University Advanced Quantum Mechanics · Midterm Examination

For a system of total angular momentum J, the time-reversal operator satisfies Θ² = (−1)^(2J), whatever phase is chosen for it. Options A to C place an ion at a crystal site with no point symmetry at all, so the crystal field removes every degeneracy that symmetry does not protect; option D is a free atom. In which case does time-reversal invariance guarantee that every level of the ground multiplet stays at least twofold degenerate?

1. AHo³⁺, whose ground multiplet has J = 8, in the crystal field alone
2. BEr³⁺, whose ground multiplet has J = 15/2, in the crystal field and a uniform static magnetic field of 0.10 T
3. CEr³⁺, whose ground multiplet has J = 15/2, in the crystal field and a uniform static electric field of 1.0 × 10⁷ V/m
4. DA free hydrogen atom in its 1s level, the proton's spin included, in a uniform static electric field of 1.0 × 10⁷ V/m

Two electrons scatter in their centre-of-mass frame through a spin-independent interaction whose direct amplitude is f(θ). Beam and target are unpolarised, and a detector at θ = 90° counts an electron whichever of the two it is. In units of |f(90°)|², what differential cross-section does the detector measure?

1. A0
2. B1
3. C2
4. D4

For a spinless particle bound to a point nucleus of charge Ze, the exact Klein–Gordon energies are E = mc²[1 + (Zα)²/(n − l − ½ + s)²]^(−1/2), with s = √((l + ½)² − (Zα)²) and α = 1/137.04. Above about what Z does the formula first fail, giving a complex energy for the 1s state?

1. A69
2. B137
3. C172
4. D206

A free Dirac particle has H = cα⋅p + βmc², where α_i and β are the 4 × 4 Dirac matrices, Σ = diag(σ, σ) is the spin matrix and L = r × p. Which statement is correct?

1. A[H, L] = 0, because a free particle experiences no torque
2. B[H, Σ_z] = 0, because H contains no magnetic field to turn the spin
3. C[H, β] = 0, so the upper and lower components of ψ evolve independently
4. D[H, Σ⋅p] = 0, so helicity is conserved, although Σ_z on its own is not

Under parity a Dirac field transforms as ψ(r, t) → γ⁰ψ(−r, t), with ψ̄ = ψ†γ⁰ and γ⁵ = iγ⁰γ¹γ²γ³. Which bilinear is unchanged by every proper Lorentz transformation but changes sign under parity?

1. Aψ̄ψ
2. Bψ̄γ⁰ψ
3. Cψ̄γ⁵ψ
4. Dψ̄γ⁰γ⁵ψ

An electron is held in a Penning trap whose magnetic field is 4.70 T. Its cyclotron frequency is f_c = eB/(2πm), with e/m = 1.75882 × 10¹¹ C kg⁻¹. In the exact Dirac Landau levels, where g = 2, the level with n quanta of cyclotron motion and spin up along B coincides with the level with n + 1 quanta and spin down. Taking g = 2(1 + a), with a = 1.15965 × 10⁻³, and neglecting relativistic shifts and the trap's electric field, at what frequency must a drive be applied to connect these two levels?

1. A152.6 MHz
2. B305.1 MHz
3. C131.72 GHz
4. D76.3 MHz

In the exact Dirac–Coulomb spectrum of a hydrogen-like ion with a point nucleus and Z < 137, which statement is correct?

1. A2S_(1/2) and 2P_(1/2) coincide only to order (Zα)⁴; the exact energies separate them at order (Zα)⁶, and that separation is the Lamb shift.
2. B3P_(3/2) and 3D_(3/2) coincide exactly, because the energies depend only on n and on |κ| = j + ½.
3. C3D_(3/2) and 3D_(5/2) coincide exactly, because the energies depend only on n and l.
4. D2S_(1/2), with κ = −1, and 2P_(1/2), with κ = +1, are separated, because the energies depend on the sign of κ.

An electron of total energy E = 2mc², so kinetic energy mc², meets an electrostatic step on which its potential energy rises by V₀. Beyond the step its momentum ℏq obeys c²ℏ²q² = (E − V₀)² − m²c⁴. For which step heights is the transmitted wave oscillatory rather than evanescent?

1. AV₀ < mc² only: any higher step is penetrated only evanescently
2. BV₀ < 2mc² only: a step below the total energy can be crossed
3. CV₀ < mc² and V₀ > 2mc²
4. DV₀ < mc² and V₀ > 3mc²

For a spin-½ particle, the time-reversal operator in the basis of S_z eigenstates |↑⟩ and |↓⟩ is Θ = exp(−iπS_y/ℏ)K = −iσ_yK, where K takes the complex conjugate of every component in that basis. Θ is antiunitary: Θ(a|u⟩ + b|v⟩) = a*Θ|u⟩ + b*Θ|v⟩ and ⟨Θu|Θv⟩ = ⟨v|u⟩. The Pauli matrices are σ_x = [[0, 1], [1, 0]], σ_y = [[0, −i], [i, 0]] and σ_z = [[1, 0], [0, −1]], and S = (ℏ/2)σ.

