University Quantum Mechanics I · Final Examination

ψ(x) is a normalised wavefunction for a particle moving along the x-axis. A second state of the particle has a position probability density of the same shape as |ψ(x)|², but compressed by a factor of 2 towards x = 0, so that every feature of the density lies at half its original distance from x = 0. Which wavefunction describes the second state?

1. Aψ(2x)
2. B2ψ(2x)
3. C√2 ψ(x/2)
4. D√2 ψ(2x)

Each option refers to a particle of mass m in one dimension. Which pair of operators commutes, so that the two observables are compatible and can share a complete set of eigenfunctions?

1. Ap̂ and Ĥ = p̂²/2m, for a free particle
2. Bx̂ and p̂
3. Cp̂ and Ĥ = p̂²/2m + ½mω²x̂², for a harmonic oscillator
4. Dx̂ and Ĥ = p̂²/2m, for a free particle

The simplest model of a wave packet for a free neutron (m = 1.675×10⁻²⁷ kg) is the sum of two plane waves, Ψ(x,t) = e^(i(k₁x − ω₁t)) + e^(i(k₂x − ω₂t)), with k₁ = 2.90×10¹⁰ m⁻¹ and k₂ = 3.10×10¹⁰ m⁻¹, and each frequency given by the free-particle dispersion relation ω = ℏk²/(2m). The probability density |Ψ(x,t)|² is then a row of fringes about 3.1 nm apart. Take ℏ = 1.055×10⁻³⁴ J s. At what speed does the fringe pattern of |Ψ(x,t)|² move?

1. A9.4×10² m s⁻¹
2. B1.9×10³ m s⁻¹
3. C3.8×10³ m s⁻¹
4. D1.2×10⁴ m s⁻¹

An F-centre is an electron trapped at the site of a missing negative ion in an alkali-halide crystal. Along one axis, model it as an electron (m = 9.109×10⁻³¹ kg) in a symmetric finite square well of total width 0.56 nm and depth V₀. In this model the centre's optical absorption lifts the electron from the ground state to the first excited bound state, so the well must bind at least two states. Take ℏ = 1.055×10⁻³⁴ J s and 1 eV = 1.602×10⁻¹⁹ J. What is the smallest depth V₀ for which the well binds a second state?

1. AAny depth greater than zero
2. B0.30 eV
3. C1.2 eV
4. D4.8 eV

An electron with kinetic energy E meets a rectangular barrier of height V₀ > E and width w. The thick-barrier estimate T ≈ exp(−2κw), with κ = √(2m(V₀ − E))/ℏ and the prefactor ignored, gives T = 1.0×10⁻³. A proton with the same kinetic energy meets the same barrier. Taking m_p/m_e = 1836, what does the same estimate give for the proton's transmission probability?

1. A2.3×10⁻⁵
2. B5.4×10⁻⁷
3. C3×10⁻¹²⁹
4. D10⁻⁵⁵⁰⁸

For a one-dimensional harmonic oscillator with energy eigenstates |n⟩, the position operator can be written x̂ = √(ℏ/2mω)(â + â†), where â|n⟩ = √n |n−1⟩ and â†|n⟩ = √(n+1) |n+1⟩. Which of the following matrix elements is non-zero?

1. A⟨3|x̂|2⟩
2. B⟨2|x̂|2⟩
3. C⟨0|x̂|2⟩
4. D⟨4|x̂|2⟩

An observable  is represented in one orthonormal basis by the Hermitian matrix A, and in a second orthonormal basis by A′ = U†AU, where U is the unitary matrix of overlaps between the two bases. Which of the following is in general different for A′ than for A?

1. AThe trace of the matrix
2. BThe set of eigenvalues
3. CThe expectation value ⟨ψ|Â|ψ⟩ for a given state, computed from that state's components in the same basis as the matrix
4. DThe first diagonal element, A₁₁

A two-state system with orthonormal basis states |1⟩ and |2⟩ is prepared in |ψ⟩ = (|1⟩ + i|2⟩)/√2. Which of the following is the same physical state, giving identical probabilities for every possible measurement?

1. A(|1⟩ − i|2⟩)/√2
2. B(i|1⟩ − |2⟩)/√2
3. C(|1⟩ + |2⟩)/√2
4. D(i|1⟩ + |2⟩)/√2

Near the centre of an optical trap, an atom of mass m moves in the potential V(x, y, z) = ½mω²(x² + y²) + ½m(2ω)²z², so that its classical oscillation frequency along z is twice that along x and y. How many independent energy eigenstates share the energy E = 5ℏω?

1. A4
2. B6
3. C10
4. D12

The lithium ion Li²⁺ (Z = 3) has a single electron. The energy levels of hydrogen are E_n = −13.6 eV/n². How much energy is needed to ionise Li²⁺ from its first excited state?

