Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 3 / 33C / ii / Solution

The supplied commutators say that the three operators $X_q$ transform as an angular-momentum-one tensor. Thus the span of $X_q|n\ell m⟩$ transforms inside the product representation $1\otimes \ell$. The angular-momentum addition rule requires $\boxed{|\ell -1|\le {\ell }^{\prime }\le \ell +1}$; specifically $\ell ={\ell }^{\prime }=0$ is forbidden. Alsoso a nonzero amplitude requires $m^{\prime }=m+q$. Since ${\hat{x}}_3=X_0$, its rule is $\Delta m=0$. The components ${\hat{x}}_1,{\hat{x}}_2$ are linear combinations of $X_{+1},X_{-1}$, so each requires $\Delta m=\pm 1$. Projection bounds $|m|\le \ell$, $|m^{\prime }|\le {\ell }^{\prime }$ remain necessary.

Hydrogen orbital states have parity $(-1{)}^{\ell }$, while the position operator is odd. Inserting parity on both sides of a position matrix element multiplies it by $(-1{)}^{{\ell }^{\prime }+\ell +1}$; a nonzero element therefore requires this factor to equal one. Combining the resulting odd ${\ell }^{\prime }+\ell$ with the angular-momentum condition leaves the electric dipole selection ruleThese are necessary conditions, not guarantees of a nonzero radial integral. Angular momentum supplies no additional selection rule on $n,n^{\prime }$ apart from the allowed orbital labels.