For example, the spin operator for spin 1 particle is

$$ \hat{S}_z\doteq\hbar\begin{pmatrix} 1&&\\&0&\\&&-1\end{pmatrix} {\qquad} \text{for state} {\qquad} \left|\psi\right> \doteq\begin{pmatrix}0\\1\\0\end{pmatrix} \,.$$


  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?

  • 3
    $\begingroup$ Yes, we could write $0|\psi\rangle=0$, where the second zero just means the zero vector in the vector space. This is actually also explained in the link you post. $\endgroup$
    – Martin
    Oct 13, 2014 at 11:58
  • $\begingroup$ If you like this question, you may also enjoy reading this Phys.SE post. $\endgroup$
    – Qmechanic
    Nov 7, 2014 at 13:41

3 Answers 3


(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.

  • $\begingroup$ I disagree with (2). The vacuum is the zero-particle sector of the Fock-Space and simply a none-zero complex number of length one. The annihilation operators map $a: \mathcal{F}^N\rightarrow \mathcal{F}^{N-1}$ Since there can't be negative particle numbers, one insists that $\hat{a}\vert 0\rangle\equiv 0 \in\mathbb{C}$. Furthermore, states of definite particle number are never eigenstates of $a$ or $a^\dagger$. The annih. operators do possess eigenstates called coherent states, but they are rather linear combinations of different particle numbers. $\endgroup$
    – Nephente
    Oct 13, 2014 at 13:02
  • 3
    $\begingroup$ Your statement that $\hat{a}|0\rangle=0\in \Bbb{C}$ is not quite correct in the sense that $\hat{a}$ is an operator $\hat{a}:\mathcal{F}\to \mathcal{F}$, so the image of $\hat{a}$ must be a vector of $\mathcal{F}$. Saying that the vacuum is a complex number of length one is an abuse of language of the representation of the Fock space as $\mathcal{F}=\bigoplus_{n=0}^\infty \mathcal{H}^{\otimes n}$, with $\mathcal{H}^{\otimes 0}=\Bbb{C}$. $\endgroup$ Oct 13, 2014 at 13:13

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:

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

  2. 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...


They are different.

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

  2. $|0\rangle$ 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).


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