Past exam of the physics A course of the University of Cambridge / 2019 / Paper 1

Write $L=10^{-9}\mathrm{m}$ and use the Heisenberg uncertainty relation separately in the three Cartesian directions. Since a position coordinate is confined to an interval of length $L$, Popoviciu's inequality on variances gives $\Delta x_i\le L/2$. Consequently $\Delta p_i\ge \hslash /L$, and the expectation value of the kinetic energy obeysThus an uncertainty-based lower estimate is $\boxed{K\gtrsim 0.11\mathrm{eV}}$. This confinement energy bound from position support is not sharp: the exact ground-state energy for a cubical infinite square well is $3{\pi }^2{\hslash }^2/(2m_e{L}^2)=1.13\mathrm{eV}$. The Heisenberg uncertainty principle establishes the confinement scale ${\hslash }^2/(m_e{L}^2)$; identifying a box size with an approximate position standard deviation changes its numerical coefficient.

The rotational kinetic energy is $K=L^2/(2I)$. For a three-dimensional linear rigid rotor, angular momentum operator replaces the squared magnitude by ${\hslash }^{2}l(l+1)$, where $l=0,1,2,\dots$. HenceThe three lowest distinct energy levels are thereforeTheir magnetic quantum degeneracy values are $1,3,5$. A single moment of inertia specifies this rigid rotor model; an arbitrary asymmetric body would require its other principal moments.

Inside the barrier, the stationary Schrödinger equation gives an exponentially decaying wavefunction with inverse penetration lengthThe amplitude falls approximately as $e^{-\kappa a}$, so the leading exponential estimate for the transmission probability isFor comparison, wavefunction matching at a finite potential step at both edges gives the rectangular-barrier tunnelling probabilityThe difference is the interface prefactor: for an opaque barrier it is $16E(V_0-E)/V_0^2=2.56$. Both give a transmission of order $10^{-6}$; the first is the requested exponential estimate.

Ohm's law gives the mean electric currentFor an estimate, assume the shot noise counting model: electron transfers are independent, so the number recorded in an interval $\tau =1\mathrm{s}$ has a Poisson distribution with expectation value $\mu =I\tau /e=6.2415\times 10^6$ and variance $\mu$. One measurement has relative standard deviation $1/\sqrt{\mu }=4.00\times 10^{-4}$. Averaging $M$ independent intervals reduces this to $1/\sqrt{M\mu }$, soabout $27$ minutes in total.

This is a one-standard deviation, shot noise estimate. A resistance and mean voltage alone do not specify the actual current fluctuations: a real resistor also has Johnson-Nyquist noise, and electron-transfer correlations can change the shot noise. Its temperature, measurement bandwidth and noise model would be needed for an unconditional acquisition time.

For independent small uncertainties interpreted as standard deviations, propagation of uncertainty givesThe angular uncertainty is in radians. Since ${\sigma }_n/n=0.075/1.35=1/18$, the allowed angular contribution to the fractional uncertainty of the numerical aperture is only $\sqrt{0.06^2-(1/18{)}^2}$. ThereforeThe uncertainty in the refractive index already contributes $5.56\%$. If the quoted errors instead mean strict worst-case bounds, first-order absolute errors must be added rather than added in quadrature; that convention would require $|\delta \alpha |\lesssim \tan 60^{\circ }(0.06-1/18)=0.441^{\circ }$. The $2.25^{\circ }$ result uses the usual independent statistical-error convention.

Write $S_i=(\hslash /2){\sigma }_i$, with ${\sigma }_i$ the Pauli matrices. Direct multiplication gives ${\sigma }_x^2={\sigma }_y^2={\sigma }_z^2=I$, and henceFor the products in the other order,Their difference gives the angular momentum commutation relationsThe first identity corresponds to the spin one-half value $s(s+1){\hslash }^2$ with $s=1/2$.

In the $S_z$ eigenbasis, let a candidate eigenvector be $(a,b{)}^{\mathsf{T}}$. The eigenvalue equation is $b=(2s_x/\hslash )a$ and $a=(2s_x/\hslash )b$, so a nonzero vector requires $(2s_x/\hslash {)}^2=1$. Choosing normalized phases givesThe two spin eigenstates are orthogonal, as required for distinct eigenvalues of a self-adjoint operator.

