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

in the context of $$a_- |0\rangle =0$$

-

Are you asking what the bra and ket formalism represent? – anna v Apr 13 '11 at 20:07

## 3 Answers

$|0\rangle$ 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:

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

-

$|0\rangle$ 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 '11 at 6:06

2
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. – Ted Bunn Apr 14 '11 at 13:16
wow! I see. thanks! – yayu Apr 14 '11 at 14:11

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.

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 15, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.952467143535614, "perplexity_flag": "head"}