Editing Wigner–Eckart theorem (section)

==Proof==

Starting with the definition of a [[spherical tensor operator]], we have

:<math>[J_\pm, T^{(k)}_q] = \hbar \sqrt{(k \mp q)(k \pm q + 1)}T_{q\pm 1}^{(k)},</math>

which we use to then calculate

:<math>
\begin{align}
    &\langle j \, m | [J_\pm, T^{(k)}_q] | j' \, m' 
\rangle = \hbar \sqrt{(k \mp q) (k \pm q + 1)} \, 
    \langle j \, m | T^{(k)}_{q \pm 1} | j' \, m' \rangle.
\end{align}
</math>

If we expand the commutator on the LHS by calculating the action of the {{math|''J''<sub>±</sub>}} on the bra and ket, then we get

:<math>
\begin{align} 
    \langle j \, m | [J_\pm, T^{(k)}_q] | j' \, m' 
\rangle ={} &\hbar\sqrt{(j \pm m) (j \mp m + 1)} \, \langle j \, (m \mp 1) | T^{(k)}_q | j' \, m' \rangle \\
   &-\hbar\sqrt{(j' \mp m')(j' \pm m' + 1)} \, \langle j \, m | T^{(k)}_q | j' \, (m' \pm 1) \rangle.
\end{align}
</math>

We may combine these two results to get

:<math>
\begin{align} 
    \sqrt{(j \pm m) (j \mp m + 1)} \langle j \, (m \mp 1) | T^{(k)}_q | j' \, m' 
\rangle = &\sqrt{(j' \mp m') (j' \pm m' + 1)} \, \langle j \, m | T^{(k)}_q | j' \, (m' \pm 1) \rangle \\
   &+\sqrt{(k \mp q) (k \pm q + 1)} \, \langle j \, m | T^{(k)}_{q \pm 1} | j' \, m' \rangle.
\end{align}
</math>

This recursion relation for the matrix elements closely resembles that of the [[Clebsch–Gordan coefficient]].  In fact, both are of the form {{math|Σ<sub>''c''</sub> ''a''<sub>''b'', ''c''</sub> ''x''<sub>''c''</sub> {{=}} 0}}. We therefore have two sets of linear homogeneous equations:

:<math>
\begin{align}
  \sum_c a_{b, c} x_c &= 0, &
  \sum_c a_{b, c} y_c &= 0.
\end{align}
</math>

one for the Clebsch–Gordan coefficients ({{math|''x<sub>c</sub>''}}) and one for the matrix elements ({{math|''y<sub>c</sub>''}}). It is not possible to exactly solve for {{math|''x<sub>c</sub>''}}. We can only say that the ratios are equal, that is

:<math>\frac{x_c}{x_d} = \frac{y_c}{y_d}</math>

or that {{math|''x<sub>c</sub>'' ∝ ''y<sub>c</sub>''}}, where the coefficient of proportionality is independent of the indices. Hence, by comparing recursion relations, we can identify the Clebsch–Gordan coefficient {{math|⟨''j''<sub>1</sub> ''m''<sub>1</sub> ''j''<sub>2</sub> (''m''<sub>2</sub> ± 1){{!}}''j m''⟩}} with the matrix element {{math|⟨''j''′ ''m''′{{!}}''T''<sup>(''k'')</sup><sub>''q'' ± 1</sub>{{!}}''j'' ''m''⟩}}, then we may write

:<math>
  \langle j' \, m' | T^{(k)}_{q \pm 1} | j \, m\rangle
  \propto \langle j \, m \, k \, (q \pm 1) | j' \, m' \rangle.
</math>