Wigner's 6-j symbols were introduced by Eugene Paul Wigner in 1940 and published in 1965. They are defined as a sum over products of four Wigner 3-j symbols,
{ j 1 j 2 j 3 j 4 j 5 j 6 } = ∑ m 1 , … , m 6 ( − 1 ) ∑ k = 1 6 ( j k − m k ) ( j 1 j 2 j 3 − m 1 − m 2 − m 3 ) × × ( j 1 j 5 j 6 m 1 − m 5 m 6 ) ( j 4 j 2 j 6 m 4 m 2 − m 6 ) ( j 4 j 5 j 3 − m 4 m 5 m 3 ) . {\displaystyle {\begin{aligned}{\begin{Bmatrix}j_{1}&j_{2}&j_{3}\\j_{4}&j_{5}&j_{6}\end{Bmatrix}}&=\sum _{m_{1},\dots ,m_{6}}(-1)^{\sum _{k=1}^{6}(j_{k}-m_{k})}{\begin{pmatrix}j_{1}&j_{2}&j_{3}\\-m_{1}&-m_{2}&-m_{3}\end{pmatrix}}\times \\&\times {\begin{pmatrix}j_{1}&j_{5}&j_{6}\\m_{1}&-m_{5}&m_{6}\end{pmatrix}}{\begin{pmatrix}j_{4}&j_{2}&j_{6}\\m_{4}&m_{2}&-m_{6}\end{pmatrix}}{\begin{pmatrix}j_{4}&j_{5}&j_{3}\\-m_{4}&m_{5}&m_{3}\end{pmatrix}}.\end{aligned}}}
The summation is over all six mi allowed by the selection rules of the 3-j symbols. They are closely related to the Racah W-coefficients, which are used for recoupling 3 angular momenta, although Wigner 6-j symbols have higher symmetry and therefore provide a more efficient means of storing the recoupling coefficients. Their relationship is given by:
{ j 1 j 2 j 3 j 4 j 5 j 6 } = ( − 1 ) j 1 + j 2 + j 4 + j 5 W ( j 1 j 2 j 5 j 4 ; j 3 j 6 ) . {\displaystyle {\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}).}
Symmetry relations The 6-j symbol is invariant under any permutation of the columns:
{ j 1 j 2 j 3 j 4 j 5 j 6 } = { j 2 j 1 j 3 j 5 j 4 j 6 } = { j 1 j 3 j 2 j 4 j 6 j 5 } = { j 3 j 2 j 1 j 6 j 5 j 4 } = ⋯ {\displaystyle {\begin{Bmatrix}j_{1}&j_{2}&j_{3}\\j_{4}&j_{5}&j_{6}\end{Bmatrix}}={\begin{Bmatrix}j_{2}&j_{1}&j_{3}\\j_{5}&j_{4}&j_{6}\end{Bmatrix}}={\begin{Bmatrix}j_{1}&j_{3}&j_{2}\\j_{4}&j_{6}&j_{5}\end{Bmatrix}}={\begin{Bmatrix}j_{3}&j_{2}&j_{1}\\j_{6}&j_{5}&j_{4}\end{Bmatrix}}=\cdots }
The 6-j symbol is also invariant if upper and lower arguments are interchanged in any two columns:
{ j 1 j 2 j 3 j 4 j 5 j 6 } = { j 4 j 5 j 3 j 1 j 2 j 6 } = { j 1 j 5 j 6 j 4 j 2 j 3 } = { j 4 j 2 j 6 j 1 j 5 j 3 } . {\displaystyle {\begin{Bmatrix}j_{1}&j_{2}&j_{3}\\j_{4}&j_{5}&j_{6}\end{Bmatrix}}={\begin{Bmatrix}j_{4}&j_{5}&j_{3}\\j_{1}&j_{2}&j_{6}\end{Bmatrix}}={\begin{Bmatrix}j_{1}&j_{5}&j_{6}\\j_{4}&j_{2}&j_{3}\end{Bmatrix}}={\begin{Bmatrix}j_{4}&j_{2}&j_{6}\\j_{1}&j_{5}&j_{3}\end{Bmatrix}}.}
These equations reflect the 24 symmetry operations of the automorphism group that leave the associated tetrahedral Yutsis graph with 6 edges invariant: mirror operations that exchange two vertices and a swap an adjacent pair of edges. The 6-j symbol
{ j 1 j 2 j 3 j 4 j 5 j 6 } {\displaystyle {\begin{Bmatrix}j_{1}&j_{2}&j_{3}\\j_{4}&j_{5}&j_{6}\end{Bmatrix}}}
is zero unless j1, j2, and j3 satisfy triangle conditions, i.e.,
j 1 = | j 2 − j 3 | , … , j 2 + j 3 {\displaystyle j_{1}=|j_{2}-j_{3}|,\ldots ,j_{2}+j_{3}}
In combination with the symmetry relation for interchanging upper and lower arguments this shows that triangle conditions must also be satisfied for the triads (j1, j5, j6), (j4, j2, j6), and (j4, j5, j3). Furthermore, the sum of the elements of each triad must be an integer. Therefore, the members of each triad are either all integers or contain one integer and two half-integers.
