University Quantum Mechanics II · Final Examination

A linear molecule attached by one end to a flat surface is modelled as a rigid rotor, of moment of inertia I about the attached end, that turns freely except that its axis cannot point into the surface. The polar angle θ of the axis, measured from the outward normal to the surface, is therefore confined to 0 ≤ θ ≤ π/2, and the surface acts as a hard wall: the wavefunction must vanish at θ = π/2, where the axis would lie in the surface. The rotor's Hamiltonian is H = L²/(2I). For m = 0, with x = cos θ and λ = 2IE/ℏ², the angular equation is Legendre's equation, and of its power-series solutions about x = 0 only the series of odd powers of x vanishes at x = 0, as the wall requires. Which values of λ are allowed?

1. Aλ = l(l + 1) with l = 0, 1, 2, …, the free rotor's values, because the wall changes a boundary condition but not Legendre's equation.
2. Bλ = l(l + 1) with l = 0, 2, 4, …, because only the even polynomials are symmetric about the equator, where the wall lies.
3. Cλ = l(l + 1) with l = 1, 3, 5, …, because the odd series must still terminate to stay finite at θ = 0, where the axis lies along the surface normal.
4. DAny λ ≥ 0, because the series need only be finite for 0 ≤ θ ≤ π/2, and the pole θ = π, where an unterminated series diverges, lies outside that range.

A crude model of a valence electron outside a closed-shell core replaces the Coulomb potential by V(r) = −k/r − β/r², where k = e²/(4πε₀) and β is a small positive constant representing the extra attraction felt by an electron that penetrates the core. Compared with hydrogen, which statement about the bound-state spectrum is correct?

1. AThe degeneracy in m is lifted, because the extra term breaks the symmetry that protected it.
2. BLevels of the same n but different l separate, with the lowest l pushed down most, while each keeps its (2l + 1)-fold degeneracy in m; the quantum Runge-Lenz vector is no longer conserved.
3. CNo degeneracy is lifted: the extra term simply adds to the centrifugal term ℏ²l(l + 1)/(2mr²), and so shifts every level of a given n by the same amount.
4. DThe l-degeneracy survives at first order in β, because ⟨1/r²⟩ is the same for every state of a given n, and is lifted only at second order.

In a nuclear magnetic resonance magnet whose field B points along +z, the spins of protons at rest precess at 400.0 MHz. A particle at rest whose magnetic moment is μ = γS has the Hamiltonian H = −γBS_z, and each spin component obeys the Heisenberg equation dS/dt = (i/ℏ)[H, S]; γ = +2.675 × 10⁸ rad s⁻¹ T⁻¹ for the proton and γ = −1.761 × 10¹¹ rad s⁻¹ T⁻¹ for the electron. A free electron is now held at rest in the same magnet. Which statement about the precession of its spin is correct?

1. AIt precesses at 400.0 MHz, in the same sense as the proton spins, because the Larmor frequency is set by the field alone.
2. BIt precesses at 263.3 GHz, in the same sense as the proton spins, because every moment precesses in the sense fixed by the direction of B.
3. CIt precesses at 131.7 GHz, in the opposite sense to the proton spins, because the two spinor components acquire phases of only ±γBt/2.
4. DIt precesses at 263.3 GHz, in the opposite sense to the proton spins, while S_z, which commutes with H, stays constant.

In hydrogen the fine-structure shifts are E_fs = −[Eₙ²/(2mc²)][4n/(j + ½) − 3], where Eₙ = −13.6 eV/n², so they are smaller than the Bohr energies by a factor of order α² and depend on n and j only. The 2p₃/₂–2p₁/₂ splitting is measured to be 10.97 GHz. What splitting does the formula then predict between the 3d₅/₂ and 3d₃/₂ levels?

1. A1.08 GHz
2. B3.25 GHz
3. C4.88 GHz
4. D0.722 GHz

A hydrogen atom in the 2p₁/₂ level is placed in a magnetic field of 0.10 T, which is weak compared with the 10.97 GHz fine-structure splitting of n = 2. By what frequency are its two m_j sublevels split? Take μ_B = 5.788 × 10⁻⁵ eV T⁻¹ and h = 4.136 × 10⁻¹⁵ eV s.

1. A2.80 GHz
2. B1.40 GHz
3. C0.933 GHz
4. D1.87 GHz

A system has exact ground and first excited energies E₀ = −1.00 eV and E₁ = −0.25 eV, with normalised eigenstates ψ₀ and ψ₁. A normalised trial state is φ = √(1 − ε²) ψ₀ + εψ₁ with ε = 0.10, so that 1% of its weight lies in the wrong state. What is ⟨φ|H|φ⟩?

1. A−1.020 eV
2. B−0.9925 eV
3. C−0.925 eV
4. D−1.000 eV

A cold atom of mass m is held along x in a harmonic trap whose angular frequency ω satisfies ω/2π = 2.0 kHz. A thin repulsive light sheet is then switched on at the trap centre and is modelled as an impenetrable wall at x = 0, so that the atom moves in V(x) = ½mω²x² for x > 0 only. WKB quantisation is applied using the connection-formula phase at the smooth turning point and the condition ψ = 0 at the wall. What energy does it predict for the first excited state of this half-trap, expressed as the frequency E/h and measured from the bottom of the trap?

1. A3.0 kHz
2. B5.0 kHz
3. C6.0 kHz
4. D7.0 kHz

A two-level system is swept at a constant rate through an avoided crossing. The probability of a diabatic passage, in which the system stays in its initial diabatic state and so jumps the gap, is measured to be 0.20. According to the Landau-Zener formula, what is this probability when the sweep rate is doubled?

1. A0.45
2. B0.40
3. C0.10
4. D0.040

A hydrogen atom is prepared in the 2s₁/₂ state, isolated and in zero field. It lives for about 0.12 s, whereas 2p lives for 1.6 ns. Which process is responsible for its decay?

1. AEmission of one electric-dipole photon to 1s, slowed because the 2s radial function has a node.
2. BEmission of one electric-dipole photon to 2p₁/₂, which lies 1.06 GHz below 2s₁/₂ because of the Lamb shift.
3. CEmission of two electric-dipole photons together, through virtual intermediate p states, sharing the 10.2 eV between them.
4. DEmission of one magnetic-dipole photon to 1s, allowed because parity is unchanged and Δl = 0.

Slow electrons scatter from atoms of a noble gas, which are modelled as spherically symmetric attractive wells. At the energies of interest only the s wave is appreciably scattered, and at one energy its phase shift passes through δ₀ = π. What is observed near that energy?