1. A3.40 eV
2. B10.2 eV
3. C30.6 eV
4. D122 eV

A single ⁸⁷Rb atom of mass m = 1.443×10⁻²⁵ kg moves along an optical box trap whose repulsive light walls are steep and high enough to be treated as an infinite square well on 0 < x < L, with L = 20.0 μm. The stationary states are ψ_n(x) = √(2/L) sin(nπx/L), with energies E_n = n²E₁. At t = 0 the atom is prepared near the left wall in the state Ψ(x,0) = (1/√2)ψ₁(x) + (1/2)ψ₂(x) + (1/2)ψ₃(x), for which ⟨x⟩ = 0.275L. Take ℏ = 1.055×10⁻³⁴ J s, h = 6.626×10⁻³⁴ J s and k_B = 1.381×10⁻²³ J K⁻¹.

1. (a)

   Calculate Calculate the atom's zero-point energy E₁, and express it as a temperature E₁/k_B.

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   1. 1

      E₁ = π²ℏ²/(2mL²) = π²(1.055×10⁻³⁴)²/(2 × 1.443×10⁻²⁵ × (20.0×10⁻⁶)²) = 9.52×10⁻³⁴ J

   2. 2

      E₁/k_B = 9.52×10⁻³⁴/1.381×10⁻²³ = 6.9×10⁻¹¹ K, about 69 pK

2. (b)

   State State the possible results of a measurement of the atom's energy and the probability of each, and state whether these probabilities change as time passes.

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   1. 1

      E₁, 4E₁ and 9E₁, with probabilities |1/√2|² = 1/2, |1/2|² = 1/4 and |1/2|² = 1/4, which sum to 1; no other energy can be obtained

   2. 2

      the probabilities do not change: each coefficient acquires only the phase factor e^(−iE_nt/ℏ), which leaves |c_n|² unchanged

3. (c)

   Calculate Every phase factor e^(−iE_nt/ℏ) in Ψ(x,t) returns to 1 at the same instant, after which the atom's state repeats exactly. Calculate the earliest such time.

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   1. 1

      Ψ(x,t) = Σ c_nψ_n(x)e^(−in²E₁t/ℏ); since every n² is an integer, all three factors first equal 1 together when E₁t/ℏ = 2π (the relative phases 3E₁t/ℏ and 8E₁t/ℏ have no common factor that would allow an earlier repeat)

   2. 2

      t = 2πℏ/E₁ = h/E₁ = 6.626×10⁻³⁴/9.52×10⁻³⁴ = 0.70 s

4. (d)

   Explain At t = h/(2E₁) the probability density is |Ψ(L − x, 0)|², the mirror image of the initial density about the centre of the trap. Explain why, and state ⟨x⟩ at that instant.

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   1. 1

      at t = h/(2E₁) each phase factor is e^(−iπn²) = (−1)^n, while reflection about the centre gives ψ_n(L − x) = (−1)^(n+1)ψ_n(x); so (−1)^nψ_n(x) = −ψ_n(L − x) for every n, and Ψ(x,t) = −Ψ(L − x, 0)

   2. 2

      an overall sign does not change |Ψ|², so the density is the initial one reflected about x = L/2: the atom is now near the right wall, with ⟨x⟩ = L − 0.275L = 0.725L, although no energy probability has changed

5. (e)

   Predict Suppose instead that the atom's energy is measured at t = 0 and measured again at t = h/(2E₁). Predict the probability that the two results agree, and whether the density at t = h/(2E₁) is then the mirror image of the initial density.

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   1. 1

      the first measurement leaves the atom in the single eigenstate ψ_n whose energy it found; that state is stationary, so the second measurement repeats the result with probability 1

   2. 2

      the density is then |ψ_n|², constant in time and symmetric about x = L/2 with ⟨x⟩ = L/2, so no mirror image of the initial density appears near the right wall: the measurement removed the other terms whose relative phases produced it

A beam of spin-½ atoms is prepared in the spin state χ = (1/√3)(1, 1 + i)ᵀ, written as a column in the S_z basis {|+z⟩, |−z⟩}. The spin operators are S_i = (ℏ/2)σ_i, where σ_x, σ_y and σ_z are the Pauli matrices.

1. (a)

   State State the three Pauli matrices σ_x, σ_y and σ_z in the S_z basis.

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   1. 1

      σ_x = [[0, 1], [1, 0]] and σ_z = [[1, 0], [0, −1]]

   2. 2

      σ_y = [[0, −i], [i, 0]], with −i in the upper right

2. (b)

   Calculate Calculate the probability that a measurement of S_z gives +ℏ/2, and the expectation value ⟨S_z⟩.

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   1. 1

      P(+ℏ/2) = |1/√3|² = 1/3 (and P(−ℏ/2) = |1 + i|²/3 = 2/3)

   2. 2

      ⟨S_z⟩ = (ℏ/2)(1/3 − 2/3) = −ℏ/6

3. (c)

   Determine Determine the probability that a measurement of S_x gives +ℏ/2.

