Past exam of the physics A course of the University of Cambridge / 2012 / Paper 1 (Physics A (1))

Subtract the two photoelectric effect equations $K_{\max }=hc/\lambda -\Phi$ to eliminate the work function:Here $e$ converts the energy difference from electronvolts to joules. The precision of the rounded stopping energies justifies about two significant figures, $6.7\times 10^{-34}\mathrm{J}\mathrm{s}$.

Let $\omega =\sqrt{k/m}$, $b=\sqrt{\hslash /(m\omega )}$ and $\xi =x/b$. With ${\psi }_0={\pi }^{-1/4}{b}^{-1/2}{e}^{-{\xi }^2/2}$, the next two normalized harmonic-oscillator wavefunctions are ${\psi }_1=\sqrt{2}\xi {\psi }_0$ and ${\psi }_2=(2{\xi }^2-1){\psi }_0/\sqrt{2}$. Their energies are $E_n=(n+1/2)\hslash \omega$, with respectively zero, one and two wavefunction nodes, alternating even and odd parity. The Schrödinger equation givesFor harmonic-oscillator wavefunction inflections, the inflections include both classical turning points and simple wavefunction wavefunction nodes. In dimensionless coordinates the complete sets areThe curvature changes sign at every listed point; none of these states has a wavefunction node coinciding with its turning points.

Resolving a length requires momentum transfer $q$ of order $\hslash /\ell$. Taking a hard relativistic scale $qc\sim 4\mathrm{TeV}$ givesThe full de Broglie wavelength convention gives $hc/E\simeq 3.1\times 10^{-19}\mathrm{m}$, differing by $2\pi$; the intended scale is roughly $10^{-19}\mathrm{m}$, far below a proton radius. Two opposing beams have a larger total collision energy, but proton energy is shared among partons: the actual resolution depends on the hard scattering momentum transfer, not simply on beam energy alone.

For radiative temperature uncertainty after background subtraction, write $P_{\mathrm{back}}=B$, so $P_{\mathrm{total}}\simeq 10B$ and $P\simeq 9B$. The two absolute uncertainties are both $0.1B$. If they are independent standard uncertainties,If the percentages instead denote absolute error bounds, linear worst-case propagation gives 0.56%. With correlation coefficient $\rho$, the standard-uncertainty result is $\sqrt{2(1-\rho )}/360$; subtraction makes positive common errors partly cancel. The independence convention is therefore an assumption, not information supplied by the percentages.

For independent electron arrivals, the shot-noise counting model gives mean transferred charge $Q=I\Delta t=10^{-15}\mathrm{C}$ and count $\bar{N}=Q/e\simeq 6.24\times 10^3$. Poisson statistics implyThis is about 79 electron charges rms. It assumes Poisson shot noise; charge discreteness by itself does not rule out correlations, thermal noise or other apparatus noise.

The canonical commutation relations are $[x,p_x]=[y,p_y]=i\hslash$, with $[x,y]=[p_x,p_y]=[x,p_y]=[y,p_x]=0$. Operators from different Cartesian modes commute, so $(p_x{p}_y{)}^{†}=p_y{p}_x=p_x{p}_y$ and $(xy{)}^{†}=yx=xy$. Thus $J_1$ is Hermitian and represents an observable on the usual oscillator domain.

Introduce annihilation operators $a_x=\sqrt{m\omega /(2\hslash )}x+i{p}_x/\sqrt{2m\hslash \omega }$ and similarly $a_y$. Set $N_x=a_x^{†}{a}_x$, $N_y=a_y^{†}{a}_y$ and $N=N_x+N_y$. The Schwinger boson representation givesFor $A=a_x^{†}{a}_y$ and $B=a_y^{†}{a}_x$, $AB=N_x(N_y+1)$ and $BA=N_y(N_x+1)$. ThereforeThe two-mode oscillator Casimir identity isSince $[N,A]=[N,B]=0$, $[H,J_1]=0$, and $J_1$ is conserved. However, $[J_1,J_2]=i\hslash J_3$ is not the zero operator, so there is no general joint eigenbasis. The exception is important: a common eigenstate of two angular-momentum components must satisfy $J_3\psi =0$, then the other commutators force $J_1\psi =J_2\psi =0$. Thus only the $j=0$ sector admits a common state; here that is the oscillator ground state.

Using $J^2={\hslash }^{2}j(j+1)$, with $j=0,1/2,1,\dots$, and selecting the positive energy branch givesEquivalently $N=2j=0,1,\dots$, so $E_N=\hslash \omega (N+1)$ and the degeneracy is $N+1$, corresponding to $N_x=0,\dots ,N$.

