Past exam of the physics A course of the University of Cambridge / 2016

Use the unitary position-to-momentum Fourier transform $\varphi (p)=(2\pi \hslash {)}^{-1/2}\int \psi (x)e^{-ipx/\hslash }dx$. Integration giveswhere $\mathrm{sinc}u=\sin u/u$ with its continuous value one at zero. The amplitude is real and even, with central maximum $\sqrt{a/(2\pi \hslash )}$, zeros at $p=2n\pi \hslash /a$ for nonzero integer $n$, and alternating-sign sidelobes decaying as $1/|p|$.

The probability density is $|\varphi {|}^2$, not the signed Sinc function amplitude. Its integral is one by the unitarity of the Fourier transform.

Nonrelativistic acceleration from rest gives $p=\sqrt{2m_{e}eV}$ and $\lambda =h/p$. The observed deflection is the angle between incident and diffracted beams, $2\theta =22^{\circ }$, so Bragg's law at first order givesHere $\lambda \simeq 0.100\mathrm{nm}$. The numerical evaluation uses the externally documented CODATA 2014 values $h=6.626070040\times 10^{-34}\mathrm{J}\mathrm{s}$, $m_e=9.10938356\times 10^{-31}\mathrm{kg}$ and $e=1.6021766208\times 10^{-19}\mathrm{C}$; it does not assert which rounded constants were in the exam's handbook. Relativistic corrections at this voltage are negligible at the precision of the angle.^[ref]

If $\hat{A}|a⟩=\lambda |a⟩$, the Hermitian operator property makes $\lambda$ real. Applying the operator identity to this eigenvector gives ${\lambda }^3=1$. Its only real root is $\lambda =1$; the other cube roots of unity are nonreal. In an orthonormal eigenbasis, $\hat{A}={\sum }_a|a⟩⟨a|=\boxed{\hat{I}}$. Equivalently, the spectral theorem places the entire spectrum of the self-adjoint observable at one, giving the same conclusion. The quantum-mechanical setting assumes the usual well-defined observable and operator domains.

At the central values, $z=\sqrt{\tan 45^{\circ }}+0.10^2=1.010$. For independent input standard errors, linear propagation givesThe angular derivative here is with respect to radians, so ${\sigma }_x=2\pi /180=0.03491$. At $x=\pi /4$, the derivative is one, andThus $\boxed{z=1.010\pm 0.035}$, with the stated measurement uncertainty interpreted as one standard error. Correlated inputs would require a covariance term; their correlation is not supplied.

For independent charge arrivals described by a Poisson distribution, the one-sided Shot noise current Power spectral density is $S_I=2eI$. Treating the stated bandwidth as the Effective noise bandwidth,The expected rms noise is $5.66\mathrm{pA}$, about $0.566\%$ of the mean current. This is the full Shot noise value with Fano factor one; correlations can suppress or enhance it. The elementary charge is from the external CODATA 2014 table.^[ref]

For the oppositely travelling components, write $u=c_1{e}^{ikx}$ and $v=c_2{e}^{-ikx}$ after removing the common time phase. Then ${\partial }_x\psi =ik(u-v)e^{-i\omega t}$. In $\mathrm{Im}({\psi }^{\ast }{\partial }_x\psi )$ the interference terms contribute no imaginary part, soA standing-wave interference pattern can therefore change density without adding a cross-term to the net current.

For barrier-top scattering, the stationary equation inside the barrier is $-{\hslash }^2{\psi }_2^{\prime \prime }/(2m)=0$. Thus its spatial part is linear: ${\psi }_2=(bx+c)e^{-i\omega t}$. Take incident amplitude one, $k=\sqrt{2mE}/\hslash$ and $\omega =E/\hslash$. Continuity of the spatial wave function and its derivative at the two finite steps givesSolving these four equations yieldsThe equal exterior velocities mean flux ratios are squared amplitude ratios:This is the threshold limit of finite-barrier scattering, not exponential decay with a nonzero interior decay constant.

The interior constants areSince $\mathrm{Im}[(bx+c{)}^{\ast }b]=\mathrm{Im}(c^{\ast }b)=k/(1+k^2{a}^2)$,It is positive and independent of position, matching the transmitted flux. Although the classical kinetic energy inside is zero, the complex linear quantum wave function carries conserved probability current; its phase varies with position.

