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Physics / Chapter 38: Photons and Matter Waves / Section 38-6

Schrödinger's Equation — Question 10

Section 38-6 · Physics pages 6370–6371

Question 10 A finite jump in potential changes curvature without breaking derivative continuity The potential has a finite step at x = 0, with no delta-function term and constant particle mass. A stationary solution is bounded near the step. Use −(ℏ2/2m)ψ′′ + V ψ = Eψ. x finite step Schematic; labels, not drawing lengths, specify the data. (a) Integrate the equation across [−ϵ, ϵ] and derive the derivative matching condition. (b) For a nonzero boundary value ψ(0), derive the jump in curvature when V (0+) −V (0−) = V0.
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Question 10 - Solution A finite jump in potential changes curvature without breaking derivative continuity x finite step (a) Method and calculation −ℏ2 2m[ψ′(ϵ) −ψ′(−ϵ)] = ∫ϵ −ϵ (E −V )ψ dx −→0. ψ′(0+) = ψ′(0−) ; ψ(0+) = ψ(0−) . (b) Interpretation and further calculation ψ′′(0±) = 2m ℏ2 [V (0±) −E]ψ(0). ψ′′(0+) −ψ′′(0−) = 2mV0 ℏ2 ψ(0) . Physical check A jump in the wave function would introduce a derivative of a delta function into its second derivative, with no balancing finite term. Singular or infinite potentials require separate boundary analysis.

Question 10

A finite jump in potential changes curvature without breaking derivative continuity

The potential has a finite step at x=0x=0, with no delta-function term and constant particle mass. A stationary solution is bounded near the step. Use −(ℏ2/2m)ψ′′+Vψ=Eψ-(\hbar^2/2m)\psi\prime\prime+V\psi=E\psi.

Diagram shown in the original SVG worksheet.

  1. Integrate the equation across [−ϵ,ϵ][-\epsilon,\epsilon] and derive the derivative matching condition.

  2. For a nonzero boundary value ψ(0)\psi(0), derive the jump in curvature when V(0+)−V(0−)=V0V(0^+)-V(0^-)=V_0.

Question 10 - Solution

A finite jump in potential changes curvature without breaking derivative continuity

Diagram shown in the original SVG worksheet.

(a) Method and calculation

−ℏ22m[ψ′(ϵ)−ψ′(−ϵ)]=∫−ϵϵ(E−V)ψdx→0.-\frac{\hbar^2}{2m}[\psi\prime(\epsilon)-\psi\prime(-\epsilon)]=\int_{-\epsilon}^{\epsilon}(E-V)\psi\,dx\longrightarrow0.ψ′(0+)=ψ′(0−);ψ(0+)=ψ(0−).\boxed{\psi\prime(0^+)=\psi\prime(0^-)};\qquad\boxed{\psi(0^+)=\psi(0^-)}.

(b) Interpretation and further calculation

ψ′′(0±)=2mℏ2[V(0±)−E]ψ(0).\psi\prime\prime(0^\pm)=\frac{2m}{\hbar^2}[V(0^\pm)-E]\psi(0).ψ′′(0+)−ψ′′(0−)=2mV0ℏ2ψ(0).\boxed{\psi\prime\prime(0^+)-\psi\prime\prime(0^-)=\frac{2mV_0}{\hbar^2}\psi(0)}.

Physical check

A jump in the wave function would introduce a derivative of a delta function into its second derivative, with no balancing finite term. Singular or infinite potentials require separate boundary analysis.

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