For observables with finite variances, the Heisenberg uncertainty principle gives $\Delta A\Delta B\ge |⟨[A,B]⟩|/2$; in particular $\Delta x\Delta p\ge \hslash /2$. It concerns the spreads of measurements in a prepared state, rather than only disturbance by a measuring device. Identical integer-spin particles have symmetric full states and Bose-Einstein statistics; half-integer-spin particles have antisymmetric full states and Fermi-Dirac statistics. The latter imply Pauli exclusion from one complete single-particle state, including spin.

Inside an infinite square well, solutions satisfying both zero boundary conditions arewith zero wavefunction outside. For the ground state, direct integration givesThe infinite-well ground-state uncertainty product follows:The ground state is not a minimum-uncertainty Gaussian. Here momentum variance is the usual kinetic-momentum second moment of the confined state; the hard-wall momentum operator has domain subtleties that do not alter this kinetic-energy result.

For spin degeneracy in filling a one-dimensional well, let $E_1={\pi }^2{\hslash }^2/(2m{L}^2)$ and assume the spin levels are degenerate and available. Four spin-zero bosons all occupy the lowest orbital, giving $4E_1$. Four spin-$1/2$ fermions fill two spin states in each of $n=1,2$, giving $10E_1$. Spin-$3/2$ fermions have four spin projections, so all four can occupy $n=1$ in different complete states, giving $4E_1$. These fillings do not impose full spin polarization.

matrix multiplication gives ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{ϵ}_{ijk}{\sigma }_k$, hence $[{\sigma }_i,{\sigma }_j]=2i{ϵ}_{ijk}{\sigma }_k$. For physical neutron spin, use $\mathbf{S}=(\hslash /2)\mathbf{\sigma }$, so $[S_i,S_j]=i\hslash {ϵ}_{ijk}{S}_k$, $S^2=3{\hslash }^2/4$, and measurement along any axis gives $\pm \hslash /2$.

For the specified field, $H=(\alpha B\hslash /2){\sigma }_y$. The Time-dependent Schrödinger equation is $i\hslash \dot{\psi }=H\psi$ with $\psi (0)=(1,0{)}^T$. The uniform-field spin precession follows from ${\sigma }_y^2=I$:

The sign of the precession is contained in the given $\alpha$. The normalized $x$-axis eigenstates are $(1,1{)}^T/\sqrt{2}$ and $(1,-1{)}^T/\sqrt{2}$.

The measurement direction is $\mathbf{n}=(0,\sin \theta ,\cos \theta )$, as confirmed by the original diagram. Write $c=\cos (\theta /2)$ and $s=\sin (\theta /2)$. The spinors ${\chi }_{+}=(c,is{)}^T$ and ${\chi }_{-}=(is,c{)}^T$ have unit norm and inner product $ics-isc=0$. Applying $\mathbf{n}\cdot \mathbf{\sigma }$ gives eigenvalues $+1$ and $-1$. At $\theta =0$ they are the usual $z$ basis; at $\pi /2$ they are the $y$ basis up to a phase; at $\pi$ the $z$ up/down roles are exchanged, with a factor $i$; and at $2\pi$ both return with a minus sign. A spinor changes sign under a full rotation, while its isolated-state measurement probabilities are unchanged.

For spin measurement in the yz plane, projection givesFor $\theta =0$ these are ${\cos }^2\varphi ,{\sin }^2\varphi$; for $\pi /2$ both are $1/2$ at all times; for $\pi$ they are exchanged; and for $2\pi$ they are the same as at zero angle. The constant $y$ probabilities also follow from conservation of $S_y$.

For the two inline operator labels, (a) $L_z$ has eigenvalue $m_{\ell }\hslash$, and (b) $L^2$ has eigenvalue $\ell (\ell +1){\hslash }^2$, with $\ell =0,1,\dots$ and $m_{\ell }=-\ell ,\dots ,\ell$. The subscript distinguishes the magnetic quantum number from the electron mass.

Put $b=a_0(\ell +1)$ and $\kappa =e^2/(4\pi {ϵ}_0)$. For the proposed radial shape,The angular-momentum term cancels the inverse-square part. The inverse-radius coefficient vanishes if ${\hslash }^2(\ell +1)/(mb)=\kappa$. ThereforeThis nodeless Coulomb radial eigenstate has principal quantum number $n=\ell +1$, not the complete set of hydrogen radial states. Since the spherical harmonic is normalized, radial normalization requiresup to an arbitrary phase.