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   1. 1

      the +ℏ/2 eigenspinor of S_x is |+x⟩ = (1/√2)(1, 1)ᵀ

   2. 2

      ⟨+x|χ⟩ = (1/√6)(1 + 1 + i) = (2 + i)/√6, with the bra's components conjugated (they are real here)

   3. 3

      P = |2 + i|²/6 = 5/6

4. (d)

   Determine The atoms pass through an S_x analyser, and only those leaving its +ℏ/2 port go on to an S_z analyser. Determine the fraction of the original atoms that leave the −ℏ/2 port of the S_z analyser, and give the reason it is not related to the 2/3 found in (b).

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   1. 1

      the atoms kept after the S_x analyser are in |+x⟩, and they are a fraction 5/6 of the original beam

   2. 2

      |⟨−z|+x⟩|² = 1/2, so the fraction leaving the −ℏ/2 port is (5/6)(1/2) = 5/12

   3. 3

      the S_x measurement replaces χ by |+x⟩ and erases the earlier S_z amplitudes; since [S_x, S_z] ≠ 0 no state has definite values of both, so the loss comes from the change of state, not from the magnet mechanically disturbing the atoms

Very slow electrons are strongly reflected by a crystal surface even though the crystal attracts them. Inside the crystal an electron's potential energy is lower than in the vacuum outside by U, the inner potential, so the surface at x = 0 is modelled as a downward potential step: V(x) = 0 for x < 0 (vacuum) and V(x) = −U for x > 0 (crystal). Electrons of energy E = 0.50 eV arrive from the vacuum, moving in the +x direction at normal incidence. Their stationary scattering state is ψ(x) = e^(ikx) − (2/3)e^(−ikx) for x < 0 and ψ(x) = (1/3)e^(iqx) for x > 0, where ℏk = √(2mₑE) and ℏq = √(2mₑ(E + U)), but the value of U is not known. The probability current is j = (ℏ/mₑ) Im(ψ* dψ/dx). Take mₑ = 9.109×10⁻³¹ kg, ℏ = 1.055×10⁻³⁴ J s and 1 eV = 1.602×10⁻¹⁹ J.

1. (a)

   Derive Stating the matching conditions that ψ must satisfy at x = 0, and why each holds here, derive the ratio q/k for this state and hence the inner potential U in electronvolts.

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   1. 1

      ψ is continuous at x = 0, because a jump in ψ would put the derivative of a delta function into ψ″, which no finite V can balance in the Schrödinger equation: the two sides give 1 − 2/3 = 1/3, so the amplitudes given satisfy this condition

   2. 2

      dψ/dx is continuous at x = 0, because the step in V is finite and so ψ″ has at most a finite jump: ik(1 + 2/3) = iq(1/3), giving q = 5k

   3. 3

      (q/k)² = (E + U)/E = 25, so U = 24E = 12.0 eV

2. (b)

   Show that Show that the probability current is the same on both sides of the surface, as conservation of probability requires in a stationary state.

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   1. 1

      for x < 0 the cross terms between e^(ikx) and e^(−ikx) contribute nothing to Im(ψ* dψ/dx), so j = (ℏk/mₑ)(1 − 4/9) = (5/9)ℏk/mₑ

   2. 2

      for x > 0, j = (ℏq/mₑ)(1/3)² = (5ℏk/mₑ)(1/9) = (5/9)ℏk/mₑ, equal to the current for x < 0, so no probability builds up at the surface

3. (c)

   Calculate A beam of these electrons carrying a current of 2.0 nA strikes the surface. Calculate the reflected and transmitted electron currents, and the speed of the transmitted electrons inside the crystal.

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   1. 1

      R = |−2/3|² = 4/9, so the reflected current is (4/9) × 2.0 nA = 0.89 nA

   2. 2

      T = (q/k)|t|² = 5 × (1/9) = 5/9, so the transmitted current is (5/9) × 2.0 nA = 1.1 nA, and R + T = 1

   3. 3

      inside the crystal v = ℏq/mₑ = 5√(2E/mₑ) = 5 × 4.19×10⁵ m s⁻¹ = 2.1×10⁶ m s⁻¹

4. (d)

   Calculate Calculate the electron energy at which the reflected fraction at this surface falls to 1.0%.

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   1. 1

      R = 0.010 requires |r| = (q − k)/(q + k) = 0.10, so q/k = 1.10/0.90 = 11/9

   2. 2

      (q/k)² = 1 + U/E = 121/81, giving E = 81U/40 = 24 eV (24.3 eV)

5. (e)

   Explain A classical electron crossing the surface would simply speed up, and none would be reflected. Explain why the quantum electron can be reflected by a drop in potential, and why far fewer electrons would be reflected if V fell from 0 to −U smoothly over several nanometres instead of abruptly.

