University Physics V · Final Examination

A single photon on an integrated-optics chip is divided equally among three waveguides, leaving it in |ψ⟩ = (|1⟩ + |2⟩ + |3⟩)/√3, where |k⟩ is the state in which the photon travels in waveguide k. A heater on waveguide 3 then multiplies the amplitude in that waveguide by e^(iπ). One of two measurements is made afterwards: M₁ asks whether the photon is in waveguide 3, with projector |3⟩⟨3|, and M₂ asks whether it is in the original state |ψ⟩, with projector |ψ⟩⟨ψ|. How does the probability of the outcome 'yes' in each measurement compare with its value when the heater is off?

1. ANeither changes: M₁ stays at 1/3 and M₂ stays at 1, because multiplying an amplitude by a phase factor cannot change a probability.
2. BM₁ stays at 1/3, and M₂ falls from 1 to 1/9.
3. CM₁ stays at 1/3, and M₂ falls from 1 to 1/3.
4. DM₁ stays at 1/3, and M₂ falls from 1 to 0.

Conduction electrons travelling in the +x direction along a single-channel nanowire meet a thin impurity layer at x = 0, modelled as the potential V(x) = αδ(x) with α > 0. The potential is zero on both sides of the layer, so the wavenumber k is the same on both sides. Matching ψ at x = 0, together with the jump in dψ/dx that the delta function requires, gives a reflected amplitude r = −iβ/(1 + iβ) and a transmitted amplitude t = 1/(1 + iβ) for unit incident amplitude, where β = mα/(ℏ²k) and m is the effective mass of the conduction electrons. For electrons with β = 2, what is the reflection coefficient R, the ratio of the reflected to the incident probability current?

1. A1/5
2. B4/9
3. C4/5
4. D2/√5

A mode of a superconducting microwave resonator behaves as a harmonic oscillator of angular frequency ω, with Hamiltonian H = ℏω(a†a + ½) and number states |n⟩ that satisfy a†|n⟩ = √(n + 1)|n + 1⟩ and a†a|n⟩ = n|n⟩. A sequence of control pulses prepares the mode in |ψ⟩ = η(1 + a†)²|0⟩, where 1 is the identity operator and η is a positive constant that normalises the state. What is the energy uncertainty ΔE in this state?

1. A(2√5/7)ℏω
2. Bℏω/√3
3. C(2√5/7 + 1/2)ℏω
4. D√(12/7) ℏω

A particle is in a simultaneous eigenstate of L² and L_z with quantum numbers l = 2 and m_l = 2. The component L_x is then measured. Which list gives every result that measurement can return?

1. A+2ℏ only, because the state already has a definite component of angular momentum
2. B0 only, because ⟨L_x⟩ = 0 in this state
3. C−ℏ, 0 and +ℏ
4. D−2ℏ, −ℏ, 0, +ℏ and +2ℏ

In the non-relativistic Coulomb model of hydrogen, with the electron's spin carried as a tensor factor but no spin-dependent term in the Hamiltonian, how many orthonormal states share the energy E₄ = −0.850 eV?

1. A7
2. B32
3. C16
4. D8

Treat the ground configuration of nickel as [Ar] 3d⁸ 4s², with the 4s² subshell closed. Using antisymmetry and Hund's rules, what is the ground term?

1. A¹G₄
2. B³F₂
3. C³G₅
4. D³F₄

A nucleus with A = 236 and binding energy per nucleon 7.59 MeV splits into two fragments: one with A = 138 and 8.39 MeV per nucleon, and one with A = 98 and 8.60 MeV per nucleon. Ignoring any emitted neutrons, what energy is released?

1. A209 MeV
2. B−209 MeV
3. C0.887 MeV
4. D1.79 × 10³ MeV

A nuclide decays by α emission with branching fraction 0.25 and by β⁻ emission with branching fraction 0.75. Its half-life, found by following the number of undecayed nuclei, is 48.0 min. A detector that records only the α particles is used instead. With what half-life does the α count rate fall?

1. A12.0 min
2. B192 min
3. C48.0 min
4. D64.0 min

Which of these proposed processes is forbidden by a conservation law of the Standard Model with massless neutrinos?

1. Aμ⁻ → e⁻ + γ
2. BK⁻ + p → Ω⁻ + K⁺ + K⁰
3. Cν̄_e + p → n + e⁺
4. Dπ⁺ → μ⁺ + ν_μ

Which of these normalised two-qubit states, each written in the computational basis |q₁q₂⟩, is entangled?

