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Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 46 / 3 / ii / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-06

Use the prepoint prescription for the separable Hamiltonian. Its regulated exponent is

\sum_{i = 1}^{N} [i p_{i} (q_{i + 1} - q_{i}) - Δ t_{i} (\frac{p _{i}^{2}}{2 m} + V (q_{i}))] .

(1)

Each momentum integration is an ordinary Gaussian integral:

\int_{- \infty}^{\infty} d p_{i} e^{- Δ t_{i} p_{i}^{2} / (2 m) + i p_{i} (q_{i + 1} - q_{i})} = \frac{2 πm}{Δ t _{i}} exp [- \frac{m ( q _{i + 1} - q _{i} ) ^{2}}{2Δ t _{i}}] .

(2)

Therefore the measure written in the question gives exactly

D q_{raw, N} = i = 1 \prod N \frac{2 πm}{Δ t _{i}} d q_{i}, S_{E, N} = i = 1 \sum N [\frac{m ( q _{i + 1} - q _{i} ) ^{2}}{2Δ t _{i}} + Δ t_{i} V (q_{i})] .

(3)

The resulting configuration integral is

\int D q_{raw, N} e^{- S_{E, N}}

. This is Gaussian momentum integration in a phase-space path integral. Formally its exponent tends to the usual Euclidean kinetic-plus-potential action, but the slice-dependent factors in the measure must be retained.

For a normalized quantum-mechanical propagator, Fourier completeness uses

d p_{i} / (2 π)

rather than

d p_{i}

. With that normalization the configuration measure is

D q_{N} = i = 1 \prod N \frac{m}{2 π Δ t _{i}} d q_{i} .

(4)

This differs from the raw measure by

(2 π)^{- N}

. It makes the free single-step kernel integrate to one and tend to a delta distribution as the interval tends to zero. In these formulas

q_{N + 1}

is fixed and

q_{1}, \dots, q_{N}

are integrated for propagation from an initial wavefunction. If both endpoints are fixed, omit

d q_{1}

while retaining its slice normalization factor. The continuum expression means the limit of these measures and exponents, not a flat product of

d q (t)

with no time-step weights.

We now address the unheaded real-time continuation. Set

m = ℏ = 1

. The normalized short-time Schrödinger kernel gives the final-slice recurrence

ψ (q, t + Δ t) = \frac{1}{2 πi Δ t} \int_{- \infty}^{\infty} d q^{'} exp [\frac{i ( q - q ^{'} ) ^{2}}{2Δ t} - i Δ t V (q^{'})] ψ (q^{'}, t) .

(5)

The square-root branch is fixed by the usual damped Fresnel integral, or continuation from the Euclidean kernel. The hint's

d q^{'} / Δ t

contains the essential time-step dependence; the constant

(2 πi)^{- 1/2}

fixes the identity limit and is included at every step.

Put

η = q^{'} - q

. The normalized oscillatory Gaussian has moments

⟨ 1 ⟩ = 1, ⟨ η ⟩ = 0, ⟨ η^{2} ⟩ = i Δ t, ⟨ η^{4} ⟩ = 3 (i Δ t)^{2} .

(6)

Taylor-expand the smooth wavefunction and the potential over one short step. Odd moments vanish, and potential-derivative corrections first contribute at order

(Δ t)^{2}

. Thus

ψ (q, t + Δ t) = ψ (q, t) + \frac{i Δ t}{2} \partial_{q}^{2} ψ (q, t) - i Δ t V (q) ψ (q, t) + O ((Δ t)^{2}) .

(7)

Subtract the initial value, divide by

Δ t

and take the limit:

i \partial_{t} ψ (q, t) = [- \frac{1}{2} \partial_{q}^{2} + V (q)] ψ (q, t) .

(8)

This proves the Time-dependent Schrodinger equation by the Schrödinger equation from a short-time path integral argument. Nonuniform partitions give the same limit when their largest time step tends to zero.

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