Wave Functions of a Trapped Electron — Question 7

Section 39-3 · Physics pages 6464–6465

Question 7 A coherent dark point disappears when the same populations are mixed incoherently Compare the pure states ψ± = (u1 ± u3)/ √ 2 with an incoherent mixture containing equal populations of modes 1 and 3. 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 L/2 Schematic; labels, not drawing lengths, specify the data. (a) Find the three probability densities at x = L/2. (b) Explain mathematically why equal mode populations do not specify the local density.
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Question 7 - Solution A coherent dark point disappears when the same populations are mixed incoherently x 0 L L/2 (a) Method and calculation u1(L/2) = √ 2/L, u3(L/2) = − √ 2/L. ρ+(L/2) = 0, ρ−(L/2) = 4/L, ρmix(L/2) = 2/L . (b) Interpretation and further calculation ρ± = 1 2(u2 1 + u2 3) ± u1u3, ρmix = 1 2(u2 1 + u2 3). ρ± −ρmix = ±u1u3 . Physical check Density at a point is not a point probability; a finite detector integrates over a finite interval. Coherence controls the cross term even when energy populations agree.

Question 7

A coherent dark point disappears when the same populations are mixed incoherently

Compare the pure states ψ±=(u1±u3)/2\psi_\pm=(u_1\pm u_3)/\sqrt2 with an incoherent mixture containing equal populations of modes 1 and 3. 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 three probability densities at x=L/2x=L/2.

  2. Explain mathematically why equal mode populations do not specify the local density.

Question 7 - Solution

A coherent dark point disappears when the same populations are mixed incoherently

Diagram shown in the original SVG worksheet.

(a) Method and calculation

u1(L/2)=2/L,u3(L/2)=−2/L.u_1(L/2)=\sqrt{2/L},\quad u_3(L/2)=-\sqrt{2/L}.ρ+(L/2)=0,ρ−(L/2)=4/L,ρmix(L/2)=2/L.\boxed{\rho_+(L/2)=0,\quad\rho_-(L/2)=4/L,\quad\rho_{\mathrm{mix}}(L/2)=2/L}.

(b) Interpretation and further calculation

ρ±=12(u12+u32)±u1u3,ρmix=12(u12+u32).\rho_\pm=\tfrac12(u_1^2+u_3^2)\pm u_1u_3,\qquad\rho_{\mathrm{mix}}=\tfrac12(u_1^2+u_3^2).ρ±−ρmix=±u1u3.\boxed{\rho_\pm-\rho_{\mathrm{mix}}=\pm u_1u_3}.

Physical check

Density at a point is not a point probability; a finite detector integrates over a finite interval. Coherence controls the cross term even when energy populations agree.