Past exam of the physics A course of the University of Cambridge / 2018

For a neutral atom, count both the nucleons and the electrons. Protons, neutrons and electrons are all fermions, and exchanging two identical intact atoms produces the sign $(-1{)}^{N_f}$ from exchanging their fermionic constituents. An odd number gives a composite fermion; an even number gives a composite boson, consistently with the Spin-statistics theorem.

A lithium atom has three protons and three electrons. Lithium-6 also has three neutrons, giving $3+3+3=9$ fermionic constituents; lithium-7 has four neutrons, giving $3+4+3=10$. ThereforeThe composite statistics of neutral atoms cannot be inferred from the mass number alone: counting only nucleons would classify the nuclei and give the opposite answers here. The atom's total spin includes its electrons as well as its nucleus.

The infinite square well has energy levels $E_n=n^2{h}^2/(8m_e{L}^2)$, with $n=1,2,\dots$. The longest absorption wavelength corresponds to the smallest available energy difference from the ground state, namely $n=1\to 2$. Its position matrix element is nonzero because the two states have opposite parity about the center of the well, so it is an allowed electric dipole transition.

Equating the photon energy to this gap givesHere the photon gap is approximately $1.55\mathrm{eV}$, not the ground-state energy itself.

In the orthonormal $S_z$ eigenbasis $(\alpha ,\beta )$, the spin one-half operator is $S_x=(\hslash /2)\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$. An eigenvector $a\alpha +b\beta$ with eigenvalue $s_x$ obeys $b=(2s_x/\hslash )a$ and $a=(2s_x/\hslash )b$. A nonzero vector requires $(2s_x/\hslash {)}^2=1$. Choosing normalized quantum phases givesThese spin eigenstates are orthogonal and have unit norm because $\alpha$ and $\beta$ do.

Substitution in the moment of inertia givesFor independent errors interpreted as standard uncertainties, logarithmic differentiation and propagation of uncertainty giveConsequentlyThis is a $0.515\%$ statistical precision requirement. If “accuracy” instead means hard bounds whose worst cases must be guaranteed, first-order absolute errors add: $0.01+0.01+4{ϵ}_R\le 0.025$, giving $\boxed{{ϵ}_R\le 0.125\%,|\delta R|\le 1.25\mathrm{mm}}$ to first order. For an exact symmetric hard bound about the nominal values, the upper extreme is slightly more restrictive: $(1.01{)}^2(1+{ϵ}_R{)}^4\le 1.025$ gives ${ϵ}_R\le 0.1199\%$, or $1.20\mathrm{mm}$. The question does not distinguish a statistical uncertainty from a worst-case tolerance, so the convention matters.

Use an unloaded RC low-pass filter: a series resistor between the source and the output node, with a capacitor from that node to ground, and measure the output across the capacitor. The components are ideal in this first-order model; “ideal” here does not mean a brick-wall frequency response.

The time constant is $\tau =RC$, henceThe capacitive electrical impedance gives the transfer function $H(i\omega )=1/(1+i\omega RC)$, soOn the magnitude Bode plot, $20{\log }_{10}|H|=-10{\log }_{10}[1+(f/f_c{)}^2]$. The gain approaches $0\mathrm{dB}$ for $f\ll f_c$, is $-3.01\mathrm{dB}$ at $f_c$, and falls with slope $-20\mathrm{dB}$ per decade for $f\gg f_c$. These results assume negligible source resistance beyond the specified $R$ and a sufficiently large load impedance.

Let $\rho ={\psi }^{\ast }\psi$ be the probability density. For a real potential energy $V(x)$, the Schrödinger equation and its complex conjugate areDifferentiate $\rho$ and substitute both equations. The potential terms cancel, leavingComparison with the probability continuity equation ${\partial }_t\rho +{\partial }_{x}J=0$ identifiesFor a plane wave of amplitude $A$ and positive wavenumber $k$, $J=\hslash k|A{|}^2/m$, the density times the velocity. Integrating the continuity equation over an interval makes its probability change equal to the incoming current minus the outgoing current.

SetAll are real and $k_0>k_1>k_2>0$. Suppressing a common time quantum phase $e^{-iEt/\hslash }$, chooseThe rightmost region has no left-moving term because no wave is incident from the right. The shifted quantum phase convention makes $F$ the transmitted amplitude at the second interface.