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   1. 1

      at an abrupt step the wavenumber jumps from k to q = 5k; a transmitted wave alone would need 1 = t from continuity of ψ and k = qt from continuity of dψ/dx, which cannot both hold when q ≠ k, so a reflected wave is required: a wave effect, like the partial reflection of light where the refractive index changes suddenly, with no counterpart for a classical particle

   2. 2

      if V changes over a distance much larger than the reduced wavelength 1/k (0.28 nm in vacuum at 0.50 eV, and 1/q = 0.055 nm inside), the local wavelength adjusts gradually and ψ follows it without a mismatch at any point, so the reflected amplitude becomes exponentially small and is negligible for a change spread over several nanometres; reflection is appreciable only if V changes within a distance of order 1/k or less, a fraction of a nanometre

An electron shared between two neighbouring semiconductor quantum dots, L and R, is modelled as a two-state system in the basis {|L⟩, |R⟩} of states localised on one dot. Energies are measured from the mean of the two on-dot energies, and in that basis the Hamiltonian is H = [[ε/2, −τ], [−τ, −ε/2]]. The detuning ε = 30 μeV is set by a gate voltage, and the tunnel coupling is τ = 20 μeV. The next single-electron level in each dot lies about 1.5 meV above these states. If the electron starts on dot L at t = 0, the probability of finding it on dot R is P_R(t) = [4τ²/(E₊ − E₋)²] sin²[(E₊ − E₋)t/(2ℏ)], where E₊ and E₋ are the eigenvalues of H. Take h = 6.626×10⁻³⁴ J s and 1 eV = 1.602×10⁻¹⁹ J.

1. (a)

   Justify Justify the two-state model of the electron, using the energies given.

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   1. 1

      the omitted levels lie about 1.5 meV away, some 30 times the 50 μeV scale of ε, τ and the level splitting, so they mix only weakly into the two localised states

   2. 2

      nothing supplies the 1.5 meV needed to reach those levels: the gate voltage must change slowly compared with h/(1.5 meV) ≈ 3 ps, and the thermal energy k_BT must be much less than 1.5 meV (either condition, stated as a comparison with the 1.5 meV spacing, earns the mark)

2. (b)

   Determine Determine the eigenvalues E₊ and E₋ of H, and hence the splitting E₊ − E₋.

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   1. 1

      det(H − EI) = 0 gives (ε/2 − E)(−ε/2 − E) − τ² = 0, so E² = ε²/4 + τ²

   2. 2

      E± = ±√(ε²/4 + τ²) = ±√(225 + 400) μeV = ±25 μeV

   3. 3

      E₊ − E₋ = √(ε² + 4τ²) = 50 μeV, larger than the 30 μeV detuning alone because the coupling pushes the levels apart

3. (c)

   Determine Determine the probability of finding the electron on dot R when it is in the ground state.

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   1. 1

      for the ground state, E₋ = −25 μeV, writing v = a|L⟩ + b|R⟩, (H − E₋I)v = 0 gives (15 + 25)a − 20b = 0 with energies in μeV

   2. 2

      so b = 2a, and the normalised ground state is (|L⟩ + 2|R⟩)/√5

   3. 3

      P_R = 4/5 = 0.80 (equivalently, the mixing angle with tan 2χ = 2τ/ε = 4/3 gives sin²χ = 0.20 on dot L)

4. (d)

   Calculate The electron is placed on dot L at t = 0. Calculate the maximum probability of later finding it on dot R, and the period of the oscillation of P_R(t).

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   1. 1

      maximum P_R = 4τ²/(E₊ − E₋)² = 1600/2500 = 0.64

   2. 2

      sin² repeats when (E₊ − E₋)t/(2ℏ) advances by π, so the period is 2πℏ/(E₊ − E₋) = h/(E₊ − E₋) = 6.626×10⁻³⁴/(50×10⁻⁶ × 1.602×10⁻¹⁹) = 8.3×10⁻¹¹ s

5. (e)

   Predict The gate voltage is adjusted until ε = 0. Predict the maximum probability of transfer to dot R, and the time after t = 0 at which it is first reached.

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   1. 1

      with ε = 0 the eigenstates are (|L⟩ ± |R⟩)/√2, with equal weight on each dot, and E₊ − E₋ = 2τ = 40 μeV, so the maximum P_R = 4τ²/(2τ)² = 1: complete transfer

   2. 2

      P_R first reaches 1 when (2τ)t/(2ℏ) = π/2, at t = πℏ/(2τ) = h/(4τ) = 5.2×10⁻¹¹ s

For orbital angular momentum with l = 1, use the ordered basis {|1,1⟩, |1,0⟩, |1,−1⟩}, in which L_z = ℏ diag(1, 0, −1). The raising and lowering operators L± = L_x ± iL_y act as L±|l,m⟩ = ℏ√(l(l+1) − m(m±1)) |l,m±1⟩. The uncertainty relation for L_x and L_y is ΔL_x ΔL_y ≥ (ℏ/2)|⟨L_z⟩|.

1. (a)

   Show that Show that the only non-zero matrix elements of L₊ in this basis are ⟨1,1|L₊|1,0⟩ = ⟨1,0|L₊|1,−1⟩ = √2 ℏ.