For the hydrogen ground-state nuclear-volume probability, use $R=2a_0^{-3/2}{e}^{-r/a_0}$. With $u=r_p/a_0\ll 1$,No numerical proton radius is supplied or required. A sharp proton boundary is an approximation to a charge distribution, and the point-Coulomb potential changes inside an extended proton. Finite proton mass, relativistic and electromagnetic corrections also limit the ideal wavefunction's precision.

Classically the kinetic energy must satisfy $E-V(r)\ge 0$. For $E=-\kappa /(2a_0)$ this demands $r\le 2a_0$. The hydrogen ground-state probability beyond the classical turning point isA substantial probability lies beyond the classical turning radius; the bound-state wavefunction decays there rather than vanishing.

angular momentum generates rotations. Its components obey $[J_i,J_j]=i\hslash {ϵ}_{ijk}{J}_k$, and $J^2$ commutes with each component. A compatible basis has $J^2={\hslash }^{2}j(j+1)$ and $J_z=m\hslash$, where $m=-j,-j+1,\dots ,j$. ladder operators change $m$ by one, terminate at $\pm j$, and give $2j+1$ magnetic substates. Different Cartesian components generally cannot all be sharp simultaneously.

orbital angular momentum is $\mathbf{L}=\mathbf{r}\times \mathbf{p}=-i\hslash \mathbf{r}\times \nabla$. Its angular eigenfunctions are spherical harmonics, with integer $\ell$ arising from their spatial rotation properties and single-valuedness. spin is intrinsic angular momentum, not literal rotation of a finite charged sphere. It may be half-integer; spin $1/2$ is represented by $\mathbf{S}=\hslash \mathbf{\sigma }/2$. For several contributions, $\mathbf{J}=\mathbf{L}+\mathbf{S}$ obeys the same algebra. Coupling angular momenta $j_1,j_2$ permits $J=|j_1-j_2|,\dots ,j_1+j_2$, with Clebsch-Gordan coefficients relating product and coupled bases. spin-orbit coupling can distinguish these total-angular-momentum sectors.

The Stern-Gerlach apparatus sends a collimated neutral atomic beam through an inhomogeneous magnetic field and records its deflection. A magnetic moment experiences $F_z\simeq {\mu }_z\partial B_z/\partial z$; field uniformity would produce precession but no analogous net separating force. A classically continuous distribution of moment orientations would broaden the beam continuously. Gerlach and Stern instead observed separated silver-beam components. Reversing or removing the field and comparing the undeflected profile controls the magnetic origin; collimation, velocity spread and finite resolution affect the spots. The original experiment demonstrated space quantization; its modern silver interpretation uses the unpaired electron spin with nearly zero orbital angular momentum, rather than a direct measurement of a free electron's trajectory.^[ref]

For two spin-$1/2$ particles, the triplet state sector has $S=1$ and symmetric spin statesThe spin singlet state has $S=0$ and antisymmetric spin stateTheir total $S^2$ values are $2{\hslash }^2$ and zero, while ${\mathbf{S}}_1\cdot {\mathbf{S}}_2$ has values ${\hslash }^2/4$ and $-3{\hslash }^2/4$. The singlet is rotationally invariant and gives anticorrelated spin measurements along a common axis. For identical fermions the complete state is antisymmetric: a symmetric spatial state pairs with a singlet spin state, whereas an antisymmetric spatial state pairs with triplet spin. Orbital, spin and coupled angular momenta share an algebra, but their physical origins and allowed representations differ.

In a finite potential barrier with $E<V$, the stationary Schrödinger equation has exponentially decaying and growing solutions, with local decay constant $\kappa =\sqrt{2m(V-E)}/\hslash$. Matching the wavefunction and its derivative at finite steps produces a transmitted probability current even though a classical particle at that energy could not cross. energy remains conserved. For a smooth thick barrier, the leading WKB transmission factor isFor a rectangular barrier of height $V_0$, width $d$ and $0<E<V_0$, exact matching gives $T=[1+V_0^2{\sinh }^2(\kappa d)/(4E(V_0-E)){]}^{-1}$. Thickness and energy therefore affect transmission exponentially in the opaque limit; prefactors and near-turning-point corrections require more than the leading exponent.

alpha decay illustrates escape from a nuclear well through a Coulomb barrier. A rough decay rate is an internal encounter frequency times barrier penetrability, multiplied by the probability of forming the outgoing cluster; penetrability alone is not a complete nuclear lifetime calculation. In a scanning tunnelling microscope, electron transmission through the vacuum gap gives an approximately exponential distance dependence $I\propto e^{-2\kappa d}$ at fixed conditions. A feedback loop adjusts tip height to maintain current, providing surface information. Bias, tip electronic structure and local electronic density of states also affect the current, so constant-current height is not purely geometric topography. Both applications exploit the same sensitivity of transmission to a classically forbidden region.

