Past exam of the quantum physics course of the University of Oxford / 2016 / 8

Hermiticity of the Cartesian angular momentum components givesThe angular momentum commutation relations $[L_i,L_j]=i\hslash {ϵ}_{ijk}{L}_k$ implyFor example, $[L_z,L_x+i{L}_y]=i\hslash L_y+\hslash L_x=\hslash L_{+}$. The two terms in $[L_x^2+L_y^2+L_z^2,L_i]$ cancel in pairs, giving the first commutator. Also $[L_{+},L_{-}]=2\hslash L_z$.

The spherical harmonics satisfyCommuting $L^2$ or $L_z$ past $L_{\pm }$ therefore givesFor a nonzero result these are the eigenvalue equations with unchanged $l$ and shifted magnetic quantum number. Taking the squared norm, using $L_{\mp }{L}_{\pm }=L^2-L_z^2\mp \hslash L_z$, fixes the conventional coefficient:At the appropriate endpoint the coefficient is zero. The zero vector is not itself an eigenstate.

There is no allowed spherical harmonic with $m=l+1$, so $L_{+}{Y}_l^l=0$. The equation $-i\hslash {\partial }_{\varphi }Y=\hslash lY$ first gives $Y=f(\theta )e^{il\varphi }$. Combining the differential expressions for $L_x,L_y$ yieldsIts annihilation of the highest-weight state becomesIntegrating $f^{\prime }/f=l\cot \theta$ gives $f=C{\sin }^l\theta$, with the constant solution included for $l=0$. HenceThe phase $e^{il\varphi }$ is single valued for integral orbital quantum number $l$; normalization fixes only $|C|$, leaving a conventional constant phase.

Direct integration over solid angle givesThus $|C|=\sqrt{(2l+1)!/(4\pi )}/(2^{l}l!)$. Choosing the usual alternating sign convention,Apply $L_{-}$ successively and divide by its ladder coefficient at each step. Explicitly,The normalization follows because the product of the squared ladder coefficients is ${\hslash }^{2(l-m)}(2l)!(l-m)!/(l+m)!$. This constructs every magnetic sublevel, including negative $m$.