1. (a)

   Determine Determine Θ|↑⟩ and Θ|↓⟩, and hence Θ². Show that replacing Θ by e^(iδ)Θ, for any real δ, leaves Θ² unchanged.

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   1. 1

      −iσ_y = [[0, −1], [1, 0]] is real, so Θ|↑⟩ = |↓⟩ and Θ|↓⟩ = −|↑⟩; then Θ²|↑⟩ = Θ|↓⟩ = −|↑⟩ and Θ²|↓⟩ = −Θ|↑⟩ = −|↓⟩, so Θ² = −1

   2. 2

      (e^(iδ)Θ)(e^(iδ)Θ) = e^(iδ)e^(−iδ)Θ² = Θ², because the antilinear Θ conjugates the phase it passes through: the sign of Θ² belongs to the spin, not to a convention

2. (b)

   Show that Show that ΘσΘ⁻¹ = −σ, so that ΘSΘ⁻¹ = −S.

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   1. 1

      with Θ⁻¹ = K(iσ_y), ΘσΘ⁻¹ = σ_yσ*σ_y: σ_x and σ_z are real and anticommute with σ_y, giving σ_yσ_xσ_y = −σ_x and σ_yσ_zσ_y = −σ_z, while σ_y* = −σ_y gives σ_y(−σ_y)σ_y = −σ_y

3. (c)

   Justify A particle, or an atom, whose only degeneracy is the 2j + 1 values of m in a rotation-invariant theory is said to have a permanent electric dipole moment d if ⟨j, m|d|j, m⟩ ≠ 0. Using the Wigner–Eckart theorem within the multiplet, justify the claim that such a moment requires both parity and time-reversal invariance to fail.

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   1. 1

      within one multiplet the Wigner–Eckart theorem makes every vector operator proportional to J: ⟨j, m′|d|j, m⟩ = k⟨j, m′|J|j, m⟩, with k a single real constant because d and J are Hermitian, so a permanent moment means k ≠ 0

   2. 2

      d is odd under P and even under T, while J is even under P and odd under T; if P is a symmetry it maps the non-degenerate multiplet into itself, and PdP⁻¹ = −d with PJP⁻¹ = J forces k = −k, and if T is a symmetry, ΘdΘ⁻¹ = d with ΘJΘ⁻¹ = −J forces k = −k again, so k ≠ 0 needs both to fail

4. (d)

   Explain The n = 2 level of hydrogen shows a linear Stark effect, a shift proportional to the applied field, as though the atom had a permanent dipole moment. Explain why this does not contradict (c).

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   1. 1

      the linear effect needs states of opposite parity at one energy, 2s and 2p, so the level is not the non-degenerate multiplet that (c) assumes; the field mixes them into states that carry a dipole, and once the field is weak beside the 2S_(1/2)–2P_(1/2) splitting the shift becomes quadratic, as it must for a non-degenerate level

5. (e)

   Show that For a Dirac spinor in the Dirac representation, γ^i = [[0, σ_i], [−σ_i, 0]] for i = 1, 2, 3 in 2 × 2 blocks, and time reversal acts as T = iγ¹γ³K. Show that T² = −1.

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   1. 1

      γ¹ and γ³ contain only σ_x and σ_z, so they are real, and K(iγ¹γ³) = (−iγ¹γ³)K; hence T² = (iγ¹γ³)(−iγ¹γ³)K² = γ¹γ³γ¹γ³

   2. 2

      γ³γ¹ = −γ¹γ³ and (γ^i)² = g^(ii) = −1, so γ¹γ³γ¹γ³ = −(γ¹)²(γ³)² = −(−1)(−1) = −1: the four-component spinor, like the two-component one in (a), has T² = −1, and a Dirac particle's levels in any T-invariant field are Kramers doublets

A spinless particle of mass m and charge −e is bound by a point nucleus of charge Ze, treated as the fixed source of the static potential energy V(r) = −Zαℏc/r. Covariant notation uses x^μ = (ct, r), the metric g = diag(1, −1, −1, −1) and p^μ = iℏ∂^μ, so that p⁰ = (iℏ/c)∂/∂t and p = −iℏ∇; minimal coupling replaces the energy E by E − V, and a stationary state has ψ ∝ e^(−iEt/ℏ). Schrödinger's hydrogen problem, −(ℏ²/2M)u″ + [ℏ²λ(λ + 1)/(2Mr²) − k/r]u = εu, has bound states ε = −Mk²/[2ℏ²(n_r + λ + 1)²], n_r = 0, 1, 2, …, for any real λ > −½, taking the solution regular at the origin, u ∝ r^(λ+1). Take α = 1/137.036.