1. A(|00⟩ + |01⟩ + |10⟩ + |11⟩)/2
2. B(|00⟩ − |01⟩ + |10⟩ − |11⟩)/2
3. C(|00⟩ + i|01⟩ + i|10⟩ − |11⟩)/2
4. D(|00⟩ + |01⟩ + |10⟩ − |11⟩)/2

An electron in a GaAs quantum well is modelled as a particle of effective mass m* = 0.0670 mₑ in a one-dimensional infinite square well of width L = 12.0 nm, with eigenstates ψ_n(x) = √(2/L) sin(nπx/L) for 0 ≤ x ≤ L and energies E_n = n²π²ℏ²/(2m*L²). At t = 0 it is prepared in Ψ(x, 0) = [3ψ₁(x) + 4ψ₂(x)]/5. Use mₑ = 9.109 × 10⁻³¹ kg, ℏ = 1.055 × 10⁻³⁴ J s, h = 6.626 × 10⁻³⁴ J s and e = 1.602 × 10⁻¹⁹ C. You may use ∫₀^L ψ₁(x) x ψ₂(x) dx = −16L/(9π²).

1. (a)

   Calculate Calculate E₁ and E₂, in meV.

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   1. 1

      E₁ = π²ℏ²/(2m*L²) = π²(1.055 × 10⁻³⁴ J s)²/(2 × 0.0670 × 9.109 × 10⁻³¹ kg × (12.0 × 10⁻⁹ m)²) = 6.25 × 10⁻²¹ J

   2. 2

      E₁ = 39.0 meV and E₂ = 4E₁ = 156 meV

2. (b)

   Show that Show that the expectation value of the energy is 73E₁/25 at every time t.

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   1. 1

      ⟨E⟩ = |c₁|²E₁ + |c₂|²E₂ = (9/25)E₁ + (16/25)(4E₁) = 73E₁/25 (= 114 meV)

   2. 2

      time evolution multiplies each c_n by e^(−iE_n t/ℏ), which leaves every |c_n|² unchanged, so ⟨E⟩ is the same at all t (equivalently, the propagator commutes with H)

3. (c)

   Determine Determine ⟨x⟩ as a function of time, giving the centre and the amplitude of its oscillation in nm.

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   1. 1

      Ψ(x, t) = [3ψ₁e^(−iE₁t/ℏ) + 4ψ₂e^(−iE₂t/ℏ)]/5, with ⟨ψ₁|x|ψ₁⟩ = ⟨ψ₂|x|ψ₂⟩ = L/2

   2. 2

      ⟨x⟩ = L/2 + 2(3/5)(4/5)x₁₂ cos(ωt), where ω = (E₂ − E₁)/ℏ and x₁₂ = −16L/(9π²)

   3. 3

      ⟨x⟩ = L/2 − [384/(225π²)]L cos(ωt) = 6.00 nm − (2.08 nm) cos(ωt)

4. (d)

   Calculate Calculate the period of the oscillation of ⟨x⟩.

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   1. 1

      T = 2πℏ/(E₂ − E₁) = h/(3E₁) = 6.626 × 10⁻³⁴ J s/(3 × 6.25 × 10⁻²¹ J) = 3.53 × 10⁻¹⁴ s (35.3 fs); 2πℏ/(3E₁) with the given ℏ gives 3.54 × 10⁻¹⁴ s, equally acceptable

5. (e)

   Predict The electron is instead prepared in [3ψ₁ + 4ψ₃]/5. Predict how ⟨x⟩ behaves in time, and give the reason.

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   1. 1

      ⟨x⟩ = L/2 at all times: it does not oscillate

   2. 2

      ψ₁ and ψ₃ are both even about x = L/2, so (x − L/2)ψ₁ψ₃ is odd about the centre and integrates to zero, which with ⟨ψ₁|ψ₃⟩ = 0 gives x₁₃ = ∫ψ₁ x ψ₃ dx = 0; the cross term vanishes even though |Ψ|² still changes shape symmetrically in time

X-ray spectra of the hot gas in galaxy clusters contain lines from O⁷⁺, an oxygen nucleus (Z = 8) with a single bound electron. Treat the ion with the non-relativistic Coulomb Hamiltonian and an infinitely heavy nucleus, for which E_n = −(13.6 eV)Z²/n². The ground-state radial probability density is P(r) = (4Z³/a₀³) r² e^(−2Zr/a₀), with a₀ = 52.9 pm. Use hc = 1239.8 eV nm.

1. (a)

   Calculate Calculate the ionisation energy of ground-state O⁷⁺, and the photon energy and wavelength of its 2p → 1s (Lyman-α) line.

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   1. 1

      ionisation energy = (13.6 eV) × 8² = 870 eV

   2. 2

      photon energy for 2p → 1s = 870.4 eV × (1 − 1/4) = 653 eV

   3. 3

      λ = hc/E = 1239.8 eV nm/652.8 eV = 1.90 nm

2. (b)

   Show that Show that the most probable distance of the electron from the nucleus in the ground state is a₀/Z, and calculate that distance for O⁷⁺.

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   1. 1

      dP/dr ∝ (2r − 2Zr²/a₀)e^(−2Zr/a₀)

   2. 2

      dP/dr = 0 for r > 0 gives r = a₀/Z, a maximum because P = 0 at r = 0 and P → 0 as r → ∞

   3. 3

      r = 52.9 pm/8 = 6.61 pm

3. (c)

   Justify The electron is excited into one of the n = 3 states. Justify which of 3s, 3p and 3d can decay to 1s by emitting a single electric-dipole photon, and state the route to 1s for each of the others.