1. AA peak in the total cross-section, reaching the unitarity limit 4π/k².
2. BA deep minimum in the total cross-section, because (4π/k²)sin²δ₀ vanishes, so the gas becomes nearly transparent to the electrons.
3. CA new bound state appearing at exactly that energy, because Levinson's theorem requires one whenever δ₀ reaches π.
4. DA cross-section equal to the geometric value πR², because a phase shift of π reflects the wave as a hard sphere of the well's radius would.

Two electrons, each held in one of a pair of neighbouring quantum dots, interact through the exchange Hamiltonian H = (J/ℏ²)S₁ ⋅ S₂, with J = 2.0 μeV. Their orbital states are frozen, so only the spins evolve. Write |↑↓⟩ for spin 1 up and spin 2 down along z, and similarly for the other product states. Take ℏ = 1.055 × 10⁻³⁴ J s and 1 eV = 1.602 × 10⁻¹⁹ J.

1. (a)

   Show that Show that the eigenvalues of H are J/4 for each of the three triplet states and −3J/4 for the singlet.

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   1. 1

      S² = (S₁ + S₂)² = S₁² + S₂² + 2S₁ ⋅ S₂, so S₁ ⋅ S₂ = (S² − S₁² − S₂²)/2 = (ℏ²/2)[s(s + 1) − 3/2] on states of total spin s

   2. 2

      s = 1 (the three triplet states) gives S₁ ⋅ S₂ = ℏ²/4 and E = J/4; s = 0 (the singlet) gives S₁ ⋅ S₂ = −3ℏ²/4 and E = −3J/4, a triplet-singlet splitting of J

2. (b)

   Determine At t = 0 the spins are prepared in |↑↓⟩. By expressing this state in the coupled basis, determine the probability P(t) that a measurement at time t finds the spins in |↓↑⟩.

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   1. 1

      |↑↓⟩ = (|T₀⟩ + |S⟩)/√2, with |T₀⟩ = (|↑↓⟩ + |↓↑⟩)/√2 and |S⟩ = (|↑↓⟩ − |↓↑⟩)/√2

   2. 2

      each coupled state evolves with its own energy: |ψ(t)⟩ = [e^(−iJt/4ℏ)|T₀⟩ + e^(3iJt/4ℏ)|S⟩]/√2, so ⟨↓↑|ψ(t)⟩ = [e^(−iJt/4ℏ) − e^(3iJt/4ℏ)]/2

   3. 3

      P(t) = |1 − e^(iJt/ℏ)|²/4 = sin²(Jt/2ℏ), oscillating between 0 and 1

3. (c)

   Calculate Calculate the shortest time after which the two spins have completely exchanged their states.

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   1. 1

      P = 1 first when Jt/2ℏ = π/2, at t = πℏ/J = π × 1.055 × 10⁻³⁴ J s/(2.0 × 10⁻⁶ × 1.602 × 10⁻¹⁹ J) = 1.03 × 10⁻⁹ s, about 1.0 ns

4. (d)

   Determine Determine the reduced density matrix of spin 1 at time t and its purity Tr ρ₁², and hence the earliest time at which spin 1 is maximally mixed.

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   1. 1

      |ψ(t)⟩ = a|↑↓⟩ + b|↓↑⟩ with |a|² = cos²(Jt/2ℏ) and |b|² = sin²(Jt/2ℏ); tracing over spin 2 removes the cross terms because they carry ⟨↓|↑⟩ = 0 for spin 2, so ρ₁ = cos²(Jt/2ℏ)|↑⟩⟨↑| + sin²(Jt/2ℏ)|↓⟩⟨↓|

   2. 2

      Tr ρ₁² = cos⁴(Jt/2ℏ) + sin⁴(Jt/2ℏ) = 1 − ½sin²(Jt/ℏ), which reaches its minimum of ½ when Jt/ℏ = π/2

   3. 3

      t = πℏ/(2J) = 0.52 ns, half the swap time: ρ₁ = I/2, its Bloch vector (0, 0, cos(Jt/ℏ)) is at the centre of the ball, and the pair is in the maximally entangled pure state (|↑↓⟩ − i|↓↑⟩)/√2, up to an overall phase, with entanglement entropy ln 2

5. (e)

   Explain At that time an experimenter measures spin 1 alone, along any axis they choose. Explain why no such measurement can distinguish it from a spin drawn from an ensemble prepared half in |↑⟩ and half in |↓⟩, and what kind of measurement would reveal the difference.

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   1. 1

      every outcome probability for a measurement on spin 1 alone is Tr(ρ₁Π) for the projector Π of that outcome, and both preparations have ρ₁ = I/2, so every axis gives ½ and ½: preparations with the same density operator cannot be told apart by any measurement on that system

   2. 2

      the difference lies only in correlations between the two spins, which need joint measurements: the entangled state has ⟨σ_x ⊗ σ_y⟩ = +1, whereas a mixture of |↑↓⟩ and |↓↑⟩ has zero transverse correlations (a measurement in the Bell basis would also distinguish them)

An electron in a cubic InAs nanocrystal is modelled as a particle of charge −e and effective mass m* = 0.023mₑ in a cubical infinite well of side L = 20 nm centred on the origin, −L/2 < x, y, z < L/2. Spin is ignored, since its Zeeman energy is the same for every orbital state. The unperturbed states are ψ_(n₁n₂n₃) = u_(n₁)(x)u_(n₂)(y)u_(n₃)(z), with energies (n₁² + n₂² + n₃²)ε, where ε = π²ℏ²/(2m*L²), u₁(x) = √(2/L) cos(πx/L) and u₂(x) = √(2/L) sin(2πx/L). A uniform magnetic field B = 1.5 T is applied along z. In the symmetric gauge, and to first order in B, it adds H′ = (μ*B/ℏ)L_z, where μ* = eℏ/(2m*) is the effective magneton and L_z = −iℏ(x ∂/∂y − y ∂/∂x); the term in B² is neglected. You may use ∫u₁(x) x u₂(x) dx = 16L/(9π²) and ∫u₁(du₂/dx) dx = 8/(3L) = −∫u₂(du₁/dx) dx, each integral running from −L/2 to L/2. Take mₑ = 9.109 × 10⁻³¹ kg, ℏ = 1.055 × 10⁻³⁴ J s, e = 1.602 × 10⁻¹⁹ C and 1 eV = 1.602 × 10⁻¹⁹ J.

1. (a)

   Calculate Calculate ε in meV, and the energies and degeneracies of the ground level and of the first excited level.

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   1. 1

      ε = π²ℏ²/(2m*L²) = π²(1.055 × 10⁻³⁴ J s)²/(2 × 0.023 × 9.109 × 10⁻³¹ kg × (20 × 10⁻⁹ m)²) = 6.55 × 10⁻²¹ J = 40.9 meV

   2. 2

      the ground level ψ₁₁₁ lies at 3ε = 123 meV and is non-degenerate; the first excited level lies at 6ε = 245 meV and is threefold degenerate, spanned by ψ₂₁₁, ψ₁₂₁ and ψ₁₁₂

2. (b)

   Calculate Calculate μ*B in meV, and its ratio to the spacing 3ε that separates the first excited level from the levels on either side of it.

