University Quantum Mechanics II · Midterm Examination

Separating the angular equation for m = 0 and writing x = cos θ gives Legendre's equation, (1 − x²)y″ − 2xy′ + λy = 0. Substituting y = Σ aₖxᵏ gives the recursion aₖ₊₂ = [k(k + 1) − λ]aₖ/[(k + 1)(k + 2)], so the even and the odd coefficients form two independent series. Take λ = 6. Which statement is correct?

1. ABoth series terminate, the even one as 1 − 3x² and the odd one as x − (2/3)x³, so λ = 6 has two independent polynomial solutions.
2. BNeither series terminates, but both converge for |x| < 1, so λ = 6 is as acceptable as any other value; λ is quantised only by the boundary condition on the φ equation.
3. CThe even series terminates as 1 − 3x²; the odd series does not, and because its coefficients fall off only as 1/k it diverges logarithmically at x = ±1, so its coefficient must be set to zero.
4. DThe even series terminates as 1 − 3x²; the odd series does not, but its coefficients alternate in sign, so it converges at x = ±1 and any combination of the two solutions is acceptable.

For hydrogen, with a₀ the Bohr radius, compare the 3s state (n = 3, l = 0) with a 3d state (n = 3, l = 2). Which expectation value is the same for the two states?

1. A⟨1/r⟩, equal to 1/(9a₀) in both, because the virial theorem fixes ⟨V⟩ = 2E₃
2. B⟨r⟩, equal to 9a₀ in both, the radius of the third Bohr orbit
3. C⟨1/r²⟩, because the centrifugal term averages to the same value in every state of a given n
4. D⟨r²⟩, because the mean-square radius is fixed by the energy through ⟨T⟩ = −E₃

A particle of charge q and mass m moves in the potentials φ and A with Hamiltonian Ĥ = (p̂ − qA)²/(2m) + qφ, where p̂ = −iℏ∇. Under the gauge transformation A → A + ∇Λ, φ → φ − ∂Λ/∂t, with Λ(r, t) any real function, the Schrödinger equation keeps its form provided ψ → e^(iqΛ/ℏ)ψ. Which expectation value is the same in every gauge, whatever Λ is?

1. A⟨p̂⟩, because p̂ = −iℏ∇ is the same operator in every gauge
2. B⟨Ĥ⟩, because the Hamiltonian generates the same physical time evolution in every gauge
3. CThe phase of ψ at a fixed point r₀, because multiplying ψ by e^(iqΛ/ℏ) leaves |ψ|² unchanged
4. D⟨p̂ − qA⟩, which equals m times the rate of change of ⟨r⟩

A source emits spin-½ particles, each prepared either in |+z⟩ or in |+x⟩ = (|+z⟩ + |−z⟩)/√2, with probability ½ each, and the record of which state was prepared is discarded. What is the purity Tr(ρ²) of the ensemble?

1. A0.50
2. B0.75
3. C0.85
4. D1.00

An electron in a 2p orbital of hydrogen is in the coupled state with l = 1, s = ½, j = ½ and m_j = ½. The stretched state is |j = 3/2, m_j = 3/2⟩ = |m_l = 1⟩|↑⟩, and J₋ = L₋ + S₋, with L₋|l, m⟩ = ℏ√(l(l + 1) − m(m − 1)) |l, m − 1⟩ and S₋|↑⟩ = ℏ|↓⟩. What is the probability that a measurement of S_z on the j = ½ state gives +ℏ/2?

1. A1/2
2. B2/3
3. C1/3
4. D1

A particle of mass m moves in the isotropic two-dimensional oscillator H₀ = (p_x² + p_y²)/(2m) + ½mω²(x² + y²). Its second excited level, at 3ℏω, is spanned by |2, 0⟩, |1, 1⟩ and |0, 2⟩, the labels counting quanta along x and along y. Here x = √(ℏ/(2mω))(a_x + a_x†), with y written likewise, a†|n⟩ = √(n + 1)|n + 1⟩ and a|n⟩ = √n|n − 1⟩. A weak perturbation H′ = εxy is added. What are the first-order shifts of this level, and which good state is left unshifted?