Put $\Omega =\lambda B$, including its sign. The initial vector is the $+\hslash /2$ eigenstate of $S_z$; it is not an energy eigenstate of $H=\Omega S_x$. In terms of the spin eigenstates found above, $|+z⟩=(|+x⟩+|-x⟩)/\sqrt{2}$. Each energy eigenstate acquires its own unitary time evolution phase:Equivalently, $e^{-iHt/\hslash }=I\cos (\Omega t/2)-i{\sigma }_x\sin (\Omega t/2)$ follows from ${\sigma }_x^2=I$. The Born rule givesThe expectation value of the spin angular momentum is $(\hslash /2)(0,-\sin \Omega t,\cos \Omega t)$, a rotation around the $x$ axis. Its direction follows the specified sign of the Hamiltonian; changing the magnetic coupling sign reverses the precession.

A Stern-Gerlach experiment uses an oven and slits to produce a collimated atomic beam, conventionally of neutral silver atoms, and sends it through an inhomogeneous magnetic field onto a position-sensitive detector. The unpaired outer electron supplies the effective spin one-half magnetic dipole moment; neutral atoms avoid the ordinary Lorentz deflection of a charged beam. For an analysing direction $z$, the spatial gradient produces a force approximately $F_z={\mu }_z{\partial }_z{B}_z$. The resulting separated beam positions measure the allowed projections of angular momentum. A spin one-half beam gives two branches, rather than the continuous distribution expected from classical dipoles with arbitrary orientations. A uniform magnetic field can rotate a spin but does not itself provide this spatial separation.

To measure the rotation, first use a $z$-directed Stern-Gerlach experiment and an aperture to transmit only the $+z$ branch. Pass this prepared spin eigenstate through a region of approximately uniform field in the $x$ direction, then use a second $z$ analyser to separate and count the two populations. For a field region of length $\ell$ and beam speed $v$, the precession angle is $\theta =\lambda B\ell /v$, and the predicted fractions areVarying $B$ or the transit time gives complementary oscillations, demonstrating coherent Larmor precession. Velocity selection and a sufficiently uniform rotation field keep different particles from accumulating different phases and washing out the contrast. Detection should compare the branch fractions after allowing for losses and analyser efficiency.

Use the nonrelativistic Coulomb potential energy $V=-Z{e}^2/(4\pi {ϵ}_{0}r)$, with the nucleus treated as fixed. Since the wavefunction has zero orbital angular momentum, its Laplacian isSubstituting into the time-independent Schrödinger equation givesFor a constant energy eigenvalue, the coefficient of $1/r$ must vanish. Therefore $\boxed{\beta =Z/a_0}$, where $a_0$ is the Bohr radius. Wavefunction normalization then requiresChoosing the irrelevant overall phase so that $\alpha$ is real and positive gives $\boxed{{\alpha }^2={\beta }^3/\pi }$. The remaining constant term is the hydrogen ground-state energyReplacing $m_e$ by the electron-nucleus reduced mass gives the usual finite-nuclear-mass refinement.

The normalized radial probability density givesConsequently the expectation value of the potential energy isHere the last equality uses $Z{e}^2/(4\pi {ϵ}_0)={\hslash }^2\beta /m_e$. Since this is an energy eigenstate, $E=⟨T⟩+⟨V⟩$, and so $\boxed{⟨T⟩=-E}$. The kinetic energy is positive because $E<0$, in agreement with the virial theorem for a Coulomb potential.

The probability of finding the electron in a spherical shell is $P(r)dr=4\pi r^2|\psi (r){|}^{2}dr$, soThis radial probability density, rather than the density per unit volume, determines the most probable radius. Its expectation value isFor its interior maximum, differentiate $\log P=\text{constant}+2\log r-2\beta r$. The stationary condition $2/r-2\beta =0$ givesThe density per unit volume $|\psi {|}^2$ is instead largest at the nucleus; including the spherical-shell volume factor is essential to the requested radial mode.

The nuclear beta decay is fast compared with the electronic orbital timescale. In the sudden approximation, the electronic wavefunction does not have time to change during the increase from $Z_i=1$ to $Z_f=2$. The subsequent probability of a particular daughter energy eigenstate is therefore its squared overlap with the initial state.

For the final $1s$ state, write ${\beta }_i=1/a_0$ and ${\beta }_f=2/a_0$. The sudden hydrogenic charge change gives the amplitudeThusThe final $2p$ states have spherical harmonics $Y_{1m}$, whereas the initial state is proportional to $Y_{00}$. Their angular overlap $\int Y_{1m}^{\ast }{Y}_{00}d\Omega$ vanishes by orthogonality. Accordingly $\boxed{P_{2p}=0}$ in the ideal spherically symmetric sudden-change model. Nuclear recoil or interactions with the emitted particles can produce small corrections. This is an overlap selection rule for a sudden change, not a photon-induced electric dipole transition.