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   1. 1

      the dipole operator is a rank-1 vector operator of odd parity, so the Wigner–Eckart triangle rule and parity together require Δl = ±1 (with Δm_l = 0, ±1)

   2. 2

      only 3p → 1s (Lyman-β) is allowed directly; 3s → 1s has Δl = 0 and 3d → 1s has Δl = 2, so both are forbidden

   3. 3

      3s and 3d each decay first to 2p (Δl = −1 from 3d, Δl = +1 from 3s) and then 2p → 1s

4. (d)

   Explain In this model the 3s, 3p and 3d levels are exactly degenerate. Explain why this degeneracy is special to the Coulomb potential, and name the effect that splits these levels in a real O⁷⁺ ion.

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   1. 1

      for a 1/r potential the Runge–Lenz vector commutes with H and, with L, closes an so(4) algebra whose multiplets contain every l from 0 to n − 1; any other central potential keeps only the (2l + 1)-fold rotational degeneracy in m_l

   2. 2

      relativistic fine structure (the kinetic-energy correction, spin–orbit coupling and the Darwin term), which scales as Z⁴α² and is about 8⁴ ≈ 4 × 10³ times larger than in hydrogen, splits the levels by j; levels of equal n and j are then separated by the much smaller Lamb shift

A water sample sits in the uniform static field B₀ = 9.40 T of an NMR magnet, directed along +z. Consider the spin of a single hydrogen nucleus. The proton's magnetic moment is μ = (g_p μ_N/ℏ)S, with g_p = 5.586 and μ_N = 5.051 × 10⁻²⁷ J T⁻¹, so the Hamiltonian is H = −μ⋅B = −ω₀S_z with ω₀ = g_p μ_N B₀/ℏ. Take ℏ = 1.055 × 10⁻³⁴ J s. A radio-frequency pulse, treated as instantaneous, leaves the spin at t = 0 in the +ℏ/2 eigenstate of n̂⋅S, where n̂ = (sin θ cos φ, sin θ sin φ, cos θ) is the unit vector with polar angle θ and azimuth φ. The spin operators are S_i = (ℏ/2)σ_i, and in the basis of the S_z eigenstates |↑⟩ and |↓⟩ the Pauli matrices are σ_x = (0, 1; 1, 0), σ_y = (0, −i; i, 0) and σ_z = (1, 0; 0, −1), with rows separated by semicolons.

1. (a)

   Determine Determine the matrix of n̂⋅S in the S_z basis, and its normalised eigenvector with eigenvalue +ℏ/2 written as a combination of |↑⟩ and |↓⟩.

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   1. 1

      n̂⋅S = (ℏ/2)(cos θ, sin θ e^(−iφ); sin θ e^(iφ), −cos θ)

   2. 2

      |n̂+⟩ = cos(θ/2)|↑⟩ + e^(iφ) sin(θ/2)|↓⟩, normalised and checked against the first row: cos θ cos(θ/2) + sin θ sin(θ/2) = cos(θ/2); any overall phase is accepted

2. (b)

   Show that Using the propagator U(t) = exp(−iHt/ℏ), show that at time t the spin is in the +ℏ/2 eigenstate of n̂(t)⋅S, where n̂(t) has the same polar angle θ and the azimuth φ − ω₀t.

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   1. 1

      H = −ω₀S_z is diagonal in the S_z basis, with energy −ℏω₀/2 for |↑⟩ and +ℏω₀/2 for |↓⟩, so U(t) multiplies |↑⟩ by e^(iω₀t/2) and |↓⟩ by e^(−iω₀t/2)

   2. 2

      U(t)|n̂+⟩ = e^(iω₀t/2) cos(θ/2)|↑⟩ + e^(iφ) e^(−iω₀t/2) sin(θ/2)|↓⟩

   3. 3

      factoring out the global phase e^(iω₀t/2), which changes no prediction, leaves cos(θ/2)|↑⟩ + e^(i(φ − ω₀t)) sin(θ/2)|↓⟩, the eigenvector of (a) with θ unchanged and φ replaced by φ − ω₀t

3. (c)

   Calculate Calculate ω₀, and, for 0 < θ < π, the shortest time after t = 0 at which the spin is again in the same physical state as at t = 0.

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   1. 1

      ω₀ = g_p μ_N B₀/ℏ = 5.586 × 5.051 × 10⁻²⁷ J T⁻¹ × 9.40 T/(1.055 × 10⁻³⁴ J s) = 2.51 × 10⁹ rad s⁻¹, a Larmor frequency f₀ = ω₀/2π = 400 MHz

   2. 2

      the azimuth first returns to φ when ω₀t = 2π, so t = 2π/ω₀ = 2.50 × 10⁻⁹ s; the ket is then −|n̂+⟩, which is the same ray

4. (d)

   Describe A receiver coil with its axis along x picks up a voltage proportional to d⟨S_x⟩/dt. Describe the signal it records, and how the amplitude of that signal depends on θ.

