University Quantum Mechanics I · Midterm Examination

The Compton shift for a photon scattered through an angle θ by a free electron at rest is Δλ = (h/mₑc)(1 − cos θ), with h/mₑc = 2.43 pm. Photons of wavelength 71.0 pm and photons of wavelength 500 nm are each scattered through 90°. Which statement best explains why the effect was established with X-rays rather than with visible light?

1. AThe shift is 2.43 pm in both cases, which is 3.4% of 71.0 pm but only 4.9 × 10⁻⁶ of 500 nm, a fractional change about 7000 times smaller and far harder to detect.
2. BA visible photon carries no momentum, because its energy is far below the electron's rest energy, so it cannot make the electron recoil.
3. CThe shift is proportional to the incident wavelength, so for 500 nm light it is so large that the scattered light leaves the visible band and goes undetected.
4. DThe shift is 2.43 pm for X-rays but smaller by the factor 71.0 pm/500 nm for visible light, about 0.35 fm, which is too small to resolve.

Each function in the options is finite and continuous for all x, and takes the form stated at large |x|, where C and k are nonzero real constants. Which one could be normalised, and so represent the state of a particle?

1. Aψ(x) ≈ C/√|x| at large |x|
2. Bψ(x) ≈ C e^(ikx) at large |x|
3. Cψ(x) ≈ C/|x| at large |x|
4. Dψ(x) ≈ C cos(kx)/|x|^(1/4) at large |x|

A particle in a time-independent potential is in the state Ψ(x,t) = [ψ₁(x)e^(−iE₁t/ℏ) + ψ₂(x)e^(−iE₂t/ℏ)]/√2, where ψ₁ and ψ₂ are real, normalised, mutually orthogonal eigenfunctions of Ĥ with eigenvalues E₁ ≠ E₂. Which quantity is independent of time whatever the forms of ψ₁ and ψ₂?

1. A|Ψ(x,t)|² at a point x where ψ₁ and ψ₂ are both nonzero
2. B⟨x⟩
3. C|∫Ψ*(x,0)Ψ(x,t) dx|², the probability of finding the particle still in its initial state
4. D⟨H⟩

Acting on square-integrable functions on −∞ < x < ∞ that vanish, together with their derivatives, as |x| → ∞, which one of these operators is Hermitian?

1. Ad/dx
2. Bx d/dx
3. Ci d²/dx²
4. Di d/dx

A particle has Hamiltonian Ĥ = p̂²/(2m) + V(x̂), and [x̂, p̂] = iℏ. What is the commutator [x̂, Ĥ]?

1. Aiℏp̂/m
2. B0, because x̂ commutes with V(x̂)
3. Ciℏp̂/(2m)
4. D−iℏp̂/m

Position is measured once on each of N electrons, all prepared in the same state ψ, with a detector whose resolution is far finer than the spread of the results. The readings have sample standard deviation s, and their mean has standard error s/√N. N is increased from 100 to 10 000. Which statement is correct?

1. As and the standard error both fall by a factor of 10, because more data reduces every uncertainty.
2. Bs stays close to σ_x of the state, while the standard error falls by a factor of 10.
3. Cs and the standard error both stay the same, because the uncertainty principle fixes both.
4. Ds grows, because each extra measurement disturbs the ensemble, while the standard error falls by a factor of 10.

A particle of mass m has V(x) = 0 everywhere on −∞ < x < ∞. Which statement about the solutions of its time-independent Schrödinger equation is correct?

1. AOnly discrete energies are allowed, because ψ must be normalisable and only certain wavelengths fit.
2. BEvery E > 0 is allowed, and each is twofold degenerate, with solutions e^(ikx) and e^(−ikx) where k = √(2mE)/ℏ.
3. CEvery E > 0 is allowed, and each is non-degenerate, because states in one dimension are never degenerate.
4. DNegative energies are also allowed, with solutions e^(κx) and e^(−κx) where κ = √(−2mE)/ℏ.

A particle is in the ground state ψ₁(x) = √(2/L) sin(πx/L) of an infinite square well occupying 0 < x < L. What is the probability that a measurement of position finds it in 0 < x < L/4?

1. A0.41
2. B0.25
3. C0.091
4. D0.045

A particle of mass m moving along the x-axis is in a stationary state of some potential V(x). Its wavefunction is ψ(x) = 0 for x < 0; ψ(x) = Ax/a for 0 ≤ x ≤ a; and ψ(x) = A e^(−(x−a)/a) for x > a, where a is a positive length and A is a positive real constant. ψ is continuous everywhere, but its slope jumps at x = 0 and at x = a.