Differentiating the expectation value and using ${\partial }_t|\psi ⟩=-i\hat{H}|\psi ⟩/\hslash$ and ${\partial }_t⟨\psi |=i⟨\psi |\hat{H}/\hslash$ givesThe second term means explicit operator time dependence, distinct from the state's evolution.

The commutator product rule follows by adding and subtracting $\hat{B}\hat{A}\hat{C}$:For the mathematical induction step, let $\hat{D}=[\hat{A},\hat{B}]$ and use $[\hat{A},\hat{D}]=0$. The case $n=1$ is immediate. If $[{\hat{A}}^m,\hat{B}]=m\hat{D}{\hat{A}}^{m-1}$, thenThus the claimed identity holds for every positive integer. Commutation of $\hat{B}$ with $\hat{D}$ is not needed for this particular step. This is the central-commutator differentiation rule. Termwise application to a convergent power series, on a common domain where the operations are justified, gives

For position and momentum, $[\hat{x},\hat{p}]=i\hslash \hat{I}$. Hence $[{\hat{p}}^2,\hat{x}]=-2i\hslash \hat{p}$, $[V(\hat{x}),\hat{x}]=0$ and $[V(\hat{x}),\hat{p}]=i\hslash V^{\prime }(\hat{x})$. Substitution gives Ehrenfest theorem:The mean momentum changes by the mean force. It does not follow for a general nonlinear Potential energy that $⟨V^{\prime }(\hat{x})⟩=V^{\prime }(⟨\hat{x}⟩)$: a closed classical trajectory for the mean position requires an additional condition, such as a quadratic Potential energy or a sufficiently localized wave packet.

Expanding with the angular momentum commutators givesFor ${\hat{L}}_z|\psi ⟩=\alpha \hslash |\psi ⟩$, commuting ${\hat{L}}_z$ past a ladder operator givesThus the raised or lowered vector, if nonzero, is an eigenvector with the shifted eigenvalue; an angular momentum ladder endpoint can instead be annihilated. Also $[{\hat{L}}^2,{\hat{L}}_{\pm }]=0$, so ${\hat{L}}^2{\hat{L}}_{\pm }|\psi ⟩=\Lambda {\hslash }^2{\hat{L}}_{\pm }|\psi ⟩$ for every nonzero ladder result.

For orbital angular momentum, ${\hat{L}}_z=-i\hslash {\partial }_{\varphi }$ gives an azimuthal factor $e^{i\alpha \varphi }$. Single-valuedness under $\varphi \mapsto \varphi +2\pi$ requires integer $\alpha$. Positivity of ladder norms givesfor normalized states. A fixed finite $\Lambda$ therefore has upper and lower endpoints. Writing the upper endpoint as $m=\ell$, its vanishing raising norm gives $\Lambda =\ell (\ell +1)$; the lower endpoint is $m=-\ell$. Consequentlywith $\ell =0,1,2,\dots$. This integer conclusion is for orbital angular momentum; spin representations can also have half-integer labels.

For the specified Gaussian-radial state, use a Cartesian p-orbital decomposition: write $x+y+z=r(\sin \theta \cos \varphi +\sin \theta \sin \varphi +\cos \theta )$. Its angular factor isOnly $\ell =1$ occurs. The squared angular coefficients are $2,2,2$ before common normalization, so each allowed $m$ has probability one third. Therefore the simultaneous outcome isThe common normalized radial factor cancels from this ratio. The standard symmetric $m$ sequence used above resolves the duplicated central positive entry in the printed list.

Direct matrix multiplication givesThe normalized $S_x$ eigenvectors are $(|\uparrow ⟩\pm |\downarrow ⟩)/\sqrt{2}$, and the $S_y$ eigenvectors are $(|\uparrow ⟩\pm i|\downarrow ⟩)/\sqrt{2}$, with eigenvalues $\pm \hslash /2$ in each case.

For $N=|a{|}^2+|b{|}^2>0$, allowing for a possibly unnormalized input pair of coefficients,For a normalized state $N=1$. The second result follows because each squared Spin operator matrix equals ${\hslash }^2\hat{I}/4$.