The Time-dependent Schrödinger equation $i\hslash {\partial }_t|\psi ⟩=H(t)|\psi ⟩$ governs unitary evolution of a closed system. If $H$ is independent of time, expand in its energy eigenstates:An individual energy eigenstate changes only by a phase, so its probability density and time-independent-observable expectations are stationary. Superpositions can have interference terms oscillating at $(E_n-E_m)/\hslash$; fixed energy populations do not imply that every observable is constant.

For general $H(t)$, $U(t,t_0)=\mathcal{T}\exp [-(i/\hslash )\int H(t^{\prime })d{t}^{\prime }]$ requires time ordering unless Quantum Hamiltonians at different times commute. The equivalent Heisenberg description evolves observables rather than the state, giving $d⟨A⟩/dt=(i/\hslash )⟨[H,A]⟩+⟨{\partial }_{t}A⟩$. Explicit time dependence can transfer energy from an applied perturbation.

For a weak perturbation $V(t)$ to a known $H_0$, the first-order transition amplitude from $n$ to $m\ne n$ isIts squared modulus gives the leading transition probability while depletion remains small. A nearly monochromatic perturbation emphasizes energy differences near its quantum of frequency; a continuum and suitable long-time limit yield Fermi's golden rule. These approximations have stated weak-coupling and time-resolution limits; they do not replace exact unitary dynamics in every regime.

For a photon scattering through angle $\theta$ from a free electron initially at rest, conserve energy and momentum. Eliminating the final electron momentum by its relativistic energy relation givesor $E^{\prime }=E/[1+(E/m{c}^2)(1-\cos \theta )]$. The recoil electron gains kinetic energy $E-E^{\prime }$. The shift is zero forward, one electron Compton wavelength at $90^{\circ }$, and twice that wavelength backward.

Compton's X-ray scattering measurements compared scattered spectra at different angles and identified an angle-dependent shifted component. This supports photon momentum $p=E/c$ and inelastic recoil, beyond the unchanged-frequency prediction of classical elastic Thomson scattering. Bound electrons and coherent scattering can also produce an unshifted component; initial momentum and binding broaden or modify the ideal free-electron result. Treating an electron as initially free and stationary is therefore a kinematic approximation appropriate when the relevant photon transfer dominates binding scales.^[ref]

A chi-squared test compares discrepancies with their expected random size. For independent Gaussian observations with known standard deviations, a fitted model givesUnder a correct regular model with $p$ fitted parameters, its reference statistical degrees of freedom are usually $\nu =n-p$. Correlated errors instead require ${\chi }^2={\mathbf{r}}^T{C}^{-1}\mathbf{r}$ with the correct covariance matrix. The reduced statistic has mean about one and standard deviation $\sqrt{2/\nu }$, so it should not be forced to equal exactly one.

For binned counts, Pearson's statistic is ${\sum }_i(O_i-E_i{)}^2/E_i$. With a fixed total and $k$ populated categories, fitting $p$ distribution parameters generally gives $\nu =k-1-p$. The approximation needs adequately populated expected bins and the appropriate dependence assumptions. Sparse data may require a Poisson likelihood, exact calculation or simulated reference distribution rather than a casual chi-squared approximation.

The upper-tail probability $p_{\mathrm{value}}=P({\chi }_{\nu }^2\ge {\chi }_{\mathrm{obs}}^2)$ measures how unusual the discrepancy would be under the null model; it is not the probability that the model is true. A large statistic can indicate a wrong model, underestimated noise or unmodelled systematic effects; an unusually small one can indicate overestimated errors or correlations. Residual plots and checks of the noise model are needed alongside the numerical test.

For an experimental-noise interpretation of Nyquist's theorem, a resistor $R$ in thermal equilibrium at temperature $T$ has one-sided open-circuit voltage-noise spectral densityin the classical range $hf\ll k_{B}T$, over an approximately rectangular bandwidth $\Delta f$. Across a matched load, half the open-circuit voltage appears, giving available noise power $k_{B}T\Delta f$. This is thermal agitation, present without a mean current, and is distinct from shot noise. A real readout weights the spectrum with its transfer function: $⟨v_{\mathrm{out}}^2⟩={\int }_0^{\infty }{S}_V(f)|H(f){|}^{2}df$. Cooling, reduced bandwidth and suitable impedance/noise matching reduce measured noise; the resistor's noise does not simply disappear in an ideal amplifier model.^[ref]