For finite potential steps and equal mass, wavefunction matching at a finite potential step requires continuity of both $\psi$ and ${\partial }_x\psi$. Explicitly,A discontinuous wavefunction would introduce an inadmissible singular kinetic term, while integrating the stationary Schrödinger equation across a finite step shows that the derivative cannot jump. These conditions also keep the probability current continuous, so a finite interface neither creates nor destroys probability. The derivative rule would need modification for a delta-function potential or a different effective mass; neither occurs here.

Zero reflection means $B=0$. Matching the first interface then givesThe specified layer quantum phase is $k_{1}a=\pi /2$, so $e^{i{k}_{1}a}=i$ and $e^{-i{k}_{1}a}=-i$. The second-interface conditions becomeSince $C+D=A$, compatibility requires $\boxed{k_1^2=k_0{k}_2}$. This is a quantum quarter-wave matching layer: the intermediate wavenumber is the geometric mean, and the two interface reflections cancel after their round trip through the layer.

In terms of the physical energies,Rearranging also givesA finite positive energy satisfying the stated ordering requires $V_2/2<V_1<V_2$. Under that condition $E-V_2=(V_2-V_1{)}^2/(2V_1-V_2)>0$, consistent with the assumed propagating transmitted wave. The width must simultaneously satisfy the given quarter-wave quantum phase condition.

From the second-edge amplitude in part (c),This uses ${\psi }_{\mathrm{R}}=F{e}^{i{k}_2(x-a)}$. If the transmitted wave is instead written $T{e}^{i{k}_{2}x}$, its coefficient is $\boxed{T/A=i\sqrt{k_0/k_2}{e}^{-i{k}_{2}a}}$; the extra factor only changes the quantum phase convention.

Using the probability current of a plane wave,Thus the flux transmission probability is one. The larger transmitted density $|F{|}^2>|A{|}^2$ compensates for the smaller transmitted velocity $\hslash k_2/m$; equal currents need not have equal wave amplitudes.

The one-dimensional time-independent Schrödinger equation isHere $m$ is the particle mass, $V(x)$ its potential energy, $E$ the stationary energy eigenvalue, and $\hslash =h/(2\pi )$ the reduced Planck constant. The spatial wavefunction is $\Psi$; the full stationary state is $\Psi (x)e^{-iEt/\hslash }$. Physical bound states satisfy the boundary conditions and wavefunction normalization $\int |\Psi {|}^{2}dx=1$. The operator on the left is the Hamiltonian operator, the sum of kinetic and potential energies.

Assume $a>0$ and define $\omega =\sqrt{2a/m}$ so that the potential is that of a quantum harmonic oscillator. The two trial forms have derivativesSubstitution into the stationary Schrödinger equation cancels the $x^2$ terms precisely when $a=2{\hslash }^2{C}^2/m$. Normalizability selects the positive root,The remaining constants giveFor later overlaps choose positive normalization constants. The Gaussian integral gives $A=(2C/\pi {)}^{1/4}$, while its second-moment integral gives $B=2\sqrt{C}A$. Thus both wavefunctions have unit norm and are orthogonal by their opposite parity.

The real Gaussian is positive everywhere and has no nodes; the odd state has exactly one node at $x=0$. The Sturm oscillation theorem orders one-dimensional bound states by their node count. Consequently these are the ground state and first excited state, respectively, also agreeing with the oscillator spectrum $E_n=(n+1/2)\hslash \omega$.

The infinite wall imposes $\boxed{\Psi (0)=0}$, and the wavefunction vanishes for $x<0$. The even full-oscillator ground state fails this boundary condition; the lowest admissible state is the positive-half restriction of the odd first excited state. Multiplying by $\sqrt{2}$ restores normalization because half its probability originally lay on either side.

With $\ell =\sqrt{\hslash /(m\omega )}$, the half-harmonic oscillator ground state isIt rises from zero at the wall, reaches its wave-amplitude maximum at $x=\ell$, and decays to zero as $x\to \infty$. It is positive throughout the interior of the allowed region, with no interior node, as required for its ground state. The wall need not make the derivative continuous with the forbidden half-line.