1. (a)

   Show that Show that A = √(6/(5a)).

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   1. 1

      splits the normalisation integral at x = a and evaluates the linear piece: ∫₀^a A²x²/a² dx = A²a/3

   2. 2

      evaluates the tail: ∫ₐ^∞ A²e^(−2(x−a)/a) dx = A²a/2

   3. 3

      sets A²(a/3 + a/2) = 5A²a/6 = 1, so A = √(6/(5a))

2. (b)

   Calculate Calculate the probability that a measurement of position finds the particle in 0 < x < a/2, and the probability that it finds the particle at x > a.

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   1. 1

      P(0 < x < a/2) = A²∫₀^(a/2) x²/a² dx = A²a/24 = (6/5)(1/24) = 0.050

   2. 2

      P(x > a) = A²a/2 = 3/5 = 0.60

3. (c)

   Explain The slope of ψ must be continuous wherever V is finite. Explain why, and explain why the jump in slope at x = 0 is nevertheless admissible.

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   1. 1

      integrating −(ℏ²/2m)ψ″ + Vψ = Eψ from x₀ − ε to x₀ + ε gives ψ′(x₀ + ε) − ψ′(x₀ − ε) = (2m/ℏ²)∫(V − E)ψ dx, which tends to zero as ε → 0 when V is finite, so ψ′ cannot jump at such a point

   2. 2

      ψ = 0 for all x < 0 is the wavefunction beside an infinite wall (V = ∞ for x < 0); V is not finite there, so the slope may jump from 0 to A/a at x = 0 while ψ itself stays continuous

4. (d)

   Determine The jump in slope at x = a requires V(x) to contain a delta-function term gδ(x − a). Determine g in terms of ℏ, m and a, and say whether the spike is attractive or repulsive.

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   1. 1

      the slope changes from +A/a to −A/a, so Δψ′ = −2A/a, while integrating the Schrödinger equation across x = a gives Δψ′ = (2m/ℏ²)gψ(a) = (2m/ℏ²)gA

   2. 2

      g = −ℏ²/(ma): the spike is attractive, V(x) containing −(ℏ²/(ma))δ(x − a)

An electron (mₑ = 9.109 × 10⁻³¹ kg) moves freely along the x-axis, V = 0. At t = 0 it is prepared as a Gaussian wave packet centred on wavenumber k₀, with position standard deviation σ₀ = 0.40 nm and σ_xσ_p = ℏ/2. The kinetic energy corresponding to the central wavenumber is E₀ = ℏ²k₀²/(2mₑ) = 1.50 eV. The width of the packet then evolves as σ_x(t) = σ₀√(1 + (ℏt/(2mₑσ₀²))²). Take ℏ = 1.055 × 10⁻³⁴ J s and 1 eV = 1.602 × 10⁻¹⁹ J.

1. (a)

   Explain Explain why no single plane wave e^(ik₀x) can describe the electron, and how a superposition of plane waves overcomes the difficulty.

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   1. 1

      |e^(ik₀x)|² = 1 at every x, so ∫|ψ|² dx over the whole line diverges: the plane wave cannot be normalised and so is not a physical state

   2. 2

      a Fourier superposition ψ(x) = (1/√(2π))∫φ(k)e^(ikx) dk with square-integrable φ(k) is normalisable, since ∫|ψ|² dx = ∫|φ|² dk (Plancherel's theorem); the price is a spread of wavenumbers, and so of momenta, about k₀

2. (b)

   Show that Using the dispersion relation of a free particle, show that the centre of the packet moves at v_g = ℏk₀/mₑ, twice the phase velocity of its central component.

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   1. 1

      for V = 0, ℏω = ℏ²k²/(2mₑ), so ω(k) = ℏk²/(2mₑ) and the phase velocity of the central component is v_p = ω/k = ℏk₀/(2mₑ)

   2. 2

      the envelope moves at v_g = dω/dk evaluated at k₀, which is ℏk₀/mₑ = 2v_p and equals the classical speed p₀/mₑ

3. (c)

   Calculate Calculate the group velocity, the time at which σ_x has grown to 2σ₀, and the distance the centre of the packet travels in that time.

