Past exam of the advanced quantum physics course of the University of Cambridge / 2018 / Section A / 1

Use $|m_1,m_2⟩$ for a product of the two individual spin-one spin eigenstates, with magnetic quantum numbers $m_i=-1,0,1$. Total $S_z=0$ requires $m_1+m_2=0$, leaving preciselyHere these are product labels, not labels of the total angular momentum. Write their coefficients in this order as $\alpha ,\beta ,\gamma$. With physical spin ladder operators $S_{\pm }=S_{1\pm }+S_{2\pm }$, each nonzero individual step has amplitude $\sqrt{2}\hslash$, soBoth must vanish for a singlet state. The independent product states therefore require $\alpha =\gamma =-\beta$. Wavefunction normalization gives the spin-one singlet stateup to an overall quantum phase. Conversely, $S_z|\psi ⟩=S_{+}|\psi ⟩=0$ implies $S^2|\psi ⟩=0$ through $S^2=S_{-}{S}_{+}+S_z^2+\hslash S_z$, so these conditions really select total spin zero. Distinguishability imposes no additional exchange restriction. The ladder hint uses dimensionless angular momentum units; restoring its common factor $\hslash$ does not change the coefficient ratios.

For a bound energy eigenstate with a nondegenerate energy eigenvalue of an even one-dimensional potential, the wavefunction can be chosen with definite parity. A uniform electric field $\mathcal{E}$ gives the electric-dipole interaction $H^{\prime }=-q\mathcal{E}x$, which is odd under reflection. Hencesince $x|{\psi }_n(x){|}^2$ is odd. Second-order nondegenerate perturbation theory then gives the leading allowed Stark effect shift:Only opposite-parity intermediate states have nonzero position matrix elements. More generally, reflection maps $H(\mathcal{E})$ to $H(-\mathcal{E})$, so an analytic nondegenerate energy level is an even function of the field. This explains the absence of all odd orders and the usual quadratic Stark effect.

A level separation changes by the difference of the two quadratic coefficients. Symmetry alone does not guarantee a nonzero differential shift: in a displaced quantum harmonic oscillator every level moves by the same $-q^2{\mathcal{E}}^2/(2m{\omega }^2)$, leaving every spacing unchanged. Degenerate states would instead require degenerate perturbation theory and can split linearly; the nondegeneracy of one-dimensional bound states applies to ordinary isolated well levels.

Neglect hyperfine structure and use the electronic weak-field Zeeman splitting regime, where ${\mu }_{B}B$ is small compared with the spin-orbit interaction splitting. The sodium levels are $3s{}^2{S}_{1/2}$, $3p{}^2{P}_{1/2}$ and $3p{}^2{P}_{3/2}$. The Landé g-factor formula givesFor example, inserting $L=1,S=1/2,J=1/2$ gives $3/2+(3/4-2)/(2\cdot 3/4)=2/3$. Each sublevel with magnetic quantum number $m_J$ shifts by $g_J{m}_J{\mu }_{B}B$. Thus the lower level has shifts $\pm {\mu }_{B}B$, the upper $J=1/2$ level has shifts $\pm {\mu }_{B}B/3$, and the upper $J=3/2$ level has shifts $\pm 2{\mu }_{B}B$ and $\pm 2{\mu }_{B}B/3$.

The electric-dipole selection rule allows $\Delta J=0,\pm 1$ except $0\leftrightarrow 0$, and $m_u-m_l=0,\pm 1$. Both fine structure branches are allowed. The emitted photon energy isAll the allowed components of the sodium D-line Zeeman pattern are listed below; the last column is the photon shift in units of ${\mu }_{B}B$:These are four components of D1 and six of D2. Their strengths need not be equal; the table and arrows specify allowed transitions, not intensities.

The $3p{}^2{P}_{3/2}$ level lies above $3p{}^2{P}_{1/2}$, so D2 has the larger zero-field photon energy. ^[ref] In the stated weak-field regime its most positive shift is $5{\mu }_{B}B/3$. Since $\lambda =hc/E_{\gamma }$, the shortest wavelength is therefore produced byThe tempting $m_u=3/2\to m_l=-1/2$ pair has a larger difference of level shifts but is forbidden because $|m_u-m_l|=2$.