Schrödinger's Equation — Question 6

Section 38-6 · Physics pages 6362–6363

Question 6 Counterpropagating waves can form density fringes while carrying constant net current A stationary free-particle state has spatial part ψ = Aeikx + Be−ikx, with real A = 2B > 0 and k > 0. The common time factor is e−iEt/ℏ. The probability current is j = (ℏ/m) Im(Ψ∗∂xΨ). x ρ j > 0 Schematic; labels, not drawing lengths, specify the data. (a) Derive the spatial probability density and its extrema. (b) Compute the current and explain why a fringed density need not imply zero transport.
Show solutionHide solution
Question 6 - Solution Counterpropagating waves can form density fringes while carrying constant net current x ρ j > 0 (a) Method and calculation ρ = A2 + B2 + 2AB cos 2kx = B2(5 + 4 cos 2kx). ρmax = 9B2, ρmin = B2 . (b) Interpretation and further calculation Im(ψ∗ψ′) = k(A2 −B2). j = ℏk m (A2 −B2) = 3ℏkB2 m > 0 . Physical check The interference term modulates density but cancels from current. Unequal rightward and leftward fluxes leave a nonzero net flow.

Question 6

Counterpropagating waves can form density fringes while carrying constant net current

A stationary free-particle state has spatial part ψ=Aeikx+Be−ikx\psi=Ae^{ikx}+Be^{-ikx}, with real A=2B>0A=2B>0 and k>0k>0. The common time factor is e−iEt/ℏe^{-iEt/\hbar}. The probability current is j=(ℏ/m)Im⁡(Ψ*∂xΨ)j=(\hbar/m)\operatorname{Im}(\Psi^*\partial_x\Psi).

Diagram shown in the original SVG worksheet.

  1. Derive the spatial probability density and its extrema.

  2. Compute the current and explain why a fringed density need not imply zero transport.

Question 6 - Solution

Counterpropagating waves can form density fringes while carrying constant net current

Diagram shown in the original SVG worksheet.

(a) Method and calculation

ρ=A2+B2+2ABcos⁡2kx=B2(5+4cos⁡2kx).\rho=A^2+B^2+2AB\cos2kx=B^2(5+4\cos2kx).ρmax=9B2,ρmin=B2.\boxed{\rho_{\max}=9B^2,\qquad\rho_{\min}=B^2}.

(b) Interpretation and further calculation

Im⁡(ψ*ψ′)=k(A2−B2).\operatorname{Im}(\psi^*\psi\prime)=k(A^2-B^2).j=ℏkm(A2−B2)=3ℏkB2m>0.\boxed{j=\frac{\hbar k}{m}(A^2-B^2)=\frac{3\hbar kB^2}{m}>0}.

Physical check

The interference term modulates density but cancels from current. Unequal rightward and leftward fluxes leave a nonzero net flow.