University Advanced Quantum Mechanics · Final Examination

A metallic grain a few nanometres across holds an odd number N of conduction electrons. Its shape is irregular, its impurities are random, and strong spin-orbit scattering mixes spin with orbital motion, so that no spatial symmetry and no component of the total spin is conserved. No magnetic field is applied, so every term in the grain's Hamiltonian H is even under time reversal, and for N spin-half particles the time-reversal operator satisfies Θ² = (−1)^N. Which statement about the energy levels of the grain is correct?

1. AEvery level is non-degenerate: once spin-orbit scattering has mixed the spin states, only a spatial symmetry could force two states to share an energy.
2. BEvery level is at least twofold degenerate: for an eigenstate |ψ⟩, Θ|ψ⟩ has the same energy, and Θ² = −1 makes it orthogonal to |ψ⟩.
3. CThe levels are twofold degenerate only while the spin-orbit scattering is weak, because the pairs are the spin-up and spin-down states of each orbital, which strong spin-orbit coupling mixes.
4. DEvery level is twofold degenerate, and stays so when one more electron is added, because each electron's states still come in time-reversed pairs.

The elastic scattering of ¹⁶O by ¹⁶O is followed at a fixed centre-of-mass angle θ = 60° while the centre-of-mass energy is stepped down from 8.0 MeV to 3.0 MeV, below the Coulomb barrier of about 10 MeV; at this angle the nuclei never come closer than about 17 fm, so only the Coulomb interaction acts. ¹⁶O nuclei are identical spin-0 bosons. The Coulomb amplitude the pair would have if the nuclei were distinguishable is, up to a phase that does not depend on θ, f(θ) = −[η/(2k sin²(θ/2))] exp[−iη ln sin²(θ/2)], where k is the relative wavenumber and the Sommerfeld parameter η = Z²α/β, with βc the relative speed, rises from 10.1 to 16.5 over the range. Let R be the measured differential cross-section divided by |f(60°)|², the Rutherford value for the projectile alone. How does R behave over the range?

1. AR stays at 1 + tan⁴30° = 10/9 at every energy: spin-0 nuclei have no spin states to average over, and the Coulomb phase cancels when each amplitude is squared.
2. BR oscillates between 0 and 4, because the amplitudes of identical bosons always add, so they can interfere completely.
3. CR oscillates between 4/9 and 16/9, passing through more than one full cycle as η ln 3 changes by about 7 rad.
4. DR oscillates between 10/9 and 16/9, because the interference of identical bosons is always constructive, so the cross-section never falls below the sum of the two Rutherford terms.

A negative kaon, K⁻, is a spinless particle of charge −e that can be captured into atomic orbits. Consider a kaon bound to a lead nucleus (Z = 82), treated as a point charge, with the strong interaction ignored. Its Klein-Gordon radial equation contains the centrifugal term ℏ²c²l(l + 1)/r² together with an attractive term −(Zαℏc)²/r² that comes from squaring the Coulomb energy, so that near the origin the radial function R(r) behaves as r^s, with s(s + 1) = l(l + 1) − (Zα)². Take α = 1/137.04. Which statement about the kaon's s, p and d orbitals (l = 0, 1 and 2) is correct?

1. AOnly the s orbitals fail: for l = 0, s is complex and the solutions oscillate without limit near the origin, a fall to the centre, while the p and d orbitals have s ≈ 0.88 and s ≈ 1.93.
2. BNone fails, because Zα = 0.60 is below 1; the s orbitals merely acquire a weak singularity, R ∝ r^s with s ≈ −0.36.
3. CEvery orbital fails, because the attractive −(Zα)²/r² term outweighs the Coulomb term near the origin whatever l is.
4. DOnly the s orbitals fail, and the p and d orbitals keep s = 1 and s = 2 exactly, as in the non-relativistic atom, because (Zα)² is negligible beside l(l + 1).

In Fermi's theory of beta decay and its generalisations, n → p e⁻ ν̄ arises from contact terms in the Lagrangian density, each the product of a bilinear in the proton and neutron fields and a bilinear in the electron and neutrino fields, with ψ̄ = ψ†γ⁰, γ⁵ = iγ⁰γ¹γ²γ³ and σ^μν = (i/2)[γ^μ, γ^ν]. Under parity each Dirac field becomes γ⁰ψ(t, −x) times an intrinsic phase, and the phases are chosen so that Fermi's vector term G_V(ψ̄_pγ^μψ_n)(ψ̄_eγ_μψ_ν) conserves parity. Which further term, with a real constant G, would make the interaction violate parity?

1. AG(ψ̄_pψ_n)(ψ̄_eψ_ν)
2. BG(ψ̄_pσ^μνψ_n)(ψ̄_eσ_μνψ_ν)
3. CG(ψ̄_pγ^μγ⁵ψ_n)(ψ̄_eγ_μγ⁵ψ_ν)
4. DG(ψ̄_pγ^μψ_n)(ψ̄_eγ_μγ⁵ψ_ν)

In a uniform static electric field of strength ℰ, the potential energy of an electron changes steadily along the field, so that a positive-energy electron state at one point has the same total energy as a negative-energy state a distance d = 2mc²/(eℰ) further along: Sauter's smooth version of the Klein step. In hole theory an electron tunnelling from the filled negative-energy states across this gap leaves a hole behind, so the field creates electron-positron pairs. Take mc² = 0.511 MeV, the reduced Compton wavelength ℏ/mc = 3.86 × 10⁻¹³ m and e = 1.602 × 10⁻¹⁹ C. At about what field strength does pair creation stop being exponentially suppressed?

1. AAbout m²c³/(eℏ) ≈ 1.3 × 10¹⁸ V m⁻¹, at which the gap d has shrunk to two reduced Compton wavelengths, so tunnelling across it is no longer suppressed.
2. BAbout 1.9 × 10¹⁶ V m⁻¹, at which the potential energy changes by 2mc² across one Bohr radius, the size of the atoms that would supply the electrons.
3. CAny field at all, because Klein's step needs only a total rise of more than 2mc², which a uniform field supplies over a long enough distance, whatever its gradient.
4. DAbout 3.6 × 10²⁰ V m⁻¹, at which the potential energy changes by 2mc² across one classical electron radius, 2.82 fm, the length at which classical electrodynamics fails.

A free complex scalar field and a free Dirac field are each quantised in two ways: with commutators and with anticommutators of their mode operators. Two requirements are imposed: a Hamiltonian bounded below on a space of states of positive norm, and local observables that commute at spacelike separation. Which statement is correct?

1. AThe Dirac field with commutators keeps a positive energy but fails microcausality, while the scalar field with anticommutators has an energy unbounded below.
2. BThe Dirac field with commutators has an energy unbounded below, or states of negative norm if creators and annihilators are relabelled, while the scalar field with anticommutators fails microcausality; only the scalar with commutators and the Dirac field with anticommutators pass both tests.
3. CBoth wrong quantisations fail only the positivity requirement, since microcausality holds for any choice because the vacuum amplitude D(x − y) vanishes outside the light cone.
4. DThe scalar field with anticommutators is excluded only because it allows at most one quantum in each mode, which contradicts the classical limit of the field.

Three fermion orbitals are ordered 1, 2, 3, and occupation-number states are defined by |n₁ n₂ n₃⟩ = (c₁†)^n₁ (c₂†)^n₂ (c₃†)^n₃ |0⟩, with {c_i, c_j†} = δ_ij and {c_i, c_j} = 0. The one-body operator T = t(c₃†c₁ + c₁†c₃) moves a fermion between orbitals 1 and 3. What is T|1 1 0⟩?

1. A+t|0 1 1⟩
2. B0, because orbital 2 lies between orbitals 1 and 3 and is occupied
3. C−t|1 0 1⟩
4. D−t|0 1 1⟩

A single mode of the quantised field is prepared first in the number state |n = 4⟩ and then in the coherent state |α⟩ with |α|² = 4. Its normalised second-order correlation at zero delay is g⁽²⁾(0) = ⟨a†a†aa⟩/⟨a†a⟩². What are the two values of g⁽²⁾(0), number state first?

1. A1 and 1
2. B0.75 and 2
3. C0.75 and 1
4. D0 and 1

A quantum dot emitting at a free-space wavelength λ = 930 nm sits at the field maximum of a GaAs micropillar cavity of refractive index n = 3.5, quality factor Q = 6000 and mode volume V = 1.5(λ/n)³. Its transition is resonant with the cavity mode and its dipole is parallel to the cavity field. The Purcell factor, the ratio of its emission rate into the cavity mode to its spontaneous emission rate in bulk GaAs, is F_P = (3/4π²)(λ/n)³(Q/V). What is F_P?

