= Solution
For the isotropic three-dimensional <quantum harmonic oscillator>, $V(r)=m\omega^2r^2/2$, hence $r^{-1}dV/dr=m\omega^2$ is constant. No radial integral remains, so the <spin-orbit energy shift in a central potential> reduces to
$$
\Delta E_{l,j}=\frac{\hbar^2\omega^2}{4mc^2}
[j(j+1)-l(l+1)-3/4].
$$
For $l=0$, the only possibility is $j=1/2$, and the bracket vanishes. For $l=1$, coupling to <spin one-half> permits $j=1/2$ and $j=3/2$, whose brackets are $-2$ and $+1$. Thus
$$
\boxed{\begin{aligned}
\Delta E_{0,1/2}&=0,\\
\Delta E_{1,1/2}&=-\frac{\hbar^2\omega^2}{2mc^2},\\
\Delta E_{1,3/2}&=+\frac{\hbar^2\omega^2}{4mc^2}.
\end{aligned}}
$$
These <spin-orbit shifts of an isotropic harmonic oscillator> are a zero line, a downward parabola, and an upward parabola of half the downward magnitude. They are independent of the radial excitation number, and their magnetic multiplicities are respectively two, two and four. Their weighted mean within the $l=1$ manifold is zero: $2\Delta E_{1,1/2}+4\Delta E_{1,3/2}=0$.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2021/paper-1-spin-orbit-shifts.png]
{title=Harmonic-oscillator spin-orbit shifts grow quadratically with trap frequency}
{description=Original plot with an arbitrary reference frequency $\omega_*$ and energy unit $E_*=\hbar^2\omega_*^2/(4mc^2)$. Only the <spin-orbit interaction> shifts are shown, not the unperturbed oscillator energies. The perturbative nonrelativistic regime requires these shifts to remain small relative to the orbital spacings.}
Back to article page
