Past exam of the quantum theory course of the University of Oxford / 2023 / 3

The angular momentum commutation relations arewhere ${\epsilon }_{ijk}$ is the Levi-Civita symbol. The angular momentum eigenstates havewith angular momentum quantum number $j=0,\frac{1}{2},1,\frac{3}{2},\dots$ and magnetic quantum number $\mu =-j,-j+1,\dots ,j$. There are $2j+1$ such states in one irreducible spin representation, so the $J^2$ eigenspace has dimension $2j+1$ within that multiplet. The use of $\mu$ avoids confusing the magnetic quantum number with the particle mass in the earlier questions.

For completeness, the ladder operators $J_{\pm }=J_1\pm i{J}_2$ satisfy $[J_3,J_{\pm }]=\pm \hslash J_{\pm }$ andThe upper and lower ends are $\mu =\pm j$; integer steps between them require $2j$ to be a nonnegative integer. This explains both the spectrum and the multiplet dimension. Pure orbital angular momentum permits only integer $j$, whereas general spin allows half-integers. If a larger Hilbert space contains several copies of the same irreducible spin representation, the full $J^2$ eigenspace includes all those copies: its dimension is $(2j+1)$ times their multiplicity, and can be infinite. The paper's dimension question is understood per irreducible multiplet.

In the ordered orthonormal basis $(|\frac{1}{2},\frac{1}{2}⟩,|\frac{1}{2},-\frac{1}{2}⟩)$, the spin one-half angular momentum operators are $J_i=(\hslash /2){\sigma }_i$, with Pauli matricesThe ladder operators areExplicit multiplication gives $J_{+}{J}_{-}={\hslash }^2\mathrm{diag}(1,0)$ and $J_{-}{J}_{+}={\hslash }^2\mathrm{diag}(0,1)$. Their commutator is therefore

Choose the ordered angular momentum eigenstates $(|1,1⟩,|1,0⟩,|1,-1⟩)$, so $J_3=\hslash \mathrm{diag}(1,0,-1)$. Since $[J_3,J_{\pm }]=\pm \hslash J_{\pm }$, a nonzero matrix element of $J_{+}$ must increase the magnetic quantum number by one, and one of $J_{-}$ must decrease it by one. Thus only entries $(1,2),(2,3)$ of $J_{+}$ and $(2,1),(3,2)$ of $J_{-}$ can be nonzero. The adjoint operator identity $J_{-}=J_{+}^{†}$ givesFor the $j=1$ irreducible spin representation, both adjacent transitions are present; the following calculation determines their sizes.

The phases of the three basis vectors can be chosen so that $A$ and $B$ are nonnegative real numbers. Matrix multiplication then givesThe angular momentum commutation relations require this to equal $2\hslash J_3=2{\hslash }^2\mathrm{diag}(1,0,-1)$. Equating diagonal entries yieldsHenceThese are also the $j=1$ special case of the normalized spin ladder operator coefficients $\hslash \sqrt{j(j+1)-\mu (\mu \pm 1)}$.

Recover the angular momentum operators from $J_1=(J_{+}+J_{-})/2$ and $J_2=(J_{+}-J_{-})/(2i)$. The spin-one angular momentum matrices in the chosen descending-$J_3$ orthonormal basis areThey are self-adjoint operators, satisfy $[J_i,J_j]=i\hslash {\epsilon }_{ijk}{J}_k$, and give $J^2=2{\hslash }^{2}I$, as required for $j=1$.

The Hamiltonian operator $H=\omega J_3$ is already diagonal in the angular momentum eigenstate basis of part (b). Its three distinct energy eigenvalues and stationary states areThus the three column eigenvectors are $(1,0,0{)}^{\mathsf{T}}$, $(0,1,0{)}^{\mathsf{T}}$ and $(0,0,1{)}^{\mathsf{T}}$, with energies $+\hslash \omega$, zero and $-\hslash \omega$, respectively. Each is nondegenerate since $\omega >0$. Any nonzero scalar multiple gives the same stationary state after normalization; a quantum superposition of different energies is generally not stationary.

Diagonalizing the $J_1$ matrix gives the normalized eigenvectorswith eigenvalues $+\hslash ,0,-\hslash$, respectively. The first projective measurement prepares $v_{+}$ up to global phase, regardless of the earlier state, conditional on obtaining $+\hslash$.

The unitary time evolution under $H=\omega J_3$ gives, with $\theta =\omega t$,The probability amplitudes for the second projective measurement areThe Born rule therefore gives the general spin-one transverse measurement after axial precession:At the specified times $t_k=\pi k/\omega$, $\theta =\pi k$, soFor even $k$, the result is certainly $+\hslash$; for odd $k$, it is certainly $-\hslash$. Although zero belongs to the spectrum of the observable, it never occurs at these times. Indeed, for even $k$ the evolved vector is $v_{+}$, while for odd $k$ it is $-v_{-}$, where the minus sign is a global phase.