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   1. 1

      v_g = √(2E₀/mₑ) = √(2 × 1.50 × 1.602 × 10⁻¹⁹ J / 9.109 × 10⁻³¹ kg) = 7.26 × 10⁵ m s⁻¹

   2. 2

      σ_x = 2σ₀ when (ℏt/(2mₑσ₀²))² = 3, so t = 2√3 mₑσ₀²/ℏ = 2√3 × 9.109 × 10⁻³¹ kg × (0.40 × 10⁻⁹ m)² / 1.055 × 10⁻³⁴ J s = 4.79 × 10⁻¹⁵ s

   3. 3

      distance = v_g t = 7.26 × 10⁵ m s⁻¹ × 4.79 × 10⁻¹⁵ s = 3.5 × 10⁻⁹ m (3.5 nm)

4. (d)

   Predict The electron is instead prepared with σ₀ = 0.80 nm, still with σ_xσ_p = ℏ/2 at t = 0. Predict, with a reason, the new time for the width to double.

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   1. 1

      the packet spreads because ω is quadratic in k, so its components travel with a spread of group velocities σ_p/mₑ; with σ_p = ℏ/(2σ₀) halved, they separate half as fast, while the width to be doubled is twice as large

   2. 2

      the doubling time scales as σ₀², so it becomes 4 times longer: 1.91 × 10⁻¹⁴ s

An electron (mₑ = 9.109 × 10⁻³¹ kg) is confined along one axis of a nanostructure modelled as an infinite square well on 0 < x < L, with L = 1.20 nm. Its energy eigenfunctions are ψₙ(x) = √(2/L) sin(nπx/L), with energies Eₙ = n²π²ℏ²/(2mₑL²). At t = 0 its state is Ψ(x,0) = √(2/3) ψ₁(x) + (i/√3) ψ₂(x). You may use ∫₀^L ψₙ x ψₙ dx = L/2 for every n, and ∫₀^L ψ₁ x ψ₂ dx = −16L/(9π²). Take ℏ = 1.055 × 10⁻³⁴ J s and 1 eV = 1.602 × 10⁻¹⁹ J.

1. (a)

   Calculate Calculate the probabilities of obtaining E₁ and E₂ in a measurement of energy, and the expectation value ⟨H⟩ in electronvolts.

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   1. 1

      P(E₁) = |√(2/3)|² = 2/3 and P(E₂) = |i/√3|² = 1/3, which sum to 1, so Ψ is normalised

   2. 2

      E₁ = π²ℏ²/(2mₑL²) = 4.19 × 10⁻²⁰ J = 0.261 eV, so ⟨H⟩ = (2/3)E₁ + (1/3)(4E₁) = 2E₁ = 0.523 eV

2. (b)

   Show that Write down Ψ(x,t), and show that ⟨x⟩(t) = L/2 − (32√2L/(27π²)) sin(ωt), where ω = (E₂ − E₁)/ℏ.

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   1. 1

      Ψ(x,t) = √(2/3)ψ₁e^(−iE₁t/ℏ) + (i/√3)ψ₂e^(−iE₂t/ℏ), with each phase factor attached to its own term

   2. 2

      ⟨x⟩ = (2/3)(L/2) + (1/3)(L/2) + 2Re[c₁*c₂e^(−iωt)]∫ψ₁xψ₂ dx = L/2 + 2Re[(i√2/3)e^(−iωt)] × (−16L/(9π²))

   3. 3

      Re[i e^(−iωt)] = sin(ωt), so ⟨x⟩ = L/2 − (2√2/3)(16L/(9π²)) sin(ωt) = L/2 − (32√2L/(27π²)) sin(ωt)

3. (c)

   Calculate Calculate the period of the oscillation of ⟨x⟩, and its amplitude in nanometres.

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   1. 1

      ω = 3E₁/ℏ = 3 × 4.19 × 10⁻²⁰ J / 1.055 × 10⁻³⁴ J s = 1.19 × 10¹⁵ rad s⁻¹, so the period is 2π/ω = 5.28 × 10⁻¹⁵ s

   2. 2

      amplitude = 32√2L/(27π²) = 0.170L = 0.204 nm

4. (d)

   Predict Suppose instead that Ψ(x,0) = √(2/3) ψ₁(x) + (1/√3) ψ₂(x). Predict which of P(E₁), P(E₂), ⟨H⟩, and the period and amplitude of ⟨x⟩(t) change, and what does change.

