Past exam of the advanced quantum physics course of the University of Cambridge / 2017 / 1

Use the isolated nuclear spin Zeeman effect implied by the supplied nuclear magnetic moment, with hyperfine structure neglected. The proton gyromagnetic ratio isThe two spin levels are separated by $g_p{\mu }_N{B}_0$. Resonant nuclear magnetic resonance therefore requires the Larmor frequency

The duration is a Rabi pi pulse: in a frame rotating with the resonant transverse magnetic field, the spin must rotate through $\pi$. If $B_1$ denotes the magnitude of a circularly rotating field with the correct sense of rotation, the Rabi angular frequency is ${\Omega }_R={\gamma }_p{B}_1$. The shortest pulse is then

There is an amplitude convention to specify. If the applied transverse magnetic field is instead the linearly oscillating $B_1\cos (\omega t)\hat{\mathbf{x}}$, its resonant circular component has magnitude $B_1/2$. The rotating-wave approximation gives ${\Omega }_R={\gamma }_p{B}_1/2$ and henceThe wording does not specify this polarization convention, so the two pulse durations should be distinguished. Odd multiples of either shortest Rabi pi pulse also invert the spin within the corresponding ideal model.

The operators associated with the two angular momenta commute with one another, and each squared angular momentum operator commutes with every component of that same angular momentum. Two useful identities areConsequently the two maximal commuting subsets of the displayed collection are

The first set gives the uncoupled eigenbasis $|j_1{m}_1⟩\otimes |j_2{m}_2⟩$. Its $J_z$ eigenvalue is already fixed by $m=m_1+m_2$. The second gives the coupled eigenbasis $|j_1{j}_2;jm⟩$ for the addition of angular momentum. Its scalar-product eigenvalue is already fixed by $j,j_1,j_2$. Thus an independent complete set of commuting observables uses four operators in either case; literal maximality within the question's list includes the dependent fifth operator.

For generic nonzero angular momenta, no further listed operator can be added. In particular,is generally nonzero, and $[J_{1z},J^2]=2[J_{1z},{\mathbf{J}}_1\cdot {\mathbf{J}}_2]$. The same obstruction applies to $J_{2z}$. Their sum $J_z$ commutes because the two commutators cancel.

Write $C=e^2/(4\pi {ϵ}_0)$. The shell theorem makes the potential energy of the Electron constant inside the charged spherical shell. Relative to the point-proton Coulomb potential, the perturbation isIt is positive: the finite charge distribution weakens the attraction close to the origin.

The first-order energy correction is the expectation value of this perturbation in the unperturbed hydrogen atom ground state. Integrating its probability density over spherical volume givesFor a proton radius much smaller than the Bohr radius, expand the exponential uniformly over this short interval. The leading integral isSince $E_{1s}=-C/(2a_0)$, the signed fractional change isEquivalently, $\Delta E/|E_{1s}|=+4b^2/(3a_0^2)$ to leading order: the energy increases and the binding-energy magnitude decreases. This is the finite proton-shell correction to hydrogen ground energy.

If no small-radius expansion is made, the same time-independent perturbation theory integral gives, with $\beta =b/a_0$,Its expansion starts with $4{\beta }^2/3$, in agreement with the leading result. The integral is exact for the first-order calculation; it is not an exact solution of the perturbed Schrödinger equation.