Wave Functions of a Trapped Electron — Question 10

Section 39-3 · Physics pages 6470–6471

Question 10 A narrow detector near a node collects probability cubically in its half-width An electron is in mode 2. Compare narrow windows of half-width ϵ ≪L centered on the node x = L/2 and the antinode x = L/4. Both windows lie inside the well. The infinite well occupies 0 < x < L; wave functions vanish outside. Its normalized modes are un(x) = √ 2/L sin(nπx/L) and energies En = n2E1. x 0 L Schematic; labels, not drawing lengths, specify the data. (a) Find the leading probability in the node-centered window. (b) Find the leading antinode-window probability and their ratio.
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Question 10 - Solution A narrow detector near a node collects probability cubically in its half-width x 0 L (a) Method and calculation x = L/2 + δ : u2 ≃− √ 2/L (2πδ/L). Pnode ≃8π2 L3 ∫ϵ −ϵ δ2dδ = 16π2ϵ3 3L3 . (b) Interpretation and further calculation x ≃L/4 : |u2|2 ≃2/L ⇒Panti ≃4ϵ/L . Pnode/Panti ≃4π2 3 (ϵ/L)2 . Physical check A finite detector centered on a node still has nonzero probability. The suppression is stronger than the linear width scaling near an antinode.

Question 10

A narrow detector near a node collects probability cubically in its half-width

An electron is in mode 2. Compare narrow windows of half-width ϵ≪L\epsilon\ll L centered on the node x=L/2x=L/2 and the antinode x=L/4x=L/4. Both windows lie inside the well. The infinite well occupies 0<x<L0<x<L; wave functions vanish outside. Its normalized modes are un(x)=2/Lsin⁡(nπx/L)u_n(x)=\sqrt{2/L}\sin(n\pi x/L) and energies En=n2E1E_n=n^2E_1.

Diagram shown in the original SVG worksheet.

  1. Find the leading probability in the node-centered window.

  2. Find the leading antinode-window probability and their ratio.

Question 10 - Solution

A narrow detector near a node collects probability cubically in its half-width

Diagram shown in the original SVG worksheet.

(a) Method and calculation

x=L/2+δ:u2≃−2/L(2πδ/L).x=L/2+\delta:\quad u_2\simeq-\sqrt{2/L}\,(2\pi\delta/L).Pnode≃8π2L3∫−ϵϵδ2dδ=16π2ϵ33L3.\boxed{P_{\mathrm{node}}\simeq\frac{8\pi^2}{L^3}\int_{-\epsilon}^{\epsilon}\delta^2d\delta=\frac{16\pi^2\epsilon^3}{3L^3}}.

(b) Interpretation and further calculation

x≃L/4:|u2|2≃2/L⇒Panti≃4ϵ/L.x\simeq L/4:\quad|u_2|^2\simeq2/L\Rightarrow\boxed{P_{\mathrm{anti}}\simeq4\epsilon/L}.Pnode/Panti≃4π23(ϵ/L)2.\boxed{P_{\mathrm{node}}/P_{\mathrm{anti}}\simeq\frac{4\pi^2}{3}(\epsilon/L)^2}.

Physical check

A finite detector centered on a node still has nonzero probability. The suppression is stronger than the linear width scaling near an antinode.