1. AAll three shifts are zero: every diagonal element ⟨n_x, n_y|H′|n_x, n_y⟩ vanishes, so the level stays threefold degenerate at first order.
2. B+εℏ/(mω), 0 and −εℏ/(mω), with |1, 1⟩ left unshifted
3. C+εℏ/(√2 mω), 0 and −εℏ/(√2 mω), with (|2, 0⟩ − |0, 2⟩)/√2 left unshifted
4. D+εℏ/(mω), 0 and −εℏ/(mω), with (|2, 0⟩ − |0, 2⟩)/√2 left unshifted

The ground configuration of oxygen is 1s²2s²2p⁴. The 2p⁴ shell has the same terms as 2p², namely ¹S, ¹D and ³P. According to Hund's rules, which level is the ground level of oxygen?

1. A³P₂
2. B³P₀
3. C¹D₂
4. D¹S₀

Three small splittings are measured in the n = 2 level of hydrogen: the fine-structure splitting 2P_(3/2)–2P_(1/2), the Lamb shift 2S_(1/2)–2P_(1/2), and the hyperfine splitting of 2S_(1/2). Using mc² = 511 keV, α = 1/137, mₑ/m_p = 1/1836, and the powers of α and of mₑ/m_p that set the size of each effect, which ordering, largest first, is correct?

1. Afine structure, then hyperfine splitting, then Lamb shift
2. Bfine structure, then Lamb shift, then hyperfine splitting
3. Chyperfine splitting, then fine structure, then Lamb shift
4. DLamb shift, then fine structure, then hyperfine splitting

On the unit sphere the angular-momentum operators act as L_z = −iℏ ∂/∂φ and L± = ±ℏe^(±iφ)(∂/∂θ ± i cot θ ∂/∂φ), and L±Yₗᵐ = ℏ√(l(l + 1) − m(m ± 1)) Yₗ^(m±1). The Condon–Shortley convention is used throughout, in which Y₁¹ = −√(3/(8π)) sin θ e^(iφ).

1. (a)

   Derive Write Y₂² = g(θ)e^(2iφ). Starting from the condition L₊Y₂² = 0, derive g(θ) up to a constant factor.

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   1. 1

      L₊ acting on g(θ)e^(2iφ) gives ℏe^(3iφ)[g′ − 2 cot θ g], since ∂/∂φ brings down 2i and i × 2i = −2, so L₊Y₂² = 0 requires dg/dθ = 2 cot θ g

   2. 2

      integrating, ln g = 2 ln sin θ + constant, so g = C sin²θ, which is finite at both poles as a spherical harmonic must be

2. (b)

   Show that Show that normalising Y₂² over the sphere, with C real and positive, gives C = (1/4)√(15/(2π)).

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   1. 1

      ∫|Y₂²|² dΩ = C² ∫₀^(2π) dφ ∫₀^π sin⁵θ dθ, and with u = cos θ, ∫₀^π sin⁵θ dθ = ∫₋₁¹ (1 − u²)² du = 2 − 4/3 + 2/5 = 16/15

   2. 2

      setting 2πC² × 16/15 = 1 gives C² = 15/(32π), so C = √(15/(32π)) = (1/4)√(15/(2π)) = 0.386

3. (c)

   Determine Determine Y₂¹ by applying L₋ to Y₂², keeping the sign that the operator produces.

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   1. 1

      L₋(sin²θ e^(2iφ)) = −ℏe^(−iφ)[2 sin θ cos θ − i cot θ (2i) sin²θ]e^(2iφ) = −4ℏ sin θ cos θ e^(iφ)

   2. 2

      dividing by ℏ√(6 − 2) = 2ℏ and multiplying by C: Y₂¹ = −(1/2)√(15/(2π)) sin θ cos θ e^(iφ) = −√(15/(8π)) sin θ cos θ e^(iφ), the minus sign being the Condon–Shortley phase that the lowering generates

4. (d)

   State State the parity of Y₂¹ and the number of values of θ strictly between 0 and π at which it vanishes.

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   1. 1

      even parity, (−1)ˡ = +1, since under θ → π − θ and φ → φ + π both sin θ cos θ and e^(iφ) change sign; and one value, θ = π/2, where cos θ = 0, a nodal plane at the equator, as l − |m| = 1 requires

5. (e)

   Explain In general the recursion for cos θ Yₗᵐ gives a combination of Yₗ₊₁ᵐ and Yₗ₋₁ᵐ. Explain why cos θ Y₁¹ is a multiple of Y₂¹ alone, find the multiple, and say what this implies for a z-polarised electric-dipole transition from an l = 1, m = 1 state.