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   1. 1

      L₊|1,0⟩ = ℏ√(2 − 0)|1,1⟩ = √2ℏ|1,1⟩ and L₊|1,−1⟩ = ℏ√(2 − (−1)(0))|1,0⟩ = √2ℏ|1,0⟩

   2. 2

      L₊|1,1⟩ = ℏ√(2 − 2) = 0, and each L₊|1,m⟩ is proportional to |1,m+1⟩, so by orthonormality every other element ⟨1,m′|L₊|1,m⟩ vanishes

2. (b)

   Determine Determine the 3 × 3 matrix that represents L_x in this basis.

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   1. 1

      L₋ = L₊†, with non-zero elements ⟨1,0|L₋|1,1⟩ = ⟨1,−1|L₋|1,0⟩ = √2ℏ

   2. 2

      L_x = (L₊ + L₋)/2 = (ℏ/√2)[[0, 1, 0], [1, 0, 1], [0, 1, 0]]

3. (c)

   Determine Determine the eigenvalues of L_x.

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   1. 1

      sets up det(L_x − λI) = 0 using the matrix from (b)

   2. 2

      expanding gives −λ³ + λℏ² = 0, that is −λ(λ² − ℏ²) = 0

   3. 3

      λ = ℏ, 0 and −ℏ, the same values as L_z, as they must be, since nothing distinguishes the x axis from the z axis

4. (d)

   Calculate A particle is prepared in |1,1⟩. Calculate the probabilities of obtaining ℏ, 0 and −ℏ in a measurement of L_x.

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   1. 1

      the normalised eigenvectors of L_x are (1, √2, 1)/2 for ℏ, (1, 0, −1)/√2 for 0, and (1, −√2, 1)/2 for −ℏ

   2. 2

      their overlaps with |1,1⟩ = (1, 0, 0) are the first components, 1/2, 1/√2 and 1/2

   3. 3

      the probabilities are 1/4, 1/2 and 1/4, which sum to 1

5. (e)

   Show that Show that in the state |1,1⟩ the product ΔL_x ΔL_y is exactly equal to the lower bound (ℏ/2)|⟨L_z⟩|.

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   1. 1

      from (d), ⟨L_x⟩ = 0 and ⟨L_x²⟩ = (1/4)ℏ² + (1/2)(0) + (1/4)ℏ² = ℏ²/2, so ΔL_x = ℏ/√2; the state is unchanged except for a phase by rotation about z, so ΔL_y = ℏ/√2 as well (or ⟨L_x²⟩ + ⟨L_y²⟩ = ⟨L²⟩ − ⟨L_z²⟩ = 2ℏ² − ℏ² = ℏ²)

   2. 2

      ΔL_x ΔL_y = ℏ²/2, and (ℏ/2)|⟨L_z⟩| = (ℏ/2)(ℏ) = ℏ²/2, so the bound is met with equality

For a hydrogen atom, the radial function u(r) = rR(r) satisfies −(ℏ²/2m) d²u/dr² + [−e²/(4πε₀r) + ℏ²l(l+1)/(2mr²)] u = Eu. A proposed solution is u(r) = N r³ e^(−r/(3a₀)), where N is a normalisation constant and a₀ = 4πε₀ℏ²/(me²) = 5.29×10⁻¹¹ m is the Bohr radius, so that e²/(4πε₀) = ℏ²/(ma₀) and ℏ²/(2ma₀²) = 13.6 eV. Treat the proton as infinitely heavy. You may use ∫₀^∞ rⁿ e^(−βr) dr = n!/βⁿ⁺¹ for a non-negative integer n and β > 0.

1. (a)

   Determine Determine the value of l for which u(r) satisfies the radial equation, and the corresponding energy E in electronvolts.

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   1. 1

      d²u/dr² = N[6r − 2r²/a₀ + r³/(9a₀²)] e^(−r/(3a₀))

   2. 2

      writes the Coulomb term as −(ℏ²/2m)(2/(a₀r))u, whose r² contribution −(ℏ²/2m)(2r²/a₀)Ne^(−r/(3a₀)) cancels the +(ℏ²/2m)(2r²/a₀)Ne^(−r/(3a₀)) from −(ℏ²/2m)u″

   3. 3

      the terms linear in r are −(ℏ²/2m)(6r) from the kinetic term and +(ℏ²/2m)l(l+1)r from the centrifugal term (each times Ne^(−r/(3a₀))), which cancel only if l(l+1) = 6, so l = 2

   4. 4

      the surviving r³ term gives −(ℏ²/2m)(1/(9a₀²))u = Eu, so E = −ℏ²/(18ma₀²) = −13.6 eV/9 = −1.51 eV

2. (b)

   State State the principal quantum number n of this state, its number of radial nodes, and its spectroscopic label.

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   1. 1

      E = −13.6 eV/n² = −1.51 eV gives n = 3

   2. 2

      u has no zero for r > 0, so there are no radial nodes (n − l − 1 = 0), and the state is 3d

3. (c)

   Determine Determine the most probable distance of the electron from the proton, in nanometres.