The tilted component is ${\hat{S}}_{z^{\prime }}=(\hslash /2)\left(\begin{array}{cc}\cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{array}\right)$. Its eigenvalues remain $\pm \hslash /2$, with one real phase convention for the normalized eigenvectorsProjecting the initial $|\uparrow ⟩$ onto them gives $P_{+}={\cos }^2(\theta /2)$ and $P_{-}={\sin }^2(\theta /2)$. ThusA smaller expectation does not imply a smaller individual spin eigenvalue; it describes the probabilities of the same two measurement outcomes.

identical particles have the same intrinsic properties, including mass, charge and spin. Their labels are bookkeeping devices rather than measurable identities. When wave functions overlap, quantum mechanics supplies no observable trajectory that permanently identifies “particle 1” independently of “particle 2”. A many-particle description must therefore treat exchange consistently. The Hamiltonian and physical observables are invariant under relabelling identical particles.

For two particles, the exchange operator obeys ${\hat{P}}_{12}^2=\hat{I}$. The usual three-dimensional symmetrization postulate selects states symmetric under exchange for bosons and antisymmetric for fermions. In relativistic quantum theory, the Spin-statistics theorem associates integer spin with bosons and half-integer spin with fermions under its locality and positivity assumptions; the connection is not derived merely from ${\hat{P}}_{12}^2=\hat{I}$ in elementary nonrelativistic mechanics. For orthonormal one-particle states $u$ and $v$, the two-particle spatial combinations areThe sign concerns the complete state, including spin and other internal variables, not necessarily the spatial factor alone.

The antisymmetric state vanishes if both fermions occupy the same one-particle state: this is the Pauli exclusion principle. For $N$ fermions in orthonormal spin orbitals ${\chi }_j$, the normalized Slater determinant $\Psi =(N!{)}^{-1/2}\det [{\chi }_j(i)]$ supplies the required exchange sign. Bosonic occupation has no corresponding upper limit, and symmetrized products describe many bosons. Fermionic occupation is zero or one per complete one-particle state; two electrons can still occupy the same spatial atomic orbital with opposite spin.

Spin and spatial symmetry are linked. A two-electron spin singlet state is antisymmetric, so its spatial factor must be symmetric. A symmetric triplet state instead requires an antisymmetric spatial factor. Exchange then changes pair correlations and the energies available when interactions depend on particle separation, even without introducing a new classical “exchange force”. It contributes to atomic shell structure, bonding and magnetic behaviour.

In an occupation-number description, bosonic creation and annihilation operators satisfy $[{\hat{a}}_i,{\hat{a}}_j^{†}]={\delta }_{ij}$, whereas fermionic operators satisfy $\{{\hat{c}}_i,{\hat{c}}_j^{†}\}={\delta }_{ij}$ and anticommute among themselves. The latter imply $({\hat{c}}_i^{†}{)}^2=0$. This formalism implements the same exchange constraints without assigning permanent particle labels.

For noninteracting particles in thermal equilibrium, the mean occupation of a state of energy $ϵ$ isFermi-Dirac statistics produces a filled Fermi sea at zero temperature and degeneracy pressure; Bose-Einstein statistics permits macroscopic Ground state occupation when the density of available states and thermodynamic conditions allow condensation. The different statistics therefore have measurable collective consequences, not just different notation.^[ref]

When phase-space occupation is dilute, both distributions reduce to the Maxwell–Boltzmann statistics limit. Likewise, well-separated localized wave packets have negligible overlap and small exchange corrections, so effectively labelled classical trajectories can become a useful approximation. In two spatial dimensions more general exchange statistics are possible; the standard boson/fermion account here concerns ordinary three-dimensional particles and their total states.

For the nonrelativistic Coulomb Hamiltonian with reduced electron-proton mass $\mu$, separation in Spherical coordinate system gives ${\psi }_{n\ell m}(r,\theta ,\varphi )=R_{n\ell }(r)Y_{\ell m}(\theta ,\varphi )$. The quantum numbers obey $n\ge 1$, $0\le \ell \le n-1$ and $-\ell \le m\le \ell$. These are joint eigenstates of energy, $L^2$ and $L_z$, with the latter two eigenvalues $\ell (\ell +1){\hslash }^2$ and $m\hslash$.