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   1. 1

      cos θ Y₁¹ = −√(3/(8π)) sin θ cos θ e^(iφ) = (1/√5)Y₂¹, using Y₂¹ = −√(15/(8π)) sin θ cos θ e^(iφ); the l − 1 partner would be Y₀ with m = 1, which does not exist because |m| cannot exceed l

   2. 2

      so ⟨l′, m′|cos θ|1, 1⟩ is nonzero only for l′ = 2, m′ = 1: a z-polarised dipole transition from a p state with m = 1 can reach d states but no s state, although Δl = −1 is allowed in general

In a crystal of cuprous oxide, an electron in the conduction band and a hole in the valence band bind into a hydrogen-like exciton. Model it as two particles of charges −e and +e with effective masses 0.99mₑ (electron) and 0.58mₑ (hole), interacting through the Coulomb potential screened by the relative permittivity ε_r = 7.5, V(r) = −e²/(4πε₀ε_r r). Once the centre-of-mass motion is separated, the relative motion is that of one particle of reduced mass μ, whose radial function u(r) = rR(r) obeys −(ℏ²/2μ)u″ + [V(r) + ℏ²l(l + 1)/(2μr²)]u = Eu. For a bound state write κ = √(−2μE)/ℏ, ρ = κr, a* = 4πε₀ε_rℏ²/(μe²) and ρ₀ = 2/(κa*). Substituting u = ρ^(l+1)e^(−ρ)v(ρ) gives ρv″ + 2(l + 1 − ρ)v′ + [ρ₀ − 2(l + 1)]v = 0. For hydrogen with an infinitely heavy nucleus, ℏ²/(2mₑa₀²) = 13.606 eV and a₀ = 0.05292 nm.

1. (a)

   Derive Substituting v(ρ) = Σⱼ cⱼρʲ, derive the recursion relation between cⱼ₊₁ and cⱼ.

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   1. 1

      with every sum re-indexed to carry ρʲ: ρv″ gives Σ j(j + 1)cⱼ₊₁ρʲ, 2(l + 1)v′ gives Σ 2(l + 1)(j + 1)cⱼ₊₁ρʲ, and −2ρv′ gives Σ (−2j)cⱼρʲ

   2. 2

      equating coefficients of ρʲ: (j + 1)(j + 2l + 2)cⱼ₊₁ = [2(j + l + 1) − ρ₀]cⱼ, so cⱼ₊₁ = [2(j + l + 1) − ρ₀]cⱼ/[(j + 1)(j + 2l + 2)]

2. (b)

   Explain The series solves the differential equation for any value of E. Explain why a physical solution must nevertheless have a terminating series, and why this condition fixes the energy.

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   1. 1

      for large j, cⱼ₊₁/cⱼ → 2/j, the ratio for e^(2ρ) = Σ(2ρ)ʲ/j!, so an unterminated v grows like e^(2ρ) times a power of ρ, and u = ρ^(l+1)e^(−ρ)v grows like e^(ρ) times a power of ρ: it diverges at large r and cannot be normalised

   2. 2

      the series must therefore stop at some c_N, which needs 2(N + l + 1) − ρ₀ = 0 for an integer N ≥ 0; ρ₀ = 2/(κa*) depends on E through κ, so only discrete energies satisfy the condition

3. (c)

   Show that Show that the bound-state energies are Eₙ = −(μ/mₑ)(13.606 eV)/(ε_r²n²), with n = N + l + 1, where N is the degree of the terminated polynomial, and that l ≤ n − 1.

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   1. 1

      termination requires ρ₀ = 2(N + l + 1) ≡ 2n, and since N ≥ 0, n ≥ l + 1, that is l ≤ n − 1

   2. 2

      κ = 2/(ρ₀a*) = 1/(na*), so E = −ℏ²κ²/(2μ) = −ℏ²/(2μa*²n²); with a* = a₀ε_r(mₑ/μ) this is −[ℏ²/(2mₑa₀²)](μ/mₑ)/(ε_r²n²) = −(μ/mₑ)(13.606 eV)/(ε_r²n²)

4. (d)

   Calculate Calculate μ as a multiple of mₑ, the exciton Rydberg energy R* = −E₁ in meV, and the effective Bohr radius a* in nm.

