Past exam of the advanced quantum physics course of the University of Cambridge / 2021 / Section A

The initial quantum state is a coherent state. Since $a|\alpha ⟩=\alpha |\alpha ⟩$ and the state is normalized, its mean occupation number is $N=⟨a^{†}a⟩=|\alpha {|}^2$. Each quantum harmonic oscillator energy eigenstate acquires the phase $e^{-i\omega (n+1/2)t}$, soThe prefactor is a global phase and does not change an expectation value. Inverting the definitions of the annihilation operator and creation operator gives $x=\sqrt{\hslash /(2m\omega )}(a+a^{†})$. If $\alpha =\sqrt{N}{e}^{i\phi }$, thenFor a classical harmonic oscillator with amplitude $A$, the conserved energy is $E=m{\omega }^2{A}^2/2$. Setting $E=\hslash \omega N$ gives exactly the same amplitude, $A=\sqrt{2\hslash N/(m\omega )}$. Thus these coherent-state oscillator expectation values follow the classical orbit.

The full quantum mean energy is nevertheless $⟨H⟩=\hslash \omega (N+1/2)$: the extra zero-point energy belongs to the nonzero position and momentum uncertainties required by the Heisenberg uncertainty relation, rather than to the motion of the packet centre. In particular, $N=0$ gives zero mean displacement but a finite ground-state width. A coherent state follows a classical mean trajectory while retaining finite position and momentum uncertainties.

Use the uncoupled angular momentum eigenstates $|m_L,\uparrow ⟩$ and $|m_L,\downarrow ⟩$, with $m_L=-1,0,1$. For the Hamiltonian operator actually written in the paper, these are the energy eigenstates, withThe initial $J=3/2$ states are not all energy eigenstates of this operator. To find their energy populations, first perform the Clebsch-Gordan decomposition. The extremal states are $|3/2,3/2⟩=|1,\uparrow ⟩$ and $|3/2,-3/2⟩=|-1,\downarrow ⟩$. Applying the total angular momentum lowering operator to the first givesSimilarly,The initial preparation is an equal incoherent mixture of the four coupled states. Each extremal state contributes population $1/4$ to its one energy eigenvalue, while each nonextremal state contributes $2/3$ and $1/3$ of its own $1/4$ population to the two displayed components. The resulting Zeeman splitting is thereforefor nonzero $B_z$. The relative populations are $\boxed{3:2:2:2:3}$ in this order; the two zero-energy states each have population $1/12$. These Zeeman populations from an unpolarized angular-momentum multiplet remain constant under subsequent evolution by the stated static Hamiltonian operator, although off-diagonal quantum coherence can evolve.

There is a distinct weak-field model in which a retained spin-orbit interaction keeps $J=3/2$ isolated and one projects the magnetic perturbation into that quartet. The Landé g-factor is then $4/3$, giving four shifts ${\mu }_B{B}_z(\pm 2,\pm 2/3)$ with equal populations. That calculation requires the additional projection or dominant spin-orbit interaction; it is not diagonalization of the full magnetic Hamiltonian operator specified here, which also mixes the $J=1/2$ sector.

A confined ground state requires $k>0$; take $\alpha >0$ so the trial wavefunction is normalizable. Its normalized form isThe Gaussian moment recurrence gives $⟨x^2⟩=1/(2\alpha )$ and $⟨x^4⟩=3/(4{\alpha }^2)$. Also, integration by parts gives $⟨p^2⟩={\hslash }^2\int |f_{\alpha }^{\prime }{|}^{2}dx={\hslash }^2\alpha /2$. The Rayleigh quotient for the quartic oscillator is consequentlyDifferentiation yields ${\mathcal{E}}^{\prime }(\alpha )={\hslash }^2/(4m)-3k/(2{\alpha }^3)$, so the unique minimum is at ${\alpha }_{\ast }=(6mk/{\hslash }^2{)}^{1/3}$. Since the quotient diverges at both endpoints $0$ and $\infty$, this is the global minimum in the trial family. The Rayleigh-Ritz variational principle givesFor the first excited energy eigenstate, choose the odd trial familyThe even confining potential has an even, nodeless ground state, by the nodeless theorem for a one-dimensional ground state. By the Sturm oscillation theorem, its first excited state is odd with one node. The proposed family has the same parity operator eigenvalue and one node, and is orthogonal to the exact ground state, not merely to its Gaussian trial approximation. Minimization in the odd linear subspace therefore bounds the first excited energy from above. For completeness, its Rayleigh quotient is $3{\hslash }^2\beta /(4m)+15k/(4{\beta }^2)$, minimized at ${\beta }^3=10mk/{\hslash }^2$, giving $E_1\le (9{\hslash }^2/8m)(10mk/{\hslash }^2{)}^{1/3}$. These are Gaussian variational bounds for a quartic oscillator.

If $k=0$, there is no normalizable free particle ground state on the whole line, though the spectral infimum is zero. If $k<0$, the Hamiltonian operator is not bounded below, so a ground-energy bound of the requested kind does not exist.