Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2018 / S3 (Quantum ideas)

The de Broglie hypothesis associates a wave phase with a material particle. Its momentum and energy setThis is not a literal extended classical lump of matter oscillating at every point. A wavefunction gives amplitudes for measurements, and a localized particle is represented by a superposition of momentum waves. Electron diffraction, for example, exhibits interference at angles set by the same wavelength relation.

Using the given one-component equipartition estimate, $p^2=M{k}_{B}T$. Substituting into the de Broglie relation givesComparison with a confinement length or interparticle spacing $d$ gives the criterion $T\lesssim h^2/(M{k}_B{d}^2)$. Numerical factors depend on the convention: the statistical thermal de Broglie wavelength is usually $h/\sqrt{2\pi M{k}_{B}T}$, while a three-component root-mean-square momentum introduces a factor $\sqrt{3}$. The question's specified one-component convention is used consistently here. Once degeneracy becomes important, classical equipartition itself ceases to describe the electrons accurately.

Set $\lambda \simeq d$ with the convention of part b. ThenThus ordinary solid-state temperatures are far below this rough overlap scale. The electrons need quantum statistics and quantum states throughout that regime. The number is an order-of-magnitude wavelength criterion, not an exact Fermi temperature: electron density, electronic band structure, effective mass and the numerical convention all matter.

At the same temperature, momenta increase as the square root of mass, so wavelengths decrease as its inverse square root. For a nucleus of mass $M_N\simeq A{m}_p$,At the electron overlap threshold, ${\lambda }_e=d$, the nuclear wavelength is about $7.0\mathrm{pm}/\sqrt{A}$, far below the $0.3$ nm spacing. Nuclear exchange effects are consequently much less important than electron exchange effects. This statement about nuclear overlap does not mean lattice vibrations are always classical.

At room temperature, electrons require quantum mechanics, including their occupied orbitals or bands and Fermi statistics. Heavy nuclei can often be treated as distinguishable particles near classical equilibrium lattice sites, because their exchange wavelengths are short. Their coupled displacements form phonons, however, and the occupation of a vibrational mode is classical only if $k_{B}T\gg \hslash \omega$ for that mode. Even at room temperature some high-frequency modes remain quantum mechanical.

At $1$ K, quantum lattice vibrations, zero-point motion and the scarcity of thermally excited phonons are essential. Many heavy nuclei still do not have overlapping exchange wavefunctions; quantum vibration and quantum exchange are distinct criteria. For the same spacing, the nuclear overlap threshold from the rough estimate is $T_N\simeq 211/A$ K, illustrating why the nuclear mass cannot be omitted from a blanket low-temperature claim.

Choose the overall phase so that $A$ is positive. Normalization requiresThereforeAn arbitrary constant phase multiplying this answer has no effect on any expectation value.

The probability density is even, so its first moment vanishes. Its second moment follows by integrating the polynomial:Using the normalization from part a givesThe variance in position is consequently $a^2/7$.

The wavefunction is real and vanishes continuously at both endpoints. HenceThe momentum second moment is most safely evaluated from its quadratic form:ThusThe derivative jumps where the wavefunction joins zero. On the whole real line the strong second-derivative operator therefore produces endpoint delta distributions, so the state is not in the usual operator domain of $p^2$. The finite momentum second moment, equivalently the kinetic-energy quadratic form or the integral of $p^2$ against the momentum probability density, remains well defined and gives the value above.

Both first moments vanish, so the standard deviations follow directly from the second moments:The Heisenberg uncertainty relation is satisfied and is not saturated. A Gaussian wave packet can saturate the position-momentum bound; this compactly supported polynomial is not Gaussian.

For a single slit of width $d$, the far-field amplitude is proportional to $\mathrm{sinc}(\pi d\sin \theta /\lambda )$. The first zeros satisfy $d\sin \theta =\pm \lambda$. At small angles the angular width between those zeros is $2\lambda /d$, or $2L\lambda /d$ on a screen a distance $L$ away.The result assumes the wavelength, propagation distance and far-field conditions stay fixed. If the slit becomes comparable to the wavelength, the exact sine relation replaces the small-angle scaling.

Quantum tunnelling is transmission through a region where the particle's energy is below the classical potential energy. The stationary wavefunction decays within that region but need not vanish at its far boundary, so matching can leave a nonzero transmitted probability current.

This does not require a particle to borrow energy temporarily. Energy is conserved for a time-independent barrier; the classically forbidden region is described by an evanescent wave, with the same stationary energy as the incident and transmitted waves.

Inside a rectangular barrier with $E<V$, the stationary Schrödinger equation givesIts independent solutions are $e^{\pm \alpha x}$ withFor an opaque barrier the amplitude is suppressed by approximately $e^{-\alpha L}$ and the probability by $e^{-2\alpha L}$. Matching contributes a prefactor, so this exponential approximation is not an exact transmission coefficient for a transparent or very thin barrier. For two moving particles the relative-coordinate mass is their reduced mass.

At a separation of $1$ fm, the proton Coulomb energy is approximately $V=e^2/(4\pi {\epsilon }_{0}r)\simeq 1.44$ MeV, whereas $k_{B}T\simeq 0.862$ keV at the stated solar-core temperature. As a deliberately crude rectangular-barrier estimate, take a width of $1$ fm and use the two-proton reduced mass $\mu =m_p/2$. ThensoTreating one proton as fixed and using $m_p$ instead gives about $0.59$. Both estimates only say that this arbitrarily shortened barrier would be fairly transparent; the opaque-barrier exponential is itself uncontrolled for such a small exponent.

There is an important conditioning issue. If the protons have already reached a separation of one femtometre, most of the long-range Coulomb suppression has already been overcome. The rectangular estimate cannot represent the probability for thermal protons starting far apart. Their classical turning separation at this energy is about $1.7\times 10^3$ fm, not one femtometre.

Two major suppressions are missing from the one-femtometre rectangular estimate. First, the Coulomb barrier extends far beyond nuclear distances and is not rectangular. A semiclassical integral through it produces the Gamow factor $\exp (-2\pi \eta )$, with $\eta ={\alpha }_{\mathrm{fs}}\sqrt{\mu c^2/(2E)}$; at the illustrative thermal energy it is already of order $10^{-11}$. A thermal average additionally weights the high-energy tail and is not the same as evaluating a single thermal energy.

Second, two protons cannot simply form a stable bound diproton. The proton–proton chain starts with a weak-interaction conversion producing a deuteron, a Positron and a neutrino. The weak transition probability supplies a further very large suppression. Thus the quoted tiny fusion probability is not a pure one-dimensional barrier transmission probability with universal energy-independent meaning.

The Sun contains of order $10^{57}$ nucleons in a hot dense core and offers enormously many encounters over astronomical times. A very small reaction probability per encounter can therefore supply its luminosity while consuming only a small fraction of the fuel per unit time. The small weak reaction rate is also crucial for the long stellar lifetime; a larger core temperature increases burning and triggers expansion and cooling.