The infinite walls impose $\psi (0)=\psi (L)=0$, and the interior time-independent Schrödinger equation is ${\psi }^{\prime \prime }+k^2\psi =0$ with $E={\hslash }^2{k}^2/(2m)$. Its general solution is $A\sin kx+B\cos kx$. The first boundary condition gives $B=0$, and the second gives $kL=n\pi$. A nonzero normalizable state needs $n=1,2,\dots$; $n=0$ would give the zero function.

Wavefunction normalization, ${\int }_0^L{\sin }^2(n\pi x/L)dx=L/2$, gives the infinite square well states and their energy eigenvalues:The wavefunctions vanish outside the well, and their full time dependence is ${\psi }_n(x)e^{-i{E}_{n}t/\hslash }$.

The first two probability densities areBoth vanish at the walls. The ground state has one maximum at $L/2$ and no interior node. The first excited state has a node at $L/2$ and equal maxima at $L/4$ and $3L/4$.

For a finite square well whose exterior potential is $V_0>E_n$, the exterior wavefunction decays as $e^{-\kappa d}$, where $\kappa =\sqrt{2m(V_0-E_n)}/\hslash$ and $d$ is distance from the relevant edge. Wavefunction matching at a finite potential step makes the wavefunction and, for constant mass, its first derivative continuous. Thus the probability density is nonzero at the edges and has evanescent tails outside, the interior wavelength increases, and the energy levels are lower than for infinite walls at the same positions. Reflection symmetry still gives an even ground state and odd first excited state about the center, when both are bound. A shallow well may not support that excited bound state at all.

Choose the two real energy eigenstates in part (a) and take their coefficients to be in phase initially. A different initial relative phase only shifts the oscillation in time. The normalized quantum superposition isOrthogonality removes the cross terms in the energy expectation value, soFor position, the diagonal matrix elements equal $L/2$ by reflection symmetry, but the off-diagonal element is nonzero. Using the product-to-sum identity and ${\int }_0^{L}x\cos (j\pi x/L)dx=L^2[(-1{)}^j-1]/(j^2{\pi }^2)$ givesTherefore the equal superposition of the first two box states hasThe amplitude is $\boxed{16L/(9{\pi }^2)\simeq 0.1801L}$, and the ordinary frequency is $\boxed{f=\Omega /(2\pi )=3h/(8m{L}^2)}$. The motion of the mean position is an interference effect between distinct energies, while their populations, and hence the mean energy, remain fixed.

The boundary force in an infinite square well is obtained from the kinetic momentum flux at each wall. With $C_1={\hslash }^2{\pi }^2/(2m{L}^3)=E_1/L$, the derivatives of the state in part (c) areThe outward force magnitudes are $P_j=({\hslash }^2/2m)|{\Psi }^{\prime }(x_j,t){|}^2$. Taking positive force to mean rightwards, the forces exerted by the particle on the walls areBoth point outwards at all times, but their magnitudes oscillate out of phase. This force formula follows directly by differentiating the mean particle momentum and integrating the interior Schrödinger equation by parts: $d⟨p⟩/dt=({\hslash }^2/2m)(|{\Psi }^{\prime }(0){|}^2-|{\Psi }^{\prime }(L){|}^2)$. As a check, the net force on the particle is $8C_1\cos \Omega t=m{d}^2⟨x⟩/d{t}^2$, the opposite of the net force on the two walls.

A classical particle of energy $E=⟨E⟩$ transfers momentum $2mv$ in each collision with a particular wall, and returns to that wall after $2L/v$. Its mean outward force on that wall isThus the time averages agree separately at each wall, although the classical force consists of collision impulses and the quantum expectation varies continuously. Constant mean energy alone does not make the instantaneous quantum wall forces constant; their unequal instantaneous values supply the oscillation of $⟨x⟩$.

Let $c=\sqrt{m\omega /(2\hslash )}$ and $d=1/\sqrt{2m\hslash \omega }$, so $a=cx+idp$ and $a^{†}=cx-idp$. The canonical commutation relation $[x,p]=i\hslash$ givesHence $\boxed{H=\hslash \omega (a^{†}a+1/2)}$. Similarly,Using the product rule for a commutator, $[AB,C]=A[B,C]+[A,C]B$, givesThus $a^{†}a$ is the number operator, while $a$ and $a^{†}$ are the annihilation operator and creation operator.

