Schrödinger's Equation — Question 4

Section 38-6 · Physics pages 6358–6359

Question 4 Amplitude decay and probability decay have different characteristic lengths In a semi-infinite region with constant V0 > E, a stationary electron has ψ(x) = Ce−κx for x ≥0 and κ > 0. The potential exceeds the energy by 1.00 eV. Use ℏ2/(2me) = .0381 eV nm2. Use −(ℏ2/2m)ψ′′ + V ψ = Eψ. x relative size |ψ| |ψ|2 Schematic; labels, not drawing lengths, specify the data. (a) Find κ and the amplitude e-folding length. (b) Find the distance over which the density decreases to one tenth of its boundary value.
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Question 4 - Solution Amplitude decay and probability decay have different characteristic lengths x relative size |ψ| |ψ|2 (a) Method and calculation κ2 = (V0 −E)/(ℏ2/2me) ⇒ κ = 5.12316 nm−1 . ℓamp = 1/κ = .195192 nm . (b) Interpretation and further calculation ρ(x)/ρ(0) = e−2κx = .100 ⇒ x = ln 10/(2κ). x = .224723 nm . Physical check The density e-folding length is half the amplitude e-folding length. This real decaying solution alone carries zero probability current.

Question 4

Amplitude decay and probability decay have different characteristic lengths

In a semi-infinite region with constant V0>EV_0>E, a stationary electron has ψ(x)=Ce−κx\psi(x)=Ce^{-\kappa x} for x≥0x\ge0 and κ>0\kappa>0. The potential exceeds the energy by 1.00 eV. Use ℏ2/(2me)=.0381eVnm2\hbar^2/(2m_e)=.0381\,\mathrm{eV\,nm^2}. Use −(ℏ2/2m)ψ′′+Vψ=Eψ-(\hbar^2/2m)\psi\prime\prime+V\psi=E\psi.

Diagram shown in the original SVG worksheet.

  1. Find κ\kappa and the amplitude e-folding length.

  2. Find the distance over which the density decreases to one tenth of its boundary value.

Question 4 - Solution

Amplitude decay and probability decay have different characteristic lengths

Diagram shown in the original SVG worksheet.

(a) Method and calculation

κ2=(V0−E)/(ℏ2/2me)⇒κ=5.12316nm−1.\kappa^2=(V_0-E)/(\hbar^2/2m_e)\quad\Rightarrow\quad\boxed{\kappa=5.12316\,\mathrm{nm}^{-1}}.ℓamp=1/κ=.195192nm.\boxed{\ell_{\mathrm{amp}}=1/\kappa=.195192\,\mathrm{nm}}.

(b) Interpretation and further calculation

ρ(x)/ρ(0)=e−2κx=.100⇒x=ln⁡10/(2κ).\rho(x)/\rho(0)=e^{-2\kappa x}=.100\quad\Rightarrow\quad x=\ln10/(2\kappa).x=.224723nm.\boxed{x=.224723\,\mathrm{nm}}.

Physical check

The density e-folding length is half the amplitude e-folding length. This real decaying solution alone carries zero probability current.