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   1. 1

      μ*B = eℏB/(2m*) = (1.602 × 10⁻¹⁹ C)(1.055 × 10⁻³⁴ J s)(1.5 T)/(2 × 0.023 × 9.109 × 10⁻³¹ kg) = 6.05 × 10⁻²² J = 3.78 meV

   2. 2

      μ*B/3ε = 3.78 meV/123 meV = 0.031 ≪ 1, so admixtures from outside the level are at most a few per cent and second-order shifts at most about (μ*B)²/3ε ≈ 0.1 meV: first-order theory within the degenerate level is adequate

3. (c)

   Determine Using the integrals given, determine the matrix of H′ in the first excited level, its eigenvalues in meV, and the good states.

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   1. 1

      the diagonal elements vanish because each ψ is real, and ψ₁₁₂ is uncoupled because L_z leaves the z factor alone and ⟨u₂|u₁⟩ = 0; the one independent element is ⟨ψ₂₁₁|L_z|ψ₁₂₁⟩ = −iℏ[⟨u₂|x|u₁⟩⟨u₁|u₂′⟩ − ⟨u₂|u₁′⟩⟨u₁|y|u₂⟩] = −iℏ × 2 × (16L/9π²)(8/3L) = −iℏ(256/27π²) = −0.961iℏ

   2. 2

      in the basis (ψ₂₁₁, ψ₁₂₁, ψ₁₁₂), H′ = 0.961μ*B [[0, −i, 0], [i, 0, 0], [0, 0, 0]], with eigenvalues +0.961μ*B = +3.63 meV, 0 and −3.63 meV

   3. 3

      the good states are (ψ₂₁₁ + iψ₁₂₁)/√2, raised by 3.63 meV, (ψ₂₁₁ − iψ₁₂₁)/√2, lowered by 3.63 meV, and ψ₁₁₂, unshifted, so the level at 245 meV splits into three levels spanning 7.26 meV

4. (d)

   Explain Explain how the rotation through 90° about the z-axis identifies these good states without any diagonalisation.

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   1. 1

      the rotation R, which sends ψ(x, y, z) to ψ(y, −x, z), maps the cube onto itself and so commutes with H₀, and it commutes with L_z, the generator of rotations about z, and so with H′; because u₁ is even and u₂ odd, it takes ψ₂₁₁ to ψ₁₂₁, ψ₁₂₁ to −ψ₂₁₁ and ψ₁₁₂ to itself

   2. 2

      within the level its eigenstates are (ψ₂₁₁ ± iψ₁₂₁)/√2, with eigenvalues ∓i, and ψ₁₁₂, with eigenvalue 1; the three eigenvalues are distinct, so H′ has no element between these states and they are the good states, the analogues of m = +1, −1 and 0

5. (e)

   Show that The field is turned, at the same strength, to lie along the body diagonal n̂ = (1, 1, 1)/√3, so that H′ = (μ*B/ℏ)n̂⋅L. Show that the first excited level splits in exactly the same way, into shifts of +3.63 meV, 0 and −3.63 meV.

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   1. 1

      the cube is unchanged by the cyclic relabelling x → y → z, so L_x couples only ψ₁₂₁ and ψ₁₁₂, and L_y only ψ₁₁₂ and ψ₂₁₁, with the same magnitude as L_z; in the basis (ψ₂₁₁, ψ₁₂₁, ψ₁₁₂) every element is (L_k)_ij = −iℏcε_kij, with c = 256/27π² = 0.961

   2. 2

      these are c times the matrices of an l = 1 angular momentum in the Cartesian basis, so n̂⋅L has the characteristic equation λ(λ² − c²ℏ²) = 0 and the eigenvalues 0 and ±cℏ for every unit vector n̂: the shifts are 0 and ±0.961μ*B = ±3.63 meV whatever the direction of B

A hydrogen atom in the n = 2 shell is placed in a uniform static electric field F along z; hyperfine structure is ignored. Without the field, 2s₁/₂ lies above 2p₁/₂ by the Lamb shift, L = h × 1057.8 MHz, and 2p₃/₂ lies h × 10.97 GHz above 2p₁/₂. The perturbation is H′ = eFz. In the uncoupled basis ⟨2s|z|2p, m_l = 0⟩ = −3a₀, with a₀ = 5.29 × 10⁻¹¹ m, and in the convention used here |2p₁/₂, m_j = +½⟩ = √(2/3)|m_l = 1⟩|↓⟩ − √(1/3)|m_l = 0⟩|↑⟩ and |2s₁/₂, m_j = +½⟩ = |2s⟩|↑⟩. The 2p lifetime is 1.60 ns, and an isolated 2s atom lives for 0.122 s. Take h = 6.626 × 10⁻³⁴ J s and e = 1.602 × 10⁻¹⁹ C.

1. (a)

   Show that Show that H′ couples |2s₁/₂, +½⟩ to |2p₁/₂, +½⟩ and to no other 2p₁/₂ state, with a matrix element of magnitude √3 ea₀F.

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   1. 1

      z does not act on spin and, being proportional to rY₁⁰, cannot change m_l, so it conserves m_j: |2s₁/₂, +½⟩ couples only to m_j = +½, and within |2p₁/₂, +½⟩ only the |m_l = 0⟩|↑⟩ component contributes

   2. 2

      ⟨2s₁/₂, +½|eFz|2p₁/₂, +½⟩ = eF × (−√(1/3)) × (−3a₀) = √3 ea₀F

2. (b)

   Explain Explain why the 2s₁/₂ level has no linear Stark shift in a weak field, although the n = 2 level of hydrogen is often said to show a linear Stark effect of ±3ea₀F.

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   1. 1

      the Lamb shift separates 2s₁/₂ from 2p₁/₂, so in a weak field the two 2s₁/₂ states (m_j = ±½) are degenerate only with each other and both have definite (even) parity; every first-order matrix element between them, ⟨2s₁/₂|eFz|2s₁/₂⟩, vanishes because z is odd, so the leading shift is second order and, dominated by 2p₁/₂, about 3e²a₀²F²/L

   2. 2

      the linear effect ±3ea₀F assumes 2s and 2p₀ exactly degenerate, so that the field mixes them at first order; with the Lamb shift present the shift becomes linear only once √3 ea₀F is large compared with L, and reaches ±3ea₀F only when the field also overwhelms the 10.97 GHz fine structure

3. (c)

   Determine Neglecting 2p₃/₂, determine the energies of the two m_j = +½ states as functions of F, measured from the midpoint between 2s₁/₂ and 2p₁/₂, and the field at which the coupling equals half the Lamb shift.