For removal of a half-oscillator wall, the sudden approximation keeps the state unchanged during the change of potential. Its probability in the new full-oscillator ground state ${\Psi }_0={\pi }^{-1/4}{\ell }^{-1/2}{e}^{-x^2/(2{\ell }^2)}$ is the squared overlap:where ${\int }_0^{\infty }x{e}^{-x^2/{\ell }^2}dx={\ell }^2/2$. ThereforeThe integral runs only over the initially occupied half-line. The initial state is not the odd full-line eigenstate: it is zero, rather than negative, on the left. Indeed its overlap with that final first excited state is $1/\sqrt{2}$, giving probability $1/2$, and the remaining probability lies in higher states. Subsequent evolution under the new fixed Hamiltonian operator changes the quantum phases of its energy components without changing these energy-measurement probabilities.

The quantum rotor has orientation angle $\varphi$ of period $2\pi$, so physical wavefunctions satisfy $\psi (\varphi +2\pi )=\psi (\varphi )$. The exponential solutions of $Hu=Eu$ are thereforeThey are normalized and orthogonal on $[0,2\pi )$. The corresponding angular momentum operator is $L_z=-i\hslash {\partial }_{\varphi }$, with eigenvalue $m\hslash$. The $m=0$ ground level is nondegenerate, and every $|m|>0$ energy level has the twofold degeneracy $m\leftrightarrow -m$. This uses ordinary periodic boundary conditions, with no flux or twisted quantum phase.

Expand the initial wavefunction in the angular momentum eigenbasis using ${\cos }^2\varphi =1/2+(e^{2i\varphi }+e^{-2i\varphi })/4$. With the normalized $u_m$ from part (a), this givesThe coefficient squares sum to one, also verifying the stated normalization. The Born rule therefore gives angular momentum outcomesFor an energy measurement the two distinct $m=\pm 2$ states belong to the same energy eigenspace. Sum their probabilities, rather than adding their amplitudes, to obtainTheir relative probabilities are $4:1:1$ for the angular momenta and $2:1$ for the energies. In particular, zero mean angular momentum does not imply zero rotational energy.

Each energy eigenstate evolves by its own quantum phase $e^{-i{E}_{m}t/\hslash }$. The two $m=\pm 2$ components have the same energy, while the $m=0$ quantum phase stays one. ThusIn angle representation this isAt $t=0$, $1+\cos 2\varphi =2{\cos }^2\varphi$ recovers the initial state. The fixed coefficient magnitudes keep the angular momentum and energy probabilities constant even though the density in angle changes.

Taking the modulus squared gives the three-mode revival of a planar quantum rotor:The angular density has period $\pi$, while both the state and its density revive after $\boxed{T=2\pi /\Omega =\pi I/\hslash }$. Its integral over $0\le \varphi <2\pi$ is one because $\int \cos 2\varphi d\varphi =0$ and $\int {\cos }^{2}2\varphi d\varphi =\pi$.

At $t=0$, the density is $4{\cos }^4\varphi /(3\pi )$, with maxima at $0,\pi$ and zeros at $\pi /2,3\pi /2$. At $t=T/2$ it becomes $4{\sin }^4\varphi /(3\pi )$, interchanging those maxima and zeros. At $t=T/4$ and $3T/4$, it is $[1+{\cos }^{2}2\varphi ]/(3\pi )$, with four equal maxima along the coordinate axes and positive minima halfway between them. At $t=T$ the original profile returns.

The changing profile is interference between the stationary $m=0$ mode and the degenerate counter-rotating $m=\pm 2$ modes. It does not translate rigidly around the circle; the standing-wave weights alternate between two orientations.

The commutator is $\boxed{[A,B]=AB-BA}$. To evaluate it for differential operators, apply it to a smooth test wavefunction $f$:Thus the canonical commutation relation is $\boxed{[x,p_x]=i\hslash I}$. The associated Heisenberg uncertainty relation is $\Delta x\Delta p_x\ge \hslash /2$. Position and momentum cannot both have arbitrarily sharp outcome distributions in the same prepared state, even with ideal measuring apparatus. This concerns quantum-state preparation, not merely a calibration error in a particular instrument.

