String Waves and Matter Waves — Question 5

Section 39-1 · Physics pages 6420–6421

Question 5 Periodic confinement admits a constant mode that a hard-wall box excludes A particle moves on a ring of circumference L. Let ψ(s) = Aeiks and require ψ(s + L) = ψ(s). There is no magnetic flux, and E = ℏ2k2/(2m). R Schematic; labels, not drawing lengths, specify the data. (a) Derive the allowed k values and normalize the traveling waves. (b) Identify the zero-energy state and the degeneracy of each positive energy.
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Question 5 - Solution Periodic confinement admits a constant mode that a hard-wall box excludes R (a) Method and calculation eikL = 1 ⇒kj = 2πj/L , j ∈Z. 1 = ∫L 0 |A|2ds = |A|2L ⇒|A| = L−1/2 . (b) Interpretation and further calculation j = 0 : ψ0 = L−1/2, E0 = 0 . Ej = 2π2ℏ2j2 mL2 = E−j ⇒gj = 2 (j̸ = 0) . Physical check The two states at positive energy circulate in opposite directions. A ring has periodic continuity rather than zero-value endpoint constraints.

Question 5

Periodic confinement admits a constant mode that a hard-wall box excludes

A particle moves on a ring of circumference L. Let ψ(s)=Aeiks\psi(s)=Ae^{iks} and require ψ(s+L)=ψ(s)\psi(s+L)=\psi(s). There is no magnetic flux, and E=ℏ2k2/(2m)E=\hbar^2k^2/(2m).

Diagram shown in the original SVG worksheet.

  1. Derive the allowed k values and normalize the traveling waves.

  2. Identify the zero-energy state and the degeneracy of each positive energy.

Question 5 - Solution

Periodic confinement admits a constant mode that a hard-wall box excludes

Diagram shown in the original SVG worksheet.

(a) Method and calculation

eikL=1⇒kj=2πj/L,j∈ℤ.e^{ikL}=1\Rightarrow\boxed{k_j=2\pi j/L},\quad j\in\mathbb Z.1=∫0L|A|2ds=|A|2L⇒|A|=L−1/2.1=\int_0^L|A|^2ds=|A|^2L\Rightarrow\boxed{|A|=L^{-1/2}}.

(b) Interpretation and further calculation

j=0:ψ0=L−1/2,E0=0.j=0:\quad\boxed{\psi_0=L^{-1/2},\quad E_0=0}.Ej=2π2ℏ2j2mL2=E−j⇒gj=2(j≠0).E_j=\frac{2\pi^2\hbar^2j^2}{mL^2}=E_{-j}\Rightarrow\boxed{g_j=2\ (j\ne0)}.

Physical check

The two states at positive energy circulate in opposite directions. A ring has periodic continuity rather than zero-value endpoint constraints.