quantum mechanics - 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:

1. 
   

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

   
   


2. 
   

   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?

   
   

Answer

(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.