Past exam of the physics A course of the University of Cambridge / 2022 / Paper 1 / Section B

Use the canonical commutation relations $[\hat{x},{\hat{p}}_x]=[\hat{y},{\hat{p}}_y]=i\hslash$, with cross-coordinate commutators zero. For either coordinate,Thus $2\alpha \beta \hslash =1$. To obtain the required Hamiltonian, the ratio of its position operator and momentum operator coefficients must be preserved:where the positive root follows from the requested signs. ConsequentlyThe annihilation operators are dimensionless; $\alpha$ has units of inverse length and $\beta$ has units of inverse momentum. Operators belonging to different coordinates commute automatically. Writing the number operators as ${\hat{N}}_j={\hat{a}}_j^{†}{\hat{a}}_j$, substitution givesThe constant is the sum of the two zero-point energies of this two-dimensional isotropic harmonic oscillator.

For either number operator $\hat{N}={\hat{a}}^{†}\hat{a}$, positivity follows fromHence an eigenvalue $\nu$ cannot be negative. The commutatorsshow that $\hat{a}|\nu ⟩$, if nonzero, is an eigenstate with eigenvalue $\nu -1$. For a normalized eigenstate, $\Vert \hat{a}|\nu ⟩{\Vert }^2=\nu$.

If $\nu$ were not an integer, successive nonzero lowerings would eventually reach an eigenstate with $0<{\nu }^{\prime }<1$. Its next lowering would have nonzero norm but a negative eigenvalue, contradicting positivity. The ladder must therefore terminate at $\nu =0$, with $\hat{a}|0⟩=0$. Conversely, successive applications of the creation operator never vanish, becauseThe normalizable bottom state exists explicitly: solving $\hat{a}{\psi }_0=0$ gives ${\psi }_0(x)\propto \exp [-m\omega x^2/(2\hslash )]$. Thus the number operator has exactly the non-negative integer eigenvalues.

Since the two coordinate number operators commute, use the product eigenstates $|n_x,n_y⟩$. Their energy eigenvalues areAt fixed $N=n_x+n_y$, the allowed pairs are $(0,N),(1,N-1),\dots ,(N,0)$. Thus the energy level $E_N=(N+1)\hslash \omega$ has quantum degeneracy $N+1$, and the ground-state energy is $\hslash \omega$.

The number operator commutators established above imply that ${\hat{a}}_x^{†}|n⟩$ is proportional to $|n+1⟩$ and ${\hat{a}}_x|n⟩$ to $|n-1⟩$. Their norms determine the magnitudes of the proportionality constants:Choose the phase of the normalized ground eigenstate once, then define recursivelyThis removes the otherwise arbitrary relative phases of the eigenstates. The Hermitian adjoint relation then fixes the lowering coefficient with the same positive sign, givingAt $n=0$, the second expression means the zero vector, ${\hat{a}}_x|0⟩=0$; it does not introduce a physical state $|-1⟩$.

Set $\hat{B}={\hat{a}}_x^{†}{\hat{a}}_y-{\hat{a}}_y^{†}{\hat{a}}_x$. Taking the Hermitian adjoint reverses the order of products, so ${\hat{B}}^{†}=-\hat{B}$. Complex conjugation also reverses the sign of $-i$, and thereforeThus $\hat{Q}$ is a self-adjoint operator on the usual oscillator states.

Using the coefficients in part (a), products containing two position operators or two momentum operators cancel in the difference. Cross-coordinate operators commute, leavingHenceIt is the component of orbital angular momentum perpendicular to the plane, with units $\mathrm{kg}{\mathrm{m}}^2{\mathrm{s}}^{-1}$ (equivalently $\mathrm{J}\mathrm{s}$). The rotational symmetry of the two-dimensional isotropic harmonic oscillator makes this a conserved observable; directly, each term in $\hat{B}$ preserves $N_x+N_y$, so $[\hat{H},\hat{Q}]=0$.