1. (a)

   Derive Derive the free Klein–Gordon equation from p_μp^μ = m²c², and show that with the Coulomb potential included a stationary state obeys [(E − V)² − m²c⁴]ψ = −ℏ²c²∇²ψ.

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   1. 1

      p_μp^μ = (p⁰)² − p⋅p, with p⁰ = (iℏ/c)∂/∂t and p = −iℏ∇, gives −(ℏ²/c²)∂²ψ/∂t² + ℏ²∇²ψ = m²c²ψ, that is (1/c²)∂²ψ/∂t² − ∇²ψ + (mc/ℏ)²ψ = 0, Lorentz invariant because p_μp^μ is a scalar

   2. 2

      for ψ ∝ e^(−iEt/ℏ), cp⁰ acting on ψ gives E; minimal coupling replaces it by E − V, so (E − V)²ψ − c²p²ψ = m²c⁴ψ, that is [(E − V)² − m²c⁴]ψ = −ℏ²c²∇²ψ

2. (b)

   Determine With ψ = [u(r)/r]Yₗᵐ(θ, φ), the equation becomes −ℏ²c²u″ + [ℏ²c²l(l + 1) − (Zαℏc)²]u/r² − 2EZαℏc u/r = (E² − m²c⁴)u. By mapping it onto Schrödinger's problem, determine the exact bound-state energies, writing n = n_r + l + 1.

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   1. 1

      dividing by 2E gives Schrödinger's form with M = E/c², k = Zαℏc, ε = (E² − m²c⁴)/(2E) and λ(λ + 1) = l(l + 1) − (Zα)², so λ = −½ + √((l + ½)² − (Zα)²), the root that tends to l as Zα → 0

   2. 2

      then (E² − m²c⁴)/(2E) = −(E/c²)(Zαℏc)²/(2ℏ²N²) = −E(Zα)²/(2N²), with N = n_r + λ + 1 = n − l − ½ + √((l + ½)² − (Zα)²)

   3. 3

      so E²[1 + (Zα)²/N²] = m²c⁴ and E = mc²[1 + (Zα)²/N²]^(−1/2), which depends on l through N, unlike the Schrödinger levels

3. (c)

   Show that Show that, to order (Zα)⁴, E = mc² − mc²(Zα)²/(2n²) − [mc²(Zα)⁴/(2n⁴)][n/(l + ½) − ¾].

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   1. 1

      N = n − δ with δ = (l + ½) − √((l + ½)² − (Zα)²) ≈ (Zα)²/(2l + 1), so (Zα)²/N² ≈ (Zα)²/n² + 2(Zα)⁴/[n³(2l + 1)]

   2. 2

      with (1 + x)^(−1/2) ≈ 1 − x/2 + 3x²/8 and x² ≈ (Zα)⁴/n⁴, E ≈ mc²[1 − (Zα)²/(2n²) − (Zα)⁴/(n³(2l + 1)) + 3(Zα)⁴/(8n⁴)], and (Zα)⁴/(n³(2l + 1)) = [(Zα)⁴/(2n⁴)]n/(l + ½), which gives the result

4. (d)

   Calculate The charged-pion mass is determined from the X-rays of the 5g → 4f transition in pionic nitrogen, a π⁻ bound to ¹⁴N (Z = 7), where the strong interaction is negligible for these orbits. Treat the pion as moving in the static field of the nucleus with its mass replaced by the reduced mass μ, using m_πc² = 139.570 MeV and a nuclear rest energy of 13040.20 MeV. Calculate the transition energy to order (Zα)², and the order-(Zα)⁴ correction to it, in eV.

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   1. 1

      μc² = (139.570 × 13040.20/13179.77) MeV = 138.092 MeV and Zα = 0.051081, so to order (Zα)² the line is [μc²(Zα)²/2](1/16 − 1/25) = 4053.66 eV

   2. 2

      the (Zα)⁴ terms are −0.272 eV for 5g (n = 5, l = 4) and −0.721 eV for 4f (n = 4, l = 3), so the correction is +0.450 eV and the line lies at 4054.11 eV; at 1.1 × 10⁻⁴ of the energy, a pion mass quoted to a few parts per million must include it

5. (e)

   Sketch On one energy-level diagram, sketch the n = 2 levels of a spinless particle of the electron's mass bound to a proton, from (c) with Z = 1, beside the n = 2 levels of hydrogen given by the Dirac equation, E = mc² − mc²α²/(2n²) − [mc²α⁴/(2n⁴)][n/(j + ½) − ¾]. Mark each level's shift from the unperturbed n = 2 level in units of mc²α⁴/128, and label the hydrogen levels with the terms of the fine-structure Hamiltonian that move them away from the Klein–Gordon levels.