1. A304
2. B1.30 × 10⁴
3. C4.00 × 10³
4. DIt cannot be found without the dot's transition dipole moment, which sets both rates.

In Hartree-Fock theory the energy per electron of the uniform electron gas, in rydbergs, is E/N = 2.21/r_s² − 0.916/r_s, where r_s is the radius, in Bohr radii, of the sphere that contains one electron, and 1 Ry = 13.61 eV. The conduction electrons of lithium have r_s = 3.25. What is E/N at lithium's density, and how does that density compare with the density at which the Hartree-Fock E/N is lowest?

1. AE/N = −0.073 Ry (−0.99 eV), and lithium's electron gas is less dense than at the Hartree-Fock minimum.
2. BE/N = −0.282 Ry (−3.84 eV), and lithium's electron gas is denser than at the Hartree-Fock minimum.
3. CE/N = +0.209 Ry (+2.85 eV), and lithium's electron gas is denser than at the Hartree-Fock minimum.
4. DE/N = −0.073 Ry (−0.99 eV), and lithium's electron gas is denser than at the Hartree-Fock minimum.

A negative muon, a spin-half Dirac particle of charge q = −e and mass m about 207 times the electron's, is captured by a lead nucleus, ²⁰⁸Pb (Z = 82), and cascades to the 1s orbit of muonic lead. That orbit is so small that the muon spends most of its time inside the nucleus, modelled as a uniformly charged sphere of radius R = 7.1 fm, where its potential energy is V(r) = −(Zαℏc/R)(3/2 − r²/2R²). In static potentials φ and A its Dirac Hamiltonian is H = cα ⋅ π + βmc² + qφ, with π = p − qA, p = −iℏ∇ and V = qφ. In the Dirac representation, writing a stationary state as e^(−i(mc² + W)t/ℏ) times the column (ξ, χ) of upper and lower two-component spinors, Hψ = (mc² + W)ψ becomes (W − V)ξ = cσ ⋅ π χ and (W − V + 2mc²)χ = cσ ⋅ π ξ. Pauli's anomalous-moment term adds to H, in a magnetic field B, the operator −a(qℏ/2m)βΣ ⋅ B, where Σ = diag(σ, σ). For vector operators a and b that commute with the Pauli matrices, (σ ⋅ a)(σ ⋅ b) = a ⋅ b + iσ ⋅ (a × b), and for any function f(r), pf − fp = −iℏ∇f. In a storage-ring measurement of the anomaly, muons of Lorentz factor γ = 29.3 circulate in a uniform field B = 1.4513 T, with no electric field acting on them, and the rate of their decay positrons oscillates at the frequency ω_a/2π = 229.08 kHz at which the spin turns relative to the momentum. Take mc² = 105.66 MeV, m = 1.8835 × 10⁻²⁸ kg, ℏc = 197.33 MeV fm, α = 1/137.036, e = 1.602 × 10⁻¹⁹ C and the muon's lifetime at rest τ = 2.197 µs.

1. (a)

   Show that Show that the Dirac equation iℏ∂ψ/∂t = Hψ keeps its form under the gauge transformation A → A + ∇Λ, φ → φ − ∂Λ/∂t, for any smooth function Λ(r, t), provided that ψ → e^(iqΛ/ℏ)ψ.

   [2]
   Ready to self-mark?Reveal the detailed answerU22 marks+

   Mark-by-mark answer

   1. 1

      p(e^(iqΛ/ℏ)ψ) = e^(iqΛ/ℏ)(p + q∇Λ)ψ, so (p − qA′)ψ′ = e^(iqΛ/ℏ)(p − qA)ψ: the kinetic momentum acting on ψ′ is the phase times π acting on ψ, and cα ⋅ π and βmc² pass through the phase unchanged

   2. 2

      iℏ∂ψ′/∂t = e^(iqΛ/ℏ)(iℏ∂ψ/∂t − q(∂Λ/∂t)ψ), and the extra term −q(∂Λ/∂t)ψ′ is exactly the change qφ′ − qφ = −q∂Λ/∂t in the potential energy, so ψ′ obeys the same equation with the new potentials

2. (b)

   Estimate Inside the nucleus V is the potential energy of a three-dimensional harmonic oscillator. Estimate, stating your assumptions, the oscillator quantum ℏω, the muon's speed in the 1s state as a fraction of c and hence the size of χ relative to ξ, and judge whether an expansion in v/c is trustworthy for muonic lead.

   [2]
   Ready to self-mark?Reveal the detailed answerU42 marks+

   Mark-by-mark answer

   1. 1

      assuming the 1s muon lies mostly inside the nucleus, V = V(0) + ½kr² with k = Zαℏc/R³, so ℏω = ℏc√(Zαℏc/(mc²R³)) = 197.33 MeV fm × √(0.5984 × 197.33 MeV fm/(105.66 MeV × 357.9 fm³)) ≈ 11.0 MeV, and the oscillator length √(ℏ/mω) ≈ 5.8 fm, less than R = 7.1 fm, supports the assumption

   2. 2

      with ⟨T⟩ = ¾ℏω ≈ 8.3 MeV, v²/c² ≈ 2⟨T⟩/mc² ≈ 0.16 and v/c ≈ 0.4; the lower equation gives χ ≈ (σ ⋅ p/2mc)ξ, so |χ|/|ξ| ≈ v/2c ≈ 0.2 and χ carries about 4% of the probability: the expansion converges, but its order-(v/c)² terms are corrections of order 15%, not the parts in 10⁵ of hydrogen

3. (c)

   Derive For a slow muon in a uniform magnetic field B, with A = ½B × r and no electric field, eliminate χ to leading order in v/c and derive the Hamiltonian for ξ to first order in B. Hence find the g-factors with which the orbital angular momentum L = r × p and the spin S = (ℏ/2)σ couple to B under minimal coupling alone, and the spin g-factor once Pauli's term is included.

   [3]
   Ready to self-mark?Reveal the detailed answerU23 marks+

   Mark-by-mark answer

   1. 1

      with V = 0 and W neglected beside 2mc², the lower equation gives χ ≈ (σ ⋅ π/2mc)ξ, so Wξ = (σ ⋅ π)²ξ/2m; (σ ⋅ π)² = π² + iσ ⋅ (π × π), and because the components of π do not commute, π × π = −q(p × A + A × p) = iqℏ∇ × A = iqℏB, so (σ ⋅ π)² = π² − qℏσ ⋅ B

   2. 2

      with A = ½B × r, ∇ ⋅ A = 0, so p ⋅ A = A ⋅ p and π² = p² − 2qA ⋅ p + q²A², where 2A ⋅ p = (B × r) ⋅ p = B ⋅ L; to first order in B, H = p²/2m − (q/2m)L ⋅ B − (q/m)S ⋅ B: an orbital moment (q/2m)L, with g_l = 1, and a spin moment (q/m)S = g(q/2m)S, with g = 2

   3. 3

      the upper block of βΣ ⋅ B is σ ⋅ B, so Pauli's term adds −a(qℏ/2m)σ ⋅ B = −a(q/m)S ⋅ B, making the spin term −(1 + a)(q/m)S ⋅ B and g = 2(1 + a) while leaving the orbital term alone; a is a parameter the Dirac equation itself does not fix

4. (d)

   Calculate For a slow muon the Pauli Hamiltonian makes the spin precess at ω_s = g|q|B/2m while the momentum turns at ω_c = |q|B/m, so the spin turns relative to the momentum at ω_a = a|q|B/m; with no electric field the Thomas-BMT equation shows that this holds at any speed. Calculate a for the muon from the storage-ring measurement, and the number of cycles of the positron-rate oscillation in one dilated lifetime γτ.

   [2]
   Ready to self-mark?Reveal the detailed answerU32 marks+

   Mark-by-mark answer

   1. 1

      a = ω_a m/(eB) = 2π × 229.08 × 10³ s⁻¹ × 1.8835 × 10⁻²⁸ kg/(1.602 × 10⁻¹⁹ C × 1.4513 T) = 1.166 × 10⁻³

   2. 2

      γτ = 29.3 × 2.197 µs = 64.4 µs, so the positron rate completes 229.08 kHz × 64.4 µs ≈ 14.7 cycles per dilated lifetime, enough to fit the frequency directly rather than measuring ω_s and ω_c separately

5. (e)

   Show that The Foldy-Wouthuysen reduction to order (v/c)² contains the even term −(1/8m²c⁴)[O, [O, V]], where O = cα ⋅ p is the odd part of H when A = 0. Show that for a static V this term equals (ℏ²/8m²c²)∇²V + (ℏ/4m²c²)Σ ⋅ (∇V × p), and that inside the uniformly charged nucleus the Darwin part is the constant 3Zα(ℏc)³/(8(mc²)²R³). Evaluate it for muonic lead.

