 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/iii/paper-48/1/a/solution

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

Write

I = I_{2}

σ^{i} σ^{j} + σ^{j} σ^{i} = 2 δ^{ij} I

(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}) .

2 I_{4}

, and the spatial anticommutators are

- 2 δ^{ij} I_{4}

. Together these prove the Clifford algebra identity

{γ^{μ}, γ^{ν}} = 2 η^{μν} I_{4} .

The negative spatial sign comes from the lower-left block of each spatial gamma matrix.

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