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   1. 1

      P(E₁), P(E₂) and ⟨H⟩ are unchanged because they depend only on |cₙ|²; the period and amplitude are unchanged because they depend only on E₂ − E₁ and on |c₁||c₂|

   2. 2

      only the phase of the oscillation changes: ⟨x⟩(t) = L/2 − (32√2L/(27π²)) cos(ωt), so ⟨x⟩ starts at its minimum, 0.600 − 0.204 = 0.396 nm, instead of at the centre of the well

5. (e)

   Describe Return to the original state. At t = 2.0 fs a measurement of energy on the electron returns E₂. Describe the state immediately afterwards, and how ⟨x⟩ and the result of a second energy measurement behave from then on.

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   1. 1

      the state collapses to ψ₂, multiplied by a phase factor of no physical consequence and normalised, with no ψ₁ component remaining

   2. 2

      it is then stationary: ⟨x⟩ = L/2 = 0.600 nm at all later times, and a repeated energy measurement returns E₂ = 4E₁ = 1.05 eV with probability 1

An electron is bound in a thin layer of material modelled as a symmetric finite square well: V(x) = 0 for |x| < a and V(x) = V₀ for |x| > a, with well width 2a = 0.80 nm and V₀ = 2.00 eV. Treat the electron as having the free-electron mass mₑ = 9.109 × 10⁻³¹ kg. For a bound state with 0 < E < V₀, write k = √(2mₑE)/ℏ, κ = √(2mₑ(V₀ − E))/ℏ, z = ka and z₀ = (a/ℏ)√(2mₑV₀). Take ℏ = 1.055 × 10⁻³⁴ J s and 1 eV = 1.602 × 10⁻¹⁹ J.

1. (a)

   Justify Justify looking for bound states that are either even or odd in x, rather than of general form.

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   1. 1

      V(−x) = V(x), so if ψ(x) solves the time-independent Schrödinger equation with energy E, then ψ(−x) solves it with the same E

   2. 2

      bound states in one dimension are non-degenerate, so ψ(−x) = λψ(x); reflecting twice gives λ² = 1, so λ = ±1 and every bound state is even or odd as a matter of theorem, not merely of choice

2. (b)

   Derive Starting from the solutions of the Schrödinger equation in each region, derive the condition that the energies of the even bound states must satisfy, written as a single equation in z and z₀ alone.

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   1. 1

      inside the well ψ = C cos(kx); for x > a, ψ = De^(−κx), mirrored for x < −a, the growing solution e^(+κx) being discarded because ψ must be normalisable

   2. 2

      continuity of ψ at x = a: C cos(ka) = De^(−κa)

   3. 3

      continuity of ψ′ at x = a, which holds because V is finite there: −kC sin(ka) = −κDe^(−κa); dividing the two conditions eliminates C and D and gives k tan(ka) = κ

   4. 4

      multiplying by a and using k² + κ² = 2mₑV₀/ℏ², so that (κa)² = z₀² − z², gives z tan z = √(z₀² − z²)

3. (c)

   Determine Calculate z₀, and hence determine how many bound states the well has and the parity of each.

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   1. 1

      z₀ = (0.40 × 10⁻⁹ m / 1.055 × 10⁻³⁴ J s) × √(2 × 9.109 × 10⁻³¹ kg × 2.00 × 1.602 × 10⁻¹⁹ J) = 2.90

   2. 2

      an even state exists for each nπ < z₀ and an odd state for each (n + ½)π < z₀, with n = 0, 1, 2, …; since π/2 < 2.90 < π there are two bound states, the even ground state and one odd state

4. (d)

   Compare The even condition has its lowest root at z = 1.159. Compare the ground-state energy with that of an infinite square well of the same width, 0.80 nm, and account for the difference.

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   1. 1

      E = V₀(z/z₀)² = 2.00 eV × (1.159/2.897)² = 0.320 eV (accept 0.319 eV from z₀ = 2.90)

   2. 2

      infinite well of width 2a: E = π²ℏ²/(2mₑ(2a)²) = 0.588 eV, so the finite-well ground state is lower, about 0.54 of the infinite-well value

   3. 3

      ψ leaks into the classically forbidden region over a decay length 1/κ = a/√(z₀² − z²) = 0.15 nm, so it fits a longer wavelength inside the well: ka = 1.159 rather than π/2 = 1.571, giving a smaller k and a lower energy

In a scanning tunnelling microscope a sharp tungsten tip is held above a clean metal surface with a small bias voltage between them, and the tunnelling current I is recorded as a piezoelectric actuator withdraws the tip in equal steps. Δd is the increase in the tip-to-surface gap from its starting value d₀, which is not known accurately but is estimated to be about 0.5 nm. The gap is modelled as a rectangular barrier of height φ above the energy of the tunnelling electrons. In the thick-barrier limit I = I₀ exp(−2κd), with κ = √(2mₑφ)/ℏ, d = d₀ + Δd, and I₀ independent of d. Take mₑ = 9.109 × 10⁻³¹ kg, ℏ = 1.055 × 10⁻³⁴ J s and 1 eV = 1.602 × 10⁻¹⁹ J.