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   1. 1

      Klein–Gordon: 2s at −13 units and 2p at −7/3 units, 2p above 2s by 32/3 units, mc²α⁴/12 or 29.2 GHz; hydrogen: 2S_(1/2) and 2P_(1/2) drawn as one level at −5 units and 2P_(3/2) at −1 unit, every level below the unperturbed one

   2. 2

      the differences labelled: the Darwin term raises 2s by 8 units to 2S_(1/2); spin–orbit coupling raises 2p by 4/3 units to 2P_(3/2) and lowers it by 8/3 units to 2P_(1/2); the Klein–Gordon levels carry only the −p⁴/(8m³c²) term, so their pattern is right only for a spinless particle, as in a pionic atom

In the Dirac representation the free Dirac Hamiltonian is H = cα⋅p + βmc², with β = [[I, 0], [0, −I]] and α_i = [[0, σ_i], [σ_i, 0]] in 2 × 2 blocks, and ψ̄ = ψ†β. An electron moves along +z with momentum p = (3/4)mc. Positive-energy spinors are normalised to u†u = 2E/c, and the matrix that carries a rest spinor into the spinor of an electron moving along +z with Lorentz factor γ is S = cosh(η/2) + α_z sinh(η/2), where cosh η = γ. Take mc² = 0.511 MeV, ℏ = 1.055 × 10⁻³⁴ J s, ℏc = 197.3 MeV fm and 1 MeV = 1.602 × 10⁻¹³ J.

1. (a)

   Derive From the Dirac equation Hu = Eu, derive the positive-energy spinor u(p, ↑), with spin up along z, in the form N(1, 0, b, 0), finding E, b and N.

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   1. 1

      the anticommutation relations make H² = (p²c² + m²c⁴)I, so the positive energy is E = mc²√(1 + 9/16) = (5/4)mc²

   2. 2

      writing u = (φ, χ), the lower block of Hu = Eu reads cσ⋅p φ − mc²χ = Eχ, so χ = [cpσ_z/(E + mc²)]φ, and with φ = (1, 0), b = (3/4)/(9/4) = 1/3

   3. 3

      N²(1 + 1/9) = 2E/c = (5/2)mc, so N² = (9/4)mc and u = √(mc)(3/2, 0, 1/2, 0)

2. (b)

   Show that Show that S applied to the rest spinor √(2mc)(1, 0, 0, 0) gives the spinor of (a), and that S is not unitary, in a way consistent with ūu staying at 2mc while u†u grows to 2E/c.

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   1. 1

      γ = 5/4, so cosh(η/2) = √((γ + 1)/2) = 3/(2√2) and sinh(η/2) = √((γ − 1)/2) = 1/(2√2); since α_z(1, 0, 0, 0) = (0, 0, 1, 0), S√(2mc)(1, 0, 0, 0) = √(2mc)(3/(2√2), 0, 1/(2√2), 0) = √(mc)(3/2, 0, 1/2, 0), the spinor of (a)

   2. 2

      S is Hermitian, so S†S = S² = cosh η + α_z sinh η ≠ I, and u†u = 2mc cosh η = (5/2)mc = 2E/c because α_z has no diagonal element in the rest spinor; but βSβ = S⁻¹, so S†βS = β and ūu = 2mc is unchanged

3. (c)

   Calculate Calculate ⟨cα_z⟩ = u†cα_zu/u†u for this spinor, and compare it with the group velocity ∂E/∂p.

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   1. 1

      α_zu = √(mc)(1/2, 0, 3/2, 0), so u†α_zu = mc(3/4 + 3/4) = (3/2)mc and ⟨cα_z⟩ = [(3/2)/(5/2)]c = 0.6c

   2. 2

      ∂E/∂p = pc²/E = [(3/4)/(5/4)]c = 0.6c, the same: in a positive-energy plane wave the velocity operator cα = (i/ℏ)[H, r] has the group velocity as its expectation value

4. (d)

   Explain Every eigenvalue of cα_z is +c or −c. Explain how this is consistent with (c).

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   1. 1

      α_z² = I gives eigenvalues ±1, but α_z does not commute with H, since it anticommutes with β, so its eigenstates are superpositions of positive- and negative-energy solutions; the electron of (c) is not in an α_z eigenstate, and 0.6c is the average of cα_z over a state of one sign of energy

5. (e)

   Estimate A wave packet built from both positive- and negative-energy solutions near this momentum shows Zitterbewegung. Estimate the angular frequency and the amplitude of the trembling, stating your assumptions, and say what happens to it in a packet built from positive-energy solutions alone.