The bare name also has a sampling interpretation. The Nyquist–Shannon sampling theorem permits exact reconstruction of a bandlimited signal from sufficiently fast uniform samples. The practical condition is $f_s>2f_{\max }$, with an anti-aliasing filter before sampling; otherwise distinct continuous frequencies produce identical samples. Ideal reconstruction uses shifted sinc functions. Critical-rate equality needs an appropriate signal class and can fail for unrestricted edge-frequency sinusoids with unknown phase. Sampling fast enough prevents aliasing, but does not remove thermal noise. This distinction covers the source's unspecified use of “Nyquist's theorem”.^[ref]

A closed conducting enclosure redistributes surface charge so that an empty interior is protected from an external static electric field. Grounding fixes a reference and drains deposited charge, but an ideal isolated closed conductor also provides inward electrostatic shielding. Real seams, apertures and cables need careful bonding, shielding and feedthrough filtering. For time-varying fields the relevant wavelength, conductivity, geometry and penetration depth matter; a static argument alone is insufficient. Differential wiring and small pickup-loop areas can complement shielding, while careless multiple ground connections create unwanted current loops.

Ordinary copper or aluminium does not generally exclude a static magnetic field. High-permeability shells instead guide magnetic flux around the protected volume. Their effectiveness depends on shape, continuity, openings and saturation; magnetic history and mechanical stress can worsen performance. Nested shields and degaussing can reduce residual fields. Superconducting shielding can exploit the Meissner effect, but trapped flux, openings and critical-field limits require attention.

At nonzero frequency, eddy currents in conducting shields oppose changing magnetic flux. The skin depth isfor a good conductor in the usual diffusion approximation. Thickness of several skin depths suppresses penetration, but the very-low-frequency limit is weak for a normal metal. High-permeability and conducting layers can therefore complement each other. Active compensation coils are an additional method, not passive shielding. Test performance by applying known fields along relevant axes and measuring the residual field or induced signal over the apparatus bandwidth; room-temperature shielding materials and electronics can themselves add noise.

The ideal operational-amplifier rules are zero input currents and, when stable negative feedback maintains unsaturated linear operation, equal input voltages. Infinite open-loop gain motivates the latter; it must not be assumed to guarantee asymptotic stability for every external network.

For the numerical sketch and usefulness discussion, use the transfer function derived in part (c). The chosen values give $K=5$ and $\tau =RC=10^{-4}\mathrm{s}$. With $x=\omega \tau$,The formal amplitude falls from 5 to zero, with high-frequency slope $-40$ dB per decade. Its unwrapped phase rises from zero through $+90^{\circ }$ at $\omega =10^4{\mathrm{s}}^{-1}$ to $+180^{\circ }$; the corresponding characteristic frequency is 1.59 kHz. At that frequency the amplitude is 2.5.

These are formal sinusoidal particular-solution curves, not a stable settled filter response. For the formal sinusoidal response of an unstable linear system, replacing $i\omega$ by the Laplace variable $s$ gives $G(s)=5/(1-s\tau {)}^2$, with a repeated pole at $s=+1/\tau$. A homogeneous output contains $(A+Bt)e^{t/\tau }$ and grows, so real perturbations lead to supply-limited nonlinear behaviour. With equal positive resistors and capacitors in the frequency-setting network, the general denominator is $1+(3-K)s\tau +s^2{\tau }^2$: the Sallen-Key equal-component stability bound is $K<3$. For example $K=3-\sqrt{2}$ gives a Butterworth filter response. Thus the Sallen-Key low-pass filter architecture is useful for active filtering with suitable gain and damping, but the stated $K=5$ component set is unsuitable for linear filtering. The component values are retained in this diagnosis.^[ref]

Zero capacitance removes the capacitor branches. No current enters the non-inverting input, so no current flows through $R_1$ or $R_2$ and $V_{+}=V_{\mathrm{in}}$. The inverting divider gives $V_{-}=V_{\mathrm{out}}{R}_3/(R_3+R_4)$. Equating the two input voltages in the linear solution gives

Let $V_B$ be the non-inverting input voltage. The feedback divider gives $V_B=V_{\mathrm{out}}/K$. With $s=i\omega$, current conservation at this wavefunction node gives $(V_A-V_B)/R_2=s{C}_2{V}_B$, because the amplifier input draws no current. Hence

Current conservation at A, with the capacitor $C_1$ connected from A to the output as verified in the PDF, givesSubstitute $V_A=(1+s{R}_2{C}_2)V_B$ and $V_{\mathrm{out}}=K{V}_B$:Therefore, at $s=i\omega$,This algebraic transfer function alone does not establish that transients decay; the pole check in the parent solution is necessary for the requested numerical choice.