Past exam of the quantum theory course of the University of Oxford / 2017 / 3 / b

Put $f_l(\theta ,\varphi )={\sin }^l\theta e^{-il\varphi }$. For the regular, single-valued orbital angular momentum states, $l$ is a nonnegative integer. Differentiation immediately givesFor $0<\theta <\pi$, the derivatives needed for $L^2$ areConsequently the angular differential expression inside the parentheses of $-L^2/{\hslash }^2$ equalsusing ${\csc }^2\theta -{\cot }^2\theta =1$. ThereforeThe expressions extend regularly through the polar coordinate singularities: $f_l$ is the restriction of the polynomial $(x-iy{)}^l$ to the unit sphere. It is proportional to the lowest-projection spherical harmonic of degree $l$; for $l=0$ it is the constant state.

Starting from $f_{-1}=\sin \theta e^{-i\varphi }$, part (i) gives $L^2{f}_{-1}=2{\hslash }^2{f}_{-1}$ and $L_3{f}_{-1}=-\hslash f_{-1}$. Apply the angular momentum raising operator explicitly:A second application givesSince $L_{+}$ commutes with $L^2$, both raised functions retain the eigenvalue $2{\hslash }^2$. Their $L_3$ eigenvalues follow either by differentiating their $\varphi$ dependence or from $[L_3,L_{+}]=\hslash L_{+}$. Thus three valid unnormalized functions, in the requested order, areFor later Born rule calculations, normalize with the solid angle measure $d\Omega =\sin \theta d\theta d\varphi$. Their squared norms are $8\pi /3$, $4\pi /3$, $8\pi /3$, respectively. For example ${\int }_0^{\pi }{\sin }^3\theta d\theta =4/3$ and ${\int }_0^{\pi }{\cos }^2\theta \sin \theta d\theta =2/3$. A convenient orthonormal basis is thereforeOrthogonality follows from the distinct Fourier factors in $\varphi$. An overall phase of each basis function is arbitrary; these choices need not impose the usual phase convention for the spherical harmonics.

Recover $L_1$ from the given angular momentum raising operator and angular momentum lowering operator:This is the generator of rotations about the first coordinate axis. A natural candidate is the angular function pointing along that axis, $f_x=\sin \theta \cos \varphi$. Direct substitution givesAlso $f_x=(f_{+1}+f_{-1})/2$, so linearity and part (ii) give $L^2{f}_x=2{\hslash }^2{f}_x$. Thus an answer is $\sin \theta \cos \varphi$. Its squared norm on the sphere isso the normalized quantum state isThe supplied starting state gives an alternative derivation with a changed quantization axis. Put $g=\cos \theta +i\sin \theta \sin \varphi$ and $K_{+}=L_2+i{L}_3$. The angular momentum commutation relations give $[L_1,K_{+}]=\hslash K_{+}$ and $[L^2,K_{+}]=0$, so $K_{+}$ raises the $L_1$ projection from $-\hslash$ to zero. Direct differentiation giveswhich is nonzero and proportional to the same ${\psi }_x$. This also connects the construction directly to the given state.

Use the normalized $L_3$ eigenstates from part (ii). The quantum state found in part (iii) has the decompositionThere is no ${\chi }_0$ component. Equivalently, its inner product with ${\chi }_0$ contains ${\int }_0^{2\pi }\cos \varphi d\varphi =0$. The Born rule therefore givesFor completeness the only outcomes are $L_3=\pm \hslash$, each with probability $1/2$. Thus $⟨L_3⟩=0$ does not mean that zero is a possible result in this state: its quantum uncertainty in $L_3$ is $\hslash$, even though $L_1$ is exactly zero.