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   1. 1

      in the basis {|2s₁/₂, +½⟩, |2p₁/₂, +½⟩} the Hamiltonian is [[L/2, V], [V, −L/2]], with V = √3 ea₀F

   2. 2

      E± = ±√(L²/4 + 3e²a₀²F²): for √3 ea₀F ≪ L the 2s-like level rises quadratically, by 3e²a₀²F²/L, and for √3 ea₀F ≫ L the shifts become linear, ±√3 ea₀F

   3. 3

      V = L/2 at F = L/(2√3 ea₀) = (6.626 × 10⁻³⁴ × 1057.8 × 10⁶ J)/(2√3 × 1.602 × 10⁻¹⁹ C × 5.29 × 10⁻¹¹ m) = 2.39 × 10⁴ V m⁻¹, about 240 V cm⁻¹

4. (d)

   Calculate The 2s-like state decays through the 2p₁/₂ component that the field mixes into it. Calculate its lifetime in a field of 2.0 V cm⁻¹.

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   1. 1

      with F = 200 V m⁻¹, V/h = √3 ea₀F/h = √3 × 1.602 × 10⁻¹⁹ × 5.29 × 10⁻¹¹ × 200/(6.626 × 10⁻³⁴) Hz = 4.43 MHz

   2. 2

      the admixed 2p probability is (V/L)² = (4.43/1057.8)² = 1.75 × 10⁻⁵, so the decay rate is 1.75 × 10⁻⁵/(1.60 × 10⁻⁹ s) = 1.10 × 10⁴ s⁻¹, beside which the field-free rate 1/(0.122 s) = 8.2 s⁻¹ is negligible

   3. 3

      τ = 1/(1.10 × 10⁴ s⁻¹) = 91 μs, about 1300 times shorter than the field-free lifetime: a field of a few V cm⁻¹ quenches the metastable state

5. (e)

   Justify Justify the neglect in (d) of the 2p₃/₂ level and of the natural width of the 2p state.

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   1. 1

      |2p₃/₂, +½⟩ = √(1/3)|m_l = 1⟩|↓⟩ + √(2/3)|m_l = 0⟩|↑⟩, orthogonal to the 2p₁/₂ state, is coupled by √6 ea₀F, twice as strongly in square, but it lies 10.97 − 1.06 = 9.91 GHz from 2s₁/₂, so its admixture is 2 × (1.058/9.91)² ≈ 0.023 of the 2p₁/₂ one: it adds about 2% to the rate

   2. 2

      the 2p natural width is Γ/2π = 1/(2π × 1.60 ns) = 99.5 MHz; including it replaces L² by L² + (ℏΓ/2)², a fractional change of (49.7/1057.8)² ≈ 0.2%, so neither neglect changes the 91 μs by more than a few per cent

A heavy quark and antiquark bound in a meson are modelled non-relativistically as one particle of reduced mass μ, with μc² = 0.75 GeV, in the linear confining potential V(r) = σr, with σ = 0.90 GeV fm⁻¹. Only l = 0 is considered. The exact ground-state energy is E₀ = 2.3381ε₀, where ε₀ = (ℏ²σ²/2μ)^(1/3) and −2.3381 is the first zero of the Airy function Ai; for any s state of this potential it can also be shown that |ψ(0)|² = μσ/(2πℏ²) exactly. Take ℏc = 0.1973 GeV fm, and ∫₀^∞ rⁿ e^(−βr) dr = n!/βⁿ⁺¹.

1. (a)

   Show that Show that for the normalised trial function ψ_β(r) = √(β³/π) e^(−βr), ⟨H⟩ = ℏ²β²/(2μ) + 3σ/(2β).

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   1. 1

      ⟨T⟩ = (ℏ²/2μ)∫|∇ψ|² d³r, and |∇ψ|² = β²|ψ|² for this function, so ⟨T⟩ = ℏ²β²/(2μ) (equivalently, from the full radial Laplacian with the integrals given)

   2. 2

      ⟨r⟩ = (β³/π) × 4π∫₀^∞ r³e^(−2βr) dr = 4β³ × 3!/(2β)⁴ = 3/(2β), so ⟨V⟩ = 3σ/(2β)

2. (b)

   Determine Determine the value of β that minimises ⟨H⟩, and show that the resulting bound is E_exp = 3(3/4)^(2/3)ε₀ = 2.476ε₀.

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   1. 1

      d⟨H⟩/dβ = ℏ²β/μ − 3σ/(2β²) = 0 gives β³ = 3μσ/(2ℏ²)

   2. 2

      at the minimum 3σ/(2β) = ℏ²β²/μ, twice the kinetic term (the virial theorem for a linear potential), so ⟨H⟩ = 3ℏ²β²/(2μ)

   3. 3

      substituting β² = (3μσ/2ℏ²)^(2/3) gives E_exp = 3(ℏ²/2μ)^(1/3)(3σ/4)^(2/3) = 3(3/4)^(2/3)ε₀ = 2.476ε₀

3. (c)

   Calculate Calculate ε₀, the exponential bound and the exact ground-state energy in GeV, and the percentage by which the bound exceeds the exact value.

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   1. 1

      ε₀ = [(ℏc)²σ²/(2μc²)]^(1/3) = [(0.1973 GeV fm)² × (0.90 GeV fm⁻¹)²/(1.50 GeV)]^(1/3) = 0.276 GeV

   2. 2

      E_exp = 2.476 × 0.276 GeV = 0.683 GeV against E₀ = 2.3381 × 0.276 GeV = 0.645 GeV: the bound is 5.9% high, and above the exact value, as it must be

4. (d)

   Explain A Gaussian trial function e^(−αr²), optimised in the same way, gives 2.3448ε₀, within 0.3% of the exact value, although an exponential is the exact ground state of hydrogen. Explain why the Gaussian does so much better in this potential.

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   1. 1

      for a potential that is finite at r = 0, the s-wave equation gives u = rψ = c₁r + c₃r³ + … with no r² term, so ψ has zero slope at the origin; the exponential's cusp, ψ ≈ 1 − βr, is right only where a 1/r singularity balances it, and here it costs kinetic energy for nothing, whereas the Gaussian is smooth at the origin

   2. 2

      the energy is decided by the shape of ψ where |ψ|²r² is large; the exact function decays as exp[−(2/3)(r/l)^(3/2)] with l = (ℏ²/2μσ)^(1/3), between the two trials, and the Gaussian's smooth, rounded peak matches the bulk of the exact function far better than the exponential's cusp

5. (e)

   Evaluate The meson's decay rate to a lepton pair is proportional to |ψ(0)|². The optimal Gaussian trial, ψ = (2α/π)^(3/4) e^(−αr²), has α^(3/2) = (μσ/3ℏ²)√(2/π). A student proposes using it to predict |ψ(0)|², arguing that an energy within 0.3% of the exact value shows the trial state to be accurate. Evaluate this proposal.

