Past exam of the physics A course of the University of Cambridge / 2013 / Paper 1 / Section B

Apply the radial momentum operator to a smooth radial test function. More generally $[f(r),p_r]=i\hslash f^{\prime }(r)$, giving the first two labelled identities. Expanding the square gives the third:The kinetic-energy decomposition is $p^2=p_r^2+L^2/r^2$. A central potential commutes with $L^2$ and $L_z$, so choose simultaneous angular eigenfunctions $Y_{lm}$, with $L^2=l(l+1){\hslash }^2$ and $l=0,1,\dots$. The radial Hamiltonian consequently has the stated centrifugal term. The radial inner product uses measure $r^{2}dr$; adjoints below are taken on compatible regular bound-state domains. The formal radial momentum is not thereby asserted to be a self-adjoint observable on a half-line.

Set $q=l+1$, $W=-q/r+1/(qa)$ and $D=i{p}_r/\hslash$. Since $D^{†}=-D$ formally and $[D,W]=W^{\prime }=q/r^2$,The hydrogen radial Hamiltonian factorization follows by matching the Coulomb coefficient, which fixes the Bohr radiusFor the second pair of labelled relations, the opposite product changes the sign of $W^{\prime }$, henceUsing $[A,A^{†}A]=[A,A^{†}]A$ in the factorization then givesThis Coulomb angular-momentum intertwining means applying $A_l$ to an energy eigenstate either gives zero or an eigenstate of $H_{l+1}$ with the same energy. For a normalized negative-energy state,At fixed $E<0$ this would become negative if raising $l$ continued indefinitely. The bound-state ladder termination by norm positivity occurs at some $L$ with $A_L\psi =0$, givingThe terminating radial function is proportional to $r^L{e}^{-r/[(L+1)a]}$, regular and normalizable; adjoint intertwiners generate lower $l$ values at fixed $n$. Positive-energy scattering states are not subject to this terminating negative-energy argument. Here $m$ is the mass in the supplied Hamiltonian; including nuclear motion replaces it by the reduced mass.

The remaining cyclic commutators are $[S_y,S_z]=i\hslash S_x$ and $[S_z,S_x]=i\hslash S_y$. In a spin-$s$ irreducible representation,For spin one-half, $S_i=\hslash {\sigma }_i/2$, appropriate to an electron's spin. Each $S_i^2={\hslash }^{2}I/4$, so this Hamiltonian is $H=A{\hslash }^{2}I/4$. The orthonormal vectors $(1,0{)}^T$ and $(0,1{)}^T$ are eigenstates with the same energy $A{\hslash }^2/4$; every normalized superposition is also an eigenstate. The $B$ term vanishes identically because spin-one-half quadratic anisotropy is constant.

For the supplied three-dimensional basis one consistent choice isThese Hermitian matrices obey the cyclic commutators with the given $S_x$, and $S^2=2{\hslash }^{2}I$, representing spin one. Their squares giveThus the three normalized Cartesian basis vectors $e_1,e_2,e_3$ have energies ${\hslash }^2(A-B)$, ${\hslash }^2(A+B)$ and $0$, respectively; degeneracies permit arbitrary orthonormal choices in the corresponding eigenspaces.

The parallel and antiparallel $S_x$ states are $|+x⟩=(0,1,-i{)}^T/\sqrt{2}$ and $|-x⟩=(0,1,i{)}^T/\sqrt{2}$. Evolution from the first givesThis spin-one reversal under quadratic anisotropy is complete whenprovided $A+B\ne 0$. If $A+B=0$, its initial state is stationary up to phase and complete reversal never occurs. Assume real $A,B$ for a Hermitian Hamiltonian.

The stationary equation isThe second derivative represents kinetic energy; the delta term is an attractive point interaction, with $U$ having dimensions energy times length. A jump in $\psi$ would produce a delta derivative in ${\psi }^{\prime \prime }$, with no matching term in this potential, so $\psi$ is continuous. Integrating through the origin givesFor $E<0$ define $\kappa =\sqrt{-2mE}/\hslash$. Each side has general solution $c_{+}{e}^{\kappa x}+c_{-}{e}^{-\kappa x}$ with independent constants. Square integrability retains $C{e}^{\kappa x}$ on the left and $C{e}^{-\kappa x}$ on the right. The jump requires $\kappa =mU/{\hslash }^2$, and wavefunction normalization givesThere is one bound state for every $U>0$.

For $E>0$ write $k=\sqrt{2mE}/\hslash$. General solutions are independent combinations of $e^{ikx}$ and $e^{-ikx}$ on each side. For incidence from the left choose ${\psi }_L=e^{ikx}+r{e}^{-ikx}$ and ${\psi }_R=t{e}^{ikx}$. Continuity gives $t=1+r$ and the jump gives $2ikr=-2(mU/{\hslash }^2)t$. With $g=mU/({\hslash }^{2}k)$,The incident and transmitted velocities are equal, so squared amplitude ratios are current ratios here. At fixed positive energy, $U\to 0$ gives full transmission and increasing $U$ gives increasing reflection, approaching unity. Even an attractive point well reflects quantum waves; energy is conserved, and the static potential does not irreversibly capture the incident particle into its bound state.

The anisotropic quantum harmonic oscillator is a sum of three commuting one-coordinate oscillator Hamiltonians. Substituting $\psi =X(x)Y(y)Z(z)$ and dividing by $XYZ$ gives a sum of three functions of independent coordinates, so each equals a separation constant. ConsequentlyFor a time-independent operator, conservation for arbitrary states follows from $[H,Q]=0$, since $d⟨Q⟩/dt=(i/\hslash )⟨[H,Q]⟩$. The first labelled operator is $i{L}_z/\hslash$; the second is $-L^2/{\hslash }^2$. The kinetic energy is rotationally invariant. The potential is invariant under rotations about $z$ precisely when (i) ${\omega }_x^2={\omega }_y^2$; full rotational invariance requires (ii) ${\omega }_x^2={\omega }_y^2={\omega }_z^2$. For positive oscillator frequencies these are equality of the frequencies themselves. Acting with the commutators on arbitrary test functions excludes anisotropic potentials in case (ii).

In the isotropic case let $b=m\omega /\hslash$. ThenThe polynomial $x$ is the first excited oscillator factor and the other two factors are ground-state Gaussians. The first excited spatial level has three independent choices, proportional to $x,y,z$. Odd parity gives $⟨x⟩=0$, and Gaussian integration yieldsFor oscillator uncertainty and allocation of classical energy, the $x$ mode has energy $3\hslash \omega /2$, whose classical amplitude is $A_x=\sqrt{3\hslash /(m\omega )}$, giving $\Delta x=A_x/\sqrt{2}$. If “same energy” instead means assigning the entire three-dimensional energy $5\hslash \omega /2$ to one classical coordinate, its amplitude is $\sqrt{5\hslash /(m\omega )}$ and the ratio is $\sqrt{3/10}$. A general three-dimensional classical orbit's $x$ amplitude is not fixed by total energy without specifying its allocation among modes.

The capacity of an isotropic oscillator shell, including the two electron spin states, is $(N+1)(N+2)$ electrons. Nine non-interacting electrons occupy shells as $2,6,1$ for $N=0,1,2$. ThusThe least positive excitation promotes an electron by one oscillator quantum. Rearrangements within the partly filled degenerate shell remain ground states and cost no energy.