---
tag: 6Z33W3D482
kind: definition
name: 6-j symbol
msc: [33C80, 81R05, "22E70"]
defines: D17XX5N9VF
uses: [YJRJVZ998D, E5P8AG6CK7, 0BF2DE2SYK]
sources:
  - url: https://en.wikipedia.org/wiki/6-j_symbol
---

## Definition

Let $V_j$ be the irreducible representation of [[SU(2)|E5P8AG6CK7]] of dimension $2j + 1$, and let $j_1, j_2, j_3, J$ be spins.
Coupling uses the Clebsch–Gordan coefficients of the Condon–Shortley convention, as in the definition of the [[3-j symbol|YJRJVZ998D]].
In the [[tensor product|0BF2DE2SYK]] $V_{j_1} \otimes V_{j_2} \otimes V_{j_3}$, let $\lvert (j_1 j_2) j_{12}, j_3; J M \rangle$ be the orthonormal vectors obtained by coupling $j_1$ and $j_2$ to $j_{12}$, then $j_{12}$ and $j_3$ to $J$.
Let $\lvert j_1, (j_2 j_3) j_{23}; J M \rangle$ be obtained by coupling $j_2$ and $j_3$ to $j_{23}$ and then $j_1$ and $j_{23}$ to $J$.
The *6-j symbol* is the number $\begin{Bmatrix} j_1 & j_2 & j_{12} \\ j_3 & J & j_{23} \end{Bmatrix}$ such that
$$\langle (j_1 j_2) j_{12}, j_3; J M \mid j_1, (j_2 j_3) j_{23}; J M \rangle = (-1)^{j_1 + j_2 + j_3 + J} \sqrt{(2 j_{12} + 1)(2 j_{23} + 1)}\, \begin{Bmatrix} j_1 & j_2 & j_{12} \\ j_3 & J & j_{23} \end{Bmatrix}.$$
The left side does not depend on $M$.
The symbol is set to $0$ when one of the couplings is impossible.

## Remarks

A symbol $\begin{Bmatrix} j_1 & j_2 & j_3 \\ j_4 & j_5 & j_6 \end{Bmatrix}$ can be nonzero only if each of the triads $(j_1, j_2, j_3)$, $(j_1, j_5, j_6)$, $(j_4, j_2, j_6)$, $(j_4, j_5, j_3)$ satisfies the triangle inequality and has an integer sum.
It is unchanged by any permutation of its columns, and by exchanging the upper and lower entries in any two columns.

Example: take $j_3 = 0$, so $J = j_{12} = c$ and $j_{23} = j_2 = b$, with $j_1 = a$.
Coupling with the trivial representation changes nothing, so both vectors equal $\lvert (a b) c; c M \rangle$ and the overlap is $1$.
Hence $\begin{Bmatrix} a & b & c \\ 0 & c & b \end{Bmatrix} = \frac{(-1)^{a + b + c}}{\sqrt{(2b+1)(2c+1)}}$.

Non-example: $\begin{Bmatrix} 1 & 1 & 3 \\ 1 & 1 & 1 \end{Bmatrix} = 0$, because the triad $(1, 1, 3)$ fails the triangle inequality.

The Racah W-coefficient is the same number with another sign and order: $\begin{Bmatrix} j_1 & j_2 & j_3 \\ j_4 & j_5 & j_6 \end{Bmatrix} = (-1)^{j_1 + j_2 + j_4 + j_5} W(j_1 j_2 j_5 j_4; j_3 j_6)$.