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   1. 1

      μ = (0.99 × 0.58/1.57)mₑ = 0.366mₑ

   2. 2

      R* = 0.366 × 13.606 eV/7.5² = 0.0885 eV = 88.5 meV

   3. 3

      a* = 0.05292 nm × 7.5/0.366 = 1.09 nm

5. (e)

   Determine For these states Kramers' relation reads ((s + 1)/n²)⟨rˢ⟩ − (2s + 1)a*⟨rˢ⁻¹⟩ + (s/4)[(2l + 1)² − s²]a*²⟨rˢ⁻²⟩ = 0. Absorption lines of the exciton P series (l = 1) have been resolved up to n = 25. Determine ⟨r⟩ for the n = 25, l = 1 exciton.

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   1. 1

      s = 0 gives ⟨1/r⟩ = 1/(n²a*); s = 1 then gives (2/n²)⟨r⟩ − 3a* + l(l + 1)a*/n² = 0, so ⟨r⟩ = (a*/2)[3n² − l(l + 1)]

   2. 2

      ⟨r⟩ = (1.09 nm/2)(3 × 625 − 2) = 1.02 × 10³ nm, about 1.0 μm

The neutron is a spin-½ particle whose magnetic moment is μ = γS, with γ = −1.832 × 10⁸ rad s⁻¹ T⁻¹, so the moment points opposite to the spin. In a magnetic field B its spin Hamiltonian is H = −μ⋅B = −γB⋅S, and the spin operators obey [S_x, S_y] = iℏS_z and its cyclic permutations. In every part the neutrons travel at 2200 m/s, and their motion through the apparatus is treated classically, so that a neutron spends a time L/v in a region of length L.

1. (a)

   Derive For a uniform static field B = Bẑ, derive from the Heisenberg equation of motion the equations of motion of S_x, S_y and S_z, written as one vector equation, and hence the angular frequency ω_L at which ⟨S⟩ precesses about B and the sense of the precession.

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   1. 1

      dS_x/dt = (i/ℏ)[H, S_x] = (i/ℏ)(−γB)[S_z, S_x] = (i/ℏ)(−γB)(iℏS_y) = γBS_y; likewise dS_y/dt = −γBS_x and dS_z/dt = 0, which together are dS/dt = γS × B

   2. 2

      taking expectation values, ⟨S_z⟩ is constant while ⟨S_x⟩ and ⟨S_y⟩ rotate about B at ω_L = |γ|B; because γ < 0, dS/dt = |γ|B × S, a right-handed rotation about B

2. (b)

   Calculate In one arm of a neutron interferometer the neutrons cross a region 2.0 cm long in which a uniform field B is applied perpendicular to their path. Calculate the time each neutron spends in the field, and the field B₂π for which its spin precesses through exactly one turn.

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   1. 1

      t = 0.020 m/(2200 m s⁻¹) = 9.09 × 10⁻⁶ s

   2. 2

      B₂π = 2π/(|γ|t) = 2π/(1.832 × 10⁸ rad s⁻¹ T⁻¹ × 9.09 × 10⁻⁶ s) = 3.77 × 10⁻³ T (3.77 mT)

3. (c)

   Predict The two arms are recombined, and the count rate at one output port is recorded as B is raised from zero; at B = 0 this port is at an interference maximum. In the field the spin state is acted on by the rotation operator exp(−iφn̂⋅σ/2), where φ is the precession angle and n̂ the field direction. Predict the count rate at B = B₂π, and the smallest nonzero field at which the port returns to its B = 0 count rate.

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   1. 1

      the recombined intensity is proportional to |χ + e^(−iφn̂⋅σ/2)χ|² = 2 + 2 cos(φ/2) for any spin state χ, since ⟨χ|n̂⋅σ|χ⟩ is real and contributes only to the imaginary part of the overlap; at φ = 2π the operator is −I, so the port is at an interference minimum, ideally dark, although every spin points as it did at B = 0

   2. 2

      the count rate first returns to its B = 0 value at φ = 4π, that is B = 2B₂π = 7.55 mT

4. (d)

   Describe A resonance spin flipper is a region with a static field B₀ along z and a field of magnitude B₁ ≪ B₀ rotating in the xy-plane at angular frequency ω, in the same sense as the precession. Describe how transforming to the frame that rotates with B₁ reduces the problem to precession about a static effective field, giving that field's components and the flip probability that follows.