   [3]
   Ready to self-mark?Reveal the detailed answerU23 marks+

   Mark-by-mark answer

   1. 1

      [O, V] = cα_j[p_j, V] = −iℏcα ⋅ ∇V, so [O, [O, V]] = −iℏc²Σ_jk[α_jp_j, α_k∂_kV]; using p_j∂_kV = (∂_kV)p_j − iℏ∂_j∂_kV this is −iℏc²Σ_jk[(α_jα_k − α_kα_j)(∂_kV)p_j − iℏα_jα_k∂_j∂_kV]

   2. 2

      with α_jα_k = δ_jk + iε_jklΣ_l the symmetric part gives −ℏ²c²∇²V and the antisymmetric part −2ℏc²Σ ⋅ (∇V × p), so −(1/8m²c⁴)[O, [O, V]] = (ℏ²/8m²c²)∇²V + (ℏ/4m²c²)Σ ⋅ (∇V × p)

   3. 3

      inside the sphere dV/dr = Zαℏcr/R³, so ∇²V = (1/r²)d(r²dV/dr)/dr = 3Zαℏc/R³ and the Darwin term is 3Zα(ℏc)³/(8(mc²)²R³) = 3 × 0.5984 × (197.33 MeV fm)³/(8 × (105.66 MeV)² × 357.9 fm³) ≈ 0.43 MeV, a uniform upward shift wherever the muon is inside the nucleus

6. (f)

   Explain The 1s and 2s states of muonic lead are orthogonal four-component spinors. Using your estimate from (b), explain why their upper components ξ₁ₛ and ξ₂ₛ cannot be eigenfunctions of any Hermitian Hamiltonian at order (v/c)², so that no two-component Schrödinger equation for ξ can be exact to that order, and what the Foldy-Wouthuysen transformation does instead.

   [2]
   Ready to self-mark?Reveal the detailed answerU12 marks+

   Mark-by-mark answer

   1. 1

      orthonormality belongs to the whole spinor, ∫(ξ₁ₛ†ξ₂ₛ + χ₁ₛ†χ₂ₛ)d³r = 0, and with χ ≈ (σ ⋅ p/2mc)ξ this gives ∫ξ₁ₛ†ξ₂ₛ d³r ≈ −⟨ξ₁ₛ|p²|ξ₂ₛ⟩/(4m²c²), which is not zero, while ∫ξ₁ₛ†ξ₁ₛ d³r ≈ 1 − ⟨p²⟩/(4m²c²) ≈ 0.96; eigenfunctions of a Hermitian operator belonging to different eigenvalues are orthogonal, so no such operator has the ξ as its eigenfunctions, and expanding 1/(W − V + 2mc²) one step further gives instead an operator that depends on the energy W

   2. 2

      the Foldy-Wouthuysen transformation is unitary, so it keeps the spinors orthonormal while removing the odd operators order by order; the upper components of the transformed spinors, Φ ≈ [1 + p²/(8m²c²)]ξ to this order, are orthonormal two-component functions obeying one energy-independent Hermitian Hamiltonian, which contains the Darwin and spin-orbit terms of (e)

The Lyman-α X-rays of hydrogen-like uranium, U⁹¹⁺ (Z = 92), emitted when bare uranium nuclei circulating in a heavy-ion storage ring capture an electron that cascades to the ground state, test the Dirac-Coulomb spectrum where Zα is not small. The two lines are Lyα1, 2p₃/₂ → 1s₁/₂, and Lyα2, 2p₁/₂ → 1s₁/₂, and spectroscopy of the stored ions places Lyα1 at 102.17 keV. In a central potential V(r) the radial functions G(r) and F(r) of the upper and lower components satisfy dG/dr = −(κ/r)G + [(E + mc² − V)/ℏc]F and dF/dr = (κ/r)F − [(E − mc² − V)/ℏc]G, where E includes mc², κ = (l − j)(2j + 1) and l is the orbital angular momentum of the upper component. For a point nucleus V = −Zαℏc/r, and the exact energies are E_nj = mc²[1 + (Zα)²/(n − |κ| + γ)²]^(−1/2), with γ = √(κ² − (Zα)²). The normalised point-nucleus 1s density, both components together, is |ψ₁ₛ(r)|² = [(2Z/a₀)^(2γ+1)/(4πΓ(2γ + 1))]r^(2γ−2)e^(−2Zr/a₀), with Γ(2γ + 1) = 1.313 for Z = 92, while the Schrödinger 1s density is (Z³/πa₀³)e^(−2Zr/a₀). The uranium nucleus has its surface at about r = 7.5 fm and is treated as infinitely heavy. Take mc² = 510.999 keV, α = 1/137.036 and a₀ = 5.2918 × 10⁴ fm.

1. (a)

   State State which conserved operator takes the place of L² in labelling the electron's states, and state κ and the orbital angular momentum l′ of the lower component for each of 1s₁/₂, 2p₁/₂ and 2p₃/₂, saying which of their components are s waves, able to reach the nucleus.

   [2]
   Ready to self-mark?Reveal the detailed answerU12 marks+

   Mark-by-mark answer

   1. 1

      L does not commute with H, because cα ⋅ p couples an upper component of orbital angular momentum l to a lower one of l′ = 2j − l; together with J², J_z and parity, K = β(Σ ⋅ L + ℏ) commutes with H, and its eigenvalue −κℏ labels the state, |κ| = j + ½ giving j and the sign of κ telling whether the upper component has l = j − ½ (κ < 0) or l = j + ½ (κ > 0)

   2. 2

      κ = −1 for 1s₁/₂, +1 for 2p₁/₂ and −2 for 2p₃/₂, with l′ = 1, 0 and 2: the s waves are the upper component of 1s₁/₂ and the lower component of 2p₁/₂, so both reach the nucleus, while 2p₃/₂, with l = 1 and l′ = 2, has none

2. (b)

   Derive Derive from the radial equations how G and F rise from the origin. For the point nucleus, writing G ≈ ar^γ and F ≈ br^γ, find γ and b/a, evaluate b/a for 1s₁/₂ and for 2p₁/₂ at Z = 92, and say which component dominates near the nucleus in each. Then find the powers of r with which G and F rise inside a nucleus of finite size, where V(r) tends to a finite V(0).

   [3]
   Ready to self-mark?Reveal the detailed answerU23 marks+

   Mark-by-mark answer

   1. 1

      for the point nucleus the Coulomb term is as singular as κ/r and dominates E ± mc², and the r^(γ−1) terms give (γ + κ)a = Zαb and (γ − κ)b = −Zαa, which have a nonzero solution only if γ² = κ² − (Zα)²; the positive root γ = √(κ² − (Zα)²) keeps the density normalisable, and b/a = (γ + κ)/(Zα)

   2. 2

      at Z = 92, Zα = 0.6714 and γ = √(1 − 0.4507) = 0.7411 for |κ| = 1, so b/a = (0.7411 + 1)/0.6714 = +2.59 for 2p₁/₂, the s-wave lower component outweighing the upper one near the nucleus, and b/a = (0.7411 − 1)/0.6714 = −0.386 for 1s₁/₂, the upper component dominating; both components share the non-integer power, so G/r and F/r diverge weakly, as r^(γ−1) = r^(−0.26)

   3. 3

      inside a finite nucleus V stays finite, so near the origin only the κ/r terms are singular: for κ < 0, G ∝ r^(−κ) = r^(l+1) and F ∝ r^(l+2) = r^(l′+1), and for κ > 0, F ∝ r^κ = r^(l′+1) and G ∝ r^(κ+1) = r^(l+1); the integer powers of the non-relativistic problem return, each component with its own orbital angular momentum, and the singularity of the point nucleus is removed

3. (c)

   Calculate Calculate, for a point nucleus, the energies of Lyα1 and Lyα2 in keV, and the 2p fine-structure splitting.

   [2]
   Ready to self-mark?Reveal the detailed answerU32 marks+

   Mark-by-mark answer

   1. 1

      E₁ₛ = γmc² = 0.74113 × 510.999 keV = 378.72 keV; for 2p₁/₂, n − |κ| + γ = 1.7411 and E = 510.999 keV/√(1 + 0.4507/3.0315) = 476.78 keV; for 2p₃/₂, γ = √(4 − 0.4507) = 1.8840 and E = 510.999 keV/√(1 + 0.4507/3.5493) = 481.35 keV

   2. 2

      Lyα1 = 481.35 keV − 378.72 keV = 102.63 keV and Lyα2 = 476.78 keV − 378.72 keV = 98.06 keV, a fine-structure splitting of 4.57 keV, where the non-relativistic theory gives one line at (3/8)(Zα)²mc² = 86.4 keV

4. (d)

   Determine Treating the 2p₃/₂ level as given correctly by the point-nucleus formula, determine from the measured Lyα1 energy the shift of the 1s level that the point-nucleus Dirac theory omits, its sign, and its size as a fraction of the 1s binding energy.