The Coulomb bound-state energies and characteristic length areFor example, ${\psi }_{100}=(\pi a_{\mu }^3{)}^{-1/2}{e}^{-r/a_{\mu }}$ is spherically symmetric and has no radial node. In general the radial part contains an exponential times a Generalized Laguerre polynomial, with $n-\ell -1$ radial nodes. The spherical harmonic specifies angular structure and azimuthal phase; real linear combinations of degenerate $m$ states give familiar directional orbitals.

The leading Coulomb energy depends on $n$ alone, giving $n^2$ spatial states for each principal shell, or $2n^2$ including electron spin. Fine structure, Lamb shift and Hyperfine structure lift parts of this ideal degeneracy. An atomic orbital is a probability amplitude, not a classical planetary path.^[ref]

Quantum tunnelling is supported by transport and decay through barriers that classical particles of the same energy cannot traverse. Three complementary examples are:

Together these test barrier penetration in nuclear decay, vacuum-gap transport and semiconductor transport. Their detailed currents depend on state supply and matching as well as the barrier transmission probability.

For the Quantum harmonic oscillator defineThe Canonical commutation relation gives $[\hat{a},{\hat{a}}^{†}]=1$, andThus the creation operator and annihilation operator raise or lower the energy by one quantum. Positivity of ${\hat{a}}^{†}\hat{a}$ and termination of downward stepping give $E_n=\hslash \omega (n+1/2)$, $n=0,1,\dots$. Normalized actions are $\hat{a}|n⟩=\sqrt{n}|n-1⟩$ and ${\hat{a}}^{†}|n⟩=\sqrt{n+1}|n+1⟩$. The annihilated ground state obeysThe nonzero Ground state energy is the zero-point energy, consistent with position-momentum quantum uncertainty. Since $\hat{x}=\sqrt{\hslash /(2m\omega )}(\hat{a}+{\hat{a}}^{†})$, a perturbation linear in position connects only neighbouring number states. The algebra is especially useful for matrix elements and selection rules.^[ref]

TODO: biology deferred by the user; all five assessed organism-growth tasks are deferred.

The ideal voltage-feedback Operational amplifier has zero input currents and, while stable negative feedback holds it in its linear operating range, equal input voltages: $i_{+}=i_{-}=0$ and $v_{+}=v_{-}$. The first rule follows from infinite input impedance; the second is the large-open-loop-gain limit of $v_o=A(v_{+}-v_{-})$. Equal input electric potential is a virtual short, not an electrically conducting connection.

For an inverting amplifier with the noninverting input grounded, $v_{-}=0$ and node-current balance gives $v_i/R_i=-v_o/R_f$, hence $v_o=-R_f{v}_i/R_i$. For a noninverting amplifier, the feedback divider gives $v_{-}=v_o{R}_g/(R_f+R_g)=v_i$, hence $v_o=(1+R_f/R_g)v_i$.

One should check output swing, input common-mode range, slew rate, Gain-bandwidth product and stability before using these limits in a real circuit. Saturation, positive feedback or operation beyond bandwidth invalidates the equal-voltage rule; finite input bias currents and offset voltage can also matter.^[ref]

Thermal agitation in a resistor produces Johnson-Nyquist noise, even without an applied mean current. In the classical low-frequency regime, its one-sided open-circuit voltage spectrum is $S_V=4k_{B}TR$ and its equivalent short-circuit current spectrum is $S_I=4k_{B}T/R$. In an Effective noise bandwidth $B$,For a matched load, available thermal noise power is $k_{B}TB$. A resistor and load at the same temperature exchange noise power without a net heat flow.

Cooling reduces the rms amplitude as $\sqrt{T}$ if Electrical resistance and bandwidth remain fixed. A cryogenic bath or refrigerator, thermal anchoring and controlled heat input can maintain a steady low temperature. Radiation shields, an insulating vacuum, thermally anchored leads and limited measurement dissipation reduce heat leaks and self-heating. Temperature sensing and stable feedback control reduce slow temperature-induced drift; keeping temperature constant does not remove microscopic thermal fluctuations.