1. (a)(i)

   Explain Explain why a plot of ln(I/nA) against Δd should give a straight line even though d₀ is not known.

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   1. 1

      ln I = (ln I₀ − 2κd₀) − 2κΔd, so ln I is linear in Δd, and the unknown d₀ only shifts the intercept

2. (a)(ii)

   State State what the gradient of that line represents.

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   1. 1

      the gradient is −2κ

3. (b)

   Determine Determine κ from the data.

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   1. 1

      computes ln(I/nA) = 1.163, 0.615, 0.122, −0.440, −0.947, −1.505

   2. 2

      gradient of the best-fit straight line of ln(I/nA) against Δd = −0.0212 pm⁻¹ (accept −0.0205 to −0.0220 pm⁻¹), from a least-squares fit or from widely separated points on a drawn line rather than from adjacent readings

   3. 3

      κ = 0.0212 pm⁻¹ / 2 = 1.06 × 10¹⁰ m⁻¹ (accept 1.03 × 10¹⁰ to 1.10 × 10¹⁰ m⁻¹)

4. (c)

   Calculate Calculate the barrier height φ in electronvolts.

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   1. 1

      φ = ℏ²κ²/(2mₑ) = (1.055 × 10⁻³⁴ J s × 1.06 × 10¹⁰ m⁻¹)² / (2 × 9.109 × 10⁻³¹ kg) = 6.9 × 10⁻¹⁹ J

   2. 2

      φ = 4.3 eV (accept 4.0 to 4.6 eV when consistent with the value of κ from (b))

5. (d)

   Evaluate Evaluate whether the thick-barrier form is adequate for analysing these readings.

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   1. 1

      at the smallest gap 2κd ≈ 2 × 1.06 × 10¹⁰ m⁻¹ × 0.5 × 10⁻⁹ m ≈ 11 ≫ 1, so the exact rectangular-barrier transmission differs from the thick-barrier form only by corrections of relative order e^(−2κd), a few parts in 10⁵, and the prefactor that form discards does not depend on d, so it cannot change the gradient

   2. 2

      the six values of ln I lie within about 0.02 of a straight line, which supports the model over this range; judgement: adequate for finding the gradient, but φ is an apparent barrier height, because the real barrier is not rectangular (image-charge forces round and lower it) and the tip geometry is not one-dimensional

6. (e)

   Estimate Estimate, stating your assumptions, the change in gap that would alter the current by 5%.

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   1. 1

      assumes I₀ stays fixed and the change is small enough to linearise the exponential, so that |ΔI|/I ≈ 2κ|Δd|

   2. 2

      |Δd| ≈ 0.05 / (2.1 × 10¹⁰ m⁻¹) ≈ 2 × 10⁻¹² m (2 pm), a small fraction of an atomic diameter

A beam of electrons of mass mₑ and energy E travels in the +x direction towards a potential step: V(x) = 0 for x < 0 and V(x) = V₀ for x > 0, with V₀ real and positive. The electrons are described by stationary scattering states of the time-dependent Schrödinger equation iℏ ∂Ψ/∂t = −(ℏ²/2mₑ)∂²Ψ/∂x² + V(x)Ψ. 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 Derive, from the time-dependent Schrödinger equation and its complex conjugate, the continuity equation ∂ρ/∂t + ∂j/∂x = 0 for ρ = |Ψ|², obtaining an expression for the probability current j and stating where the reality of V is used.