   [2]
   Ready to self-mark?Reveal the detailed answerU32 marks+

   Mark-by-mark answer

   1. 1

      the measured line lies 102.63 keV − 102.17 keV = 0.46 keV below the point-nucleus value, so with 2p₃/₂ fixed the 1s level lies about 0.46 keV higher, less bound, than the point-nucleus Dirac prediction

   2. 2

      the 1s binding energy is mc² − E₁ₛ = 510.999 keV − 378.72 keV = 132.28 keV, so the shift is 0.46/132.28 = 0.35% of it, the combined effect of the nucleus's finite size and of the radiative corrections of quantum electrodynamics, which together raise the s level

5. (e)

   Compare Compare the Dirac 1s density at the nuclear surface, r = 7.5 fm, with the Schrödinger 1s density there, and say what the comparison implies for the finite-size part of the shift found in (d).

   [2]
   Ready to self-mark?Reveal the detailed answerU32 marks+

   Mark-by-mark answer

   1. 1

      the exponentials cancel and the ratio is [2/Γ(2γ + 1)](2Zr/a₀)^(2γ−2); with 2Zr/a₀ = 2 × 92 × 7.5 fm/(5.2918 × 10⁴ fm) = 0.0261 and 2γ − 2 = −0.518 it is (2/1.313) × 0.0261^(−0.518) ≈ 10

   2. 2

      the weak singularity r^(2γ−2) puts about ten times more of the Dirac electron's density inside the nucleus than the Schrödinger estimate, so the finite-size shift is about ten times what a non-relativistic calculation gives and makes up a substantial part of the 0.46 keV

6. (f)

   Predict In the point-nucleus Dirac theory 2s₁/₂ and 2p₁/₂ are exactly degenerate. Predict which of the two the finite size of the nucleus raises more, and so which lies higher.

   [1]
   Ready to self-mark?Reveal the detailed answerU11 mark+

   Mark-by-mark answer

   1. 1

      2s₁/₂, whose s wave is the large upper component, overlaps the nucleus far more than 2p₁/₂, whose s wave is the lower component that is small over most of the orbit, so 2s₁/₂ is raised more and lies above 2p₁/₂, in the same direction as the radiative Lamb shift

A free complex scalar field φ(x, t), with φ and φ† treated as independent fields, describes spinless particles of mass m and charge q through the Lagrangian density ℒ = (1/c²)(∂φ†/∂t)(∂φ/∂t) − ∇φ† ⋅ ∇φ − κ²φ†φ, where κ = mc/ℏ. In a box of volume V with periodic boundary conditions the quantised field is φ(x, t) = Σ_k N_k[a_k e^(i(k⋅x − ω_k t)) + b_k† e^(−i(k⋅x − ω_k t))], with ω_k = c√(k² + κ²), N_k = √(ℏc²/(2ω_kV)), [a_k, a_k′†] = [b_k, b_k′†] = δ_kk′ and all other commutators zero. After normal ordering the Hamiltonian is H = Σ_k ℏω_k(a_k†a_k + b_k†b_k). Take ℏc = 197.3 MeV fm.

1. (a)

   Derive Apply the Euler-Lagrange equation for φ† to derive the field equation for φ, and use it to show that the density ρ = (iq/ℏc²)(φ†∂φ/∂t − φ∂φ†/∂t) and the current j = −(iq/ℏ)(φ†∇φ − φ∇φ†) satisfy ∂ρ/∂t + ∇ ⋅ j = 0.

   [3]
   Ready to self-mark?Reveal the detailed answerU23 marks+

   Mark-by-mark answer

   1. 1

      ∂ℒ/∂(∂_tφ†) = (1/c²)∂_tφ, ∂ℒ/∂(∇φ†) = −∇φ and ∂ℒ/∂φ† = −κ²φ, so ∂_t[∂ℒ/∂(∂_tφ†)] + ∇ ⋅ [∂ℒ/∂(∇φ†)] − ∂ℒ/∂φ† = 0 gives (1/c²)∂²φ/∂t² − ∇²φ + κ²φ = 0, the Klein-Gordon equation, with its conjugate for φ†

   2. 2

      ∂ρ/∂t = (iq/ℏc²)(φ†∂²_tφ − φ∂²_tφ†), the products of first time derivatives cancelling, and likewise ∇ ⋅ j = −(iq/ℏ)(φ†∇²φ − φ∇²φ†)

   3. 3

      substituting ∂²_tφ = c²(∇²φ − κ²φ) and its conjugate, the κ² terms cancel and ∂ρ/∂t = (iq/ℏ)(φ†∇²φ − φ∇²φ†) = −∇ ⋅ j: the current is conserved, the Noether current of the symmetry φ → e^(−iθ)φ

2. (b)

   Show that By substituting the mode expansion, show that after normal ordering the charge Q = ∫ρ d³x is qΣ_k(a_k†a_k − b_k†b_k), and hence that a_k† creates a quantum of charge +q and b_k† one of charge −q, each with energy ℏω_k.

   [3]
   Ready to self-mark?Reveal the detailed answerU23 marks+

   Mark-by-mark answer

   1. 1

      ∫e^(i(k − k′)⋅x) d³x = Vδ_kk′ pairs each operator with one of the same mode or of the opposite mode, and the terms in a_k b_(−k) e^(−2iω_k t) and in a_k†b_(−k)† e^(2iω_k t) enter φ†∂_tφ and φ∂_tφ† with the same coefficient and cancel, so Q is independent of time

   2. 2

      each mode then contributes (iq/ℏc²)(−iω_k)N_k²V(a_k†a_k + a_k a_k† − b_k b_k† − b_k†b_k) = (q/2)(a_k†a_k + a_k a_k† − b_k b_k† − b_k†b_k), since ω_kN_k²V/ℏc² = ½; normal ordering gives Q = qΣ_k(a_k†a_k − b_k†b_k)

   3. 3

      [Q, a_k†] = qa_k† and [Q, b_k†] = −qb_k†, so a_k†|0⟩ has charge +q and b_k†|0⟩ charge −q, while [H, a_k†] = ℏω_k a_k† and [H, b_k†] = ℏω_k b_k† give both the energy ℏω_k: the antiparticle has the opposite charge and a positive energy

3. (c)

   Explain The field commutator is [φ(x), φ†(y)] = D(x − y) − D(y − x), where D(x − y) = ⟨0|φ(x)φ†(y)|0⟩ = Σ_k N_k² e^(i(k⋅(x − y) − ω_k(t_x − t_y))). Explain why the commutator vanishes when x − y is spacelike although D(x − y) does not, and what would go wrong if the field contained the a_k terms alone.

   [2]
   Ready to self-mark?Reveal the detailed answerU12 marks+

   Mark-by-mark answer

   1. 1

      for spacelike x − y, D depends only on the invariant interval (at equal times it is even in x − y, by k → −k), and a proper Lorentz transformation can carry a spacelike x − y into −(x − y); so D(x − y) = D(y − x) and the commutator vanishes, although each term is nonzero

   2. 2

      the second term comes from the b-operators: the amplitude for an antiparticle to travel from x to y cancels that for a particle to travel from y to x; with the a-terms alone, [φ(x), φ†(y)] = D(x − y), nonzero at spacelike separation, so a measurement at one point could affect another outside its light cone: a relativistic charged field needs its antiparticle

4. (d)

   Estimate At equal times and separation r, D is proportional to (1/r)K₁(κr), which for κr ≫ 1 behaves as r^(−3/2)e^(−κr). For a π⁺, with mc² = 139.6 MeV, estimate the length 1/κ over which the amplitude extends outside the light cone, and the factor by which D falls between r = 5 fm and r = 10 fm, stating your assumptions.

   [2]
   Ready to self-mark?Reveal the detailed answerU42 marks+

   Mark-by-mark answer

   1. 1

      1/κ = ℏc/mc² = 197.3 MeV fm/139.6 MeV = 1.41 fm, the pion's reduced Compton wavelength

   2. 2

      assuming the asymptotic form already holds at κr = 5/1.41 = 3.5, D(10 fm)/D(5 fm) ≈ (1/2)^(3/2) e^(−5/1.41) = 0.354 × 0.029 = 0.010 (the Bessel functions themselves give 0.0098): a hundredfold fall, but never zero

5. (e)

   Compare Compare the density ρ of this question, as an operator of the quantised field, with the same expression read as a probability density in the one-particle Klein-Gordon theory.

