Past exam of the physics A course of the University of Cambridge / 2011 / Paper 2 / Section B

For each mode in Cartesian coordinates, $[a_i,a_j^{†}]={\delta }_{ij}$ and $a_i^{†}{a}_i=H_i/(\hslash \omega )-1/2$. Consequently the two-dimensional isotropic harmonic oscillator hasThe normalized actions of the creation and annihilation operators areInvert their definitions as $x=\sqrt{\hslash /(2m\omega )}(a_x+a_x^{†})$ and $p_x=i\sqrt{m\hslash \omega /2}(a_x^{†}-a_x)$, and similarly for $y$. linear operators belonging to different coordinates commute, so substitution into $L_z=x{p}_y-y{p}_x$ givesEach term transfers one excitation between the two modes: $[N_x+N_y,a_x^{†}{a}_y]=(1-1)a_x^{†}{a}_y=0$, and likewise for its adjoint. Thus $[H,L_z]=0$ and, with no explicit time dependence, $d{L}_z/dt=(i/\hslash )[H,L_z]=0$. This is conservation of angular momentum under planar rotations.

Finally $L_z|1,0⟩=i\hslash |0,1⟩$ and $L_z|0,1⟩=-i\hslash |1,0⟩$. HenceThe angular momentum is $+\hslash$; division by $\sqrt{2}$ performs wavefunction normalization on this state without changing its eigenvalue. It is one excitation of the positive circular mode.

A pure stationary state of a time-independent Hamiltonian operator is $\Psi (\mathbf{r},t)=\psi (\mathbf{r})e^{-iEt/\hslash }$. Its probability density and all expectations of time-independent observables are constant. This does not require zero probability current.

For real scalar potential energy and no vector potential, substitute $i\hslash \dot{\Psi }=-({\hslash }^2/2m){\nabla }^2\Psi +V\Psi$ and its complex conjugate into $\dot{\rho }={\Psi }^{\ast }\dot{\Psi }+{\dot{\Psi }}^{\ast }\Psi$. The potential terms cancel:Thus the probability current is $\mathbf{j}=(\hslash /m)\mathrm{Im}({\Psi }^{\ast }\nabla \Psi )$. In a stationary state $\nabla \cdot \mathbf{j}=0$: flow may circulate without changing the density.

Here only the azimuthal phase contributes an imaginary logarithmic derivative. Since $\nabla \varphi =\hat{\mathbf{\varphi }}/(r\sin \theta )$,There is no radial or polar flow; the azimuthal current is independent of $\varphi$, so its divergence vanishes. The apparent coordinate singularity at $\sin \theta =0$ cancels in the final expression.

The z-component of orbital angular momentum follows from the moment of the mass current:Indeed $A^2(24a_0^5)(4/3)(2\pi )=1$. This agrees with $L_z=-i\hslash {\partial }_{\varphi }$ acting on the wavefunction.

Write $\gamma =ge/(2m)$, retaining the printed magnetic moment convention. The displayed two-dimensional matrices are Pauli matrices, and the spin Hamiltonian operator isFor an x-directed field let $\omega =\gamma B_x/2$. The Schrödinger equation gives $i{\dot{c}}_{+}=-\omega c_{-}$ and $i{\dot{c}}_{-}=-\omega c_{+}$. With $c_{+}(0)=1,c_{-}(0)=0$, these equations yieldEquivalently, $U(t)=e^{i\omega t{\sigma }_x}=I\cos \omega t+i{\sigma }_x\sin \omega t$. If $e$ denotes a positive elementary charge magnitude, a free electron is conventionally assigned a minus sign in its magnetic moment; that alternative reverses the rotation sense. The expressions here consistently use the moment specified in the paper.

Put $c=\cos \omega t$, $s=\sin \omega t$. The real part of the off-diagonal coherence vanishes:This also follows from $[H,S_x]=0$ and its zero initial expectation. Individual measurements still give $\pm \hslash /2$, each with probability $1/2$.

The populations giveTogether with $⟨S_y⟩=(\hslash /2)\sin (\Omega t)$, this describes Larmor precession of a spin initially perpendicular to the field. The positive angular frequency is $|\Omega |$; the sign specifies the rotation sense through $d⟨\mathbf{S}⟩/dt=\gamma ⟨\mathbf{S}⟩\times \mathbf{B}$. The factor of two between the spinor amplitudes and the measured spin is essential. The model with quantum coherence predicts no decay; spin relaxation or dephasing in a real solid can damp the oscillation.

Put $\kappa =e^2/(4\pi {ϵ}_0)$. Substitution of $\psi =R(r)\Phi (\varphi )$ into the time-independent Schrödinger equation, followed by division by $R\Phi$ and multiplication by $r^2$, separates the angular dependence:Thus $\Phi =e^{im\varphi }/\sqrt{2\pi }$ andThe single-valued wavefunction condition, $\Phi (\varphi +2\pi )=\Phi (\varphi )$, requires $m\in \mathbb{Z}$. The regular boundary condition at the origin selects $R\sim r^{|m|}$. Accordingly use $s=|m|$ in the radial ansatz; the printed $x^m$ notation describes the $m\ge 0$ representatives.

For the two-dimensional Coulomb bound state, substitution of $R=x^s{e}^{-x/2}G$ with the stipulated radial and energy scales givesThe coefficient of $x^n$ in the power series isThe regular solution otherwise has an exponentially growing contribution at large $x$, incompatible with being normalizable. Termination at degree $n_r\ge 0$ therefore requiresThe resulting $G$ is proportional to a Generalized Laguerre polynomial.

Define $R_{\mathrm{y}}=\mu {\kappa }^2/(2{\hslash }^2)$, the hydrogenic Rydberg energy for the stipulated mass. The first three distinct two-dimensional energies are $-4R_{\mathrm{y}},-4R_{\mathrm{y}}/9,-4R_{\mathrm{y}}/25$, whereas in three dimensions they are $-R_{\mathrm{y}},-R_{\mathrm{y}}/4,-R_{\mathrm{y}}/9$. Both series approach zero from below, but the planar ground state is four times more tightly bound. Without spin the respective quantum degeneracies are $1,3,5$ and $1,4,9$; if individual degenerate states are counted, the second and third planar states already share the first excited energy. This comparison retains the supplied $1/r$ interaction; it does not replace it by the logarithmic electrostatic potential of a genuinely two-dimensional electromagnetic theory.