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   1. 1

      |ψ(0)|² = (2α/π)^(3/2) = (2/π)^(3/2)(μσ/3ℏ²)√(2/π) = 4μσ/(3π²ℏ²)

   2. 2

      against the exact μσ/(2πℏ²) the ratio is 8/(3π) = 0.85, so the Gaussian predicts a decay rate 15% too low; the exponential trial, with β³ = 3μσ/(2ℏ²), gives β³/π = 3μσ/(2πℏ²), three times too high

   3. 3

      the proposal should be rejected: ⟨H⟩ is stationary under variations of ψ, so a first-order error in ψ gives only a second-order error in E, while |ψ(0)|² has no such protection and is wrong at first order; an accurate energy certifies an upper bound on E, not the wavefunction, least of all its value at one point, where the weight r²|ψ|² contributes nothing to E

For spontaneous emission by an electron from an upper state |b⟩ to a lower state |a⟩ at angular frequency ω, Einstein's detailed-balance argument with the Planck spectrum gives A = ω³|⟨a|er|b⟩|²/(3πε₀ℏc³). The absorption oscillator strength from a non-degenerate state |a⟩ to a level b is f = (2mₑω/3ℏ)Σ|⟨b|r|a⟩|², summed over the degenerate states of b, and for one electron the Thomas-Reiche-Kuhn sum rule gives Σf = 1 over all levels b. As a model, take an electron bound in the isotropic three-dimensional harmonic potential ½mₑω²r², with states |n_x, n_y, n_z⟩ and ⟨0|x|1⟩ = √(ℏ/2mₑω) for each Cartesian direction. Take e = 1.602 × 10⁻¹⁹ C, mₑ = 9.109 × 10⁻³¹ kg, ε₀ = 8.854 × 10⁻¹² F m⁻¹, c = 2.998 × 10⁸ m s⁻¹, k_B = 1.381 × 10⁻²³ J K⁻¹ and 1 u = 1.661 × 10⁻²⁷ kg.

1. (a)

   Explain Explain why an atom treated by perturbation theory in a classical field would, in an excited state with no applied field, never decay, and how Einstein's argument nevertheless fixes A.

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   1. 1

      with no applied field the perturbation er ⋅ E is zero, and the excited state is an eigenstate of the atomic Hamiltonian, so it is stationary: the semiclassical theory has no spontaneous emission

   2. 2

      in equilibrium with thermal radiation, absorption at the rate Bρ(ω) must balance stimulated emission plus spontaneous emission, A + Bρ(ω), between Boltzmann-populated levels; with Planck's ρ(ω) this forces A = (ℏω³/π²c³)B, and B is computed from the golden rule in a classical field (the quantum-electrodynamic derivation, which gives the same A, is quoted)

2. (b)

   Show that Show that for the oscillator the rate from |1, 0, 0⟩ to |0, 0, 0⟩ is A = e²ω²/(6πε₀mₑc³), and comment on the absence of ℏ.

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   1. 1

      only the x component connects the two states: |⟨0,0,0|r|1,0,0⟩|² = |⟨0|x|1⟩|² = ℏ/(2mₑω), the y and z components vanishing because they would have to change n_y or n_z

   2. 2

      A = ω³e²(ℏ/2mₑω)/(3πε₀ℏc³) = e²ω²/(6πε₀mₑc³); ℏ cancels, and the result is exactly the classical radiation-damping rate of a charged oscillator, as the correspondence principle requires

3. (c)

   Determine Determine the absorption oscillator strength from |0, 0, 0⟩ to the first excited level, and state what the Thomas-Reiche-Kuhn sum rule then implies for every other transition from the ground state.

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   1. 1

      the first excited level has three states, |1,0,0⟩, |0,1,0⟩ and |0,0,1⟩, each with |⟨b|r|0,0,0⟩|² = ℏ/(2mₑω), so f = (2mₑω/3ℏ) × 3ℏ/(2mₑω) = 1

   2. 2

      this one transition exhausts the sum rule, so every other transition from the ground state has f = 0, consistent with the dipole operator changing a single quantum number by one

4. (d)

   Calculate The sodium resonance lines near 589.0 nm (3s to 3p) have statistical weights in the same 1 : 3 ratio as the oscillator's, so for them A = f × e²ω²/(6πε₀mₑc³). The measured 3p lifetime is 16.2 ns. Calculate the oscillator's lifetime at 589.0 nm, and hence the oscillator strength of the sodium resonance transition.

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   1. 1

      ω = 2πc/λ = 2π × 2.998 × 10⁸ m s⁻¹/(589.0 × 10⁻⁹ m) = 3.198 × 10¹⁵ rad s⁻¹

   2. 2

      A = e²ω²/(6πε₀mₑc³) = 6.41 × 10⁷ s⁻¹, so the oscillator lifetime is 15.6 ns

   3. 3

      f = A_Na/A_osc = 15.6 ns/16.2 ns = 0.96: the single 3s–3p transition carries about 96% of the one-electron sum, the rest going to 3s–np with n ≥ 4 and to the continuum

5. (e)

   Compare Compare the natural linewidth of the sodium line with its Doppler width in a vapour at 500 K, for which the full width at half maximum is Δν_D = (ν/c)√(8k_BT ln 2/M), with M = 22.99 u.

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   1. 1

      natural width Δν = 1/(2πτ) = 1/(2π × 16.2 × 10⁻⁹ s) = 9.8 MHz, a Lorentzian full width

   2. 2

      Δν_D = [1/(589.0 × 10⁻⁹ m)] × √(8 × 1.381 × 10⁻²³ J K⁻¹ × 500 K × ln 2/(22.99 × 1.661 × 10⁻²⁷ kg)) = (1.698 × 10⁶ m⁻¹)(1.00 × 10³ m s⁻¹) = 1.70 GHz, about 170 times wider, so the vapour line is Doppler-dominated and the natural width shows only in the far wings or in Doppler-free or cold-atom measurements

A superconducting qubit is driven close to resonance. In the frame rotating with the drive its Hamiltonian is H(φ) = (ℏ/2)[Δσ_z + Ω(cos φ σ_x + sin φ σ_y)], where the detuning is Δ/2π = 3.0 MHz, the Rabi frequency is Ω/2π = 4.0 MHz, and φ is the phase of the drive, which can be ramped. Equivalently H = (ℏΩ_eff/2) n̂ ⋅ σ, with Ω_eff = √(Δ² + Ω²) and n̂ = (sin θ cos φ, sin θ sin φ, cos θ). In the experiment φ is ramped at a constant rate from 0 to 2π in a time T = 4.0 μs, so that n̂ traces a circle on a cone about the z-axis.