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   1. 1

      the spin stays in the +ℏ/2 eigenstate of n̂(t)⋅S, so ⟨S⟩ = (ℏ/2)n̂(t) keeps its length and its angle θ to the field while precessing about +z; ⟨S_x⟩ = (ℏ/2) sin θ cos(φ − ω₀t), so the coil records a sinusoidal voltage at f₀ = 400 MHz

   2. 2

      the amplitude is proportional to sin θ: greatest for θ = π/2, a spin tipped into the xy-plane, and zero for θ = 0 or π, when the state is an eigenstate of H and nothing precesses

5. (e)

   Show that For operators A and B the Robertson relation reads ΔA ΔB ≥ ½|⟨[A, B]⟩|. Apply it to H and S_x, and show that at every time t this spin obeys ΔE ΔS_x ≥ (ℏ/2)|d⟨S_x⟩/dt|, where ΔE is the uncertainty in its energy.

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   1. 1

      [H, S_x] = −ω₀[S_z, S_x] = −iℏω₀S_y, so the Robertson relation gives ΔE ΔS_x ≥ (ℏω₀/2)|⟨S_y⟩|

   2. 2

      ⟨S_x⟩ = (ℏ/2) sin θ cos(φ − ω₀t) and ⟨S_y⟩ = (ℏ/2) sin θ sin(φ − ω₀t), so d⟨S_x⟩/dt = (ℏω₀/2) sin θ sin(φ − ω₀t) = ω₀⟨S_y⟩

   3. 3

      substituting ω₀|⟨S_y⟩| = |d⟨S_x⟩/dt| gives ΔE ΔS_x ≥ (ℏ/2)|d⟨S_x⟩/dt|: the energy spread limits how fast ⟨S_x⟩ can change

The magnesium line at λ = 517.3 nm is the transition from the upper level 3s4s ³S₁ to the lower level 3s3p ³P₁. The source is placed in a uniform magnetic field B = 1.20 T, weak enough that LS coupling holds and each level shifts by ΔE = g_J μ_B B M_J, with g_J = 1 + [J(J + 1) + S(S + 1) − L(L + 1)]/[2J(J + 1)]. Use μ_B = 5.788 × 10⁻⁵ eV T⁻¹ and hc = 1239.8 eV nm. In the lower term, the ³P₁ and ³P₂ levels are 5.05 meV apart.

1. (a)

   Calculate Calculate the Landé g factors of the upper and lower levels.

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   1. 1

      ³S₁ has L = 0, S = 1, J = 1: g = 1 + (2 + 2 − 0)/4 = 2

   2. 2

      ³P₁ has L = 1, S = 1, J = 1: g = 1 + (2 + 2 − 2)/4 = 3/2

2. (b)

   Justify Justify that this line is an allowed electric-dipole transition.

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   1. 1

      a single electron jumps 4s → 3p with Δl = 1, so the configuration parity changes from even to odd, as the odd-parity dipole operator requires

   2. 2

      ΔS = 0, ΔL = 1 and ΔJ = 0 between two J = 1 levels are all allowed; the only forbidden ΔJ = 0 case is J = 0 → J = 0, which this is not

3. (c)

   Determine Determine how many Zeeman components the line splits into, and the shift of each from the zero-field photon energy in units of μ_B B.

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   1. 1

      photon energy shift = μ_B B(g_u M_u − g_l M_l) = μ_B B(2M_u − 1.5M_l), with ΔM = M_u − M_l = 0 or ±1, giving seven (M_u, M_l) pairs

   2. 2

      the M_u = 0 → M_l = 0 component vanishes because ΔJ = 0: its Clebsch–Gordan coefficient ⟨1 0; 1 0|1 0⟩ is zero (the 3j symbol with all projections zero vanishes when 1 + 1 + 1 is odd)

   3. 3

      six components remain

   4. 4

      the π components (ΔM = 0) are shifted by ±0.5μ_B B, and the σ components (ΔM = ±1) by ±1.5μ_B B and ±2.0μ_B B

4. (d)

   Calculate Calculate the wavelength separation between the two outermost components.

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   1. 1

      the outermost components are 4.0μ_B B apart: 4 × 5.788 × 10⁻⁵ eV T⁻¹ × 1.20 T = 2.78 × 10⁻⁴ eV

   2. 2

      Δλ = λ²ΔE/(hc) = (517.3 nm)² × 2.78 × 10⁻⁴ eV/(1239.8 eV nm) = 0.0600 nm (60.0 pm)

5. (e)

   Estimate Estimate the field above which this anomalous Zeeman pattern gives way to the Paschen–Back regime, stating the criterion you use.