   [2]
   Ready to self-mark?Reveal the detailed answerU12 marks+

   Mark-by-mark answer

   1. 1

      in the one-particle theory the same bilinear (without the factor q) is negative on negative-frequency solutions and can be given either sign by the initial data, so it cannot be a probability density, and the negative-energy solutions have no consistent meaning

   2. 2

      in the field theory it is the charge density: its integral has eigenvalues q(N_a − N_b), negative values counting antiparticles, which have positive energy and states of positive norm; the indefinite sign becomes the two signs of charge, not of probability

Two electrons occupy a molecule of two sites, 1 and 2, each with one orbital, described by the Hubbard Hamiltonian H = −t Σ_σ (c₁σ†c₂σ + c₂σ†c₁σ) + U(n₁↑n₁↓ + n₂↑n₂↓), with hopping t = 0.40 eV and on-site repulsion U = 3.2 eV. The operators obey {c_iσ, c_jσ′†} = δ_ij δ_σσ′ and {c_iσ, c_jσ′} = 0. In the sector with S_z = 0 a basis is |D₁⟩ = c₁↑†c₁↓†|0⟩ and |D₂⟩ = c₂↑†c₂↓†|0⟩, with both electrons on one site, and |A⟩ = c₁↑†c₂↓†|0⟩ and |B⟩ = c₂↑†c₁↓†|0⟩, with one electron on each.

1. (a)

   Show that Show that the hopping term T sends both |A⟩ and |B⟩ to −t(|D₁⟩ + |D₂⟩), and hence that in the basis |S⟩ = (|A⟩ + |B⟩)/√2, |D₊⟩ = (|D₁⟩ + |D₂⟩)/√2 the Hamiltonian is the matrix [[0, −2t], [−2t, U]], while (|A⟩ − |B⟩)/√2 and (|D₁⟩ − |D₂⟩)/√2 are eigenstates with energies 0 and U.

   [3]
   Ready to self-mark?Reveal the detailed answerU23 marks+

   Mark-by-mark answer

   1. 1

      on |A⟩ only c₂↑†c₁↑ and c₁↓†c₂↓ act: c₂↑†c₁↑c₁↑†c₂↓†|0⟩ = c₂↑†c₂↓†|0⟩ = |D₂⟩, and c₁↓†c₂↓c₁↑†c₂↓†|0⟩ = −c₁↓†c₁↑†|0⟩ = +|D₁⟩, two anticommutation signs cancelling; likewise c₁↑†c₂↑|B⟩ = |D₁⟩ and c₂↓†c₁↓|B⟩ = |D₂⟩, so T|A⟩ = T|B⟩ = −t(|D₁⟩ + |D₂⟩)

   2. 2

      the U term acts only on |D₁⟩ and |D₂⟩, each with eigenvalue U, and |A⟩, |B⟩ have energy 0; by Hermiticity T|D₁⟩ = T|D₂⟩ = −t(|A⟩ + |B⟩), so T annihilates (|A⟩ − |B⟩)/√2 and (|D₁⟩ − |D₂⟩)/√2, which are eigenstates with energies 0 and U

   3. 3

      ⟨S|T|D₊⟩ = ½(⟨A| + ⟨B|)(−2t)(|A⟩ + |B⟩) = −2t, so on {|S⟩, |D₊⟩} H = [[0, −2t], [−2t, U]]; |S⟩ = (c₁↑†c₂↓† − c₁↓†c₂↑†)|0⟩/√2 is the spin singlet and (|A⟩ − |B⟩)/√2 the triplet with m = 0

2. (b)

   Calculate Calculate the ground-state energy and the singlet-triplet splitting, and compare the splitting with the superexchange coupling 4t²/U.

   [2]
   Ready to self-mark?Reveal the detailed answerU32 marks+

   Mark-by-mark answer

   1. 1

      E₀ = [U − √(U² + 16t²)]/2 = [3.2 − √(10.24 + 2.56)]/2 eV = (3.2 − 3.578)/2 eV = −0.189 eV, a spin singlet

   2. 2

      the triplet lies at 0, so the splitting is 0.189 eV, against 4t²/U = 4(0.40)²/3.2 eV = 0.200 eV, 6% high because t/U = 0.125 is not very small; the coupling is antiferromagnetic

3. (c)

   Determine Determine the probability of finding both electrons on the same site in the ground state.

   [2]
   Ready to self-mark?Reveal the detailed answerU32 marks+

   Mark-by-mark answer

   1. 1

      the first row of (H − E₀) acting on the amplitudes s of |S⟩ and d of |D₊⟩ gives −2td = E₀s, so d/s = −E₀/2t = 0.189/0.80 = 0.236

   2. 2

      P = d²/(s² + d²) = 0.0557/1.0557 = 0.053, about 5%, against 50% for two electrons that ignore each other

4. (d)

   Evaluate A student describes the ground state by the restricted Hartree-Fock determinant, with both electrons in the bonding orbital, created by (c₁σ† + c₂σ†)/√2. Evaluate this description: find its energy and its probability of double occupation, and say what it gets wrong as U/t grows.

   [3]
   Ready to self-mark?Reveal the detailed answerU13 marks+

   Mark-by-mark answer

   1. 1

      the determinant is ½(c₁↑† + c₂↑†)(c₁↓† + c₂↓†)|0⟩ = (|S⟩ + |D₊⟩)/√2, with energy ½(0 + U) − 2t = U/2 − 2t = +0.80 eV, 0.99 eV above the exact −0.19 eV: the correlation energy, −0.99 eV, is about five times the magnitude of the exact energy itself

   2. 2

      its probability of double occupation is 50% whatever U is, against the exact 5%: one determinant of delocalised orbitals cannot keep the two electrons apart

   3. 3

      as U/t grows the restricted energy rises as U/2 while the exact energy tends to −4t²/U, approaching zero from below: restricted Hartree-Fock fails worst exactly where the spins localise and couple antiferromagnetically, the dissociation problem of H₂

5. (e)

   Explain Unrestricted Hartree-Fock lets the up-spin and down-spin electrons occupy different orbitals, and its best determinant has energy −2t²/U = −0.10 eV. Explain why it recovers only about half of the exact binding, and what symmetry of H it breaks to do so.

   [2]
   Ready to self-mark?Reveal the detailed answerU12 marks+

   Mark-by-mark answer

   1. 1

      for U ≫ t the unrestricted determinant approaches |A⟩ = (|S⟩ + |T₀⟩)/√2, with |T₀⟩ = (|A⟩ − |B⟩)/√2: an equal mixture of the singlet, near −4t²/U, and the triplet, at 0, so its energy is about their average, −2t²/U

   2. 2

      to localise the spins it breaks the spin-rotation symmetry, since the determinant is not an eigenstate of S², and the symmetry exchanging the sites; the singlet (|A⟩ + |B⟩)/√2, which the exact ground state 0.973|S⟩ + 0.230|D₊⟩ approaches as U/t grows, is a sum of two determinants, which no single determinant can represent

In Bethe's non-relativistic calculation a hydrogen atom in a state |n⟩ is coupled to the transverse vacuum field through H′ = (e/m)A ⋅ p in the dipole approximation, photon energies being cut off at K. Summing the second-order energy over photon modes and polarisations gives W_n = −(2α/(3πm²c²)) Σ_m |⟨m|p|n⟩|² ∫₀^K k dk/(E_m − E_n + k), where k = ℏω is the photon energy and the sum runs over all bound and continuum states |m⟩. For a free electron of momentum p the same calculation gives W_free = −(2α/(3πm²c²))p²K. Write p_mn = ⟨m|p|n⟩. Take mc² = 0.510999 MeV, α = 1/137.036, 1 Ry = 13.6057 eV and h = 4.1357 × 10⁻¹⁵ eV s.

1. (a)

   Show that Show that, for K large compared with every |E_m − E_n| that matters, W_n grows linearly with K, and that the coefficient of K is the free-electron result averaged over the state |n⟩.

   [2]
   Ready to self-mark?Reveal the detailed answerU22 marks+

   Mark-by-mark answer

   1. 1

      with Δ = E_m − E_n, ∫₀^K k dk/(Δ + k) = K − Δ ln|(Δ + K)/Δ| ≈ K − Δ ln(K/|Δ|) for K ≫ |Δ|, so W_n = −(2α/3πm²c²)[K Σ_m|p_mn|² − Σ_m|p_mn|²Δ ln(K/|Δ|)]

   2. 2

      by closure Σ_m|p_mn|² = ⟨n|p²|n⟩, so the linear term is −(2αK/3πm²c²)⟨n|p²|n⟩ = ⟨n|W_free|n⟩: the bound electron's self-energy diverges exactly as a free electron's with the same momentum distribution

2. (b)

   Explain Explain why the linear term is removed by mass renormalisation, and why what remains depends on K only through a logarithm.

