Schrödinger's Equation — Question 8

Section 38-6 · Physics pages 6366–6367

Question 8 A finite potential step requires flux weighting of transmitted amplitude A particle approaches a finite upward step with E > V0. On the left ψ = eik1x + re−ik1x; on the right ψ = teik2x, where k1, k2 > 0. Both ψ and ψ′ are continuous at zero. The probability current is j = (ℏ/m) Im(Ψ∗∂xΨ). x V E V0 0 Schematic; labels, not drawing lengths, specify the data. (a) Derive r and t from the two matching equations. (b) For k2 = k1/2, find R and T from currents and verify conservation.
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Question 8 - Solution A finite potential step requires flux weighting of transmitted amplitude x V E V0 0 (a) Method and calculation 1 + r = t, k1(1 −r) = k2t. r = k1 −k2 k1 + k2 , t = 2k1 k1 + k2 . (b) Interpretation and further calculation R = |r|2 = 1/9 , T = k2 k1 |t|2 = 1 2 (4 3 )2 = 8/9 . R + T = 1 , |t|2 = 16/9̸ = T. Physical check An amplitude larger than one does not violate probability conservation. The transmitted density must be multiplied by the transmitted velocity to obtain flux.

Question 8

A finite potential step requires flux weighting of transmitted amplitude

A particle approaches a finite upward step with E>V0E>V_0. On the left ψ=eik1x+re−ik1x\psi=e^{ik_1x}+re^{-ik_1x}; on the right ψ=teik2x\psi=te^{ik_2x}, where k1,k2>0k_1,k_2>0. Both ψ\psi and ψ′\psi\prime are continuous at zero. The probability current is j=(ℏ/m)Im⁡(Ψ*∂xΨ)j=(\hbar/m)\operatorname{Im}(\Psi^*\partial_x\Psi).

Diagram shown in the original SVG worksheet.

  1. Derive r and t from the two matching equations.

  2. For k2=k1/2k_2=k_1/2, find R and T from currents and verify conservation.

Question 8 - Solution

A finite potential step requires flux weighting of transmitted amplitude

Diagram shown in the original SVG worksheet.

(a) Method and calculation

1+r=t,k1(1−r)=k2t.1+r=t,\qquad k_1(1-r)=k_2t.r=k1−k2k1+k2,t=2k1k1+k2.\boxed{r=\frac{k_1-k_2}{k_1+k_2}},\qquad\boxed{t=\frac{2k_1}{k_1+k_2}}.

(b) Interpretation and further calculation

R=|r|2=1/9,T=k2k1|t|2=12(43)2=8/9.\boxed{R=|r|^2=1/9},\qquad\boxed{T=\frac{k_2}{k_1}|t|^2=\frac12\left(\frac43\right)^2=8/9}.R+T=1,|t|2=16/9≠T.\boxed{R+T=1},\qquad |t|^2=16/9\ne T.

Physical check

An amplitude larger than one does not violate probability conservation. The transmitted density must be multiplied by the transmitted velocity to obtain flux.