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   1. 1

      assuming the packet is narrow in momentum about p, the interference between components at +E and −E oscillates at ω = 2E/ℏ = 2 × 0.639 × 1.602 × 10⁻¹³ J/(1.055 × 10⁻³⁴ J s) = 1.9 × 10²¹ s⁻¹, with amplitude of order ℏc/(2E) = 197.3 MeV fm/1.28 MeV ≈ 1.5 × 10⁻¹³ m, close to ℏ/(2mc) = 1.9 × 10⁻¹³ m, a fraction of the Compton wavelength

   2. 2

      the trembling is an interference between the two signs of energy, so it vanishes identically in a packet of positive energy only, where ⟨r⟩ moves uniformly at the group velocity of (c); confining a packet within about ℏ/(mc) forces in negative-energy components, and the trembling with them

An electron of mass m is bound to a point nucleus of charge Ze, with Z < 137, so that V(r) = −Zαℏc/r. In a central field the Dirac Hamiltonian commutes with J, with the Dirac parity operator and with K = β(Σ⋅L + ℏ), whose eigenvalue is written −κℏ, so that κ = −(j + ½) for j = l + ½ and κ = +(j + ½) for j = l − ½, l being the orbital angular momentum of the upper component. Writing the upper component as [G(r)/r] times a spinor harmonic of orbital angular momentum l, and the lower as i[F(r)/r] times one of orbital angular momentum l′, the radial equations are ℏc(dG/dr + κG/r) = (E + mc² − V)F and ℏc(dF/dr − κF/r) = −(E − mc² − V)G. Take α = 1/137.036 and mc² = 511.0 keV.

1. (a)

   State State κ for each state of the n = 2 shell, 2S_(1/2), 2P_(1/2) and 2P_(3/2), together with l′ and the parity of each state.

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   1. 1

      2S_(1/2): κ = −1; 2P_(1/2): κ = +1; 2P_(3/2): κ = −2

   2. 2

      l′ = 2j − l: 1 for 2S_(1/2), 0 for 2P_(1/2) and 2 for 2P_(3/2); the parity is (−1)^l of the upper component, even for 2S_(1/2) and odd for both 2P states, the β in the Dirac parity operator compensating the opposite (−1)^(l′) of the lower component

2. (b)

   Derive For a solution with no radial nodes, try G = Ar^γ e^(−λr) and F = Br^γ e^(−λr), with λ > 0. Derive from the radial equations γ, the ratio B/A, and λ in terms of E.

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   1. 1

      the 1/r terms of the first equation give ℏc(γ + κ)A = ZαℏcB, so B/A = (γ + κ)/(Zα); those of the second give (γ − κ)B = −ZαA, so B/A = −Zα/(γ − κ)

   2. 2

      equating the two ratios, γ² − κ² = −(Zα)², so γ = √(κ² − (Zα)²), the positive root, since the negative one gives G and F that diverge at the origin, as r^(−|κ|) when Zα is small, and cannot be normalised; for |κ| = 1, γ is less than 1

   3. 3

      the r-independent terms give −ℏcλA = (E + mc²)B and −ℏcλB = −(E − mc²)A; multiplying them, (ℏcλ)² = m²c⁴ − E², so λ = √(m²c⁴ − E²)/(ℏc), real for a bound state with |E| < mc²

3. (c)

   Show that Show that E = mc²γ/|κ|, and that such a nodeless solution exists only for κ < 0.

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   1. 1

      from the constant terms B/A = −ℏcλ/(E + mc²) = −√((mc² − E)/(mc² + E)), which is negative; with B/A = (γ + κ)/(Zα) this needs γ + κ < 0, impossible for κ > 0 because γ > 0, so κ = −|κ| and only states with j = l + ½, 1S_(1/2), 2P_(3/2), 3D_(5/2) and so on, can be nodeless

   2. 2

      squaring, (|κ| − γ)²/(Zα)² = (mc² − E)/(mc² + E); since (Zα)² = (|κ| − γ)(|κ| + γ) the left side is (|κ| − γ)/(|κ| + γ), so E/mc² = γ/|κ| = √(1 − (Zα)²/κ²)

4. (d)

   Calculate For hydrogen-like lead, Pb⁸¹⁺ (Z = 82), calculate the 1s binding energy mc² − E, comparing it with the value to order (Zα)⁴, mc²[(Zα)²/2 + (Zα)⁴/8], and the fraction of the 1s probability carried by the small component.

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   1. 1

      Zα = 0.59838 and γ = √(1 − 0.35806) = 0.80121, so mc² − E = 511.0 keV × 0.19879 = 101.6 keV, against 99.7 keV to order (Zα)⁴ and 91.5 keV non-relativistically

   2. 2

      G and F share the radial factor, so the small component carries B²/(A² + B²) = (1 − γ)/2 = 0.099, about 10% of the probability, against α²/4 = 1.3 × 10⁻⁵ in hydrogen

5. (e)

   Predict Predict how the 1s solution behaves as Z approaches 137 and beyond, for a point nucleus, and say what a nucleus of finite size changes.