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   1. 1

      criterion: the magnetic energy μ_B B becomes comparable with the fine-structure interval of the ³P term, 5.05 meV, so that the field decouples L from S

   2. 2

      B ≈ 5.05 × 10⁻³ eV/5.788 × 10⁻⁵ eV T⁻¹ ≈ 90 T (of order 10² T), far above 1.20 T, which confirms that the weak-field treatment used here is valid

A π⁻ beam strikes a liquid-hydrogen target, and the reaction π⁻ + p → K⁰ + Λ⁰ is studied. The Λ⁰ then decays as Λ⁰ → p + π⁻ with mean lifetime 2.63 × 10⁻¹⁰ s. Rest energies: m_π c² = 139.57 MeV, m_p c² = 938.27 MeV, m_K c² = 497.61 MeV and m_Λ c² = 1115.68 MeV. The quark contents are K⁰ = ds̄ and Λ⁰ = uds, and a strange quark carries strangeness S = −1.

1. (a)

   State State the total charge, baryon number and strangeness on each side of the production reaction, and state which interaction can mediate it.

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   1. 1

      charge: −1 + 1 = 0 and 0 + 0 = 0; baryon number: 0 + 1 = 1 and 0 + 1 = 1; strangeness: 0 + 0 = 0 and (+1) + (−1) = 0

   2. 2

      all three balance, strangeness included, so the strong interaction can mediate the production

2. (b)

   Derive Derive, from four-momentum conservation, an expression for the minimum laboratory energy E_π of the pion, with the target proton at rest, for which the reaction can occur.

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   1. 1

      the invariant s = (P_π + P_p)⋅(P_π + P_p) is the same in every frame, and in the laboratory s = m_π²c⁴ + m_p²c⁴ + 2E_π m_p c²

   2. 2

      at threshold the products are at rest in the centre-of-momentum frame, so s = (m_K + m_Λ)²c⁴

   3. 3

      equating the two: E_π = [(m_K + m_Λ)² − m_π² − m_p²]c²/(2m_p)

3. (c)

   Calculate Calculate the threshold kinetic energy and momentum of the pion.

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   1. 1

      E_π = (1613.29² − 139.57² − 938.27²)/(2 × 938.27) MeV = 907.5 MeV, so the kinetic energy is E_π − m_π c² = 767.9 MeV

   2. 2

      p_π c = √(E_π² − m_π²c⁴) = 896.7 MeV, so p_π = 897 MeV/c

4. (d)

   Compare The Σ⁰ baryon, also uds, decays as Σ⁰ → Λ⁰ + γ with mean lifetime 7.4 × 10⁻²⁰ s. Compare its lifetime with that of the Λ⁰, relating the difference to the interaction responsible for each decay.

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   1. 1

      Λ⁰ → p + π⁻ changes strangeness from −1 to 0, which neither the strong nor the electromagnetic interaction allows, so it is a weak decay; Σ⁰ → Λ⁰ + γ conserves strangeness and emits a photon, so it is electromagnetic

   2. 2

      the lifetimes differ by a factor 2.63 × 10⁻¹⁰ s/7.4 × 10⁻²⁰ s ≈ 4 × 10⁹; by the golden rule the width Γ = ℏ/τ scales as the square of the transition amplitude, so the ratio reflects how much weaker the weak amplitude is (phase space differs as well, but cannot account for nine orders of magnitude)

5. (e)

   Calculate Calculate the momentum of the proton in the rest frame of the Λ⁰ for the decay Λ⁰ → p + π⁻.

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   1. 1

      momentum conservation makes the two momenta equal and opposite, of magnitude p*, and energy conservation gives √(p*²c² + m_p²c⁴) + √(p*²c² + m_π²c⁴) = m_Λ c²

   2. 2

      solving: p*c = √([m_Λ² − (m_p + m_π)²][m_Λ² − (m_p − m_π)²]) c²/(2m_Λ)

   3. 3

      p* = 100.6 MeV/c; the proton's kinetic energy is then 5.4 MeV and the pion's 32.5 MeV, sharing Q = 37.8 MeV

A source sends one qubit of each pair to Alice and the other to Bob, in the state |Φ⁺⟩ = (|00⟩ + |11⟩)/√2. Alice measures the observable A(a) = cos a Z + sin a X on her qubit and Bob measures B(b) = cos b Z + sin b X on his, each with outcomes ±1, where X and Z are Pauli matrices. The correlation is E(a, b) = ⟨Φ⁺|A(a) ⊗ B(b)|Φ⁺⟩, and the CHSH combination is S = E(a, b) + E(a, b′) + E(a′, b) − E(a′, b′). Every local hidden-variable model obeys |S| ≤ 2.

1. (a)

   Show that Show that E(a, b) = cos(a − b).

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   1. 1

      expands A ⊗ B = cos a cos b Z⊗Z + cos a sin b Z⊗X + sin a cos b X⊗Z + sin a sin b X⊗X

   2. 2

      Z⊗Z and X⊗X leave |Φ⁺⟩ unchanged, so ⟨Z⊗Z⟩ = ⟨X⊗X⟩ = 1, while Z⊗X and X⊗Z map |Φ⁺⟩ to ±(|01⟩ − |10⟩)/√2, which is orthogonal to it, so those expectation values are 0

   3. 3

      E(a, b) = cos a cos b + sin a sin b = cos(a − b)

2. (b)

   Calculate Calculate S for the settings a = 0, a′ = π/2, b = π/4 and b′ = −π/4.

