Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 3 / 33C (Principles of Quantum Mechanics)

Addition of angular momenta one and $j$ gives $\boxed{J=|j-1|,|j-1|+1,\dots ,j+1}$, with projection condition $\boxed{M=q+m}$. The usual bounds $q\in \{-1,0,1\}$, $-j\le m\le j$, $-J\le M\le J$ must also hold. For $j=0$ only $J=1$ occurs; for $j=1/2$ the values are $3/2,1/2$.

For $j=1/2$, abbreviate the product state $|1,q⟩|1/2,m⟩$ by $|q,\uparrow ⟩$ or $|q,\downarrow ⟩$. The highest-weight state is $|3/2,3/2⟩=|1,\uparrow ⟩$. Applying the total lowering operator, the sum of the two individual lowering operators, and dividing by the required ladder coefficient gives the whole quartet:The normalized vector orthogonal to the $M=1/2$ quartet state is annihilated by total raising, hence is the highest state of the remaining spin-$1/2$ representation. Lowering it givesThese six orthonormal states span the six-dimensional product space, and are therefore all the states. Their relative signs are consistent with the lowering convention; a common phase for either multiplet is immaterial. The coefficients are the corresponding Clebsch-Gordan coefficients.

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.