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   1. 1

      as Zα → 1, γ → 0 and E = mc²γ → 0, so the 1s level reaches zero energy, not −mc², while the density near the origin, ∝ r^(2γ − 2), grows ever more singular; for Zα > 1, γ is imaginary, the solutions oscillate as r^(±i|γ|) with no limit at the origin, and the point-nucleus problem has no ground state, the Dirac analogue of the Klein–Gordon fall to the centre at Zα = ½

   2. 2

      a finite nucleus removes the singularity, so the 1s level continues smoothly past Z = 137 and reaches the top of the negative-energy continuum, E = −mc², near Z ≈ 172; beyond that the one-particle theory fails, and an empty 1s level would be filled by spontaneous pair creation

Identical nuclei, each in its ground state of spin s, scatter in their centre-of-mass frame through the Coulomb interaction alone, which does not act on spin; every run is far enough below the Coulomb barrier that the nuclear force can be neglected. A detector of angular half-width 0.50° centred on θ = 90° counts N₉₀ nuclei, and N_R is the number it would have counted in the same run if |f(θ)|² and |f(π − θ)|² added without interference, computed from the Rutherford cross-section and the measured beam and target. For the exact Coulomb amplitude f(θ), the interference term is proportional to cos[η ln tan²(θ/2)], where η = Z²α/β, βc is the relative speed, β² = 2E_cm/(μc²) and μ is the reduced mass. Take α = 1/137.036 and 1 u = 931.5 MeV/c², treat N₉₀ as a Poisson count and N_R as exact.

1. (a)

   Show that The (2s + 1)² spin states of the pair divide into (s + 1)(2s + 1) that are symmetric under exchange of the two spins and s(2s + 1) that are antisymmetric. Show that for an unpolarised beam and target dσ/dΩ = |f(θ)|² + |f(π − θ)|² + [2(−1)^(2s)/(2s + 1)]Re[f(θ)f*(π − θ)], and hence that R = N₉₀/N_R should equal 1 + (−1)^(2s)/(2s + 1).

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   1. 1

      the whole state must be symmetric for integer s and antisymmetric for half-integer s, so for bosons the symmetric spin states, a fraction (s + 1)/(2s + 1), go with |f + f′|² and the antisymmetric fraction s/(2s + 1) with |f − f′|², where f′ = f(π − θ), and for fermions the two fractions are exchanged; the cross terms ±2Re(ff′*) then combine with weight [(s + 1) − s]/(2s + 1) = 1/(2s + 1) and sign (−1)^(2s)

   2. 2

      at θ = 90°, f(π − θ) = f(θ), phase included, so Re(ff′*) = |f|² and dσ/dΩ = 2|f|²[1 + (−1)^(2s)/(2s + 1)], while the non-interfering sum is 2|f|², giving R = 1 + (−1)^(2s)/(2s + 1)

2. (b)

   Calculate Calculate R and its standard uncertainty for each pair.

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   1. 1

      R = N₉₀/N_R: 1.9916 for ¹²C, 0.5044 for ¹³C, 1.3384 for ¹⁴N, 1.1375 for ¹⁰B, 0.7462 for ¹¹B and 0.8371 for ¹⁷O

   2. 2

      σ_R = √N₉₀/N_R: 0.0200, 0.0079, 0.0122, 0.0097, 0.0086 and 0.0091 respectively

3. (c)

   Determine Determine the spin of each nucleus, and the significance with which the nearest alternative value is excluded. Say which assignments are secure at three standard deviations.

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   1. 1

      R > 1 marks bosons and R < 1 fermions, and |R − 1| = 1/(2s + 1): ¹²C gives 1.992 ± 0.020, consistent with s = 0, with s = 1 (R = 1.333) excluded at 33σ, and so a check on the normalisation of N_R; ¹³C gives 2s + 1 = 2.0, s = ½, with s = 3/2 (R = 0.750) excluded at 31σ

   2. 2

      ¹⁴N: 2s + 1 = 2.95, s = 1, with s = 2 (R = 1.200) excluded at 11σ; ¹¹B: 2s + 1 = 3.94, s = 3/2, with s = 5/2 (R = 0.833) excluded at 10σ; ¹⁷O: 2s + 1 = 6.1, s = 5/2, with s = 7/2 (R = 0.875) excluded at 4.1σ and s = 3/2 at 9.5σ

   3. 3

      ¹⁰B: 2s + 1 = 7.3, so s = 3 (R = 1.1429, 0.6σ away) is favoured and s = 2 is excluded at 6.4σ, but s = 4 (R = 1.1111) is only 2.7σ away, so this assignment is not secure at 3σ; separating 3 from 4 at 5σ needs σ_R ≈ 0.0318/5 = 0.0064, about 2.3 times as many counts

4. (d)

   Evaluate Near 90°, ln tan²(θ/2) ≈ 2(θ − 90°), with the angle in radians, so the detector averages cos[2η(θ − 90°)] over its acceptance. Evaluate whether this averaging biases any spin assignment, using the pair with the largest η.