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   1. 1

      the radial probability density is P(r) = |u|² ∝ r⁶ e^(−2r/(3a₀)), and dP/dr = 0 gives 6/r = 2/(3a₀)

   2. 2

      r = 9a₀ = 0.476 nm

4. (d)

   Calculate Calculate the expectation value ⟨r⟩ in nanometres.

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   1. 1

      ⟨r⟩ = ∫₀^∞ r⁷ e^(−βr) dr / ∫₀^∞ r⁶ e^(−βr) dr with β = 2/(3a₀), which is (7!/β⁸)/(6!/β⁷) = 7/β

   2. 2

      ⟨r⟩ = 21a₀/2 = 10.5a₀ = 0.555 nm, larger than the most probable distance because P(r) has a long tail at large r

5. (e)

   Compare Compare the degeneracy of this energy level in hydrogen with the degeneracy an l = 2 level would have in a generic central potential, such as a finite spherical well, giving the cause of each.

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   1. 1

      in hydrogen E depends on n alone, so the level contains every n = 3 state, 3s, 3p and 3d: 1 + 3 + 5 = n² = 9 spatial states (18 with spin); the extra degeneracy between different l belongs to the 1/r form of the Coulomb potential, not to spherical symmetry

   2. 2

      in a generic central potential such as a finite spherical well, E depends on l as well as on the radial quantum number, so only the 2l + 1 = 5 states with m = −2 to 2 share the energy; that degeneracy is guaranteed by rotational symmetry, because m does not appear in the radial equation

Gate leakage in a transistor is caused by electrons tunnelling directly through the thin silicon dioxide layer between the gate and the silicon. To study it, an engineer measures the gate current I of six test capacitors made with different oxide thicknesses t_ox and different gate areas A, all at the same small gate voltage. The model to be tested is J = J₀ exp(−2κt_ox), where J = I/A is the current density, J₀ is the same for every capacitor, and κ = √(2m*φ_B)/ℏ, with φ_B the height of the oxide barrier above the energy of the tunnelling electrons and m* the effective mass of an electron in the oxide. Internal photoemission measurements, made independently, give φ_B = 3.1 eV. Take mₑ = 9.109×10⁻³¹ kg, ℏ = 1.055×10⁻³⁴ J s, 1 eV = 1.602×10⁻¹⁹ J and 1 μm² = 1×10⁻¹² m².

1. (a)

   Describe Describe how the readings should be processed and plotted so that the model predicts a straight line, including why the currents themselves should not be plotted and what the gradient measures.

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   1. 1

      divide each current by its gate area, converted to m², to obtain J = I/A: the capacitors have different areas and the model predicts current per unit area, so ln I would contain a different ln A for each capacitor and would not follow the model

   2. 2

      plot ln(J / A m⁻²) against t_ox; the model gives ln J = ln J₀ − 2κt_ox, a straight line of gradient −2κ, with J₀ affecting only the intercept

2. (b)

   Determine Determine κ from the readings, and the reduction in oxide thickness that multiplies the leakage current density by ten.

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   1. 1

      J = 2.00×10⁴, 2.10×10³, 203, 21.8, 2.14 and 0.225 A m⁻², so ln(J / A m⁻²) = 9.90, 7.65, 5.31, 3.08, 0.76 and −1.49 for capacitors 1 to 6

   2. 2

      the best-fit gradient of ln J against t_ox is −1.14×10¹⁰ m⁻¹ (accept −1.10×10¹⁰ to −1.18×10¹⁰ m⁻¹, or −11.4 nm⁻¹), so κ = 5.7×10⁹ m⁻¹ (accept 5.5×10⁹ to 5.9×10⁹ m⁻¹)

   3. 3

      a tenfold rise in J needs a thinning of Δt_ox = ln 10/(2κ) = 2.303/(1.14×10¹⁰ m⁻¹) = 2.0×10⁻¹⁰ m, so each 0.20 nm removed from the oxide raises the leakage tenfold

3. (c)

   Estimate Estimate the effective mass m* of the tunnelling electrons in the oxide, as a multiple of mₑ, using φ_B = 3.1 eV, and state the assumptions on which your estimate rests.

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   1. 1

      rearranges κ = √(2m*φ_B)/ℏ to m* = ℏ²κ²/(2φ_B)

   2. 2

      m* = (1.055×10⁻³⁴)² × (5.7×10⁹)²/(2 × 3.1 × 1.602×10⁻¹⁹) = 3.6×10⁻³¹ kg ≈ 0.40mₑ (accept 0.37mₑ to 0.43mₑ)

   3. 3

      states at least two assumptions, for example a rectangular barrier of fixed height φ_B that the small gate voltage does not tilt appreciably, a single effective mass independent of energy and thickness, the same J₀ for every capacitor, or oxide thicknesses that are accurate and uniform across each gate

4. (d)

   Evaluate A colleague analyses the same readings taking the electron in the oxide to have the free-electron mass mₑ, and reports the barrier height that follows. Evaluate this analysis.