Bandwidth restriction, suitable impedance and a low-noise amplifier complement cooling. Warm leads, contacts and the amplifier can otherwise dominate the observed noise. At fixed mean current, shot noise is not generally reduced merely by cooling, and low-frequency excess noise need not follow the Johnson law. At frequencies with $hf\gtrsim k_{B}T$, quantum noise expressions replace the classical approximation.^[ref]

Uniform signal sampling at rate $f_s$ multiplies a continuous signal by a time-domain Dirac comb, giving frequency-domain replicas spaced by $f_s$:If a baseband signal is restricted to $|f|\le B$ and $f_s>2B$, these replicas can be separated by an ideal reconstruction filter. Frequencies differing by integer multiples of $f_s$ have the same sampled phase progression; for real signals they fold into the interval from zero to $f_s/2$. This ambiguity is aliasing.

For example, $0.8f_s$ and $0.2f_s$ cosines produce identical sample values. Apply an analogue Anti-aliasing filter before digitization and choose enough sample-rate margin for its finite transition band. A digital filter applied afterward cannot distinguish components already folded onto one another. Out-of-band noise also aliases unless filtered. The sample rate sets temporal/spectral coverage; quantization resolution sets amplitude precision and is a separate limitation. Bandpass Signal sampling has additional conditions beyond the simple baseband rule.^[ref]

Original source: Original released paper, physical PDF pages 120–127. All questions and all three D13 alternatives are covered.

For a lossless antireflection coating, cancellation requires both equal reflected wave amplitudes and an odd half-cycle difference in their round-trip wave phase. At normal incidence the interface amplitude coefficients are $r_{01}=(n_0-n)/(n_0+n)$ and $r_{12}=(n-n_s)/(n+n_s)$. Choosing $n_0<n<n_s$ makes both negative: both reflections acquire the same $\pi$ phase reversal. A quarter-wave optical thickness supplies their remaining relative phase $\pi$.

Equal amplitudes require $r_{01}=r_{12}$, which reduces to $n^2=n_0{n}_s$. ThusThe exact multilayer reflection amplitude is zero for these choices, not merely the sum of the first two reflected rays: its numerator is $r_{01}+r_{12}{e}^{2i\beta }=0$, with $\beta =2\pi nd/{\lambda }_0=\pi /2$. Any positive odd multiple of this thickness has the same cancellation at the design wavelength; the smallest is the thin-layer choice. absorption, index dispersion, angular spread and deviations from the selected wavelength limit the practical cancellation.

A Fresnel zone plate replaces a conventional refracting surface by concentric alternating transmitting and absorbing annuli. Successive annuli span half a wavelength of optical-path excess to the chosen focus; blocking alternate zones prevents their approximately opposite wave phases from cancelling. For a normally incident plane wave,The paraxial radii grow as $\sqrt{j}$, so the zones become narrower outwards.

Using the photon energy relation $\lambda =hc/E$ gives $\lambda =4.133\times 10^{-11}\mathrm{m}=0.04133\mathrm{nm}$. For $f=10^{-2}\mathrm{m}$,The first three boundaries are approximately $0.6429$, $0.9092$ and $1.1135\mu \mathrm{m}$. This fixes the zone pattern but does not fix a unique outside diameter: the number of zones is a design choice. For example, $N=100$ gives radius $6.429\mu \mathrm{m}$, diameter $12.86\mu \mathrm{m}$ and outermost zone width about $32.2\mathrm{nm}$. A larger usable diameter requires finer outer zones.

One could pattern absorbing metal annuli on a thin X-ray-transparent support, choosing absorber thickness from its attenuation at the selected photon energy; that thickness cannot be determined from energy and focal length alone. The design is chromatic: for fixed radii $f\propto 1/\lambda$, so a narrow energy band is needed for a sharp common focus.

