角運動量を合成することとテンソルを既約表現に分解すること

スピン1の波動関数をpsiで定義したとき，
$|x\rangle \equiv (|1,1\rangle+|1,-1\rangle)/\sqrt{2}$
$|y\rangle \equiv (|1,1\rangle-|1,1\rangle)/\sqrt{2}i$
$|z\rangle \equiv |1,0\rangle$
と定義すると， $\langle x|\psi\rangle, \langle y|\psi\rangle, \langle z|\psi\rangle$ は回転に対し，通常のベクトルと同じ変換則に従う．
これを $\langle x_i|\psi \rangle$ を略記する．

ベクトルA,Bの直積により作った2階の直行テンソルは $A_i B_j$ と書けるが，
これを先に議論した記法で書くと， $\langle x_i|\psi \rangle \langle x_j|\phi \rangle$ となる．

$\langle 1,1;2,2|\psi \rangle|\phi \rangle = \langle 1,1|\psi\rangle\langle 1,1 |\phi\rangle$
$\langle 1,1;2,1|\psi \rangle|\phi \rangle = (\langle 1,1|\psi\rangle\langle 1,0 |\phi\rangle + \langle 1,0|\psi\rangle\langle 1,1 |\phi\rangle)/\sqrt{2}$
$\langle 1,1;2,0|\psi \rangle|\phi \rangle = (\langle 1,1|\psi\rangle\langle 1,-1 |\phi\rangle + 2\langle 1,0|\psi\rangle\langle 1,0 |\phi\rangle + \langle 1,-1|\psi\rangle\langle 1,1 |\phi\rangle)/\sqrt{6}$
$\langle 1,1;2,-1|\psi \rangle|\phi \rangle = (\langle 1,-1|\psi\rangle\langle 1,0 |\phi\rangle + \langle 1,0|\psi\rangle\langle 1,-1 |\phi\rangle)/\sqrt{2}$
$\langle 1,1;2,-2|\psi \rangle|\phi \rangle = \langle 1,-1|\psi\rangle\langle 1,-1 |\phi\rangle$

$\langle 1,1;1,1|\psi \rangle|\phi \rangle = (\langle 1,1|\psi\rangle\langle 1,0 |\phi\rangle - \langle 1,0|\psi\rangle\langle 1,1 |\phi\rangle)/\sqrt{2}$
$\langle 1,1;1,1|\psi \rangle|\phi \rangle = (\langle 1,1|\psi\rangle\langle 1,-1 |\phi\rangle - \langle 1,-1|\psi\rangle\langle 1,1 |\phi\rangle)/\sqrt{2}$
$\langle 1,1;1,-1|\psi \rangle|\phi \rangle = -(\langle 1,-1|\psi\rangle\langle 1,0 |\phi\rangle - \langle 1,0|\psi\rangle\langle 1,-1 |\phi\rangle)/\sqrt{2}$

$\langle 1,1;2,0|\psi \rangle|\phi \rangle = (\langle 1,1|\psi\rangle\langle 1,-1 |\phi\rangle - 2\langle 1,0|\psi\rangle\langle 1,0 |\phi\rangle + \langle 1,-1|\psi\rangle\langle 1,1 |\phi\rangle)/\sqrt{6}$

この変換則は，一般の2階テンソルに応用できる．
$T_{2}^{(2)} = A_1 B_1$
$T_{1}^{(2)} = (A_1 B_0 + A_0 B_1)/\sqrt{2}$
$T_{0}^{(2)} = (A_1 B_{-1} + 2A_0 B_0 + A_{-1} B_1)/\sqrt{6}$
$T_{-1}^{(2)} = (A_{-1} B_0 + A_0 B_{-1})/\sqrt{2}$
$T_{-2}^{(2)} = A_{-1} B_{-1}$

$T_{1}^{(1)} = (A_1 B_0 - A_0 B_1)/\sqrt{2}$
$T_{0}^{(1)} = (A_1 B_{-1} - A_{-1} B_1)/\sqrt{2}$
$T_{-1}^{(1)} = -(A_{-1} B_0 - A_0 B_{-1})/\sqrt{2}$

$T_{0}^{(0)} = (A_1 B_{-1} - A_0 B_0 + A_{-1} B_1)/\sqrt{6}$

直交成分に直せば，
$T_{2}^{(2)} = (A_x B_x - A_y B_y + i (A_x B_y + A_y B_x))/2$
$T_{1}^{(2)} = (A_x B_z + A_z B_x + i(A_y B_z + A_z B_y) )/2$
$T_{0}^{(2)} = (A_x B_x + A_y B_y + 2A_z B_z)/\sqrt{6}$
$T_{-1}^{(2)} = (A_x B_z + A_z B_x - i(A_y B_z + A_z B_y) )/2$
$T_{-2}^{(2)} = (A_x B_x - A_y B_y - i (A_x B_y + A_y B_x))/2$

$T_{1}^{(1)} = (A_x B_z - A_z B_x + i (A_y B_z - A_z B_y) )/2$
$T_{1}^{(0)} = i(A_y B_x - B_y A_x)/2$
$T_{1}^{(-1)} = -(A_x B_z - A_z B_x - i (A_y B_z - A_z B_y) )/2$