 Solution

ID: past-exam-of-the-advanced-quantum-field-theory-for-particle-physics-course-of-the-university-of-oxford/2019/2/b/i/solution

First derive the contracted gamma matrices identities from the Clifford algebra in four dimensions. Moving the outer gamma matrix through two factors gives

γ_{ρ} γ_{α} γ_{β} γ^{ρ} = 4 g_{α β} 1 .

Moving it through three factors, and using this two-factor result, gives

γ_{ρ} γ_{α} γ_{β} γ_{γ} γ^{ρ} = 2 γ_{β} γ_{γ} γ_{α} - 4 g_{β γ} γ_{α} = - 2 γ_{γ} γ_{β} γ_{α} .

Consequently

γ^{ν} \neq b γ_{μ} \neq c γ_{ν} = - 2 \neq c γ_{μ} \neq b

and

γ_{μ} \neq b \neq d γ^{μ} = 4 (b \cdot d) 1

. Apply these in succession:

T = tr (\neq a γ^{ν} \neq b γ_{μ} \neq c γ_{ν} \neq d γ^{μ}) = - 2 tr (\neq a \neq c γ_{μ} \neq b \neq d γ^{μ}) = - 8 (b \cdot d) tr (\neq a \neq c) = - 32 (a \cdot c) (b \cdot d) .

The final step uses

tr (γ^{α} γ^{β}) = 4 g^{α β}

. This is a genuinely four-dimensional instance of the gamma matrix trace identities; using it unchanged inside a

D

-dimensional loop calculation would require separate justification.

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