't we use $|0\rangle$ for that zero vector? Short answer: because we already use that ket for other purposes, which brings us to your second question:
(2) The eigenvalue of the operator $\hat{a}^{\dagger}\hat{a}$ corresponding to the vector $|0\rangle$ is a scalar: $0$. This happens because when $\hat{a}$, an operator, acts on $|0\rangle$, a vector, the output is the $0$ vector.
Wait, what?
As you may have noticed (and this is most likely what got you confused in the first place) the $0$ and the $|0\rangle$ vectors are two different things. The first is the zero vector of the vector space—the one you study in your linear algebra courses, the one which is the identity element of the vector addition operation. The second, on the other hand, is completely different: the zero inside the ket is just a label of the corresponding eigenvalue (of the number operator). So it's just a notational ambiguity that somebody should have pointed out to you.
Keep these things in mind and you'll do fine.
A: I believe I found the answer. On the book Modern Quantum Mechanics by Sakurai chapter 1.2 there is a definition for null ket:

$c|\psi\rangle=|\psi\rangle c$, when $c$ is zero the resulting ket is a null ket

So 0 times any ket produce a null ket, in order not to mixed with vacuum state $|0\rangle$, I would like to denote it $0|\psi\rangle\doteq\vec{0}$, because null ket will be represented as a full zero vector in some representation.

For question (2), as is known to all, we have $\hat{a}|0\rangle=0|0\rangle\doteq\vec{0}$, but vacuum state is not a null ket, here is their difference:


*

*you can recreate particles from vacuum state:
$\hat{a}^\dagger|0\rangle=|1\rangle$;

*but you cannot create anything but null ket from a null state:
$\hat{a}^\dagger(0|0\rangle)=0|1\rangle\doteq\vec{0}$.
This is an interesting phenomenon, once you annihilate past the $|0\rangle$, you cannot recreate anything again...
A: They are different. 


*

*$\hat{S}_z|\psi\rangle= 0|\psi\rangle$ means a state with eigenvalue 0.

*$|0\rangle$ state is a state with no particles, called vacuum state.  If annihilation operator $\hat a$ act on vacuum, you will get nothing but 0 ("0" is not a state). 