The specified energy requires $n_x+n_y=2$, so the relevant energy eigenspace has the orthonormal basisApplying the creation and annihilation operators givesThus its matrix in this ordered basis isWriting $q=\hslash \lambda$, the characteristic polynomial gives $\lambda ({\lambda }^2-4)=0$. The possible angular momentum eigenvalues are thereforeFor completeness, normalized simultaneous energy and angular momentum eigenstates areEach value occurs once in this energy eigenspace. Equivalently, circular ladder operators of a two-dimensional oscillator give $\hat{H}=\hslash \omega ({\hat{N}}_{+}+{\hat{N}}_{-}+1)$ and ${\hat{L}}_z=\hslash ({\hat{N}}_{+}-{\hat{N}}_{-})$. With $N_{+}+N_{-}=2$, the difference can only be $2,0,-2$.

For a particle in a ring, the arc-length coordinate is $s=R\varphi$. The tangential kinetic energy isThe constraint removes radial motion; the constant potential energy along the ring can be chosen as zero. The tangential momentum operator is ${\hat{p}}_s=-i\hslash {\partial }_s$, or equivalently the angular momentum operator is ${\hat{L}}_z=-i\hslash {\partial }_{\varphi }$. ThusIts physical domain has periodic boundary conditions, $\psi (\varphi +2\pi )=\psi (\varphi )$ and ${\psi }^{\prime }(\varphi +2\pi )={\psi }^{\prime }(\varphi )$. These make the Hamiltonian self-adjoint and reflect that $\varphi$ and $\varphi +2\pi$ describe the same point. No magnetic flux or additional angle-dependent potential energy is present.

Solving the Time-independent Schrödinger equation with $\psi \propto e^{ik\varphi }$ gives $E={\hslash }^2{k}^2/(2m{R}^2)$. The periodic boundary condition requires $e^{2\pi ik}=1$, hence $k=n\in \mathbb{Z}$. With the angular integration measure $d\varphi$, wavefunction normalization givesso a convenient orthonormal eigenbasis isThe $n=0$ ground state is nondegenerate. For each $|n|>0$, $u_n$ and $u_{-n}$ span a two-dimensional energy eigenspace; real sine and cosine combinations are an equally valid basis. These opposite-circulation eigenstates have angular momentum $n\hslash$ and $-n\hslash$.

If instead the wavefunction is normalized with respect to physical arc length $ds=Rd\varphi$, its expression is $e^{in\varphi }/\sqrt{2\pi R}$. The two conventions describe the same probability, since $|u_n{|}^{2}d\varphi =|u_n/\sqrt{R}{|}^{2}ds$. The remaining parts use $d\varphi$.

Put $ϵ={\hslash }^2/(2m{R}^2)$. The two-particle product energy eigenstates areBecause the particles are distinguishable, $(n_1,n_2)$ and $(n_2,n_1)$ are distinct when $n_1\ne n_2$. Also, the signs of nonzero integers distinguish opposite angular momenta. The quantum degeneracy is therefore the number of ordered integer pairs at each sum of squares.

The first distinct energy levels are $0,ϵ,2ϵ,4ϵ,5ϵ,\dots$. This ordering matters: $4ϵ$ is the third excited level, while $5ϵ$ is the fourth excited level.

These counts concern energy eigenspaces; arbitrary quantum superpositions within each one do not create additional independent states.

Identical spin one-half particles are fermions. By fermionic exchange symmetry, the complete spatial-and-spin wavefunction must change sign when the two particle labels are exchanged. Define the normalized antisymmetric spin-one-half singlet stateand the three symmetric spin-one-half triplet statesThe Hamiltonian is independent of spin, so these spin functions carry no additional energy.

Both particles can occupy $u_0$ if their spin factor is antisymmetric. The unique ground state wavefunction isTwo equal-spin particles could not share this orbital, by the Pauli exclusion principle.

At the first excited energy, $E=ϵ={\hslash }^2/(2m{R}^2)$, one orbital is $u_0$ and the other is $u_s$ with $s=+1$ or $-1$. Define symmetric and antisymmetric normalized spatial functionsFor each $s$, all allowed wavefunctions are spanned byThere are $2(1+3)$ orthonormal states, hence first-excited-state degeneracy $g=8$. In each product, exactly one factor is antisymmetric, ensuring the required total fermionic exchange symmetry.