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   1. 1

      with m* = mₑ the gradient gives φ = ℏ²κ²/(2mₑ) = 2.0×10⁻¹⁹ J ≈ 1.2 eV (accept 1.15 to 1.33 eV), less than half the 3.1 eV measured independently by photoemission

   2. 2

      the gradient fixes only the product m*φ_B, so the readings alone cannot decide between the two analyses; judged against the independent barrier height the free-electron analysis must be rejected, and the data require an effective mass near 0.4mₑ rather than a barrier near 1.2 eV

A particle of mass m moves in the potential V(x) = ½mω²x², so that Ĥ = p̂²/2m + ½mω²x̂². Define â = √(mω/2ℏ)(x̂ + ip̂/(mω)) and its adjoint ↠= √(mω/2ℏ)(x̂ − ip̂/(mω)), and write N̂ = â†â. The canonical commutator is [x̂, p̂] = iℏ. Once [â, â†] = 1 has been established, you may use its consequences [N̂, â] = −â and [N̂, â†] = â†.

1. (a)

   Derive Derive [â, â†] = 1 from the canonical commutator.

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   1. 1

      [â, â†] = (mω/2ℏ)[x̂ + ip̂/(mω), x̂ − ip̂/(mω)], and [x̂, x̂] = [p̂, p̂] = 0 leaves only the cross terms

   2. 2

      these are −(i/(mω))[x̂, p̂] + (i/(mω))[p̂, x̂] = −(2i/(mω))[x̂, p̂]

   3. 3

      substituting [x̂, p̂] = iℏ gives (mω/2ℏ)(2ℏ/(mω)) = 1

2. (b)

   Show that Show that Ĥ = ℏω(N̂ + ½).

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   1. 1

      â†â = (mω/2ℏ)[x̂² + p̂²/(m²ω²) + (i/(mω))(x̂p̂ − p̂x̂)] = (mω/2ℏ)[x̂² + p̂²/(m²ω²)] − ½, using x̂p̂ − p̂x̂ = iℏ

   2. 2

      multiplying by ℏω gives ℏωâ†â = p̂²/2m + ½mω²x̂² − ½ℏω = Ĥ − ½ℏω, so Ĥ = ℏω(N̂ + ½)

3. (c)

   Derive Derive the energy spectrum E_n = (n + ½)ℏω with n = 0, 1, 2, …, using only the commutators and the fact that no vector has a negative squared norm. Do not evaluate any integral.

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   1. 1

      if N̂|ν⟩ = ν|ν⟩, then N̂(â|ν⟩) = (âN̂ − â)|ν⟩ = (ν − 1)â|ν⟩, so â|ν⟩ is either zero or an eigenstate with eigenvalue ν − 1; likewise â†|ν⟩ has eigenvalue ν + 1

   2. 2

      for normalised |ν⟩, ν = ⟨ν|â†â|ν⟩ = ‖â|ν⟩‖² ≥ 0, so no eigenvalue of N̂ is negative

   3. 3

      repeated lowering would reach a negative eigenvalue unless it ends on a state |0⟩ with â|0⟩ = 0, and that state has ν = ‖â|0⟩‖² = 0; a non-integer ν would step past zero without ending, so ν must be a non-negative integer

   4. 4

      ‖â†|n⟩‖² = ⟨n|ââ†|n⟩ = n + 1 > 0, so raising from ν = 0 never terminates and reaches every ν = n = 0, 1, 2, …; then Ĥ = ℏω(N̂ + ½) gives E_n = (n + ½)ℏω, with ground-state energy ½ℏω

4. (d)

   Determine Determine ⟨x̂²⟩ in the eigenstate |n⟩ using ladder operators alone.

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   1. 1

      x̂ = √(ℏ/2mω)(â + â†), so x̂² = (ℏ/2mω)(â² + ↲ + â↠+ â†â)

   2. 2

      â² and ↲ change n by −2 and +2, so ⟨n|â²|n⟩ = ⟨n|↲|n⟩ = 0

   3. 3

      â↠= â†â + 1 gives ⟨n|â↠+ â†â|n⟩ = 2n + 1, so ⟨x̂²⟩ = (2n + 1)ℏ/(2mω) = E_n/(mω²)

5. (e)

   Predict The particle is in the ground state |0⟩ when the potential is suddenly tightened to V′(x) = ½m(2ω)²x², so quickly that the wavefunction has no time to change. Predict which energy eigenstates of the new oscillator the particle can then be found in, and its mean energy ⟨Ĥ′⟩ immediately after the change, in terms of ℏω.

