String Waves and Matter Waves — Question 3

Section 39-1 · Physics pages 6416–6417

Question 3 Mixed boundary conditions shift the spatial spectrum by half an index Consider a mode on 0 < x < L satisfying ψ(0) = 0 and ψ′(L) = 0. Treat these as supplied mathematical boundary conditions, as for a string fixed at one end and free at the other. x L Schematic; labels, not drawing lengths, specify the data. (a) Derive the positive allowed wave numbers. (b) Using E = ℏ2k2/(2m), compare the first two mode energies for a particle with these boundary conditions.
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Question 3 - Solution Mixed boundary conditions shift the spatial spectrum by half an index x L (a) Method and calculation ψ = A sin kx, ψ′(L) = Ak cos kL = 0. kn = (n + 1 2)π/L , n = 0, 1, 2, . . . . (b) Interpretation and further calculation En = ℏ2π2 2mL2 (n + 1 2)2. E1/E0 = (3/2)2/(1/2)2 = 9 . Physical check These are not the two impenetrable-wall boundary conditions of the ordinary infinite well. The different boundary condition is what changes the spectrum.

Question 3

Mixed boundary conditions shift the spatial spectrum by half an index

Consider a mode on 0<x<L0<x<L satisfying ψ(0)=0\psi(0)=0 and ψ′(L)=0\psi\prime(L)=0. Treat these as supplied mathematical boundary conditions, as for a string fixed at one end and free at the other.

Diagram shown in the original SVG worksheet.

  1. Derive the positive allowed wave numbers.

  2. Using E=ℏ2k2/(2m)E=\hbar^2k^2/(2m), compare the first two mode energies for a particle with these boundary conditions.

Question 3 - Solution

Mixed boundary conditions shift the spatial spectrum by half an index

Diagram shown in the original SVG worksheet.

(a) Method and calculation

ψ=Asin⁡kx,ψ′(L)=Akcos⁡kL=0.\psi=A\sin kx,\quad\psi\prime(L)=Ak\cos kL=0.kn=(n+12)π/L,n=0,1,2,….\boxed{k_n=(n+\tfrac12)\pi/L},\quad n=0,1,2,\ldots.

(b) Interpretation and further calculation

En=ℏ2π22mL2(n+12)2.E_n=\frac{\hbar^2\pi^2}{2mL^2}(n+\tfrac12)^2.E1/E0=(3/2)2/(1/2)2=9.\boxed{E_1/E_0=(3/2)^2/(1/2)^2=9}.

Physical check

These are not the two impenetrable-wall boundary conditions of the ordinary infinite well. The different boundary condition is what changes the spectrum.