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

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.