Past exam of the physics A course of the University of Cambridge / 2022 / Paper 1 / Section B / B6

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 $2,0,-2$.