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   1. 1

      in the frame rotating at ω about z, B₁ is static along an axis x′, and the transformation adds −ωS_z to the Hamiltonian, which acts as a fictitious field ω/|γ| opposing B₀: B_eff = (B₀ − ω/|γ|)ẑ + B₁x̂′

   2. 2

      the spin precesses about B_eff at Ω = √(Δ² + ω₁²), with ω₀ = |γ|B₀, ω₁ = |γ|B₁ and Δ = ω − ω₀, so a spin starting along z flips with probability P = (ω₁²/Ω²) sin²(Ωt/2); on resonance B_eff = B₁x̂′ and P = sin²(ω₁t/2)

5. (e)

   Calculate The flipper is 5.0 cm long and B₀ = 10.0 mT. Calculate the frequency ω/2π at which B₁ must rotate, and the smallest value of B₁ that turns a neutron spin polarised along +z into one polarised along −z.

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   1. 1

      on resonance, ω/2π = |γ|B₀/2π = 1.832 × 10⁸ rad s⁻¹ T⁻¹ × 0.0100 T/2π = 2.92 × 10⁵ Hz (292 kHz)

   2. 2

      t = 0.050 m/(2200 m s⁻¹) = 2.27 × 10⁻⁵ s, and a flip needs ω₁t = π, so B₁ = π/(|γ|t) = 7.55 × 10⁻⁴ T (0.755 mT)

Two atoms of mass m are held in a narrow one-dimensional box trap, modelled as an infinite square well on 0 < x < L. The single-particle states are ψₙ(x) = √(2/L) sin(nπx/L), with energies Eₙ = n²E₁, where E₁ = π²ℏ²/(2mL²). The atoms interact weakly through a short-range repulsion H′ = gδ(x₁ − x₂), with g > 0, which is treated as a perturbation of H₀ = H₁ + H₂. P₁₂ exchanges every label of the two atoms, coordinates and internal states alike. You may use ∫₀^L ψ₁(x)⁴ dx = 3/(2L).

1. (a)

   Explain P₁₂ commutes with H₀ + H′. Explain why this alone does not require the two-atom state to be symmetric or antisymmetric under exchange, and state what does.

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   1. 1

      [P₁₂, H] = 0 means only that exchange symmetry, if a state has it, is conserved in time; states with no exchange symmetry, such as ψ₁(x₁)ψ₂(x₂), are perfectly good states of the two-particle space, as they are for distinguishable atoms

   2. 2

      the symmetrisation postulate: identical particles must be in states symmetric under P₁₂ (bosons, integer spin) or antisymmetric (fermions, half-integer spin), the link to spin being the spin-statistics connection, quoted from relativistic field theory rather than derived

2. (b)

   Show that Show that ∫₀^L ψ₁(x)²ψ₂(x)² dx = 1/L.

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   1. 1

      with u = πx/L, sin²u sin²2u = ¼(1 − cos 2u)(1 − cos 4u) = ¼[1 − ½cos 2u − cos 4u + ½cos 6u], using cos 2u cos 4u = ½(cos 2u + cos 6u)

   2. 2

      each cosine integrates to zero over 0 < u < π, leaving (2/L)² × ¼ × L = 1/L

3. (c)

   Determine For two distinguishable atoms the unperturbed level 5E₁ is twofold degenerate, spanned by |12⟩ = ψ₁(x₁)ψ₂(x₂) and |21⟩ = ψ₂(x₁)ψ₁(x₂). Determine the matrix of H′ in this subspace, its first-order energy shifts and the corresponding good states.

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   1. 1

      diagonal (direct) elements: ⟨12|H′|12⟩ = ⟨21|H′|21⟩ = g∫ψ₁(x)²ψ₂(x)² dx = g/L, the delta function setting x₁ = x₂ = x

   2. 2

      off-diagonal (exchange) element: ⟨12|H′|21⟩ = g∫ψ₁ψ₂ψ₂ψ₁ dx = g/L, the same integral, because a contact interaction samples both orderings at the same point

   3. 3

      the matrix (g/L)[[1, 1], [1, 1]] has eigenvalues 2g/L and 0, with good states (|12⟩ + |21⟩)/√2 and (|12⟩ − |21⟩)/√2, the eigenstates of P₁₂ with eigenvalues +1 and −1

4. (d)

   Calculate Take g = 0.10E₁L. Calculate, to first order in g and in units of E₁, the ground-state energy of two identical spinless bosons, and of two identical fermions in the same internal state.

