Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2020 / 2 / b / iii

Solution

(1/ N_{c}) \sum_{i, j} ∣ δ_{ji} ∣^{2} = 1

1/ (2 \cdot 2) = 1/4

. The Ward identities permit use of

- g_{μν}

\sum_{r} u_{r} (p) \overset{u}{ˉ}_{r} (p) = \neq p

Let

A_{μν} = γ_{ν} \frac{\neq q _{s}}{s} γ_{μ} + γ_{μ} \frac{\neq q _{u}}{u} γ_{ν} .

Reversing the gamma matrices order gives its Dirac-conjugate matrix

\overset{ˉ}{A}^{μν}

⟨ ∣ M ∣^{2} ⟩ = \frac{e ^{4} Q _{u}^{4}}{4} tr (\neq p_{3} A_{μν} \neq p_{1} \overset{ˉ}{A}^{μν}) .

For clarity evaluate the four traces before dividing by their denominators. The Mandelstam variables obey

s + t + u = 0

, with

t = (p_{1} - p_{3})^{2}

p_{1} \cdot p_{2} = s /2

p_{1} \cdot p_{4} = u (- 1/2)

and

p_{1} \cdot p_{3} = - t /2

. For the squared

s

-channel contribution, the gamma matrix trace identities imply

T_{ss} = tr (\neq p_{3} γ_{ν} \neq q_{s} γ_{μ} \neq p_{1} γ^{μ} \neq q_{s} γ^{ν}) = - 2 tr (\neq p_{3} γ_{ν} \neq q_{s} \neq p_{1} \neq q_{s} γ^{ν}) = - 2 s tr (\neq p_{3} γ_{ν} \neq p_{2} γ^{ν}) = 4 s tr (\neq p_{3} \neq p_{2}) = - 8 s u .

Here

\neq q_{s} \neq p_{1} \neq q_{s} = 2 (q_{s} \cdot p_{1}) \neq q_{s} - q_{s}^{2} \neq p_{1} = s \neq p_{2}

and

p_{3} \cdot p_{2} = - u /2

. For the squared

u

-channel contribution,

\neq q_{u} \neq p_{1} \neq q_{u} = - u \neq p_{4}

gives

T_{uu} = tr (\neq p_{3} γ_{μ} \neq q_{u} γ_{ν} \neq p_{1} γ^{ν} \neq q_{u} γ^{μ}) = - 16 u (p_{3} \cdot p_{4}) = - 8 s u .

One interference trace reduces to

T_{s u} = tr (\neq p_{3} γ_{ν} \neq q_{s} γ_{μ} \neq p_{1} γ^{ν} \neq q_{u} γ^{μ}) = - 2 tr (\neq p_{3} γ_{ν} \neq q_{s} \neq q_{u} γ^{ν} \neq p_{1}) = - 32 (q_{s} \cdot q_{u}) (p_{3} \cdot p_{1}) = 0,

since

q_{s} \cdot q_{u} = (s + t + u) /2 = 0

. The other interference trace is its complex conjugate and also vanishes. Therefore

⟨ ∣ M ∣^{2} ⟩ = - 2 e^{4} Q_{u}^{4} (\frac{u}{s} + \frac{s}{u}) = - \frac{32 e ^{4}}{81} \frac{s ^{2} + u ^{2}}{s u}, C = \frac{32 e ^{4}}{81} .

Equivalently

C = 512 π^{2} α_{em}^{2} /81

in terms of the fine-structure constant. In the physical noncollinear region

s > 0

u < 0

, so the result is positive. The singular massless endpoint

u = 0

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