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   1. 1

      E(0, π/4) = E(0, −π/4) = E(π/2, π/4) = cos(π/4) = 0.707 and E(π/2, −π/4) = cos(3π/4) = −0.707

   2. 2

      S = 3 × 0.707 + 0.707 = 2√2 = 2.83, above the local bound of 2

3. (c)

   Determine Noise in the source replaces |Φ⁺⟩⟨Φ⁺| by the density operator ρ = p|Φ⁺⟩⟨Φ⁺| + (1 − p)I/4, with 0 ≤ p ≤ 1. Determine S for the same settings as a function of p, and the range of p for which these settings show a violation.

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   1. 1

      E_ρ(a, b) = Tr[ρ A ⊗ B] = p cos(a − b) + (1 − p)Tr(A)Tr(B)/4

   2. 2

      A and B are traceless, so the noise term vanishes and S = 2√2 p

   3. 3

      a violation needs 2√2 p > 2, so p > 1/√2 = 0.707

4. (d)

   Evaluate An experiment using these settings reports S = 2.38 ± 0.05. Evaluate what this result establishes and what it does not.

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   1. 1

      S exceeds 2 by (2.38 − 2)/0.05 = 7.6 standard uncertainties, so the correlations are incompatible with every local hidden-variable model under the test's assumptions; in the noise model it implies p = 2.38/2.83 = 0.84, above the threshold of 0.707

   2. 2

      it excludes local hidden variables only if the loopholes are closed (settings chosen at spacelike separation, and detection efficient enough that the recorded pairs are not a biased sample), and it neither proves that exactly |Φ⁺⟩ was prepared nor allows any signal to pass between Alice and Bob

A radiopharmacy confirms the identity of a batch of fluorine-18 by measuring its half-life. A sample is placed in a well counter and the counts in 100 s are recorded at intervals; t is the time at the middle of each 100 s count, measured from the middle of the first. With no sample present, the counter records 1200 counts in 1000 s. The site's quality-control rule accepts the batch if the measured half-life lies between 105 min and 115 min. The accepted half-life of fluorine-18 is 109.8 min.

1. (a)

   Explain Explain why the background must be subtracted from each reading before logarithms are taken, and why the uncertainty in the background itself can be neglected here.

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   1. 1

      the background adds a constant 120 counts per 100 s to an exponential, so ln(gross counts) is not linear in t: the late readings sit above the straight line and a fit would give too small a λ, that is, too long a half-life

   2. 2

      the background uncertainty is √1200/10 ≈ 3.5 counts per 100 s, which shifts ln(net counts) by at most 3.5/5003 ≈ 7 × 10⁻⁴ at t = 360 min, far below that reading's own Poisson uncertainty of about 0.014

2. (b)

   Determine Determine the decay constant λ from the data, using either a straight-line fit of ln(net counts) against t or two well-separated readings.

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   1. 1

      subtracts 120 from each reading, for example net counts 48132 at t = 0 and 5003 at t = 360 min, giving ln(net counts) = 10.782 and 8.518

   2. 2

      gradient = (8.518 − 10.782)/(360 min) = −6.29 × 10⁻³ min⁻¹ (a least-squares fit to all seven points gives −6.30 × 10⁻³ min⁻¹)

   3. 3

      λ = 6.3 × 10⁻³ min⁻¹ = 1.05 × 10⁻⁴ s⁻¹; accept values from 6.27 × 10⁻³ to 6.33 × 10⁻³ min⁻¹

3. (c)

   Calculate Calculate the half-life and the mean lifetime.

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   1. 1

      T½ = ln 2/λ = 0.6931/(6.29 × 10⁻³ min⁻¹) = 110 min (110.2 min from the two end readings, 109.9 min from the least-squares fit)

   2. 2

      τ = 1/λ = 159 min

4. (d)

   Estimate Estimate the uncertainty in λ that arises from counting statistics, assuming each reading is Poisson-distributed and using the two readings from part (b).

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   1. 1

      σ[ln(net counts)] = √(gross counts)/(net counts): √48252/48132 = 0.0046 at t = 0 and √5123/5003 = 0.0143 at t = 360 min

   2. 2

      δλ ≈ √(0.0046² + 0.0143²)/(360 min) = 4.2 × 10⁻⁵ min⁻¹, a fractional uncertainty of 0.7 per cent

   3. 3

      δT½ = T½ δλ/λ ≈ 0.7 min, so T½ = 110.2 ± 0.7 min (a weighted fit to all seven readings narrows this to about ±0.5 min)

5. (e)

   Evaluate Evaluate whether the batch passes the quality-control rule, and whether the data show any sign of a longer-lived radionuclidic impurity.

