 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2007/iii/paper-50/1/a/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

σ_{i} σ_{j} = δ_{ij} I_{2} + i ϵ_{ijk} σ_{k}

, so

{σ_{i}, σ_{j}} = 2 δ_{ij} I_{2}

(γ^{0})^{2} = I_{4}, γ^{0} γ^{i} = (- σ_{i} 0 0 σ_{i}), γ^{i} γ^{0} = (σ_{i} 0 0 - σ_{i}),

and

γ^{i} γ^{j} = (- σ_{i} σ_{j} 0 0 - σ_{i} σ_{j}) .

Thus the mixed anticommutators vanish and the spatial ones are

- 2 δ_{ij} I_{4}

. With the Minkowski metric convention

η = diag (1, - 1, - 1, - 1)

{γ^{μ}, γ^{ν}} = 2 η^{μν} I_{4} .

These matrices therefore represent the required Clifford algebra.

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