 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-342/1/a/solution

j_{I}^{α}

and add the minimal-coupling term

L_{src} = - a_{I α} j_{I}^{α} .

\frac{1}{2 π ℏ} K_{I J} ϵ^{α β γ} \partial_{β} a_{J γ} = j_{I}^{α} .

For a static quasiparticle of integer gauge-charge vector

q

, integration over a small surrounding disk yields the Chern--Simons flux attachment relation

Φ_{I} := \int d^{2} x ϵ^{0 ij} \partial_{i} a_{I j} = 2 π ℏ (K^{- 1} q)_{I} .

q^{'}

carried once around this flux acquires the Aharonov--Bohm phase

\frac{1}{ℏ} q^{' T} Φ = 2 π q^{' T} K^{- 1} q = 2 θ_{q^{'} q},

θ_{q^{'} q} = π q^{' T} K^{- 1} q, e^{2 i θ_{q^{'} q}} = e^{2 πi q^{' T} K^{- 1} q} .

The constituent-particle current has charge vector

t = (1, \dots, 1)^{T}

. Its integral over the same disk is

N = \frac{1}{2 π ℏ} t^{T} Φ = t^{T} K^{- 1} q .

For

K = I_{n}

and

q = e_{J}

, this gives

N = 1

. The self-exchange angle is

θ_{qq} = π e_{J}^{T} e_{J} = π

, so the excitation is a fermion with one unit of microscopic particle number and may be identified with an electron.

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