Expand the proposed product rule without interchanging any operator factors:Applying this commutator identity with $A=x$ and $B=C=p_x$ givesFor a differentiable multiplication potential, apply the other commutator to a test function:ThereforeThese operator equalities are understood on a common domain where the indicated products and derivatives are defined, with the boundary conditions appropriate to the state.

For a normalized state, the expectation value isFor a time-independent operator, differentiation givesThe Schrödinger equation says $|\dot{\psi }⟩=-iH|\psi ⟩/\hslash$. The self-adjointness of $H$ gives $⟨\dot{\psi }|=i⟨\psi |H/\hslash$, soIn the integral derivation, the same self-adjointness step is integration by parts with vanishing boundary terms. An explicitly time-dependent observable would also contribute $⟨\partial Q/\partial t⟩$. These are the expectation-value equations underlying the Heisenberg equation of motion.

The Hamiltonian operator isSince multiplication by $x$ commutes with multiplication by $V(x)$, part (b) gives $[x,H]=i\hslash p_x/m$. Similarly $[p_x,H]=-i\hslash V^{\prime }(x)$. Substituting into part (c) gives Ehrenfest theorem:Thus the mean momentum changes according to the mean force, and the derivative of mean position is mean velocity. These exact quantum equations resemble Hamilton's classical equations, but $⟨V^{\prime }(x)⟩$ generally differs from $V^{\prime }(⟨x⟩)$. The classical point-particle equation for the center is exact for potentials at most quadratic and is an approximation for a localized packet when the force varies little across it. A spatially broad packet in a nonlinear potential need not follow a classical trajectory. Even where its center follows classical motion, its width and measurement uncertainties remain quantum properties.

The postulates of quantum mechanics give a framework for predicting observations from a prepared state. The experiments below motivate different features of that framework; no one experiment proves the whole set of postulates or specifies every Hamiltonian operator.

A pure quantum state is a normalized vector in a complex Hilbert space, with vectors differing by an overall quantum phase representing the same physical preparation. In position representation, $\int |\psi (\mathbf{r}){|}^2{d}^{3}r=1$. The linear state space permits quantum superposition, while the Born rule assigns probability ${\int }_{\Omega }|\psi {|}^2{d}^{3}r$ to a position outcome in a region $\Omega$. The wavefunction is a probability amplitude; it is not a classical distribution of definite particle positions and momenta.

An observable is represented by a self-adjoint operator. If $Q$ has spectral projectors $P_q$, its possible measurement outcomes are its spectral values and $\mathrm{Pr}(q)=⟨\psi |P_q|\psi ⟩$. The mean is $⟨Q⟩=⟨\psi |Q|\psi ⟩$. For an ideal repeatable projective measurement, obtaining $q$ leaves the normalized conditional state $P_q\psi /\sqrt{\mathrm{Pr}(q)}$. This projects onto an entire eigenspace if the outcome is degenerate. General detectors need more general measurement in quantum mechanics models; the projective update is an idealization. Noncommuting observables need not have simultaneous sharp values, as the canonical commutation relation and Heisenberg uncertainty relation illustrate.

Between measurements, an isolated system has unitary time evolution generated by its Hamiltonian operator:This preserves normalization and superpositions. Composite systems use a tensor product of state spaces, allowing entangled states as well as product states; identical particles additionally require the bosonic or fermionic exchange symmetry appropriate to their species. Statistical mixtures are described by a density matrix, with $\mathrm{Pr}(q)=\mathrm{tr}(\rho P_q)$. These rules separate preparation, coherent evolution and measurement.

The photoelectric effect exposes discrete energy transfer by light. Illuminate a clean metal cathode in an evacuated tube and measure the collected Electron current while varying the light frequency, intensity and a retarding voltage. The stopping potential $V_s$ probes the largest emitted Electron kinetic energy, whereas the photocurrent measures the rate at which charge is collected. In the ordinary single-photon regime there is a threshold frequency; above it the maximum energy grows with frequency, rather than with intensity. Changing intensity at fixed frequency mainly changes the number of emitted electrons. These observations are summarized by the Einstein photoelectric equationHere $\Phi$ is the surface work function, and the stopping voltage is understood after any photoelectric contact-potential correction. A photon delivers $h\nu$ in an absorption event; emission is allowed only when that energy can pay the escape cost. This accounts for the frequency threshold and the lack of the classical waiting time associated with slowly accumulating energy in a single Electron. The experiment supports energy quantization of light; it does not imply that every incident photon emits an Electron or that every emitted Electron has the maximum energy. Surface losses and collection geometry affect the full current-voltage curve.

