# What is the difference between $|0\rangle$ and $0$?

What is the difference between $|0\rangle$ and $0$ in the context of $$a_- |0\rangle =0~?$$

$|0\rangle$ is just a quantum state that happens to be labeled by the number 0. It's conventional to use that label to denote the ground state (or vacuum state), the one with the lowest energy. But the label you put on a quantum state is actually kind of arbitrary. You could choose a different convention in which you label the ground state with, say, 5, and although it would confuse a lot of people, you could still do physics perfectly well with it. The point is, $|0\rangle$ is just a particular quantum state. The fact that it's labeled with a 0 doesn't have to mean that anything about it is actually zero.

In contrast, $0$ (not written as a ket) is actually zero. You could perhaps think of it as the quantum state of an object that doesn't exist (although I suspect that analogy will come back to bite me... just don't take it too literally). If you calculate any matrix element of some operator $A$ in the "state" $0$, you will get 0 as a result because you're basically multiplying by zero:

$$\langle\psi| A (a_-|0\rangle) = 0$$

for any state $\langle\psi|$. In contrast, you can do this for the ground state without necessarily getting zero:

$$\langle\psi| A |0\rangle = \text{can be anything}$$

• Maybe it is not very important, but it should be pointed out that the "actual zero" is still a vector from the Hilbert space and not the scalar zero from the field over which the Hilbert space is defined. So, the R.H.S. zero in $a_{-}|{0}\rangle=0$ is not the complex number $0$. Rather, it is precisely the null vector of the vector space that the Hilbert space is.
– ACat
Dec 17, 2018 at 19:46
• I think you made it clear that it is a vector and not a number when you mentioned calculating matrix elements of an operator in the ""state" $0$", but I just thought it might be explicitly pointed out.
– ACat
Dec 17, 2018 at 19:49

$|0\rangle$ is a particular nonzero vector in the Hilbert space associated with this system. That vector is nonzero -- in fact, it's usually normalized to have magnitude 1. The 0 on the right refers to the zero vector in the Hilbert space. So they're quite different. For one thing, $|0\rangle$ is a possible state for a particle to be in. 0 isn't (since only unit-magnitude vectors are possible states).

• @Tedd Bunn one question: can't we have a state $|0\rangle$ where the ket represents a column vector in a particular basis where all components are zero? for an analogy in 3-space.. take a point with finite coordinates and shift the origin to that point, and in this new basis the point is represented as a 0-component vector.
– yayu
Apr 14, 2011 at 6:06
• I think you're misunderstanding what a change of basis is. Shifting the origin of a vector space is not the same thing as a change of basis. A change of basis is an invertible linear transformation (i.e., multiplication by a nonsingular matrix for finite-dimensional spaces). One consequence of this: In any vector space, any nonzero vector is nonzero in all bases. Apr 14, 2011 at 13:16

You may consider 0 as an eigenvalue and write $a|0\rangle = 0|0\rangle$.

Any eigenvector $a|\alpha \rangle = \alpha |\alpha \rangle$ is of different "length" than the corresponding normalized vector $|\alpha \rangle$. In your particular case the vector $0|0\rangle$ is of zeroth length.

$$0$$ is the additive identity of the vector space, i.e. the element of the vector space which satisfies $$0 + v = v$$ for all $$v$$ in the vector space. No system can ever be in this state since it is patently unnormalizable.

$$|0\rangle$$ is the name of an energy eigenstate of some Hamiltonian operator $$H$$ with the lowest eigenvalue in its spectrum. For example, for the harmonic oscillator $$|0\rangle$$ corresponds to the gaussian function $$~e^{-x^2}$$ while "$$0$$" actually corresponds to the real number zero. For a two-state system "$$0$$" would correspond to the column vector $$\begin{bmatrix}0\\0\\ \end{bmatrix}$$.

• As the other answers point out $0$ is not the real number zero when appearing as the result of applying an operator to a state, rather it is also a vector. In position representation it is an $L^2$ function, $0 \colon \mathbb{R}^3 \to \mathbb{C}$, $0(\vec r) = 0 \in \mathbb{C}$. Feb 2 at 15:58