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   1. 1

      T½ = 110.2 ± 0.7 min lies inside 105 min to 115 min and agrees with the accepted 109.8 min to within about 0.6 standard uncertainties, so the batch passes

   2. 2

      the intermediate readings lie on both sides of a least-squares line, each within about one Poisson standard uncertainty (within two of the line through the end readings), with no systematic upward curvature at late times; a longer-lived impurity would lift the late readings above the line, so none is detectable above a few per cent of the count rate at 360 min

In a deuterium–tritium plasma the reaction ²H + ³H → ⁴He + n releases the energy that fusion devices aim to harvest. Atomic masses: m(²H) = 2.014102 u, m(³H) = 3.016049 u and m(⁴He) = 4.002603 u; the neutron mass is 1.008665 u; 1 u = 931.494 MeV/c². For a Maxwellian plasma at temperature T, the non-resonant reaction rate per pair of nuclei is ⟨σv⟩ ∝ ∫₀^∞ S(E) exp[−E/(kT) − √(E_G/E)] dE, where E is the centre-of-mass energy, S(E) varies slowly with E, e^(−E/kT) is the Maxwell–Boltzmann factor and e^(−√(E_G/E)) is the probability of tunnelling through the Coulomb barrier. For D–T the Gamow energy is E_G = 2μc²(παZ₁Z₂)² = 1182 keV.

1. (a)

   Calculate Calculate the Q value of the reaction.

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   1. 1

      Δm = 2.014102 u + 3.016049 u − 4.002603 u − 1.008665 u = 0.018883 u, where atomic masses may be used because the two electrons on the left are matched by the two in ⁴He

   2. 2

      Q = 0.018883 u × 931.494 MeV/u = 17.59 MeV

2. (b)

   Sketch On one set of axes against E, sketch the Maxwell–Boltzmann factor, the tunnelling factor and their product, marking the energy E₀ at which the product peaks.

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   1. 1

      the Maxwell–Boltzmann factor falls monotonically from 1 at E = 0, and the tunnelling factor rises monotonically from 0 at E = 0, very flat near the origin, towards 1 at large E

   2. 2

      their product, drawn on an expanded vertical scale and labelled as such, is a single peak at E₀, several times kT, and is smaller than both factors at every energy

3. (c)

   Derive Starting from the rate integral given in the stem, derive an expression for E₀ in terms of E_G and kT by maximising the exponent of its integrand.

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   1. 1

      writes f(E) = −E/(kT) − E_G^(1/2)E^(−1/2), since S(E) varies slowly and the exponential fixes where the integrand peaks

   2. 2

      df/dE = −1/(kT) + ½E_G^(1/2)E^(−3/2) = 0

   3. 3

      E₀^(3/2) = ½E_G^(1/2)kT, so E₀ = [E_G(kT)²/4]^(1/3)

4. (d)

   Show that By expanding the exponent to second order about E₀, show that the integrand is approximately proportional to exp[−(E − E₀)²/(Δ/2)²] with Δ = 4√(E₀kT/3).

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   1. 1

      f″(E) = −¾E_G^(1/2)E^(−5/2), and the stationary condition gives E_G^(1/2) = 2E₀^(3/2)/(kT), so f″(E₀) = −3/(2E₀kT)

   2. 2

      f(E) ≈ f(E₀) + ½f″(E₀)(E − E₀)² = f(E₀) − 3(E − E₀)²/(4E₀kT), the first-order term vanishing at the maximum

   3. 3

      matching 3/(4E₀kT) = 1/(Δ/2)² gives (Δ/2)² = 4E₀kT/3, so Δ = 4√(E₀kT/3)

5. (e)

   Calculate Calculate E₀ and Δ for kT = 10.0 keV, and comment on how sharply the Gaussian form localises the reacting energies at this temperature.

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   1. 1

      E₀ = [1182 keV × (10.0 keV)²/4]^(1/3) = 30.9 keV, about 3.1kT

   2. 2

      Δ = 4√(30.9 keV × 10.0 keV/3) = 40.6 keV

   3. 3

      Δ ≈ 1.3E₀, so the window is broad rather than sharp and noticeably asymmetric, extending further above E₀ than below it: the Gaussian gives the position and width of the window but not its exact shape (a numerical integration shows the Gaussian estimate of the integral is about 4 per cent low)

6. (f)

   Evaluate The Coulomb barrier between a deuteron and a triton at nuclear contact is about 0.44 MeV. A student argues that since kT = 10.0 keV is some 44 times smaller, fusion in this plasma must be negligible. Evaluate this argument.