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   1. 1

      the largest η is for ¹⁷O + ¹⁷O: μc² = 8.5 × 931.5 MeV = 7918 MeV, β = √(8.00/7918) = 0.0318 and η = 64/(137.036 × 0.0318) = 14.7; averaging over ±0.50° = ±0.00873 rad multiplies the interference term by sin x/x with x = 2η × 0.00873 = 0.256, that is by 0.989

   2. 2

      the expected R for s = 5/2 moves from 0.8333 to 0.8352, a shift of 0.0018, that is 0.2σ and about 4% of the 0.042 gap to the s = 7/2 value; the fractional loss of interference, 1 − sin x/x ≈ x²/6, grows as η², so every other pair is moved by under 2% of the gap to its neighbouring spin and by at most 0.22σ, and the averaging biases no assignment; a detector three times wider would move ¹⁷O by 1.7σ and begin to pull the high-spin values towards 1

5. (e)

   Explain Explain why the measurement is made at 90° rather than at another angle.

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      only at 90° does f(π − θ) equal f(θ) exactly, so the Coulomb phases cancel and the interference term has its full size whatever η and E_cm are; at any other angle it carries cos[η ln tan²(θ/2)], which near 90° oscillates with a period of 180°/η, between 12° and 32° for these pairs, and the spin would be entangled with η

An electron of charge q = −e and mass m moves in static potentials with Dirac Hamiltonian H = cα⋅π + βmc² + V(r), where π = p − qA, p = −iℏ∇ and V is the electrostatic potential energy. In the Dirac representation write ψ = (φ, χ), with φ the large and χ the small two-component spinor, and E = mc² + E′. You may use (σ⋅a)(σ⋅b) = a⋅b + iσ⋅(a × b) for vector operators a and b that commute with σ, and [p, f(r)] = −iℏ∇f. Take α = 1/137.036 and mc² = 0.5110 MeV.

1. (a)

   Derive Derive the two coupled equations for φ and χ. For |E′ − V| ≪ mc², eliminate χ to leading order and derive the Pauli Hamiltonian, showing that its spin term is −(q/m)S⋅B, so that g = 2.

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   1. 1

      the blocks of Hψ = Eψ are (E′ − V)φ = cσ⋅πχ and (E′ + 2mc² − V)χ = cσ⋅πφ, so χ = [c/(2mc² + E′ − V)]σ⋅πφ ≈ [σ⋅π/(2mc)]φ, smaller than φ by about v/(2c)

   2. 2

      substituting, E′φ = [(σ⋅π)²/(2m) + V]φ, and (σ⋅π)² = π² + iσ⋅(π × π), the cross product surviving because the components of π do not commute

   3. 3

      [π_x, π_y] = iqℏ(∂A_y/∂x − ∂A_x/∂y) = iqℏB_z and cyclically, so π × π = iqℏB and (σ⋅π)² = π² − qℏσ⋅B; with S = (ℏ/2)σ, H_P = π²/(2m) − (q/m)S⋅B + V, a moment (q/m)S = g(q/2m)S with g = 2, fixed by minimal coupling rather than put in by hand

2. (b)

   Estimate Estimate the ratio |χ|/|φ| and the fraction of the probability carried by χ in the ground state of hydrogen, stating your assumptions.

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   1. 1

      assuming the typical momentum is ℏ/a₀ = αmc, so that v/c ≈ α, |χ|/|φ| ≈ p/(2mc) ≈ α/2 = 3.6 × 10⁻³

   2. 2

      the probability in χ is then about (α/2)² = 1.3 × 10⁻⁵, in agreement with the exact Dirac 1s value (1 − √(1 − α²))/2 = 1.33 × 10⁻⁵; the norm of φ alone therefore falls short of 1 by about 10⁻⁵, the same relative order as the fine structure, which is why the next order needs care

3. (c)

   Explain Now set A = 0. Keeping the next term in χ gives E′φ = [p²/(2m) + V − (1/(4m²c²))(σ⋅p)(E′ − V)(σ⋅p)]φ. Explain why this cannot be read as a Schrödinger equation for φ with a Hermitian Hamiltonian at order (v/c)².