1. (a)

   State State the eigenvalues of H, and the angle θ between n̂ and the z-axis.

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   1. 1

      E± = ±ℏΩ_eff/2 = ±(ℏ/2)√(Δ² + Ω²), that is ±h × 2.5 MHz, a gap of h × 5.0 MHz; the two levels meet only at Δ = Ω = 0

   2. 2

      cos θ = Δ/Ω_eff = 3/5 (sin θ = 4/5), so θ = 53.1°

2. (b)

   Show that Show that the adiabatic condition for the ramp becomes T ≫ πΩ/(Δ² + Ω²), and evaluate the right-hand side.

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   1. 1

      dH/dt = (2π/T)∂H/∂φ = (2π/T)(ℏΩ/2)(−sin φ σ_x + cos φ σ_y); the unit vector (−sin φ, cos φ, 0) is perpendicular to n̂, so its matrix element between the eigenstates along ±n̂ has magnitude 1, and |⟨−|dH/dt|+⟩| = πℏΩ/T

   2. 2

      the condition ℏ|⟨−|dH/dt|+⟩| ≪ (E₊ − E₋)² becomes πℏ²Ω/T ≪ ℏ²(Δ² + Ω²), so T ≫ πΩ/(Δ² + Ω²) = π(2π × 4.0 × 10⁶ s⁻¹)/(2π × 5.0 × 10⁶ s⁻¹)² = 80 ns, and the 4.0 μs ramp is fifty times longer

3. (c)

   Calculate Calculate the geometric phase acquired in one circuit by each of the two eigenstates, and the difference between their dynamic phases.

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   1. 1

      n̂ sweeps a cone of half-angle θ, enclosing a solid angle 2π(1 − cos θ) = 2π × 2/5 = 0.8π sr

   2. 2

      the geometric phase is minus half the solid angle, −0.4π rad (−72°), for the state along +n̂ (the upper level) and +0.4π rad for the state along −n̂, so the two differ by 0.8π

   3. 3

      the dynamic phases differ by Ω_eff T = 2π × 5.0 MHz × 4.0 μs = 40π rad, twenty full turns, proportional to both T and Ω_eff: a 1% error in Ω_eff shifts it by 0.4π, as much as the geometric phase itself

4. (d)

   Predict In a spin-echo version, the qubit starts in an equal superposition of the two eigenstates at φ = 0, is taken once round the circuit, has its two eigenstates exchanged by a π pulse, and is then taken round the circuit in the opposite sense in the same time T. Predict the relative phase between the two components at the end, and how it changes if T is doubled or if Δ/2π is raised to 6.0 MHz with Ω unchanged.

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   1. 1

      each component spends time T in each eigenstate, so the dynamic phases cancel, while exchanging the eigenstates and reversing the sense of the circuit each reverse the sign of a component's geometric phase, so the geometric phases add: the relative phase is 2 × 0.8π = 1.6π rad (288°, equivalently −0.4π), unchanged when T is doubled as long as the ramp stays adiabatic

   2. 2

      with Δ/2π = 6.0 MHz, cos θ = 6/√52 = 0.832 and the relative phase becomes 4π(1 − 0.832) = 0.67π rad (121°): it depends only on the geometry of the circuit

5. (e)

   Explain The same circuit also bounds the rest of the sphere, a cap of solid angle 4π minus the one used in (c). Explain why that cap gives the same phase factor, and what this has to do with the degeneracy of H at Δ = Ω = 0.

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   1. 1

      seen from the other cap the circuit runs in the opposite sense, so for the +n̂ state γ = +½(4π − 0.8π) = 2π − 0.4π, which equals −0.4π modulo 2π: e^(iγ) is the same whichever surface is used

   2. 2

      the Berry curvature is the field of a monopole at the degeneracy, the origin of the space of (Ω cos φ, Ω sin φ, Δ), where the two eigenvalues meet; its total flux for a spin-half is 2π in magnitude, a multiple of 2π, which is why the phase of a closed circuit, the flux through it, is unambiguous

A hydrogen maser oscillates on the transition from F = 1, m_F = 0 to F = 0 in the hyperfine structure of the ground state, whose zero-field frequency is ν_hfs. A small static field B along z is applied by a solenoid around the storage bulb, but the ambient field is not fully cancelled. The field inside the bulb is measured by the Zeeman frequency f_Z, the low-frequency transition between neighbouring F = 1 sublevels, which in these weak fields is f_Z = μ_B B/h. For each solenoid current the maser's output is compared with a reference oscillator, giving a frequency offset δν that also contains a constant but unknown contribution from the reference and from collisions with the bulb wall. With g_e = 2 and the proton's magnetic moment neglected, the ground-state Hamiltonian is H = A S_e ⋅ S_p/ℏ² + (2μ_B B/ℏ)S_ez, where A = hν_hfs. Take μ_B = 9.274 × 10⁻²⁴ J T⁻¹ and h = 6.626 × 10⁻³⁴ J s.

1. (a)

   Show that Working in the m_F = 0 subspace spanned by |↑⇓⟩ and |↓⇑⟩ (electron spin first), show that the frequency of the F = 1, m_F = 0 to F = 0 transition is ν_c = √(ν_hfs² + 4f_Z²) ≈ ν_hfs + 2f_Z²/ν_hfs.

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   1. 1

      S_e ⋅ S_p = S_ez S_pz + ½(S_e₊ S_p₋ + S_e₋ S_p₊) has diagonal elements −ℏ²/4 and off-diagonal elements ℏ²/2 in this subspace, while the Zeeman term adds +μ_B B to |↑⇓⟩ and −μ_B B to |↓⇑⟩

   2. 2

      the matrix [[−A/4 + μ_B B, A/2], [A/2, −A/4 − μ_B B]] has eigenvalues −A/4 ± √(A²/4 + μ_B²B²), which at B = 0 are A/4 (F = 1) and −3A/4 (F = 0)

   3. 3

      hν_c = 2√(A²/4 + μ_B²B²), so ν_c = √(ν_hfs² + 4f_Z²) ≈ ν_hfs + 2f_Z²/ν_hfs when f_Z ≪ ν_hfs: the clock transition has no term linear in B

2. (b)

   Describe Describe how the readings should be plotted to test this model and to extract ν_hfs, and explain why the solenoid current should not be used as the measure of the field.

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   1. 1

      plot δν against f_Z²: the model predicts a straight line δν = c + (2/ν_hfs)f_Z², the unknown constant offset c appearing only in the intercept and the gradient being 2/ν_hfs

   2. 2

      B is not proportional to I, because the uncancelled ambient field adds to the solenoid's (here f_Z = 0.10 kHz + 0.30 kHz mA⁻¹ × I, a residual of about 7 nT), and the solenoid's calibration inside the bulb is not known; f_Z measures the field the atoms actually experience

3. (c)

   Determine Determine the gradient of your graph and hence a value of ν_hfs, and comment on its agreement with the accepted 1420.406 MHz.

