 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-301/3/i/solution

η^{μν} = diag (1, - 1, - 1, - 1)

, the defining Clifford algebra relation is

{γ^{μ}, γ^{ν}} = γ^{μ} γ^{ν} + γ^{ν} γ^{μ} = 2 η^{μν} 1_{4} .

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