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   1. 1

      the bracket contains the eigenvalue E′, so each state obeys a different operator and the equation is not an eigenvalue problem for one fixed Hamiltonian; removing E′ with the lowest-order relation (E′ − V)φ ≈ [p²/(2m)]φ means commuting E′ − V through σ⋅p, and the commutator leaves the term (iℏ/(4m²c²))∇V⋅p, which is not Hermitian, although (σ⋅p)V(σ⋅p) itself is

   2. 2

      the reason is that probability is shared: Dirac eigenstates satisfy ∫(φₙ†φₘ + χₙ†χₘ) d³r = δₙₘ, so the large components of different states are neither normalised nor orthogonal, ∫φₙ†φₙ d³r ≈ 1 − ⟨p²⟩/(4m²c²), and no Hermitian operator can have them as its eigenfunctions until φ is renormalised

4. (d)

   Show that Show that Φ = [1 + p²/(8m²c²)]φ is normalised to order (v/c)², and that it obeys E′Φ = H_FWΦ with H_FW = p²/(2m) + V − p⁴/(8m³c²) + (ℏ/(4m²c²))σ⋅(∇V × p) + (ℏ²/(8m²c²))∇²V.

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   1. 1

      ∫χ†χ d³r ≈ ∫φ†[p²/(4m²c²)]φ d³r, so ∫(φ†φ + χ†χ) d³r = ∫φ†[1 + p²/(4m²c²)]φ d³r = ∫Φ†Φ d³r to this order, since [1 + p²/(8m²c²)]² ≈ 1 + p²/(4m²c²) and p² is Hermitian

   2. 2

      substituting φ = [1 − p²/(8m²c²)]Φ and multiplying on the left by 1 − p²/(8m²c²), the left side becomes E′[1 − p²/(4m²c²)]Φ, whose E′p² term cancels the −E′p²/(4m²c²) inside (σ⋅p)(E′ − V)(σ⋅p), leaving E′Φ = [p²/(2m) + V − p⁴/(8m³c²) − (1/(8m²c²))(p²V + Vp²) + (1/(4m²c²))(σ⋅p)V(σ⋅p)]Φ

   3. 3

      (σ⋅p)V = Vσ⋅p − iℏσ⋅∇V, so (σ⋅p)V(σ⋅p) = Vp² − iℏ∇V⋅p + ℏσ⋅(∇V × p) by the given identity, and p²V = Vp² − ℏ²∇²V − 2iℏ∇V⋅p, so p²V + Vp² = 2Vp² − ℏ²∇²V − 2iℏ∇V⋅p

   4. 4

      combining, the Vp² terms cancel and so do the two ∇V⋅p terms, neither Hermitian on its own, (1/(4m²c²))(−iℏ∇V⋅p) − (1/(8m²c²))(−2iℏ∇V⋅p) = 0, leaving (ℏ/(4m²c²))σ⋅(∇V × p) + (ℏ²/(8m²c²))∇²V: H_FW, Hermitian, with every term traced to a step

5. (e)

   Determine For a central potential V(r), and then for V = −Zαℏc/r, determine the spin–orbit term in the form ξ(r)S⋅L and the Darwin term, and evaluate the Darwin shift of the 1s level of hydrogen in eV, given |ψ₁₀₀(0)|² = 1/(πa₀³) with a₀ = ℏ/(αmc).

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   1. 1

      ∇V = (dV/dr)r/r, so σ⋅(∇V × p) = (1/r)(dV/dr)σ⋅L, and with σ = 2S/ℏ the spin–orbit term is [1/(2m²c²)](1/r)(dV/dr)S⋅L

   2. 2

      for V = −Zαℏc/r, ∇²V = −Zαℏc∇²(1/r) = 4πZαℏcδ³(r), so the Darwin term is [πℏ²/(2m²c²)]Zαℏc δ³(r), which acts only on s states

   3. 3

      for hydrogen 1s, ⟨H_D⟩ = [πℏ²αℏc/(2m²c²)]/(πa₀³) = α⁴mc²/2 = 7.25 × 10⁻⁴ eV

6. (f)

   Compare In a semiclassical picture the electron, moving with velocity v through the electric field ℰ = ∇V/e, sees in its rest frame the magnetic field B′ = −v × ℰ/c². Compare the spin–orbit coefficient that −μ⋅B′ gives, with μ = −g(e/2m)S, with the one in (e), account for the difference, and say what changes if g = 2(1 + a).

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   1. 1

      B′ = [1/(emc²)](1/r)(dV/dr)L, so −μ⋅B′ = [g/(2m²c²)](1/r)(dV/dr)S⋅L, which for g = 2 has coefficient 1/(m²c²), twice the Dirac result

   2. 2

      the rest frame of an accelerated electron rotates relative to the laboratory, the Thomas precession, which adds −[1/(2m²c²)](1/r)(dV/dr)S⋅L; the total, [(g − 1)/(2m²c²)](1/r)(dV/dr)S⋅L, equals the Dirac result for g = 2, where it emerges with no frame argument at all, and becomes (1 + 2a)/(2m²c²) for an anomalous moment, since the kinematic part does not depend on g