The commutators implyThus any nonzero resulting vector is the adjacent energy eigenstate. Since $a^{†}a|n⟩=n|n⟩$ and $a{a}^{†}=a^{†}a+1$, their squared norms areChoosing consistent phases for the eigenstates yieldsThe vanishing last expression terminates the lowering sequence at the ground state, rather than introducing a negative-energy oscillator state.

In the position representation $p=-i\hslash d/dx$, so $a{\psi }_0=0$ becomesIntegrating and using the Gaussian integral for wavefunction normalization givesThe creation operator gives $|1⟩=a^{†}|0⟩$, and thereforeIts norm is one since $\int x^2|{\psi }_0{|}^{2}dx=\hslash /(2m\omega )$. Moreover ${\psi }_0$ is even and ${\psi }_1$ is odd, so the overlap integrand ${\psi }_0^{\ast }{\psi }_1$ is odd and integrable. Integration over the whole real line gives $\boxed{⟨0|1⟩=0}$, establishing orthogonality.

Inverting the definitions of the creation operator and annihilation operator gives $x=\sqrt{\hslash /(2m\omega )}(a+a^{†})$. Their actions and the orthogonality of the oscillator energy eigenstates giveThus the electric dipole transition selection rule iswith the lower alternative absent for $n_1=0$. The field exchanges one oscillator quantum in first-order dipole coupling. For a long weak monochromatic drive, its frequency must also be resonant with the level separation, here $\omega$, to give a substantial transition probability. Higher-order multiphoton processes are beyond the stated single-position-matrix element approximation.

A localized quantum particle is described by a wave packet: a coherent quantum superposition of momentum components, rather than a single delocalized plane wave. The Born rule assigns the position probability density $|\psi (x,t){|}^2$. A useful normalized initial Gaussian wave packet isIts mean position and momentum are $x_0$ and $p_0$, and its standard deviations satisfy ${\sigma }_x(0)={\sigma }_0$, ${\sigma }_p=\hslash /(2{\sigma }_0)$. This unchirped initial state saturates the Heisenberg uncertainty relation.

For a free particle, the momentum-space Schrödinger equation multiplies each amplitude by $e^{-i{p}^{2}t/(2m\hslash )}$. The inverse Fourier transform is a Gaussian integral, giving the free Gaussian wave packet evolutionOne way to obtain this expression explicitly is to write the initial momentum amplitude as proportional to $\exp [-{\sigma }_0^2(p-p_0{)}^2/{\hslash }^2-ip{x}_0/\hslash ]$, multiply by the free evolution phase and complete the square in $p$. Taking the modulus squared removes the position-dependent phase and leavesThe expectation value follows uniform classical motion while the density spreads. A narrower initial packet needs a wider momentum distribution and therefore spreads more rapidly. At long times its width grows linearly with $|t|$; its variance grows quadratically, unlike the linear variance growth of ordinary diffusion. The probability continuity equation preserves total probability despite the decreasing peak height. The mean momentum and the mean energy $(p_0^2+{\sigma }_p^2)/(2m)$ remain constant, since the momentum populations do not change.

The same result has a wave interpretation. For a free nonrelativistic particle $\omega (k)=\hslash k^2/(2m)$, so the group velocity is $d\omega /dk=\hslash k/m$, equal to $p_0/m$ at the packet center. The curvature $d^2\omega /d{k}^2=\hslash /m$ makes different momentum components travel with different group velocities, producing dispersion. The phase velocity is $\omega /k=\hslash k/(2m)$ for this energy convention and does not describe the motion of the envelope. The evolving quadratic spatial phase is essential to the spreading; a density snapshot alone does not specify the full quantum state.

A freely spreading Gaussian wave packet remains a pure Gaussian state but develops position-momentum correlations. Its symmetrized covariance is $C_{xp}=⟨\{x-⟨x⟩,p-⟨p⟩\}⟩/2=t{\sigma }_p^2/m$. ConsequentlyEquality in the Robertson-Schrödinger uncertainty relation survives at every time, whereas the simpler product ${\sigma }_x{\sigma }_p$ equals $\hslash /2$ only initially. A chirped Gaussian wave packet can already have $C_{xp}(0)\ne 0$. More generally, $x(t)=x(0)+p(0)t/m$ gives ${\sigma }_x(t{)}^2={\sigma }_x(0{)}^2+2t{C}_{xp}(0)/m+t^2{\sigma }_p(0{)}^2/m^2$. A sufficiently negative initial covariance permits initial contraction before subsequent spreading.

