Wave Functions of a Trapped Electron — Question 1

Section 39-3 · Physics pages 6452–6453

Question 1 Normalization fixes a box amplitude while leaving its overall phase free An electron has spatial wave function ψ = A sin(3πx/L) for 0 < x < L and zero outside. x 0 L Schematic; labels, not drawing lengths, specify the data. (a) Find the magnitude of A and its physical units. (b) Find the probability in the left third and explain whether normalization fixes the phase of A.
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Question 1 - Solution Normalization fixes a box amplitude while leaving its overall phase free x 0 L (a) Method and calculation 1 = |A|2 ∫L 0 sin2(3πx/L)dx = |A|2L/2. |A| = √ 2/L , [A] = length−1/2. (b) Interpretation and further calculation P(0 < x < L/3) = 2 L ∫L/3 0 sin2(3πx/L)dx = 1/3 . A = √ 2/L eiα for arbitrary real α. Physical check The left third contains one complete density lobe. A global phase leaves all position probabilities unchanged.

Question 1

Normalization fixes a box amplitude while leaving its overall phase free

An electron has spatial wave function ψ=Asin⁡(3πx/L)\psi=A\sin(3\pi x/L) for 0<x<L0<x<L and zero outside.

Diagram shown in the original SVG worksheet.

  1. Find the magnitude of A and its physical units.

  2. Find the probability in the left third and explain whether normalization fixes the phase of A.

Question 1 - Solution

Normalization fixes a box amplitude while leaving its overall phase free

Diagram shown in the original SVG worksheet.

(a) Method and calculation

1=|A|2∫0Lsin⁡2(3πx/L)dx=|A|2L/2.1=|A|^2\int_0^L\sin^2(3\pi x/L)dx=|A|^2L/2.|A|=2/L,[A]=length−1/2.\boxed{|A|=\sqrt{2/L}},\qquad[A]=\mathrm{length}^{-1/2}.

(b) Interpretation and further calculation

P(0<x<L/3)=2L∫0L/3sin⁡2(3πx/L)dx=1/3.P(0<x<L/3)=\frac2L\int_0^{L/3}\sin^2(3\pi x/L)dx=\boxed{1/3}.A=2/Leiαfor arbitrary real α.\boxed{A=\sqrt{2/L}\,e^{i\alpha}}\quad\text{for arbitrary real }\alpha.

Physical check

The left third contains one complete density lobe. A global phase leaves all position probabilities unchanged.