Editing Wigner–Eckart theorem (section)

== Example ==

Consider the position expectation value {{math|⟨''n j m''{{!}}''x''{{!}}''n j m''⟩}}. This matrix element is the expectation value of a Cartesian operator in a spherically symmetric hydrogen-atom-eigenstate [[Basis (linear algebra)|basis]], which is a nontrivial problem. However, the Wigner–Eckart theorem simplifies the problem. (In fact, we could obtain the solution quickly using [[Parity (physics)|parity]], although a slightly longer route will be taken.)

We know that {{math|''x''}} is one component of {{math|'''r'''}}, which is a vector.  Since vectors are rank-1 spherical tensor operators, it follows that {{math|''x''}} must be some linear combination of a rank-1 spherical tensor {{math|''T''<sup>(1)</sup><sub>''q''</sub>}} with {{math|''q'' ∈ {−1, 0, 1}}}.  In fact, it can be shown that

:<math>x = \frac{T^{(1)}_{-1} - T^{(1)}_1}{\sqrt{2}},</math>

where we define the spherical tensors as<ref name="J. Sakurai 1994">J. J. Sakurai: "Modern quantum mechanics" (Massachusetts, 1994, Addison-Wesley).</ref>

:<math>T^{(1)}_{q} = \sqrt{\frac{4 \pi}{3}} r Y_1^q</math>

and {{math|''Y''<sub>''l''</sub><sup>''m''</sup>}} are [[spherical harmonic]]s, which themselves are also spherical tensors of rank {{math|''l''}}.  Additionally, {{math|''T''<sup>(1)</sup><sub>0</sub> {{=}} ''z''}}, and

:<math>T^{(1)}_{\pm 1} = \mp \frac{x \pm i y}{\sqrt{2}}.</math>

Therefore,

:<math>
\begin{align}
\langle n \, j \, m | x | n' \, j' \, m'
\rangle
  & = \left\langle n \, j \, m \left| \frac{T^{(1)}_{-1} - T^{(1)}_1}{\sqrt{2}} \right| n' \, j' \, m'
\right\rangle \\
  & = \frac{1}{\sqrt{2}} \langle n \, j \| T^{(1)} \| n' \, j'\rangle \,
    \big(\langle j' \, m' \, 1 \, (-1) | j \, m \rangle - \langle j' \, m' \, 1 \, 1 | j \, m \rangle\big).
\end{align}
</math>

The above expression gives us the matrix element for {{math|''x''}} in the {{math|{{!}}''n j m''⟩}} basis.  To find the expectation value, we set {{math|''n''′ {{=}} ''n''}}, {{math|''j''′ {{=}} ''j''}}, and {{math|''m''′ {{=}} ''m''}}.  The selection rule for {{math|''m''′}} and {{math|''m''}} is {{math|''m'' ± 1 {{=}} ''m''′}} for the {{math|''T''<sup>(1)</sup><sub>±1</sub>}} spherical tensors.  As we have {{math|''m''′ {{=}} ''m''}}, this makes the Clebsch–Gordan Coefficients zero, leading to the expectation value to be equal to zero.