A potential energy changes this behavior. A quadratic potential preserves the Gaussian form because the evolution of position and momentum is linear. For a quantum harmonic oscillator, the Gaussian covariance in a harmonic potential isA coherent state has ${\sigma }_x^2=\hslash /(2m\omega )$, ${\sigma }_p^2=m\hslash \omega /2$ and zero centered covariance. Its width is constant, and its center follows $x_c(t)=x_0\cos \omega t+p_0\sin \omega t/(m\omega )$. In oscillator notation its coherent state label rotates as $\alpha (t)=\alpha (0)e^{-i\omega t}$, apart from the common ground-energy phase of the state vector. A squeezed coherent state generally has a width that breathes at $2\omega$. Thus being Gaussian does not imply either free spreading or constant width; the Hamiltonian operator and initial covariance determine the result.

For a general nonquadratic potential, different parts of the packet experience different nonlinear forces, and an initially Gaussian profile need not remain Gaussian. Ehrenfest theorem gives the exact relations $d⟨x⟩/dt=⟨p⟩/m$ and $d⟨p⟩/dt=-⟨V^{\prime }(x)⟩$. Replacing the last mean by $-V^{\prime }(⟨x⟩)$ is exact for a quadratic potential and approximate for a packet narrow compared with the scale on which the force varies. This connects packet motion to the correspondence principle without treating a quantum state as a definite classical path. Scattering can split a packet into reflected and transmitted components, while confinement can produce oscillations and interference between discrete energy eigenstates. Their relative phases evolve even when their populations remain constant. Localization, spreading and coherent motion all follow from the same linear unitary time evolution.

The orbital angular momentum of a particle is the operator $\mathbf{L}=\mathbf{r}\times \mathbf{p}$, with $\mathbf{p}=-i\hslash \nabla$ in the position representation. Its angular momentum commutation relations areThus one may specify $L^2$ and one component, conventionally $L_z$, simultaneously. One generally cannot assign sharp values to all three components, because they do not commute. The common angular eigenfunctions are the spherical harmonics:For a scalar orbital wavefunction, single-valuedness in the azimuthal angle gives integer $m$, and regularity on the sphere and termination of the ladder give integer $l\ge |m|$. Half-integer angular momenta describe intrinsic spin rather than scalar orbital motion.

The ladder operators $L_{\pm }=L_x\pm i{L}_y$ obey $[L_z,L_{\pm }]=\pm \hslash L_{\pm }$. Their normalized action isThe coefficient vanishes at the top or bottom of the multiplet. Its squared magnitude follows from $L_{\mp }{L}_{\pm }=L^2-L_z^2\mp \hslash L_z$, and positivity of the resulting norm constrains the allowed $m$ values.

For a central potential, the Hamiltonian commutes with $L^2$ and every component of $\mathbf{L}$. Stationary states can be separated as $R_{nl}(r)Y_l^m(\theta ,\varphi )$, and rotational symmetry makes the energy independent of $m$, giving a $2l+1$ quantum degeneracy at fixed radial label and $l$. Additional degeneracy between distinct $l$ values requires additional structure, as in the Coulomb potential. The parity of $Y_l^m$ is $(-1{)}^l$. These results organize atomic orbitals and their selection rules. They describe quantized rotation and symmetry, without assigning the electron a classical orbit or a simultaneously definite three-component angular momentum vector.

In the Schrödinger picture, a time-independent Hamiltonian gives $|{\psi }_S(t)⟩=U(t)|\psi (0)⟩$, where $U=e^{-iHt/\hslash }$. The Heisenberg picture transfers this evolution to each observable:Both pictures give the same probabilities and expectation values. Differentiating yields the Heisenberg equation of motionThe last term is an explicit change of the measured operator. If it is absent and $[H,A]=0$, the observable is conserved.

The corresponding classical equation is $dA/dt=\{A,H{\}}_{\mathrm{PB}}+\partial A/\partial t$. The correspondence principle relates the Poisson bracket to $[A,H]/(i\hslash )$, with suitable operator ordering and a classical limit. For $H=p^2/(2m)+V(x)$, the canonical commutation relation givesTaking means gives Ehrenfest theorem. Its second equation contains $-⟨V^{\prime }(x)⟩$, which need not equal $-V^{\prime }(⟨x⟩)$. A narrow wave packet in a slowly varying force field can follow the classical equations approximately; a quadratic potential makes the equality exact for the means.