Special cases When j6 = 0 the expression for the 6-j symbol is:
{ j 1 j 2 j 3 j 4 j 5 0 } = δ j 2 , j 4 δ j 1 , j 5 ( 2 j 1 + 1 ) ( 2 j 2 + 1 ) ( − 1 ) j 1 + j 2 + j 3 { j 1 j 2 j 3 } . {\displaystyle {\begin{Bmatrix}j_{1}&j_{2}&j_{3}\\j_{4}&j_{5}&0\end{Bmatrix}}={\frac {\delta _{j_{2},j_{4}}\delta _{j_{1},j_{5}}}{\sqrt {(2j_{1}+1)(2j_{2}+1)}}}(-1)^{j_{1}+j_{2}+j_{3}}{\begin{Bmatrix}j_{1}&j_{2}&j_{3}\end{Bmatrix}}.}
The triangular delta {j1 j2 j3} is equal to 1 when the triad (j1, j2, j3) satisfies the triangle conditions, and zero otherwise. The symmetry relations can be used to find the expression when another j is equal to zero. Values for j6 = e = 0, 1/2, 1, 3/2 & 2 can be straightforwardly obtained from the Racah W-coefficients (Brink & Satchler 1994, Table 4, j6 = e = 0, 1/2, & 1; Biedenharn, Blatt, & Rose, 1952, j6 = e = 0, 1/2, 1, 3/2 & 2. Values for j6 = 1/2 and 1 are given below. The formulae for the recouplings for other values of j6 can be easily inferred from these by using the symmetry of the 6j-symbols and appropriate substitution.
{ j 1 j 2 j 3 j 2 + 1 2 j 1 + 1 2 1 2 } = ( − 1 ) j 1 + j 2 + j 3 + 1 [ ( j 1 + j 2 + j 3 + 2 ) ( j 1 + j 2 − j 3 + 1 ) ( 2 j 1 + 1 ) ( 2 j 1 + 2 ) ( 2 j 2 + 1 ) ( 2 j 2 + 2 ) ] 1 / 2 {\displaystyle {\begin{Bmatrix}j_{1}&j_{2}&j_{3}\\j_{2}+{\frac {1}{2}}&j_{1}+{\frac {1}{2}}&{\frac {1}{2}}\end{Bmatrix}}=(-1)^{j_{1}+j_{2}+j_{3}+1}\left[{\frac {(j_{1}+j_{2}+j_{3}+2)(j_{1}+j_{2}-j_{3}+1)}{(2j_{1}+1)(2j_{1}+2)(2j_{2}+1)(2j_{2}+2)}}\right]^{1/2}}
{ j 1 j 2 j 3 j 2 + 1 2 j 1 − 1 2 1 2 } = ( − 1 ) j 1 + j 2 + j 3 [ ( j 3 + j 1 − j 2 ) ( j 2 + j 3 − j 1 + 1 ) ( 2 j 1 ) ( 2 j 1 + 1 ) ( 2 j 2 + 1 ) ( 2 j 2 + 2 ) ] 1 / 2 {\displaystyle {\begin{Bmatrix}j_{1}&j_{2}&j_{3}\\j_{2}+{\frac {1}{2}}&j_{1}-{\frac {1}{2}}&{\frac {1}{2}}\end{Bmatrix}}=(-1)^{j_{1}+j_{2}+j_{3}}\left[{\frac {(j_{3}+j_{1}-j_{2})(j_{2}+j_{3}-j_{1}+1)}{(2j_{1})(2j_{1}+1)(2j_{2}+1)(2j_{2}+2)}}\right]^{1/2}}
{ j 1 j 2 j 3 j 2 − 1 j 1 − 1 1 } = ( − 1 ) j 1 + j 2 + j 3 [ ( j 1 + j 2 + j 3 ) ( j 1