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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-4/34e/c/solution

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Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 4 / 34E / c / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-03

The proposed potentials give

\nabla \times A = (0, 0, B)

and

- \nabla ϕ = (0, E, 0)

. Substituting

ψ = e^{ik x} e^{- i E t /ℏ} φ (y)

gives

[- \frac{ℏ ^{2}}{2 m} \frac{d ^{2}}{d y ^{2}} + \frac{( ℏ k - e B y ) ^{2}}{2 m} + e E y] φ = E φ .

This is a harmonic oscillator with cyclotron frequency and centre

ω_{1} = \frac{∣ e B ∣}{m}, y_{0} = \frac{ℏ k}{e B} - \frac{m E}{e B ^{2}}

for

e B > 0

as written. Completing the square yields

E = ℏ ω_{1} (n + \frac{1}{2}) + W, W = \frac{ℏ k E}{B} - \frac{m E ^{2}}{2 B ^{2}} .

Thus the Landau levels in crossed electric and magnetic fields retain their oscillator spacing but acquire a linear dependence on $k$ , lifting their guiding-centre degeneracy.

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