Two- and Three-Dimensional Electron Traps — Question 5

Section 39-6 · Physics pages 6520–6521

Question 5 A superposition within a degenerate level can create a stationary diagonal node In a square well, let u12 = (2/L) sin(πx/L) sin(2πy/L) and u21 = (2/L) sin(2πx/L) sin(πy/L). They are orthonormal and share energy 5C/L2. Consider ψ = (u12 −u21)/ √ 2. x y Lx Ly node x = y Schematic; labels, not drawing lengths, specify the data. (a) Check normalization and show that the diagonal x = y is a node. (b) Find the time evolution and determine whether the density changes.
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Question 5 - Solution A superposition within a degenerate level can create a stationary diagonal node x y Lx Ly node x = y (a) Method and calculation ⟨ψ|ψ⟩= 1 2(1 + 1 −0 −0) = 1 . u12(x, x) = u21(x, x) ⇒ψ(x, x) = 0 . (b) Interpretation and further calculation Hψ = (5C/L2)ψ ⇒Ψ(x, y, t) = e−i5Ct/(ℏL2)ψ(x, y) . ∂t|Ψ|2 = 0 . Physical check A superposition is not necessarily time-dependent in its density. States within the same degenerate energy share one time phase.

Question 5

A superposition within a degenerate level can create a stationary diagonal node

In a square well, let u12=(2/L)sin⁡(πx/L)sin⁡(2πy/L)u_{12}=(2/L)\sin(\pi x/L)\sin(2\pi y/L) and u21=(2/L)sin⁡(2πx/L)sin⁡(πy/L)u_{21}=(2/L)\sin(2\pi x/L)\sin(\pi y/L). They are orthonormal and share energy 5C/L25C/L^2. Consider ψ=(u12−u21)/2\psi=(u_{12}-u_{21})/\sqrt2.

Diagram shown in the original SVG worksheet.

  1. Check normalization and show that the diagonal x=yx=y is a node.

  2. Find the time evolution and determine whether the density changes.

Question 5 - Solution

A superposition within a degenerate level can create a stationary diagonal node

Diagram shown in the original SVG worksheet.

(a) Method and calculation

⟨ψ|ψ⟩=12(1+1−0−0)=1.\langle\psi|\psi\rangle=\tfrac12(1+1-0-0)=\boxed{1}.u12(x,x)=u21(x,x)⇒ψ(x,x)=0.u_{12}(x,x)=u_{21}(x,x)\Rightarrow\boxed{\psi(x,x)=0}.

(b) Interpretation and further calculation

Hψ=(5C/L2)ψ⇒Ψ(x,y,t)=e−i5Ct/(ℏL2)ψ(x,y).H\psi=(5C/L^2)\psi\Rightarrow\boxed{\Psi(x,y,t)=e^{-i5Ct/(\hbar L^2)}\psi(x,y)}.∂t|Ψ|2=0.\boxed{\partial_t|\Psi|^2=0}.

Physical check

A superposition is not necessarily time-dependent in its density. States within the same degenerate energy share one time phase.