 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ib/paper-2/15d/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

i ℏ \frac{\partial ψ}{\partial t} = - \frac{ℏ ^{2}}{2 m} \nabla^{2} ψ + U ψ, - i ℏ \frac{\partial ψ ^{*}}{\partial t} = - \frac{ℏ ^{2}}{2 m} \nabla^{2} ψ^{*} + U ψ^{*} .

Because the potential

U

is real, its two contributions cancel when differentiating the probability density

ρ = ψ^{*} ψ

. Therefore

\frac{\partial ρ}{\partial t} = ψ^{*} \frac{\partial ψ}{\partial t} + ψ \frac{\partial ψ ^{*}}{\partial t} = \frac{i ℏ}{2 m} (ψ^{*} \nabla^{2} ψ - ψ \nabla^{2} ψ^{*}) = - \nabla \cdot [- \frac{i ℏ}{2 m} (ψ^{*} \nabla ψ - ψ \nabla ψ^{*})] .

J

, so

\frac{\partial ρ}{\partial t} + \nabla \cdot J = 0 .

This probability continuity equation says that probability can leave a region only through the current across its boundary.

 New to topics? Read the docs here!