Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 4 / 34E (Applications of Quantum Mechanics)

Table of contents
- a 34E
  - Solution a
- b 34E
  - Solution b
- c 34E
  - Solution c

a

Solution

Under the stated potential transformation, set

ψ^{'} (x, t) = e^{- i e f (x, t) /ℏ} ψ (x, t) .

Then

(\nabla + \frac{i e}{ℏ} (A + \nabla f)) ψ^{'} = e^{- i e f /ℏ} (\nabla + \frac{i e}{ℏ} A) ψ

and

(\partial_{t} - \frac{i e}{ℏ} (ϕ - \partial_{t} f)) ψ^{'} = e^{- i e f /ℏ} (\partial_{t} - \frac{i e}{ℏ} ϕ) ψ .

Applying the spatial identity twice proves the Gauge covariance of the Schrödinger equation.

b

Solution

D_{A} = \nabla + i e A /ℏ

, the current is

J = - \frac{i ℏ}{2 m} (ψ^{*} D_{A} ψ - ψ (D_{A} ψ)^{*}) .

The two covariant factors acquire opposite phases under a gauge transformation, so

J

is invariant.

Multiply the Schrodinger equation by

ψ^{*}

, subtract its complex conjugate multiplied by

ψ

, and use

ψ^{*} D_{A}^{2} ψ - ψ (D_{A}^{2} ψ)^{*} = \nabla \cdot {ψ^{*} D_{A} ψ - ψ (D_{A} ψ)^{*}} .

The real scalar-potential terms cancel, leaving

\partial_{t} ∣ ψ ∣^{2} + \nabla \cdot J = 0.

c

Solution

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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