Wave Functions of a Trapped Electron — Question 8

Section 39-3 · Physics pages 6466–6467

Question 8 Sudden expansion preserves the initial wave function but changes its energy decomposition An electron starts in the ground state of 0 < x < L. The right wall suddenly moves to 2L. Immediately afterward the old wave function is unchanged on [0, L] and zero on (L, 2L). New modes are vj = L−1/2 sin(jπx/2L). x L 2L Schematic; labels, not drawing lengths, specify the data. (a) Calculate the overlap with new mode j = 2. (b) Calculate the overlap with new mode j = 1 and the probability outside those two modes.
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Question 8 - Solution Sudden expansion preserves the initial wave function but changes its energy decomposition x L 2L (a) Method and calculation c2 = √ 2 L ∫L 0 sin2(πx/L)dx = 1/ √ 2 , P2 = 1/2 . c1 = √ 2 L ∫L 0 sin(πx/L) sin(πx/2L)dx. (b) Interpretation and further calculation c1 = 4 √ 2 3π , P1 = 32 9π2 = .360253 . Pother = 1 −P1 −P2 = .139747 . Physical check The electron does not instantly become the new ground state. Sudden changes preserve the wave function at the instant of change, while the new Hamiltonian has different eigenstates.

Question 8

Sudden expansion preserves the initial wave function but changes its energy decomposition

An electron starts in the ground state of 0<x<L0<x<L. The right wall suddenly moves to 2L2L. Immediately afterward the old wave function is unchanged on [0,L][0,L] and zero on (L,2L)(L,2L). New modes are vj=L−1/2sin⁡(jπx/2L)v_j=L^{-1/2}\sin(j\pi x/2L).

Diagram shown in the original SVG worksheet.

  1. Calculate the overlap with new mode j=2j=2.

  2. Calculate the overlap with new mode j=1j=1 and the probability outside those two modes.

Question 8 - Solution

Sudden expansion preserves the initial wave function but changes its energy decomposition

Diagram shown in the original SVG worksheet.

(a) Method and calculation

c2=2L∫0Lsin⁡2(πx/L)dx=1/2,P2=1/2.c_2=\frac{\sqrt2}{L}\int_0^L\sin^2(\pi x/L)dx=\boxed{1/\sqrt2},\quad\boxed{P_2=1/2}.c1=2L∫0Lsin⁡(πx/L)sin⁡(πx/2L)dx.c_1=\frac{\sqrt2}{L}\int_0^L\sin(\pi x/L)\sin(\pi x/2L)dx.

(b) Interpretation and further calculation

c1=423π,P1=329π2=.360253.\boxed{c_1=\frac{4\sqrt2}{3\pi}},\quad\boxed{P_1=\frac{32}{9\pi^2}=.360253}.Pother=1−P1−P2=.139747.\boxed{P_{\mathrm{other}}=1-P_1-P_2=.139747}.

Physical check

The electron does not instantly become the new ground state. Sudden changes preserve the wave function at the instant of change, while the new Hamiltonian has different eigenstates.