The Wigner-Eckart theorem

From the commutation relations eq:tensor_eq it can be shown that the matrix element $\langle \gamma jm \vert T^{(k)}_q \vert \gamma' j'm' \rangle$ is proportional to a 3j-symbol

$$\langle \gamma jm \vert T^{(k)}_q \vert \gamma' j'm' \rangle ... ...ngle \gamma j\vert\vert{\bf T}^{(k)}\vert\vert\gamma'j'\rangle$$ (129)

This relation is known as the Wigner-Eckart theorem. The coefficient $\langle \gamma j\vert\vert{\bf T}^{(k)}\vert\vert\gamma'j'\rangle$is called the reduced matrix element of ${\bf T}^{(k)}$ and is independent of the magnetic quantum numbers m, m' and q.

From the properties of the 3j-symbol we obtain the following selection rules for the quantum numbers that have to be fulfilled for the matrix element $\langle \gamma jm \vert T^{(k)}_q \vert \gamma' j'm' \rangle$to be non-zero.

m=q+m' (130)

and

$$\vert j-j'\vert \leq q \leq j+j'$$ (131)

As an example we consider the matrix elements of the renormalized spherical harmonic ${\bf C}^{(k)}$From the Wigner-Eckart theorem we obtain

$$\langle lm \vert C^{(k)}_q \vert l'm' \rangle = (-1)^{l-m} \... ...y}\right) \langle l\vert\vert{\bf C}^{(k)}\vert\vert l'\rangle$$ (132)

Noting that

$$\langle lm \vert C^{(k)}_q \vert l'm' \rangle = {\sqrt \frac{4\pi... ...}\int_0^{\pi}\int_0^{2\pi}Y_{lm}^{*}Y_{kq}Y_{l'm'}\sin\theta d\theta d\varphi =$$

$$(-1)^{-m}[l,l']^{1/2} \left(\begin{array}{rrr} l & k & l' \\ 0 &... ...}\right) \left(\begin{array}{rrr} l & k & l' \\ -m & q & m' \end{array}\right)$$ (133)

it is seen that

$$\langle l\vert\vert{\bf C}^{(k)}\vert\vert l'\rangle = (-1)^{... ...begin{array}{rrr} l & k & l' \\ 0 & 0 & 0 \end{array}\right)$$ (134)

Another important example is the matrix elements of the angular momentum operator ${\bf j}$. For the j⁽¹⁾₀ = j_z component we have

$$\langle \gamma jm \vert j^{(1)}_0\vert\gamma'j'm' \rangle = ... ..._z\vert\gamma'j'm' \rangle = m\delta_{ \gamma jm ,\gamma'j'm'}$$ (135)

Thus,

$$\langle \gamma jm \vert j^{(1)}_0\vert\gamma'j'm' \rangle =$$

$$(-1)^{j-m} \left(\begin{array}{rrr} j & 1 & j' \\ -m & 0 & m' \e... ...ert{\bf j}^{(1)}\vert\vert\gamma'j' \rangle = m\delta_{ \gamma jm ,\gamma'j'm'}$$ (136)

and

$$\langle \gamma j \vert\vert{\bf j}^{(1)}\vert\vert\gamma'j' \rangle = \delta_{ \gamma j,\gamma'j'} [j(j+1)(2j+1)]^{1/2}$$ (137)