Let $x$ be mirror displacement from equal optical paths in a Michelson interferometer, so the optical path difference is $2x$. Different spectral components add detected light intensities; each component interferes with its own delayed copy. For equal line strengths and ideally balanced arms, writeThe trigonometric identity for the sum of cosines givesThus the sketch consists of rapid fringes within a slowly varying contrast envelope. With $\bar{\lambda }$ the mean wavelength and $|\Delta \lambda |\ll \bar{\lambda }$, the mirror-displacement fringe spacing is $\bar{\lambda }/2$. The interference visibility isThe first contrast zero is at $x=1/(4\Delta s)$; successive visibility maxima are separated byBetween successive contrast zeros the signed envelope changes sign, shifting the rapid fringes by half a cycle. Unequal line strengths leave a nonzero minimum contrast. Finite linewidth adds a longer-range coherence envelope, while arm imbalance and polarization mismatch can reduce contrast without indicating a different spectral spacing.

Use periodic boundary conditions temporarily to count the allowed wavevectors. Each state occupies $(2\pi {)}^3/V$ in reciprocal space. For one acoustic polarization, the spherical shell between $k$ and $k+dk$ containsBoth positive and negative wavevector directions are already included in that sphere; no further factor of two is needed. With the linear dispersion relation $k=\omega /v_s$,A three-dimensional monatomic crystal has three acoustic branches. Under the stated common-speed approximation,This is a phonon density of states per unit angular frequency, not per unit ordinary frequency. If longitudinal and transverse speeds differ, the isotropic expression becomes $g=V{\omega }^2(v_L^{-3}+2v_T^{-3})/(2{\pi }^2)$. In the Debye model, imposing ${\int }_0^{{\omega }_D}g(\omega )d\omega =3N$ sets the cutoff; the linear approximation is not continued to arbitrarily high frequency in a finite lattice.

The effective mass follows from the curvature of the electronic band structure. Differentiating twice givesAt either first Brillouin zone boundary $k=\pm \pi /a$,The boundary is a band maximum, with negative curvature and zero group velocity. In the semiclassical single-band description, an applied force changes $\hslash k$; the resulting acceleration is opposite the force near that maximum. This is a band-response coefficient, not a negative microscopic electron rest mass. A nearly filled band is often described more conveniently through positively charged electron holes near its maximum.

Put ${\omega }_0=\sqrt{k/m}$ and $\gamma =b/(2m)$. The characteristic roots are $-\gamma \pm \sqrt{{\gamma }^2-{\omega }_0^2}$. For release from rest at displacement $x_0>0$, the three regimes areunderdamping produces decaying oscillations; critical damping is the repeated-root boundary; overdamping has two real decay rates. The critical and overdamped curves from rest approach equilibrium without crossing it. All three initially have zero slope and negative acceleration. A sufficiently strong initial kick can cause a single crossing even without an oscillatory mode.

For the critical oscillator with initial velocity $-v_0$, the repeated-root solution isIts exponential is positive at finite time. Therefore the bracket has a positive-time zero precisely whenEquality gives a pure positive exponential and no finite-time crossing.

For the specified numerical oscillator, ${\omega }_0=1{\mathrm{s}}^{-1}$ and $\gamma =2/\sqrt{3}{\mathrm{s}}^{-1}>1{\mathrm{s}}^{-1}$, verifying overdamping. Its two roots are $-1/\sqrt{3}$ and $-\sqrt{3}$ in inverse seconds. The initial conditions givewith $t$ in seconds. The slow component has a negative coefficient; the fast positive component initially dominates.

The values for the requested plot areThe curve crosses near $0.56\mathrm{s}$ and then returns toward zero from below. The exact expression provides a check on the graphical estimate:

For a uniform taut sheet in the small-displacement approximation, the two-dimensional wave equation isSeek a harmonic solution $Y(y)e^{i\omega t-i{k}_{x}x}$. The clamped edges impose Dirichlet boundary conditions $Y(0)=Y(b)=0$. Hence $Y^{\prime \prime }+k_y^{2}Y=0$ has nontrivial solutions only forTaking the real part gives the stated sinusoidal transverse profile. There is no nonzero $n=0$ mode satisfying both clamped boundaries.

Substitution gives the dispersion relationThe three branches have cutoffs ${\omega }_{c,n}=n\pi v/b$ at $k_x=0$ and approach $\omega =v|k_x|$ at large $|k_x|$. Only frequencies above a branch's cutoff give traveling waves along an infinitely long guide; below it, the axial dependence is evanescent. For a forward propagating mode,The phase velocity exceeds $v$, while the group velocity is smaller: the carrier phase speed is not the signal speed.