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ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2015/paper-1/section-b/b6/solution

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Past exam of the physics A course of the University of Cambridge / 2015 / Paper 1 / Section B / B6 / Solution by

Codex  0 2026-10-20

Single travelling wave. Differentiation of the plane wave gives

\partial_{x} ψ = ik ψ

, so its probability current is

J = Re (ψ^{*} \frac{ℏ}{im} \partial_{x} ψ) = \frac{ℏ k}{m} ∣ A ∣^{2} .

This is probability density times the free-particle velocity

ℏ k / m

.

Counterpropagating waves. Remove the common time phase and set

z = B^{*} A e^{2 ik x}

. Direct multiplication gives

ψ^{*} \partial_{x} ψ = ik (∣ A ∣^{2} - ∣ B ∣^{2} + z - z^{*}) .

Consequently

J = \frac{ℏ k}{m} Re (∣ A ∣^{2} - ∣ B ∣^{2} + z - z^{*}) = \frac{ℏ k}{m} (∣ A ∣^{2} - ∣ B ∣^{2}) .

The interference term

z - z^{*}

is purely imaginary in the last real-part operation. The probability density can have interference fringes even though this net probability current is spatially constant.

Scattering amplitudes. Locate the step at

x = 0

and first take

E > V

. With

k_{1} = 2 m E /ℏ

and

k_{2} = 2 m (E - V) /ℏ

, use

ψ_{1} = e^{i k_{1} x} + r_{a} e^{- i k_{1} x}, ψ_{2} = t_{a} e^{i k_{2} x} .

The common factor

e^{- i Et /ℏ}

has been suppressed. A finite step with the same mass on both sides requires continuity of

ψ

and

ψ^{'}

1 + r_{a} = t_{a}

and

k_{1} (1 - r_{a}) = k_{2} t_{a}

. Hence

r_{a} = \frac{k _{1} - k _{2}}{k _{1} + k _{2}} = \frac{E - E - V}{E + E - V}, t_{a} = \frac{2 k _{1}}{k _{1} + k _{2}} .

If

0 < E < V

, choose

k_{2} = iκ

,

κ = 2 m (V - E) /ℏ

, so the transmitted wave decays. The same amplitude formulae continue to hold, but the semi-infinite forbidden region carries no outgoing probability current:

R = 1

and

T = 0

.

Flux balance. In the propagating case,

R = ∣ r_{a} ∣^{2}, T = \frac{k _{2}}{k _{1}} ∣ t_{a} ∣^{2} = \frac{4 k _{1} k _{2}}{( k _{1} + k _{2} ) ^{2}}, R + T = 1.

In particular,

J_{1} = \frac{ℏ k _{1}}{m} (1 - R) = \frac{ℏ k _{2}}{m} ∣ t_{a} ∣^{2} = J_{2} .

The velocity ratio in

T

is essential: the squared transmission amplitude alone is not the transmission probability when the two velocities differ.

Numerical transmission. The mass cancels from the ratio, and the energies can remain in the same electronvolt units:

T = \frac{4 3}{( 3 + 1 ) ^{2}} = 0.92820 .

Thus about

92.8%

is transmitted and

7.18%

reflected, despite the incident energy exceeding the step.

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