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Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 4 / 4C / Solution

The one-dimensional Time-dependent Schrodinger equation for a wavefunction in a real potential is

i ℏ \frac{\partial Ψ}{\partial t} = - \frac{ℏ ^{2}}{2 m} \frac{\partial ^{2} Ψ}{\partial x ^{2}} + U Ψ.

Multiplying this equation by

Ψ^{*}

, multiplying its complex conjugate by

Ψ

, and subtracting gives the probability continuity equation

\frac{\partial ∣Ψ ∣ ^{2}}{\partial t} + \frac{\partial j}{\partial x} = 0, j = \frac{ℏ}{2 mi} (Ψ^{*} \frac{\partial Ψ}{\partial x} - Ψ \frac{\partial Ψ ^{*}}{\partial x}) .

If

Ψ

and its first derivative decay sufficiently rapidly as

x \to \pm \infty

, then the probability current

j

vanishes at both ends. Integrating the continuity equation and using the fundamental theorem of calculus yields

\frac{d}{d t} \int_{- \infty}^{\infty} ∣Ψ (x, t) ∣^{2} d x = - j (\infty, t) + j (- \infty, t) = 0.

This conservation of quantum probability is required by the Born rule: once a normalizable wavefunction has total probability one, its time evolution must preserve that normalization.

For the stated Gaussian wave packet, put

β = ℏ t / m

. Since

f (t) = \frac{1}{α + i β}, ∣ f (t) ∣ = \frac{1}{α ^{2} + β ^{2}}, Re f (t) = \frac{α}{α ^{2} + β ^{2}},

its squared modulus is

∣Ψ (x, t) ∣^{2} = \frac{C ^{2}}{α ^{2} + β ^{2}} exp (- \frac{α x ^{2}}{α ^{2} + β ^{2}}) .

The Gaussian integral then gives

\int_{- \infty}^{\infty} ∣Ψ (x, t) ∣^{2} d x = \frac{C ^{2}}{α ^{2} + β ^{2}} \frac{π ( α ^{2} + β ^{2} )}{α} = C^{2} \frac{π}{α},

which is independent of time, as required.

Solved by gpt-5.6-sol high.

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