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   1. 1

      bosons: both atoms in ψ₁, so E = 2E₁ + g∫ψ₁⁴ dx = 2E₁ + 3g/(2L) = 2.15E₁

   2. 2

      fermions: antisymmetry forbids both in ψ₁, so the lowest state is (|12⟩ − |21⟩)/√2 at 5E₁ with shift J − K = 0, giving E = 5.00E₁, which is exact to all orders because every antisymmetric spatial state vanishes at x₁ = x₂, where the delta function acts

5. (e)

   Compare Two spin-½ fermions, for example atoms in two internal states labelled ↑ and ↓, occupy ψ₁ and ψ₂ in the same trap, H′ does not act on spin, and g = 0.10E₁L as in (d). Compare the first-order energies of the states with total spin S = 0 and S = 1, and account for the difference.

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   1. 1

      the whole state must be antisymmetric, so S = 0 (antisymmetric spin singlet) goes with the symmetric spatial state, at 5E₁ + 2g/L = 5.20E₁, and S = 1 (symmetric spin triplet) goes with the antisymmetric spatial state and stays at 5.00E₁: the triplet lies lower by 2g/L = 0.20E₁

   2. 2

      the splitting is an exchange correlation, not a spin-dependent force: the triplet's spatial wavefunction vanishes where the atoms meet, so the repulsion never acts, just as orthohelium lies below parahelium and as Hund's first rule favours maximum S

A source emits pairs of spin-½ particles intended to be in the singlet state. Alice measures the spin component of one particle along a direction at angle a in the xz-plane and Bob that of the other along angle b, each recording +1 or −1. For each setting the correlation E is the mean of the product of the two results; for the ideal singlet E(a, b) = −cos(a − b). Every local hidden-variable model obeys |S| ≤ 2, where S = E(a, b) − E(a, b′) + E(a′, b) + E(a′, b′). The run used a = 0°, a′ = 90°, b = 45° and b′ = 135°, and the table gives the number of detected pairs with each combination of results, Alice's sign first. Treat each detected pair as an independent trial.

1. (a)

   Calculate Calculate E for each of the four settings.

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   1. 1

      E = (N₊₊ + N₋₋ − N₊₋ − N₋₊)/N, with N the total for that setting: E(a, b) = −327/513 = −0.637 and E(a, b′) = +294/482 = +0.610

   2. 2

      E(a′, b) = −300/520 = −0.577 and E(a′, b′) = −288/482 = −0.598

2. (b)

   Determine Determine S, and the fraction of the ideal-singlet value that it reaches.

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   1. 1

      S = −0.637 − 0.610 − 0.577 − 0.598 = −2.42, the minus sign in front of the positive E(a, b′) making all four terms negative

   2. 2

      for the ideal singlet each |E| = cos 45° = 0.707 and S = −2√2 = −2.83, so the data reach 2.42/2.83 = 0.86 of it, as for a singlet mixed with uncorrelated noise, E = −V cos(a − b) with visibility V ≈ 0.86

3. (c)

   Estimate Estimate the standard uncertainty in S from counting statistics, stating the assumptions you make.

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   1. 1

      assumes that each pair is an independent trial whose product ±1 has mean E and variance 1 − E², and that the four settings are sampled independently, so σ_E = √((1 − E²)/N) ≈ √((1 − 0.37)/500) ≈ 0.036 for each setting

   2. 2

      σ_S = √(Σσ_E²) ≈ 2 × 0.036 = 0.071

4. (d)

   Evaluate Evaluate whether these data rule out every local hidden-variable model, and say what else a conclusive test must secure.