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   1. 1

      f_Z² = 0.160, 0.490, 1.00, 1.69, 2.56 and 3.61 kHz², and the six points lie on a straight line with intercept c ≈ 3.50 mHz

   2. 2

      gradient = 1.41 mHz kHz⁻² (accept 1.39 to 1.42 mHz kHz⁻²), which is 1.41 × 10⁻⁹ Hz⁻¹

   3. 3

      ν_hfs = 2/gradient = 2/(1.41 × 10⁻⁹ Hz⁻¹) = 1.42 × 10⁹ Hz (accept 1.41 to 1.44 GHz), agreeing with 1420.406 MHz within the half-per-cent uncertainty of the gradient, which confirms the quadratic Zeeman model; the neglected proton moment would change the gradient by only about 0.6%

4. (d)

   Estimate The maser is normally operated at f_Z = 0.80 kHz. Estimate how constant f_Z, and hence the field, must be kept if field fluctuations are to change the maser frequency by less than 1 part in 10¹⁵, stating your assumptions.

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   1. 1

      dν_c/df_Z = 4f_Z/ν_hfs = 4 × 800 Hz/(1.4204 × 10⁹ Hz) = 2.25 × 10⁻⁶, and the tolerance 10⁻¹⁵ × 1.42 × 10⁹ Hz = 1.4 × 10⁻⁶ Hz allows a change in f_Z of about 0.63 Hz

   2. 2

      that is a fractional stability of δf_Z/f_Z ≈ 8 × 10⁻⁴: the field of about 57 nT must be held to about 0.05 nT, assuming the quadratic law of (a) holds and every other frequency shift stays constant

A particle of mass m and energy E = ℏ²k²/2m is scattered elastically by a potential V(r) that vanishes at large r. To apply Fermi's golden rule, the system is placed in a cubic box of side L with periodic boundary conditions, so that the plane-wave states ⟨r|k⟩ = e^(ik⋅r)/L^(3/2) are normalised and the incident state carries a probability current density ℏk/(mL³). The golden rule gives the rate Γ = (2π/ℏ)|⟨f|V|i⟩|²ρ(E) into a group of final states whose density per unit energy is ρ(E). Write Ṽ(q) = ∫e^(−iq⋅r)V(r) d³r. Take mₑ = 9.109 × 10⁻³¹ kg, e = 1.602 × 10⁻¹⁹ C, c = 2.998 × 10⁸ m s⁻¹ and α = 1/137.0.

1. (a)

   Derive Derive, from the periodic boundary conditions, the density of final states per unit energy ρ(E) for plane waves whose wavevectors point into a small solid angle dΩ.

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   1. 1

      periodic boundary conditions allow k = (2π/L)(n_x, n_y, n_z) with integer n_x, n_y and n_z, one state for each volume (2π/L)³ = (2π)³/L³ of k-space

   2. 2

      the states with wavevectors in a shell k to k + dk inside dΩ number dN = L³k² dk dΩ/(2π)³

   3. 3

      with E = ℏ²k²/2m, dE = ℏ²k dk/m, so ρ(E) = dN/dE = L³mk dΩ/(8π³ℏ²)

2. (b)

   Derive Using the golden rule, derive the differential cross-section dσ/dΩ for elastic scattering in terms of Ṽ(q), where q = k_f − k_i, and confirm that it does not depend on L.

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   1. 1

      ⟨k_f|V|k_i⟩ = L⁻³∫e^(−i(k_f − k_i)⋅r)V(r) d³r = Ṽ(q)/L³, with |k_f| = |k_i| = k by energy conservation

   2. 2

      the rate into dΩ is dΓ = (2π/ℏ)(|Ṽ|²/L⁶)(L³mk dΩ/8π³ℏ²) = mk|Ṽ(q)|² dΩ/(4π²ℏ³L³)

   3. 3

      dividing by the incident current density ℏk/(mL³) gives dσ/dΩ = (m/2πℏ²)²|Ṽ(q)|², in which L³ has cancelled between the matrix element, the density of states and the flux

3. (c)

   Show that For the screened Coulomb potential V(r) = (g/r)e^(−r/a), show that Ṽ(q) = 4πg/(q² + 1/a²).

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   1. 1

      for a central potential, integrating over the angle between q and r with ∫₋₁¹ e^(−iqru) du = 2 sin(qr)/(qr) gives Ṽ(q) = (4π/q)∫₀^∞ rV(r) sin(qr) dr

   2. 2

      Ṽ(q) = (4πg/q)∫₀^∞ e^(−r/a) sin(qr) dr = (4πg/q) × q/(q² + 1/a²) = 4πg/(q² + 1/a²)

4. (d)

   Determine Take the limit a → ∞ with g = Z₁Z₂e²/(4πε₀). Determine dσ/dΩ as a function of the scattering angle θ, and compare it with Rutherford's classical result, dσ/dΩ = g²/(16E² sin⁴(θ/2)).

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   1. 1

      for elastic scattering through θ, q = |k_f − k_i| = 2k sin(θ/2), so dσ/dΩ = (m/2πℏ²)²(4πg)²/q⁴ = 4m²g²/(ℏ⁴q⁴) = m²g²/(4ℏ⁴k⁴ sin⁴(θ/2))

   2. 2

      with ℏ²k² = 2mE this is g²/(16E² sin⁴(θ/2)), identical to Rutherford's result; the total cross-section diverges at small θ, which a finite screening length a removes

5. (e)

   Explain The first Born amplitude, f(θ) = −(m/2πℏ²)Ṽ(q), is real for this potential, so Im f(0) = 0, yet the total cross-section it gives is positive. Explain why this does not contradict the optical theorem, σ = (4π/k) Im f(0).

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   1. 1

      the Born cross-section is of second order in V, so the optical theorem relates it to the imaginary part of the forward amplitude at second order; the first-order amplitude is real for a real potential, and Im f(0) first appears in the second Born term

   2. 2

      the theorem holds order by order in V: the first Born approximation conserves probability only to first order, and its failure to satisfy the optical theorem on its own is a statement about the order of the approximation, not a violation of unitarity

6. (f)

   Estimate For a Coulomb-like potential of strength g = Ze²/(4πε₀), the first Born approximation is reliable when g/(ℏv) = Zαc/v ≪ 1, where v is the speed of the incident particle. Estimate whether the Born cross-section can be trusted for electrons of kinetic energy 2.0 keV scattered by neutral helium (Z = 2) and by neutral iron (Z = 26), stating your assumptions.