   [2]
   Ready to self-mark?Reveal the detailed answerU12 marks+

   Mark-by-mark answer

   1. 1

      W_free = −(2αK/3πm²c²)p² changes the coefficient of p²: p²/2m₀ + W_free = p²/2(m₀ + δm) with δm = 4αK/(3πc²), an electromagnetic mass; the measured mass already contains it, so writing H with the observed mass subtracts W_free and removes the linear term from every state (the A² term, the same for every state in the dipole approximation, drops out of all intervals)

   2. 2

      what remains, ΔE_n = (2α/3πm²c²) Σ_m|p_mn|²(E_m − E_n) ln(K/|E_m − E_n|), exists only because the electron is bound, vanishing for a free one, and keeps K only in a logarithm, because photons much more energetic than the binding cannot tell a bound electron from a free one

3. (c)

   Show that The sum rule Σ_m|p_mn|²(E_m − E_n) = (ℏ²/2)⟨n|∇²V|n⟩ holds for V = −e²/(4πε₀r). Defining the mean excitation energy K₀ by ln K₀ = Σ_m|p_mn|²(E_m − E_n) ln|E_m − E_n| / Σ_m|p_mn|²(E_m − E_n), show that ΔE_n = (4α⁵mc²/(3πn³)) ln(K/K₀) for s states, and that the term in ln K vanishes in every other state. You may use |ψ_ns(0)|² = 1/(πn³a₀³), a₀ = ℏ/(αmc) and e²/(4πε₀) = αℏc.

   [3]
   Ready to self-mark?Reveal the detailed answerU23 marks+

   Mark-by-mark answer

   1. 1

      ∇²(1/r) = −4πδ³(r) gives ∇²V = (e²/ε₀)δ³(r) = 4παℏc δ³(r), so the sum rule gives 2παℏ³c|ψ_n(0)|², which vanishes unless l = 0: the term in ln K acts on s states only

   2. 2

      with the definition of K₀, ΔE_n = (2α/3πm²c²) × 2παℏ³c|ψ_n(0)|² × ln(K/K₀) = (4α²ℏ³/3m²c)|ψ_n(0)|² ln(K/K₀)

   3. 3

      with |ψ_ns(0)|² = 1/(πn³a₀³) = α³m³c³/(πn³ℏ³), ΔE_n = (4α⁵mc²/(3πn³)) ln(K/K₀): of order α⁵mc²/n³ times a large logarithm

4. (d)

   Calculate With the cutoff at K = mc² and the quoted mean excitation energies K₀(2s) = 16.64 Ry and K₀(1s) = 19.77 Ry, calculate the shifts of 2s and of 1s in frequency units, and compare them with the measured 2s₁/₂–2p₁/₂ splitting of 1057.8 MHz and 1s shift of 8.17 GHz.

   [3]
   Ready to self-mark?Reveal the detailed answerU33 marks+

   Mark-by-mark answer

   1. 1

      for n = 2, 4α⁵mc²/(3π × 8) = 5.610 × 10⁻⁷ eV, which is 135.6 MHz

   2. 2

      ln(mc²/K₀) = ln(510999 eV/226.4 eV) = 7.722, so the 2s shift is 135.6 MHz × 7.722 = 1047 MHz, 1% below the measured 1057.8 MHz, 2p₁/₂ receiving no logarithmic shift

   3. 3

      for 1s the prefactor is 8 × 135.6 MHz = 1085 MHz and ln(510999 eV/269.0 eV) = 7.549, giving 8.19 GHz against the measured 8.17 GHz

5. (e)

   Evaluate The non-relativistic calculation fixes the cutoff K only to within a factor of about two. Evaluate whether the agreement found in (d) shows that Bethe's choice K = mc² is correct.

   [2]
   Ready to self-mark?Reveal the detailed answerU42 marks+

   Mark-by-mark answer

   1. 1

      a factor of two in K changes the 2s shift by 135.6 MHz × ln 2 = 94 MHz, about 9%, and the 1s shift by 0.75 GHz; agreement at the 1% level is therefore much closer than the method can guarantee, and is partly fortunate

   2. 2

      what the calculation does establish is robust against the cutoff: a shift of order α⁵mc²/n³ times a logarithm, confined to s states, with a 1s to 2s ratio of 7.8 rather than 8; removing the cutoff needs the covariant QED calculation, which also adds vacuum polarisation and the anomalous moment, each worth tens of MHz

A single InGaAs quantum dot at 4 K is driven resonantly by a narrow-band laser, and its fluorescence is recorded as the laser frequency is scanned across the transition. For each laser power P at the dot, the full width at half maximum Δν of the fluorescence line, as a frequency, is recorded. A separate time-resolved measurement gives the population relaxation time T₁ = 0.80 ns. For a two-level emitter with population relaxation time T₁ and coherence time T₂, driven at Rabi frequency Ω and detuning Δ (both angular frequencies), the steady state of the optical Bloch equations has the excited-state population ρ_ee = ½Ω²T₁T₂/(1 + Δ²T₂² + Ω²T₁T₂), and Ω² is proportional to P. Without pure dephasing T₂ = 2T₁; in general 1/T₂ = 1/(2T₁) + 1/T_φ, where T_φ is the pure dephasing time.

1. (a)

   Show that Show that Δν = (1/πT₂)√(1 + Ω²T₁T₂), and hence that Δν² = Δν₀²(1 + P/P_sat), where Δν₀ = 1/(πT₂) and P_sat is the power at which Ω²T₁T₂ = 1.

   [2]
   Ready to self-mark?Reveal the detailed answerU22 marks+

   Mark-by-mark answer

   1. 1

      ρ_ee is a Lorentzian in Δ that falls to half its peak value when Δ²T₂² = 1 + Ω²T₁T₂, so its full width in angular frequency is (2/T₂)√(1 + Ω²T₁T₂)

   2. 2

      dividing by 2π, Δν = (1/πT₂)√(1 + Ω²T₁T₂); with Ω²T₁T₂ = P/P_sat this gives Δν² = Δν₀²(1 + P/P_sat), linear in P

2. (b)

   Describe Describe how the readings should be plotted to test this model and to extract Δν₀ and P_sat, and explain why the narrowest line measured is not itself Δν₀.

   [2]
   Ready to self-mark?Reveal the detailed answerU42 marks+

   Mark-by-mark answer

   1. 1

      plot Δν² against P: the model predicts a straight line whose intercept at P = 0 is Δν₀² and whose gradient is Δν₀²/P_sat, so P_sat = intercept/gradient; a plot of Δν against P curves and cannot be extrapolated reliably

   2. 2

      even the lowest power, 2.0 nW, is about a sixth of P_sat, so the 289 MHz line is already power-broadened by about 8%; the zero-power width has to be found by extrapolation

3. (c)

   Determine Determine the gradient and intercept of your graph, and hence Δν₀ and P_sat.

   [3]
   Ready to self-mark?Reveal the detailed answerU43 marks+

   Mark-by-mark answer

   1. 1

      Δν² = 0.835, 0.986, 1.303, 1.840, 3.069 and 5.358 × 10⁵ MHz², which lie on a straight line to within 2%

   2. 2

      gradient = 5.8 × 10³ MHz² nW⁻¹ (accept 5.7 to 6.0 × 10³ MHz² nW⁻¹) and intercept = 7.1 × 10⁴ MHz² (accept 6.9 to 7.3 × 10⁴ MHz²)

   3. 3

      Δν₀ = √(intercept) = 266 MHz (accept 263 to 270 MHz) and P_sat = intercept/gradient = 12 nW (accept 11.5 to 12.8 nW)

4. (d)

   Evaluate Evaluate whether this emitter's line is lifetime-limited, finding T₂ and, if it is needed, the pure dephasing time T_φ.

   [3]
   Ready to self-mark?Reveal the detailed answerU43 marks+

   Mark-by-mark answer

   1. 1

      T₂ = 1/(πΔν₀) = 1/(π × 266 × 10⁶ Hz) = 1.20 ns (accept 1.18 to 1.21 ns)

   2. 2

      a lifetime-limited line would have T₂ = 2T₁ = 1.60 ns and Δν₀ = 1/(2πT₁) = 199 MHz; the measured 266 MHz is a third wider, so the line is not lifetime-limited

   3. 3

      1/T_φ = 1/T₂ − 1/(2T₁) = (0.836 − 0.625) ns⁻¹ = 0.211 ns⁻¹, so T_φ = 4.7 ns (accept 4.5 to 5.0 ns): pure dephasing, from phonons or fluctuating charges near the dot, which a measurement of T₁ cannot see

5. (e)

   Predict Predict the rate at which the dot scatters photons when it is driven on resonance at P = 40 nW.