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   1. 1

      |S| − 2 = 0.42, about 5.9σ with σ_S = 0.071: statistically decisive, far beyond what counting fluctuations about any |S| ≤ 2 could produce

   2. 2

      but the conclusion also needs a large enough fraction of pairs detected, or a fair-sampling assumption (the detection loophole), setting choices made at random and at space-like separation from the other side's measurement (the locality loophole), and a check of systematic errors such as analyser-angle calibration, which counting statistics do not include

5. (e)

   Justify Quantum mechanics predicts that Alice's results alone cannot reveal which setting Bob chose. Justify this from the singlet's reduced density operator, and test it against the data for Alice's setting a.

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      tracing Bob's particle out of the singlet gives ρ_A = I/2, and Bob's measurement, averaged over his results, which Alice does not know, leaves ρ_A unchanged, so Alice obtains +1 with probability ½ at every a, whatever b is

   2. 2

      from the data, P_A(+1) = (43 + 215)/513 = 0.503 with Bob at b and (192 + 48)/482 = 0.498 with Bob at b′; each has standard uncertainty √(0.25/N) ≈ 0.022, so both agree with ½ and with each other: no signal

A hydrogen-like ion has one electron of mass m bound to a point nucleus of charge Ze, treated as infinitely heavy, so that V(r) = −Ze²/(4πε₀r) and the unperturbed energies Eₙ = −(Zα)²mc²/(2n²) are degenerate in l, m_l and m_s. With a = a₀/Z, so that ℏ²/(ma) = Ze²/(4πε₀) and Eₙ = −Ze²/(8πε₀an²), the unperturbed states give ⟨1/r⟩ = 1/(n²a), ⟨1/r²⟩ = 1/[n³(l + ½)a²], ⟨1/r³⟩ = 1/[n³l(l + ½)(l + 1)a³] for l ≥ 1, and |ψₙ₀₀(0)|² = 1/(πn³a³). The fine-structure Hamiltonian is H_rel + H_SO + H_D, with H_rel = −p⁴/(8m³c²); H_SO = [1/(2m²c²)](1/r)(dV/dr) L⋅S, which already contains the Thomas factor of ½, quoted from relativistic kinematics; and the Darwin term H_D = [πℏ²/(2m²c²)][Ze²/(4πε₀)]δ³(r), quoted from the Dirac equation. Take mc² = 0.5110 MeV, α = 1/137.04 and h = 4.136 × 10⁻¹⁵ eV s.

1. (a)

   Justify Each level n is degenerate, yet each fine-structure term can be evaluated with the non-degenerate formula E^(1) = ⟨H′⟩ in the coupled basis |n, l, j, m_j⟩. Justify this, naming the operators each term commutes with.

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   1. 1

      H_rel and H_D are spin-independent scalars: they commute with L², L_z, S² and S_z, so they have no matrix elements between states of different l, m_l or m_s, and are diagonal in the uncoupled basis and equally in the coupled one

   2. 2

      H_SO ∝ L⋅S = (J² − L² − S²)/2 commutes with J², J_z, L² and S² but not with L_z or S_z, so it is diagonal only in the coupled basis; there all three terms are diagonal, the degenerate states are already the good states, and E^(1) = ⟨H′⟩ holds state by state

2. (b)

   Show that Starting from p²/(2m) = H₀ − V, show that the relativistic shift is E_rel = −[Eₙ²/(2mc²)][4n/(l + ½) − 3].

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   1. 1

      H_rel = −(1/(2mc²))(p²/2m)², and because p² is Hermitian ⟨p⁴⟩ is the squared norm of p²|ψ⟩, so E_rel = −(1/(2mc²))⟨(Eₙ − V)²⟩ = −(1/(2mc²))[Eₙ² − 2Eₙ⟨V⟩ + ⟨V²⟩]

   2. 2

      ⟨V⟩ = −[Ze²/(4πε₀)]⟨1/r⟩ = −Ze²/(4πε₀n²a) = 2Eₙ, and ⟨V²⟩ = [Ze²/(4πε₀)]²⟨1/r²⟩ = [Ze²/(4πε₀a)]²/[n³(l + ½)] = 4nEₙ²/(l + ½), using Ze²/(4πε₀a) = −2n²Eₙ

   3. 3

      so E_rel = −(1/(2mc²))[Eₙ² − 4Eₙ² + 4nEₙ²/(l + ½)] = −[Eₙ²/(2mc²)][4n/(l + ½) − 3]

3. (c)

   Derive Derive the first-order spin–orbit shift E_SO of a state with l ≥ 1, expressed in terms of Eₙ, mc², n, l and j.