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   1. 1

      v = √(2E/mₑ) = √(2 × 2.0 × 10³ × 1.602 × 10⁻¹⁹ J/9.109 × 10⁻³¹ kg) = 2.65 × 10⁷ m s⁻¹, so v/c = 0.088 and non-relativistic kinematics are adequate

   2. 2

      Zαc/v = Z × (1/137.0)/0.0885 = 0.0825Z, which is 0.17 for helium and 2.1 for iron

   3. 3

      the Born cross-section is adequate for helium, with corrections of order Zαc/v, about 17%, but fails for iron, where partial waves or an exact treatment are needed; the estimate assumes that the scattering that matters happens close to the nucleus, where screening is weak, and ignores exchange with, and excitation of, the atomic electrons

A particle of mass m moves in one dimension in the potential V(x) = ½mω²x²(1 − x/L), the model used for the phase of a current-biased Josephson junction. V has a minimum at x = 0 and a barrier beyond which, for x > L, it falls without limit, so the ground state of the well is metastable and escapes by tunnelling. For the numerical parts take ω/2π = 8.0 GHz, with L such that the barrier height is V_b = 3.0ℏω. A full treatment, matching the Gaussian ground state of the well onto the decaying WKB solution under the barrier, gives the escape rate Γ = (ω/2π)√(864πV_b/ℏω) exp(−36V_b/5ℏω), which is quoted here. You may use ∫₀¹ u√(1 − u) du = 4/15. Take ℏ = 1.055 × 10⁻³⁴ J s and k_B = 1.381 × 10⁻²³ J K⁻¹.

1. (a)

   Show that Show that the top of the barrier lies at x = 2L/3 and that V_b = (2/27)mω²L².

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   1. 1

      dV/dx = mω²(x − 3x²/2L) = mω²x(1 − 3x/2L), which vanishes at x = 0, the minimum, and at x = 2L/3, the maximum

   2. 2

      V(2L/3) = ½mω²(4L²/9)(1/3) = (2/27)mω²L²

2. (b)

   Sketch Sketch V(x) from x = −L/3 to x = 4L/3, marking V_b, the ground-state energy ½ℏω and the classical turning points at that energy, and on the same axes sketch the magnitude of the metastable ground-state wavefunction.

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   1. 1

      V: a parabolic minimum at x = 0 with V rising for x < 0, a maximum V_b at x = 2L/3, back to zero at x = L and falling without limit beyond; the line E = ½ℏω = V_b/6 meets V at two points inside the well, near x ≈ −0.15L and x ≈ 0.17L, and once more on the outer slope just inside x = L (near 0.97L)

   2. 2

      |ψ|: a Gaussian-like hump centred in the well, decaying roughly exponentially through the barrier by several orders of magnitude, then emerging beyond the outer turning point as a small outgoing oscillatory wave that does not decay, with a wavelength that shortens as V falls

3. (c)

   Derive Starting from the WKB tunnelling exponent 2S/ℏ = (2/ℏ)∫√(2m(V − E)) dx taken across the barrier, derive its value for an energy E → 0 at the bottom of the well, and express it in terms of V_b/ℏω.

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   1. 1

      at E → 0 the turning points bounding the barrier are x = 0 and x = L, where V = 0, and between them √(2mV) = mωx√(1 − x/L)

   2. 2

      ∫₀^L mωx√(1 − x/L) dx = mωL²∫₀¹ u√(1 − u) du = (4/15)mωL², so 2S/ℏ = (8/15)mωL²/ℏ

   3. 3

      from (a), mω²L² = (27/2)V_b, so mωL²/ℏ = (27/2)V_b/ℏω and 2S/ℏ = (8/15)(27/2)V_b/ℏω = 36V_b/(5ℏω) = 7.2V_b/ℏω, the exponent of the quoted result

4. (d)

   Calculate Calculate, for V_b = 3.0ℏω, the naive escape rate Γ₀ = (ω/2π)exp(−2S/ℏ), which pictures the particle striking the barrier once per classical period, and the quoted rate Γ.

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   1. 1

      2S/ℏ = 7.2 × 3.0 = 21.6 and e^(−21.6) = 4.2 × 10⁻¹⁰, so Γ₀ = 8.0 × 10⁹ s⁻¹ × 4.2 × 10⁻¹⁰ = 3.3 s⁻¹

   2. 2

      √(864π × 3.0) = 90, so Γ = 90 × 3.3 s⁻¹ = 3.0 × 10² s⁻¹, a lifetime of 3.3 ms: the naive estimate is too low by a factor of about 90

5. (e)

   Evaluate Evaluated between its turning points at the actual ground-state energy E = ½ℏω, the barrier integral gives 2S/ℏ = 17.15 for these parameters. Evaluate the naive estimate in the light of this: identify which of its approximations causes most of the discrepancy, and explain why the difficulty is worst for the lowest state.

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   1. 1

      (ω/2π)e^(−17.15) = 8.0 × 10⁹ s⁻¹ × 3.6 × 10⁻⁸ = 2.9 × 10² s⁻¹, within about 5% of the quoted 3.0 × 10² s⁻¹

   2. 2

      so most of the factor of 90 comes from setting E = 0 in the exponent: E₀ = ½ℏω = V_b/6 is not small compared with V_b, and lowering the exponent by 4.45 multiplies the rate by about 86; the √(V_b/ℏω) in the quoted prefactor is this energy dependence of the exponent, expanded about E = 0

   3. 3

      near the bottom of the well the classically allowed region is only about one zero-point width across, so neither the WKB wavefunction nor the classical attempt frequency is valid there, and the slope of the exponent, d(2S/ℏ)/dE, grows logarithmically as E falls towards the bottom of the well, so the exponent cannot simply be taken at E = 0; for highly excited states the Gamow form with the true energy works, but the ground state needs its Gaussian matched onto the WKB tail

6. (f)

   Estimate Thermal activation over the barrier occurs at a rate of order (ω/2π)exp(−V_b/k_BT). Estimate the temperature below which escape is dominated by tunnelling, stating your assumptions, and say what an experiment would observe below it.

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      comparing exponents, and assuming the prefactors are of comparable size, tunnelling wins when V_b/k_BT > 7.2V_b/ℏω, that is below k_BT* = ℏω/7.2

   2. 2

      ℏω = 1.055 × 10⁻³⁴ J s × 2π × 8.0 × 10⁹ s⁻¹ = 5.30 × 10⁻²⁴ J, so ℏω/k_B = 0.384 K and T* ≈ 53 mK (a fuller treatment gives ℏω/(2πk_B) = 61 mK, the same order)

   3. 3

      below T* the escape rate stops falling with temperature and levels off at the tunnelling rate, the signature of macroscopic quantum tunnelling; at 53 mK, k_BT ≈ 0.14ℏω, so the excited levels of the well are scarcely populated, justifying the ground-state treatment, and dissipation is neglected throughout