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   1. 1

      writes the conjugate equation −iℏ ∂Ψ*/∂t = −(ℏ²/2mₑ)∂²Ψ*/∂x² + VΨ*, which uses V* = V

   2. 2

      ∂ρ/∂t = Ψ*∂Ψ/∂t + Ψ∂Ψ*/∂t; on substituting both equations the potential terms −(i/ℏ)V|Ψ|² and +(i/ℏ)V|Ψ|² cancel, because V is real

   3. 3

      ∂ρ/∂t = (iℏ/2mₑ)(Ψ*∂²Ψ/∂x² − Ψ∂²Ψ*/∂x²) = (iℏ/2mₑ)∂/∂x(Ψ*∂Ψ/∂x − Ψ∂Ψ*/∂x), the terms in (∂Ψ*/∂x)(∂Ψ/∂x) cancelling

   4. 4

      so ∂ρ/∂t + ∂j/∂x = 0 with j = (ℏ/2imₑ)(Ψ*∂Ψ/∂x − Ψ∂Ψ*/∂x) = (ℏ/mₑ)Im(Ψ*∂Ψ/∂x)

2. (b)

   Show that For E > V₀, take ψ(x) = e^(ik₁x) + r e^(−ik₁x) for x < 0 and ψ(x) = t e^(ik₂x) for x > 0, where k₁ = √(2mₑE)/ℏ and k₂ = √(2mₑ(E − V₀))/ℏ. Explaining why there is no e^(−ik₂x) term, show that the reflection and transmission probabilities are R = ((k₁ − k₂)/(k₁ + k₂))² and T = 4k₁k₂/(k₁ + k₂)².

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   1. 1

      there is no e^(−ik₂x) term because no wave comes in from x = +∞: electrons are sent in from the left only

   2. 2

      continuity of ψ at x = 0 gives 1 + r = t, and continuity of ψ′, which holds because V is finite, gives ik₁(1 − r) = ik₂t

   3. 3

      solving: r = (k₁ − k₂)/(k₁ + k₂) and t = 2k₁/(k₁ + k₂)

   4. 4

      using the current from (a): j = (ℏk₁/mₑ)(1 − |r|²) for x < 0, with no interference term, and j = (ℏk₂/mₑ)|t|² for x > 0, so R = |r|² = ((k₁ − k₂)/(k₁ + k₂))² and T = (k₂/k₁)|t|² = 4k₁k₂/(k₁ + k₂)²

3. (c)

   Calculate For electrons with E = 6.0 eV incident on a step of V₀ = 4.5 eV, calculate R, T and |t|², and confirm that R + T = 1.

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   1. 1

      k₂/k₁ = √((E − V₀)/E) = √(1.5/6.0) = 0.50, so R = (0.5/1.5)² = 1/9 = 0.11 and T = 4(0.5)/(1.5)² = 8/9 = 0.89, giving R + T = 1

   2. 2

      t = 2/(1 + 0.5) = 4/3, so |t|² = 16/9 = 1.78, and T = (k₂/k₁)|t|² = 0.89 is less than 1

4. (d)

   Sketch The electrons now have E = 3.0 eV, below the step. Sketch |ψ(x)|² against x from about x = −1 nm to x = +0.5 nm, for an incident wave of unit amplitude, marking the spacing of the maxima for x < 0, the value of |ψ|² at x = 0, and the length over which |ψ|² falls by a factor of e for x > 0.

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   1. 1

      for x < 0: a standing-wave pattern oscillating between 0 and 4, with true zeros because |r| = 1, and maxima spaced π/k₁ = 0.35 nm apart (k₁ = 8.87 × 10⁹ m⁻¹)

   2. 2

      at x = 0: matching gives t = 1 + r = 2k₁/(k₁ + iκ), so |ψ(0)|² = 4k₁²/(k₁² + κ²) = 4E/V₀ = 2.7, with the two sides joining with no jump in value or in slope

   3. 3

      for x > 0: a smooth decay |ψ(0)|²e^(−2κx) with no oscillation, falling by a factor of e in 1/(2κ) = 0.080 nm, where κ = √(2mₑ(V₀ − E))/ℏ = 6.27 × 10⁹ m⁻¹

5. (e)

   Explain Explain why every electron is reflected when E < V₀, even though |ψ|² is not zero inside the step.

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   1. 1

      for x > 0, ψ = te^(−κx) has the same complex phase at every x, so Ψ*∂Ψ/∂x = −κ|t|²e^(−2κx) is real and j = (ℏ/mₑ)Im(Ψ*∂Ψ/∂x) = 0: no probability flows into the step, so T = 0

   2. 2

      in a stationary state ∂ρ/∂t = 0, so by the continuity equation j is the same at every x; zero on the right means (ℏk₁/mₑ)(1 − |r|²) = 0 on the left, so R = |r|² = 1, and the density in the step is a standing evanescent penetration rather than a flow of particles through it