The Davisson–Germer experiment demonstrates a complementary wave property of matter. An Electron gun uses a heated emitter, accelerating voltage and apertures to send a narrow beam onto a nickel single crystal in vacuum. A movable collector measures scattered electric current versus direction; rotating the crystal also tests its azimuthal symmetry. The crystal gives sharp angular beams whose positions depend on voltage and crystal orientation. A prominent feature occurs near $54\mathrm{V}$ accelerating voltage and roughly $50^{\circ }$ detector colatitude. The nonrelativistic de Broglie wavelength predicted at that voltage isDiffraction wavelengths inferred using the known crystal structure agree with this voltage dependence. These angular peaks arise from coherent addition of scattering amplitudes from the crystal, not from classical specular reflection of independent point projectiles. Bragg's law is $2d\sin {\theta }_B=r\lambda$ for lattice-plane spacing $d$ and angle to those planes; a detector colatitude is a different angle and must not be substituted without its geometry. The historical readout was current, not a photograph of individual Electron trajectories.^[ref]

The double-slit experiment makes the role of amplitudes especially direct. A coherent source illuminates two narrow slits; an observation screen a distance $D$ away records the pattern. For slit spacing $d$ and small screen displacement $y$, the path difference is approximately $dy/D$. The two amplitudes add before probabilities are formed:For equal coherent illumination the interference term varies as $\cos (2\pi dy/(\lambda D))$, giving fringe spacing $\lambda D/d$ beneath the single-slit diffraction envelope. Blocking one slit removes that cross term. Young's optical experiment established wave interference; later particle-resolved realizations with photons or electrons show localized detection events whose accumulated distribution has the interference pattern. Low arrival rates demonstrate that this pattern does not require different incident particles to collide with one another.

If the paths become correlated with distinguishable detector states $|D_1⟩$ and $|D_2⟩$, the cross term acquires their overlap $⟨D_2|D_1⟩$. Perfectly distinguishable path records make that overlap zero, suppressing interference even if the spatial beams still overlap. This connects quantum entanglement, measurement and loss of observable coherence. Without a path measurement, the two alternatives are amplitudes for one preparation, rather than classical mutually exclusive events whose probabilities may simply be added.

Together these experiments motivate a description in which energy transfer and detector outcomes can be discrete, yet propagation retains complex quantum phases and interference. The Born rule connects those amplitudes to repeatable frequencies of outcomes. The Schrödinger equation supplies the dynamics once a Hamiltonian operator is specified, while observable operators, state preparation and the measurement model specify what the apparatus can read.

Indistinguishable particles have the same intrinsic properties and cannot be assigned physically observable permanent labels. Exchanging their mathematical coordinates leaves all physical predictions unchanged. In ordinary three-dimensional quantum mechanics the total state, including spin, obeysThe Spin-statistics theorem associates integer spin with bosons and half-integer spin with fermions. Exchange does not mean physically transporting two particles along some prescribed trajectories; it is the symmetry of the many-particle state under permutation of its arguments.

For orthonormal one-particle states $u,v$, normalized two-particle spatial combinations are $[u(1)v(2)\pm v(1)u(2)]/\sqrt{2}$. They illustrate coherent exchange amplitudes. If both fermions have the same complete one-particle state, the antisymmetric combination vanishes. This is the Pauli exclusion principle: a fermionic mode has occupation zero or one. Bosonic modes can have arbitrarily large occupation. A many-fermion product basis is antisymmetrized by a Slater determinant.

It is the full spin-spatial state that must have the appropriate symmetry. For two electrons, an antisymmetric spin singlet state can accompany a symmetric spatial orbital, so opposite-spin electrons can occupy the same spatial orbital. A symmetric spin triplet instead needs an antisymmetric spatial state. Thus exclusion concerns identical complete states, not an unconditional prohibition on sharing a position or spatial orbital.

