Schrödinger's Equation — Question 9

Section 38-6 · Physics pages 6368–6369

Question 9 An energy superposition has an evolving density even though mean energy is constant Two real orthonormal eigenfunctions u1, u2 of the same time-independent Hamiltonian have energies E1, E2, with E2 > E1. A normalized state is Ψ = (u1e−iE1t/ℏ+ u2e−iE2t/ℏ)/ √ 2. E1 E2 ∆E Schematic; labels, not drawing lengths, specify the data. (a) Derive the density and its beat period at points where u1u2̸ = 0. (b) Find the mean energy and explain why total probability remains constant.
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Question 9 - Solution An energy superposition has an evolving density even though mean energy is constant E1 E2 ∆E (a) Method and calculation ρ = 1 2(u2 1 + u2 2) + u1u2 cos[(E2 −E1)t/ℏ] . Tbeat = 2πℏ E2 −E1 = h E2 −E1 . (b) Interpretation and further calculation ⟨H⟩= 1 2(E1 + E2) . ∫ u1u2 dx = 0 ⇒ ∫ ρ dx = 1 2(1 + 1) = 1 . Physical check Local probability oscillates through interference between different energies. Orthonormality cancels the integrated cross term, preserving total probability.

Question 9

An energy superposition has an evolving density even though mean energy is constant

Two real orthonormal eigenfunctions u1,u2u_1,u_2 of the same time-independent Hamiltonian have energies E1,E2E_1,E_2, with E2>E1E_2>E_1. A normalized state is Ψ=(u1e−iE1t/ℏ+u2e−iE2t/ℏ)/2\Psi=(u_1e^{-iE_1t/\hbar}+u_2e^{-iE_2t/\hbar})/\sqrt2.

Diagram shown in the original SVG worksheet.

  1. Derive the density and its beat period at points where u1u2≠0u_1u_2\ne0.

  2. Find the mean energy and explain why total probability remains constant.

Question 9 - Solution

An energy superposition has an evolving density even though mean energy is constant

Diagram shown in the original SVG worksheet.

(a) Method and calculation

ρ=12(u12+u22)+u1u2cos⁡[(E2−E1)t/ℏ].\boxed{\rho=\tfrac12(u_1^2+u_2^2)+u_1u_2\cos[(E_2-E_1)t/\hbar]}.Tbeat=2πℏE2−E1=hE2−E1.\boxed{T_{\mathrm{beat}}=\frac{2\pi\hbar}{E_2-E_1}=\frac h{E_2-E_1}}.

(b) Interpretation and further calculation

⟨H⟩=12(E1+E2).\boxed{\langle H\rangle=\tfrac12(E_1+E_2)}.∫u1u2dx=0⇒∫ρdx=12(1+1)=1.\int u_1u_2\,dx=0\quad\Rightarrow\quad\boxed{\int\rho\,dx=\tfrac12(1+1)=1}.

Physical check

Local probability oscillates through interference between different energies. Orthonormality cancels the integrated cross term, preserving total probability.