   [1]
   Ready to self-mark?Reveal the detailed answerU31 mark+

   Mark-by-mark answer

   1. 1

      with P/P_sat = 40/12 = 3.3, ρ_ee = ½ × 3.3/4.3 = 0.38, and the scattering rate ρ_ee/T₁ = 0.38/(0.80 ns) = 4.8 × 10⁸ s⁻¹, about 77% of the saturated maximum 1/(2T₁) = 6.3 × 10⁸ s⁻¹

In a cubic box of volume V with periodic boundary conditions, the transverse vector potential in Coulomb gauge is, in the Heisenberg picture, A(r, t) = Σ_(k,λ) √(ℏ/(2ε₀ω_kV)) [a_(kλ) ε_(kλ) e^(i(k⋅r − ω_k t)) + a_(kλ)† ε_(kλ)* e^(−i(k⋅r − ω_k t))], with ω_k = ck, two unit polarisation vectors ε_(kλ) perpendicular to k for each k, and [a_(kλ), a_(k′λ′)†] = δ_kk′ δ_λλ′. A one-electron atom or ion, with an infinitely heavy nucleus at the origin, has an excited state |b⟩ and a lower state |a⟩ separated by ℏω₀. It couples to the field through H′ = (e/m)A(0) ⋅ p, the dipole approximation, where −e and m are the electron's charge and mass, and ⟨a|p|b⟩ = −imω₀⟨a|r|b⟩. For a hydrogen-like ion of nuclear charge Ze, |⟨1s|r|2p, m⟩|² = (2¹⁵/3¹⁰)(a₀/Z)² for each m, and the 2p–1s energy difference is (3/4)Z² × 13.606 eV. Take ℏ = 1.055 × 10⁻³⁴ J s, e = 1.602 × 10⁻¹⁹ C, ε₀ = 8.854 × 10⁻¹² F m⁻¹, c = 2.998 × 10⁸ m s⁻¹ and a₀ = 5.292 × 10⁻¹¹ m.

1. (a)

   Derive Derive the transverse electric field operator from A, and show that in the vacuum each mode contributes ℏω_k/(2ε₀V) to ⟨E²⟩, so that its mean electric energy is half its zero-point energy.

   [2]
   Ready to self-mark?Reveal the detailed answerU22 marks+

   Mark-by-mark answer

   1. 1

      in Coulomb gauge the transverse field is E = −∂A/∂t = Σ_(k,λ) i√(ℏω_k/(2ε₀V)) [a_(kλ)ε_(kλ)e^(i(k⋅r − ω_k t)) − a_(kλ)†ε_(kλ)*e^(−i(k⋅r − ω_k t))]

   2. 2

      in ⟨0|E²|0⟩ only the product a a† of each mode survives, with ⟨0|a a†|0⟩ = 1 and |ε|² = 1, so each mode contributes ℏω_k/(2ε₀V); its electric energy (ε₀/2)⟨E²⟩V = ℏω_k/4 is half the zero-point energy ℏω_k/2, the magnetic field carrying the other half

2. (b)

   Show that Show that the matrix element for emission of a photon into a mode (k, λ) already holding n photons is ⟨a; n + 1|H′|b; n⟩ = (e/m)√(ℏ/(2ε₀ω_kV)) √(n + 1) ε_(kλ)* ⋅ ⟨a|p|b⟩, and explain why this makes an excited atom in the vacuum emit, though an atom in a classical field of zero amplitude never does.

   [2]
   Ready to self-mark?Reveal the detailed answerU22 marks+

   Mark-by-mark answer

   1. 1

      only the a† term of A raises the photon number, and a†|n⟩ = √(n + 1)|n + 1⟩ gives the stated element, with atomic factor ε_(kλ)* ⋅ ⟨a|p|b⟩

   2. 2

      for n = 0 the factor is √1, not zero: the 1 in n + 1 has no classical counterpart, since a classical field of zero amplitude gives H′ = 0 and leaves |b⟩ stationary; stimulated emission, proportional to n, and spontaneous emission, the 1, come from the same matrix element

3. (c)

   Derive Using Fermi's golden rule, and summing over all modes, directions and both polarisations, derive A = ω₀³e²|⟨a|r|b⟩|²/(3πε₀ℏc³) for spontaneous emission from |b⟩ to |a⟩.

   [4]
   Ready to self-mark?Reveal the detailed answerU24 marks+

   Mark-by-mark answer

   1. 1

      Γ = (2π/ℏ)Σ_(k,λ)|M_(kλ)|²δ(ℏω_k − ℏω₀), with |M_(kλ)|² = (e²/m²)(ℏ/(2ε₀ω_kV))m²ω₀²|ε_(kλ)* ⋅ r_ab|² for n = 0, using p_ab = −imω₀r_ab

   2. 2

      the mode sum becomes an integral, Σ_k → (V/(2π)³)∫k² dk dΩ = (V/(8π³c³))∫ω² dω dΩ, in which V cancels the 1/V of |M|², and δ(ℏω − ℏω₀) = δ(ω − ω₀)/ℏ sets ω = ω₀

   3. 3

      the two polarisations perpendicular to k give Σ_λ|ε_λ* ⋅ r_ab|² = |r_ab|² − |k̂ ⋅ r_ab|², and since |k̂ ⋅ r_ab|² averages to |r_ab|²/3 over directions, the angular integral is 4π|r_ab|² − (4π/3)|r_ab|² = (8π/3)|r_ab|² (for a real r_ab this is ∫|r_ab|²sin²Θ dΩ, Θ the angle between k and r_ab)

   4. 4

      Γ = (2π/ℏ)(e²ℏω₀/2ε₀)|r_ab|²(ω₀²/(8π³c³ℏ))(8π/3) = ω₀³e²|r_ab|²/(3πε₀ℏc³) = A, obtained from the quantised field without any appeal to Planck's law

4. (d)

   Calculate Calculate A, the lifetime and the natural linewidth Γ/2π of the 2p state of the hydrogen-like helium ion He⁺ (Z = 2).

   [3]
   Ready to self-mark?Reveal the detailed answerU33 marks+

   Mark-by-mark answer

   1. 1

      ℏω₀ = (3/4)(4)(13.606 eV) = 40.82 eV, so ω₀ = (40.82 × 1.602 × 10⁻¹⁹ J)/(1.055 × 10⁻³⁴ J s) = 6.198 × 10¹⁶ rad s⁻¹ (a wavelength of 30.4 nm), and |r_ab|² = 0.5549 × (2.646 × 10⁻¹¹ m)² = 3.885 × 10⁻²² m²

   2. 2

      A = (6.198 × 10¹⁶ s⁻¹)³(1.602 × 10⁻¹⁹ C)²(3.885 × 10⁻²² m²)/[3π(8.854 × 10⁻¹² F m⁻¹)(1.055 × 10⁻³⁴ J s)(2.998 × 10⁸ m s⁻¹)³] = 1.00 × 10¹⁰ s⁻¹

   3. 3

      τ = 1/A = 100 ps and Γ/2π = A/2π = 1.59 GHz; A scales as Z⁴, ω₀³ growing as Z⁶ and |r_ab|² falling as Z⁻², so it is sixteen times hydrogen's 6.26 × 10⁸ s⁻¹

5. (e)

   Explain The atom now sits in thermal radiation at temperature T, whose modes near ω₀ hold n̄ photons on average. Explain how the √n and √(n + 1) factors of the field operators lead to Planck's distribution, rather than assuming it as Einstein did.

   [2]
   Ready to self-mark?Reveal the detailed answerU12 marks+

   Mark-by-mark answer

   1. 1

      because |⟨n + 1|a†|n⟩|² = n + 1 and |⟨n − 1|a|n⟩|² = n, the emission rate from |b⟩ is A(n̄ + 1), stimulated plus spontaneous, and the absorption rate from |a⟩ is An̄ for non-degenerate levels: the B coefficients and A come from the same matrix element

   2. 2

      in equilibrium N_bA(n̄ + 1) = N_aAn̄ with N_b/N_a = e^(−ℏω₀/k_BT), the Boltzmann factor being the one thermal input, so n̄ = 1/(e^(ℏω₀/k_BT) − 1): Planck's distribution is an output of the quantised field, and Einstein's relations follow with it

6. (f)

   Justify Justify the dipole approximation, and the use of the golden rule leading to exponential decay at the rate A, for the He⁺ 2p state.

