 Solution

ID: past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2018/paper-1/section-b/b6/d/solution

From the second-edge amplitude in part (c),

\frac{F}{A} = i \frac{k _{0}}{k _{1}} = i \frac{k _{0}}{k _{2}} .

ψ_{R} = F e^{i k_{2} (x - a)}

. If the transmitted wave is instead written

T e^{i k_{2} x}

T / A = i k_{0} / k_{2} e^{- i k_{2} a}

; the extra factor only changes the quantum phase convention.

J_{in} = \frac{ℏ k _{0}}{m} ∣ A ∣^{2}, J_{trans} = \frac{ℏ k _{2}}{m} ∣ F ∣^{2} = \frac{ℏ k _{2}}{m} \frac{k _{0}}{k _{2}} ∣ A ∣^{2} = J_{in} .

Thus the flux transmission probability is one. The larger transmitted density

∣ F ∣^{2} > ∣ A ∣^{2}

compensates for the smaller transmitted velocity

ℏ k_{2} / m

; equal currents need not have equal wave amplitudes.

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