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   1. 1

      (1/r)dV/dr = [Ze²/(4πε₀)]/r³, so H_SO = [Ze²/(4πε₀)] L⋅S/(2m²c²r³)

   2. 2

      in the coupled basis L⋅S = (J² − L² − S²)/2 has eigenvalue (ℏ²/2)[j(j + 1) − l(l + 1) − ¾], and ⟨1/r³⟩ = 1/[n³l(l + ½)(l + 1)a³]

   3. 3

      E_SO = [Ze²/(4πε₀)](ℏ²/(ma))[j(j + 1) − l(l + 1) − ¾]/[4mc²a²n³l(l + ½)(l + 1)], and with ℏ²/(ma) = Ze²/(4πε₀) and [Ze²/(4πε₀a)]² = 4n⁴Eₙ² this is [Eₙ²/(mc²)] n[j(j + 1) − l(l + 1) − ¾]/[l(l + ½)(l + 1)]

4. (d)

   Show that Show that, once the Darwin term is included for l = 0, the three terms sum to E_fs = [Eₙ²/(2mc²)][3 − 4n/(j + ½)] for every state, so that the energy depends on n and j only.

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   1. 1

      j = l + ½: the bracket in E_SO is l, so E_SO = nEₙ²/[mc²(l + ½)(l + 1)], and E_rel + E_SO = [Eₙ²/(2mc²)][3 − 4n/(l + ½) + 2n/((l + ½)(l + 1))] = [Eₙ²/(2mc²)][3 − 4n/(l + 1)], with l + 1 = j + ½

   2. 2

      j = l − ½: the bracket is −(l + 1), so E_SO = −nEₙ²/[mc²l(l + ½)], and E_rel + E_SO = [Eₙ²/(2mc²)][3 − 4n/(l + ½) − 2n/(l(l + ½))] = [Eₙ²/(2mc²)][3 − 4n/l], with l = j + ½

   3. 3

      l = 0: H_SO contributes nothing, E_D = [πℏ²/(2m²c²)][Ze²/(4πε₀)]/(πn³a³) = 2nEₙ²/(mc²), and E_rel = −[Eₙ²/(2mc²)](8n − 3), so the sum is [Eₙ²/(2mc²)](3 − 4n), the formula with j = ½

5. (e)

   Sketch For He⁺ (Z = 2), sketch the fine structure of the n = 2 level: draw the unperturbed level and the fine-structure levels labelled 2S_(1/2), 2P_(1/2) and 2P_(3/2), marking each shift in μeV and the 2P_(3/2)–2P_(1/2) splitting in GHz.

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   1. 1

      E₂ = −(2α)²mc²/8 = −13.6 eV and E₂²/(2mc²) = 181.1 μeV, so 2S_(1/2) and 2P_(1/2) are drawn as one level 5 × 181.1 = 906 μeV below the unperturbed level and 2P_(3/2) as a level 181 μeV below it: every state lies below E₂

   2. 2

      the 2P_(3/2)–2P_(1/2) splitting is marked as 4 × 181.1 μeV = 724 μeV, which is 724 × 10⁻⁶ eV/(4.136 × 10⁻¹⁵ eV s) = 175 GHz, sixteen times hydrogen's 10.9 GHz

6. (f)

   Evaluate The measured 2S_(1/2)–2P_(1/2) Lamb shifts are 1.058 GHz in hydrogen and 14.04 GHz in He⁺. Welton's argument gives, for an nS level, ΔE ∝ α(Zα)⁴mc² ln[1/(Zα)]/n³, with a prefactor, and an argument of the logarithm, that it fixes only roughly. Evaluate how well this scaling accounts for the ratio of the two shifts.

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   1. 1

      the scaling predicts a ratio 2⁴ ln[1/(2α)]/ln(1/α) = 16 × 4.23/4.92 = 13.7, below the Z⁴ = 16 of the fine structure because the logarithm falls with Z; the measured ratio is 14.04/1.058 = 13.3

   2. 2

      judgement: agreement to about 4%, so the scaling captures the physics, a correction proportional to Z|ψ(0)|² ∝ Z⁴ that acts on s states only; but the argument of the logarithm is fixed only to an order of magnitude, so agreement at a few per cent is partly fortunate, since the ratio cancels the prefactor but not that uncertainty