Counting allowed occupations gives Bose-Einstein statistics and Fermi-Dirac statistics. For an ideal gas in thermal equilibrium,Bosonic accumulation permits Bose-Einstein condensation; exclusion gives a filled Fermi sea and degeneracy pressure even at zero temperature. In the dilute classical limit the occupations are small and both expressions approach the Maxwell-Boltzmann distribution. Exchange effects become important when wavepackets overlap appreciably, often characterized by number density times the cube of the thermal wavelength.

Composite particles follow their total exchange statistics while remaining intact. The composite statistics of neutral atoms depend on electrons as well as nucleons, as in the two lithium isotopes above. Symmetrization can produce bunching or antibunching in joint detection probabilities even in the absence of an interaction potential; it is not an additional classical force.

A diatomic molecule has one internal bond-stretching vibrational coordinate. Near its equilibrium separation, approximate the bond by a spring of stiffness $K$ and use the reduced mass $\mu =m_1{m}_2/(m_1+m_2)$. Its frequency is $\omega =\sqrt{K/\mu }$ and its quantum harmonic oscillator levels are ${ϵ}_r=(r+1/2)\hslash \omega$.

For independent molecules in thermal equilibrium, set $x=\hslash \omega /(k_{B}T)={\theta }_v/T$. The single-molecule canonical partition function is a geometric series:Using $U=-\partial \log (z_v^N)/\partial \beta$, where $\beta =1/(k_{B}T)$, givesPer mole replace $N{k}_B$ by $R$. The zero-point energy is present in $U_v$ but is temperature-independent and contributes nothing to the heat capacity.

For $T\ll {\theta }_v$, occupation of the first excited vibrational state is exponentially suppressed: $C_{V,v}\simeq N{k}_B{x}^2{e}^{-x}\to 0$. The vibrational degree of freedom is frozen out. For $T\gg {\theta }_v$, $C_{V,v}\to N{k}_B$. This recovers classical equipartition: the quadratic kinetic term and quadratic elastic term each contribute $k_B/2$ per molecule.

The vibrational heat capacity of a diatomic gas must be distinguished from the total heat capacity. In a dilute gas with rotations already active and vibration still frozen, translation contributes $3R/2$ and rotation $R$, so molar $C_V\simeq 5R/2$. Once vibration is active it adds $R$, giving $C_V\simeq 7R/2$. For an ideal gas of a fixed number of intact molecules, $C_P=C_V+R$, so the corresponding values are $7R/2$ and $9R/2$. Very low temperatures also freeze rotations, while sufficiently high temperatures bring anharmonicity, dissociation or electronic excitations; the harmonic plateau is restricted to the range where those effects are negligible.

Quantum tunnelling is transmission through a finite classically forbidden region, where $E<V(x)$. The stationary Schrödinger equation permits an evanescent wavefunction inside such a barrier rather than forcing it to vanish at its entrance. Matching to propagating states beyond a barrier of finite width gives a nonzero transmitted probability current.

For a rectangular barrier of height $V_0$ and width $a$, with equal exterior potentials and $0<E<V_0$, write $\kappa =\sqrt{2m(V_0-E)}/\hslash$. The exact rectangular-barrier tunnelling probability isFor an opaque barrier its dominant dependence is $e^{-2\kappa a}$. For a slowly varying barrier, the WKB approximation generalizes this to $T\sim \exp [-(2/\hslash ){\int }_{x_1}^{x_2}\sqrt{2m(V(x)-E)}dx]$, apart from matching prefactors. Barrier width, particle mass and energy therefore have a strong influence. Energy is conserved in a static barrier: tunnelling does not require the particle to borrow energy or acquire a classical path with negative kinetic energy.

In alpha decay, a nuclear alpha-particle configuration is confined by the short-range nuclear interaction but separated from the exterior by a Coulomb barrier. Its energy can be positive relative to infinity while still lying below the barrier maximum. Escape by tunnelling produces alpha particles at that conserved energy, with an exponentially small decay probability per nuclear encounter. The pronounced dependence of half-life on alpha energy is a characteristic consequence of the barrier exponent, unlike a simple classical escape threshold.