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   1. 1

      V and V′ are both even, so the new eigenstates have parity (−1)ⁿ; the unchanged state |0⟩ is even, so its overlap with every odd-n state of the new oscillator vanishes, and only n = 0, 2, 4, … can be found, with energies (n + ½)(2ℏω) = ℏω, 5ℏω, 9ℏω, …

   2. 2

      the state is still the old |0⟩, so from (d) ⟨x̂²⟩ = ℏ/(2mω); then ⟨V⟩ = ½mω²⟨x̂²⟩ = ¼ℏω is half of E₀ = ½ℏω, so the kinetic energy is ⟨p̂²⟩/2m = ¼ℏω, and neither expectation changes at the instant of the change

   3. 3

      ⟨Ĥ′⟩ = ⟨p̂²⟩/2m + ½m(2ω)²⟨x̂²⟩ = ¼ℏω + ℏω = (5/4)ℏω, above the new ground energy ℏω, so the particle is not certain to be found in the new ground state

A simple model of the deuteron treats the neutron–proton pair as a single particle of reduced mass μ = 8.37×10⁻²⁸ kg moving in a spherical square well, with V(r) = −V₀ for r < a and V(r) = 0 for r > a, where a = 2.0×10⁻¹⁵ m. Only states with l = 0 are considered, so ψ depends on r alone and ∇²ψ = (1/r²) d/dr (r² dψ/dr). Primes denote d/dr. Take ℏ = 1.055×10⁻³⁴ J s and 1 MeV = 1.602×10⁻¹³ J. The measured binding energy of the deuteron is 2.22 MeV, and it has no bound excited state.

1. (a)

   Derive Starting from the time-independent Schrödinger equation in three dimensions, derive the equation −(ℏ²/2μ)u″ + V(r)u = Eu satisfied by u(r) = rψ(r).

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   1. 1

      writes −(ℏ²/2μ)∇²ψ + V(r)ψ = Eψ, with ∇²ψ = (1/r²)(r²ψ′)′ for l = 0

   2. 2

      with ψ = u/r, ψ′ = u′/r − u/r², so r²ψ′ = ru′ − u and (r²ψ′)′ = u′ + ru″ − u′ = ru″, giving ∇²ψ = u″/r

   3. 3

      substituting and multiplying through by r gives −(ℏ²/2μ)u″ + V(r)u = Eu

2. (b)

   Justify Justify the boundary condition u(0) = 0. Note that a wavefunction behaving as 1/r near the origin is square-integrable in three dimensions.

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   1. 1

      if u(0) ≠ 0 then ψ ≈ u(0)/r near r = 0, and the volume element r² dr keeps ∫|ψ|² r² dr finite there, so normalisability alone does not exclude it

   2. 2

      but ∇²(1/r) = −4πδ³(r), so such a ψ satisfies the Schrödinger equation at the origin only if V contains a delta function there; the finite well contains none, so u(0) = 0 is required

3. (c)

   Show that Show that a bound state with energy E = −E_B, where 0 < E_B < V₀, must satisfy k cot(ka) = −κ, where ℏk = √(2μ(V₀ − E_B)) and ℏκ = √(2μE_B).

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   1. 1

      inside, u″ = −k²u with u(0) = 0 gives u = A sin(kr); outside, u″ = κ²u and normalisability removes e^(+κr), leaving u = B e^(−κr)

   2. 2

      u and u′ are continuous at r = a because V is finite there: A sin(ka) = B e^(−κa) and Ak cos(ka) = −κB e^(−κa)

   3. 3

      dividing the second condition by the first gives k cot(ka) = −κ

4. (d)

   Derive Derive, in terms of μ, a and ℏ, the minimum depth V₀,min for which the well has any bound state at all.

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   1. 1

      with k > 0 and κ > 0, k cot(ka) = −κ needs cot(ka) < 0, so the lowest state requires ka > π/2; for ka ≤ π/2 there is no solution

   2. 2

      the bound state first appears at threshold, E_B → 0, where κ → 0 and the condition becomes cot(ka) = 0, with ka = π/2

   3. 3

      at E_B = 0, ℏk = √(2μV₀), so V₀,min = π²ℏ²/(8μa²)

5. (e)

   Calculate Calculate V₀,min in MeV for the values given.

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   1. 1

      V₀,min = π²(1.055×10⁻³⁴)²/(8 × 8.37×10⁻²⁸ × (2.0×10⁻¹⁵)²) = 4.10×10⁻¹² J

   2. 2

      V₀,min = 25.6 MeV

6. (f)

   Sketch A well depth of 36.6 MeV reproduces the measured binding energy, giving ka = 1.82 and κa = 0.46. Sketch u(r) for the deuteron's ground state from r = 0 to r = 5a, marking r = a on the axis.

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   1. 1

      for r < a, u rises from u(0) = 0 as a sine, reaches its maximum just inside the wall (at kr = π/2, r ≈ 0.86a) and has begun to fall by r = a

   2. 2

      at r = a, u and its slope join smoothly onto a slowly decaying exponential with decay length 1/κ ≈ 2.2a, still clearly non-zero at r = 5a, so that most of the area under u² lies outside the well