   [2]
   Ready to self-mark?Reveal the detailed answerU42 marks+

   Mark-by-mark answer

   1. 1

      the dipole approximation needs ka ≪ 1 across the ion: k = ω₀/c = 2.07 × 10⁸ m⁻¹ and the orbit size a₀/Z = 2.65 × 10⁻¹¹ m give ka = 5.5 × 10⁻³, so e^(ik⋅r) ≈ 1 to better than 1%

   2. 2

      the golden rule and the Markov step need a weak coupling and a mode density smooth across the line: Γ/ω₀ = (1.00 × 10¹⁰ s⁻¹)/(6.20 × 10¹⁶ s⁻¹) = 1.6 × 10⁻⁷, so the free-space modes are flat over many linewidths and the decay is exponential, with departures only at times of order 1/ω₀ and at very long times, both negligible

A uniform gas of N spinless bosons of mass m in a box of volume V, with periodic boundary conditions, interacts through a short-range potential represented at low energy by the contact coupling g = 4πℏ²a_s/m, where a_s is the s-wave scattering length. In second quantisation H = Σ_k ε_k a_k†a_k + (g/2V)Σ_(k,k′,q) a_(k+q)†a_(k′−q)†a_(k′)a_k, with ε_k = ℏ²k²/2m. Bogoliubov's approximation replaces a₀ and a₀† by √N₀, where N₀ is the number of bosons in the condensate, and keeps the terms of H up to second order in the operators a_k with k ≠ 0, the sums over such k being written Σ′. Write n = N/V. Take ℏ = 1.055 × 10⁻³⁴ J s, 1 u = 1.661 × 10⁻²⁷ kg and k_B = 1.381 × 10⁻²³ J K⁻¹.

1. (a)

   Show that Show that, to this order and with N₀ eliminated in favour of N, H ≈ gN²/(2V) + Σ′[(ε_k + gn)a_k†a_k + (gn/2)(a_k†a_(−k)† + a_k a_(−k))].

   [3]
   Ready to self-mark?Reveal the detailed answerU23 marks+

   Mark-by-mark answer

   1. 1

      the term with all four momenta zero gives (g/2V)N₀²; terms with one nonzero momentum vanish by momentum conservation, and those with three or four are dropped as of higher order

   2. 2

      with two nonzero momenta, the four ways of placing one condensate operator among the creators and one among the annihilators each give N₀a_k†a_k, and a_q†a_(−q)†a₀a₀ and a₀†a₀†a_q a_(−q) give N₀(a_q†a_(−q)† + a_q a_(−q)), so the interaction is (g/2V)[N₀² + 4N₀Σ′a_k†a_k + N₀Σ′(a_k†a_(−k)† + a_k a_(−k))]

   3. 3

      with N₀ = N − Σ′a_k†a_k, N₀² ≈ N² − 2NΣ′a_k†a_k cancels half of the 4N₀ term, leaving gN/V = gn for each particle outside the condensate; elsewhere N₀ may be replaced by N to this order, which gives the stated H

2. (b)

   Derive Writing a_k = u_k b_k − v_k b_(−k)† with u_k and v_k real, derive the condition on u_k and v_k that keeps [b_k, b_k†] = 1, the choice that removes the terms b_k b_(−k) and b_k†b_(−k)†, and the excitation energy E_k = √(ε_k(ε_k + 2gn)).

   [4]
   Ready to self-mark?Reveal the detailed answerU24 marks+

   Mark-by-mark answer

   1. 1

      [a_k, a_k†] = u_k²[b_k, b_k†] − v_k²[b_(−k), b_(−k)†] = u_k² − v_k² must equal 1, so u_k = cosh θ_k and v_k = sinh θ_k: a hyperbolic rotation rather than an ordinary one

   2. 2

      for the pair k, −k, with X_k = ε_k + gn and Y = gn, H_k = X_k(a_k†a_k + a_(−k)†a_(−k)) + Y(a_k†a_(−k)† + a_k a_(−k)); after substitution the coefficient of b_k b_(−k) + b_k†b_(−k)† is −2X_ku_kv_k + Y(u_k² + v_k²) = −X_k sinh 2θ_k + Y cosh 2θ_k

   3. 3

      setting it to zero gives tanh 2θ_k = gn/(ε_k + gn), which has a solution for every k ≠ 0 because gn is less than ε_k + gn

   4. 4

      the coefficient of b_k†b_k + b_(−k)†b_(−k) is X_k cosh 2θ_k − Y sinh 2θ_k = √(X_k² − Y²), so H = constant + Σ′E_k b_k†b_k with E_k = √((ε_k + gn)² − (gn)²) = √(ε_k(ε_k + 2gn))

3. (c)

   Sketch Sketch E_k against k, with ε_k on the same axes, marking the behaviour at small and at large k and the scale k ≈ 1/ξ, where ξ = ℏ/√(2mgn), at which it changes.

   [2]
   Ready to self-mark?Reveal the detailed answerU12 marks+

   Mark-by-mark answer

   1. 1

      both curves start at the origin, E_k rising linearly with slope ℏc, where c = √(gn/m), a phonon branch, while ε_k starts as a parabola of zero slope; both curves are labelled and E_k lies above ε_k everywhere

   2. 2

      the change from linear to quadratic is marked near k ≈ 1/ξ, where ε_k ≈ gn, and at large k E_k approaches from below the parabola ε_k + gn, a free particle displaced upward by the mean field gn

4. (d)

   Calculate For a condensate of ⁸⁷Rb (m = 87 u, a_s = 5.3 nm) of density n = 2.0 × 10²⁰ m⁻³, calculate gn/k_B in nanokelvin, the speed of sound c, the healing length ξ and n a_s³.

   [3]
   Ready to self-mark?Reveal the detailed answerU33 marks+

   Mark-by-mark answer

   1. 1

      m = 87 × 1.661 × 10⁻²⁷ kg = 1.445 × 10⁻²⁵ kg and g = 4πℏ²a_s/m = 4π(1.055 × 10⁻³⁴ J s)²(5.3 × 10⁻⁹ m)/(1.445 × 10⁻²⁵ kg) = 5.13 × 10⁻⁵¹ J m³, so gn = 1.03 × 10⁻³⁰ J and gn/k_B = 74 nK

   2. 2

      c = √(gn/m) = √(1.03 × 10⁻³⁰ J/1.445 × 10⁻²⁵ kg) = 2.66 × 10⁻³ m s⁻¹ and ξ = ℏ/√(2mgn) = 1.9 × 10⁻⁷ m, about 0.19 μm

   3. 3

      n a_s³ = 2.0 × 10²⁰ m⁻³ × (5.3 × 10⁻⁹ m)³ = 3.0 × 10⁻⁵ ≪ 1, so the gas is dilute and Bogoliubov's quadratic theory, an expansion in √(n a_s³), is well justified

5. (e)

   Determine Determine ⟨a_k†a_k⟩ for k ≠ 0 in the ground state, the vacuum of every b_k, and hence, given that the sum over k gives a depletion (N − N₀)/N = (8/(3√π))√(n a_s³), the fraction of the atoms of (d) that lie outside the condensate at zero temperature.

   [2]
   Ready to self-mark?Reveal the detailed answerU32 marks+

   Mark-by-mark answer

   1. 1

      a_k = u_k b_k − v_k b_(−k)† gives ⟨a_k†a_k⟩ = v_k²⟨b_(−k)b_(−k)†⟩ = v_k² = ½[(ε_k + gn)/E_k − 1], nonzero for every k: the pair term of the interaction holds atoms out of the condensate in pairs of opposite momentum even at T = 0

   2. 2

      (N − N₀)/N = 1.505 × √(3.0 × 10⁻⁵) = 8.2 × 10⁻³: less than 1% of the atoms are outside the condensate, consistent with keeping H to quadratic order

6. (f)

   Predict A tightly focused repulsive laser beam is dragged through the condensate of (d) at 1.5 mm s⁻¹. Predict, using Landau's criterion, whether it heats the gas, and whether an obstacle moving at the same speed through an ideal Bose gas would.

   [2]
   Ready to self-mark?Reveal the detailed answerU12 marks+

   Mark-by-mark answer

   1. 1

      an obstacle moving at v can create an excitation of momentum p only if E(p) − p ⋅ v ≤ 0, which needs v ≥ E(p)/p for some p; for the Bogoliubov spectrum E/p = √(c² + (p/2m)²) is least as p → 0, so the critical speed is c = 2.66 mm s⁻¹, and at 1.5 mm s⁻¹ no excitation can be made: the flow is frictionless and the gas is not heated (vortex nucleation, beyond this theory, is neglected)

   2. 2

      for the ideal gas E = p²/2m, so E/p = p/2m can be made as small as desired and the critical speed is zero: the obstacle creates excitations at any speed and heats the gas, so superfluidity comes from the interaction, which turns the low-energy spectrum into phonons