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   1. 1

      the barrier height is not a threshold: nuclei tunnel through it, and at E₀ the tunnelling factor is e^(−√(1182/30.9)) = e^(−6.18) ≈ 2 × 10⁻³, small but far from zero

   2. 2

      the rate is set where this factor overlaps the Maxwellian tail (about 10 per cent of nuclei have E > E₀ ≈ 3kT), so the reaction proceeds at an appreciable rate; the student's argument holds only classically, where no pair below 0.44 MeV could react

In the electroweak model the Higgs field is a complex doublet φ with hypercharge ½ and potential V(φ) = −μ²φ†φ + λ(φ†φ)², with μ² > 0 and λ > 0. Its covariant derivative is D_μ = ∂_μ − ig(σ^a/2)W^a_μ − i(g′/2)B_μ, summed over a = 1, 2, 3, where σ^a are the Pauli matrices, and the Lagrangian contains (D_μφ)†(D^μφ) − V(φ). In unitary gauge the vacuum is φ₀ = (0, v/√2)ᵀ. A real vector field V_μ of mass M contributes ½M²V_μV^μ to the Lagrangian, and the charged fields W^± = (W¹ ∓ iW²)/√2 of mass M_W contribute M_W²W⁺_μW^−μ. Take g = 0.652, g′ = 0.357, v = 246 GeV and λ = 0.129.

1. (a)

   Derive Derive the vacuum value v in terms of μ and λ. Then, writing φ = (0, (v + h)/√2)ᵀ, derive the mass M_H of the physical scalar h.

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   1. 1

      with ρ = φ†φ, dV/dρ = −μ² + 2λρ = 0 gives ρ = μ²/(2λ) = v²/2, so v = μ/√λ

   2. 2

      substituting φ†φ = (v + h)²/2 gives V = −½μ²(v + h)² + ¼λ(v + h)⁴, whose h² coefficient is −½μ² + (3/2)λv² = λv²

   3. 3

      matching λv²h² to ½M_H²h² gives M_H = √(2λ) v = √2 μ

2. (b)

   State State how many real field components the doublet has, how many generators of SU(2)_L × U(1)_Y the vacuum breaks, and what becomes of the corresponding field components.

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   1. 1

      four real components; the vacuum is left invariant only by the U(1) generated by the charge Q = T₃ + Y, so 4 − 1 = 3 generators are broken

   2. 2

      the three would-be Goldstone modes become the longitudinal polarisations of W⁺, W⁻ and Z, leaving one physical scalar, the Higgs boson h

3. (c)

   Show that Show that the kinetic term evaluated on φ₀ gives M_W = gv/2, and that the neutral fields acquire the mass-squared matrix M² = (v²/4)(g², −gg′; −gg′, g′²) in the basis (W³, B), with rows separated by semicolons.

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   1. 1

      the gauge-field matrix is ½(gW³ + g′B, g(W¹ − iW²); g(W¹ + iW²), −gW³ + g′B), so D_μφ₀ = −(i/2)(v/√2)(g(W¹ − iW²), −gW³ + g′B)ᵀ

   2. 2

      (D_μφ₀)†(D^μφ₀) = (v²/8)[g²(W¹W¹ + W²W²) + (gW³ − g′B)²]

   3. 3

      W¹W¹ + W²W² = 2W⁺W⁻, so the charged term is (g²v²/4)W⁺W⁻ and M_W = gv/2

   4. 4

      (v²/8)(gW³ − g′B)² = ½(W³, B)M²(W³, B)ᵀ with M² = (v²/4)(g², −gg′; −gg′, g′²)

4. (d)

   Determine Determine the eigenvalues and normalised eigenvectors of M², identify the photon and the Z boson, and express the Weinberg angle θ_W and the ratio M_W/M_Z in terms of g and g′.

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   1. 1

      det M² = (v²/4)²(g²g′² − g²g′²) = 0, so one eigenvalue is zero; its normalised eigenvector is the massless photon A = (g′W³ + gB)/√(g² + g′²)

   2. 2

      the other eigenvalue equals the trace, M_Z² = (v²/4)(g² + g′²), with normalised eigenvector the Z boson, Z = (gW³ − g′B)/√(g² + g′²)

   3. 3

      writing Z = cos θ_W W³ − sin θ_W B gives tan θ_W = g′/g, and M_W/M_Z = g/√(g² + g′²) = cos θ_W

5. (e)

   Calculate Calculate M_W, M_Z, sin²θ_W and M_H.

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   1. 1

      M_W = 0.652 × 246 GeV/2 = 80.2 GeV and M_Z = (246 GeV/2)√(0.652² + 0.357²) = 91.4 GeV

   2. 2

      sin²θ_W = g′²/(g² + g′²) = 0.231 and M_H = √(2 × 0.129) × 246 GeV = 125 GeV

6. (f)

   Evaluate The measured values are M_W = 80.37 GeV, M_Z = 91.19 GeV and M_H = 125.1 GeV. Evaluate which of your results are genuine predictions of the model and which only return the numbers put in.

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   1. 1

      g, g′, v and λ are inputs fitted to data, so the individual values of M_W, M_Z and M_H are not independent predictions; M_H in particular only restates λ, which the model leaves free

   2. 2

      the genuine tree-level predictions are structural: a massless photon, and M_W = M_Z cos θ_W, which follows from the doublet's quantum numbers and tests the model once θ_W is measured independently of the masses, in neutral-current couplings; the measured ratio 80.37/91.19 = 0.881 agrees with cos θ_W = 0.877 at the half-per-cent level that loop corrections account for