Schrödinger's Equation — Question 5

Section 38-6 · Physics pages 6360–6361

Question 5 A specified Gaussian stationary wave determines the potential that supports it A nodeless stationary state is ψ(x) = A exp[−x2/(4a2)] with a > 0. Choose the potential-energy zero so that V (0) = 0. Use −(ℏ2/2m)ψ′′ + V ψ = Eψ. x V ψ Schematic; labels, not drawing lengths, specify the data. (a) Infer V (x) and E by differentiating the prescribed state. (b) Express the resulting potential as mΩ2x2/2 and relate E to Ω.
Show solutionHide solution
Question 5 - Solution A specified Gaussian stationary wave determines the potential that supports it x V ψ (a) Method and calculation ψ′′ ψ = x2 4a4 − 1 2a2 , V = E + ℏ2 2m ψ′′ ψ . E = ℏ2 4ma2 , V (x) = ℏ2x2 8ma4 . (b) Interpretation and further calculation 1 2mΩ2 = ℏ2 8ma4 ⇒ Ω= ℏ 2ma2 . E = 1 2ℏΩ. Physical check This is an inverse check of one supplied state. It does not assume or derive the full trapped-particle spectrum, which belongs to the following chapter.

Question 5

A specified Gaussian stationary wave determines the potential that supports it

A nodeless stationary state is ψ(x)=Aexp⁡[−x2/(4a2)]\psi(x)=A\exp[-x^2/(4a^2)] with a>0a>0. Choose the potential-energy zero so that V(0)=0V(0)=0. Use −(ℏ2/2m)ψ′′+Vψ=Eψ-(\hbar^2/2m)\psi\prime\prime+V\psi=E\psi.

Diagram shown in the original SVG worksheet.

  1. Infer V(x)V(x) and E by differentiating the prescribed state.

  2. Express the resulting potential as mΩ2x2/2m\Omega^2x^2/2 and relate E to Ω\Omega.

Question 5 - Solution

A specified Gaussian stationary wave determines the potential that supports it

Diagram shown in the original SVG worksheet.

(a) Method and calculation

ψ′′ψ=x24a4−12a2,V=E+ℏ22mψ′′ψ.\frac{\psi\prime\prime}{\psi}=\frac{x^2}{4a^4}-\frac1{2a^2},\quad V=E+\frac{\hbar^2}{2m}\frac{\psi\prime\prime}{\psi}.E=ℏ24ma2,V(x)=ℏ2x28ma4.\boxed{E=\frac{\hbar^2}{4ma^2}},\qquad\boxed{V(x)=\frac{\hbar^2x^2}{8ma^4}}.

(b) Interpretation and further calculation

12mΩ2=ℏ28ma4⇒Ω=ℏ2ma2.\frac12m\Omega^2=\frac{\hbar^2}{8ma^4}\quad\Rightarrow\quad\boxed{\Omega=\frac{\hbar}{2ma^2}}.E=12ℏΩ.\boxed{E=\tfrac12\hbar\Omega}.

Physical check

This is an inverse check of one supplied state. It does not assume or derive the full trapped-particle spectrum, which belongs to the following chapter.