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 Incoming links: Past exam of the advanced quantum field theory course of the University of Oxford / 2022 / 2 / b / i / Solution

Past exam of the advanced quantum field theory course of the University of Oxford / 2022 / 2 / b / iii / Solution 2026-10-25

Both

s

-channel annihilation and

t

-channel gluon exchange are now allowed. Their spin sum factors are the Bhabha scattering factors with

e \mapsto g

. Their color charge tensors, with the external indices of Solution, are

c_{s} = (T^{a})_{ji} (T^{a})_{k l}, c_{t} = (T^{a})_{ki} (T^{a})_{j l} .

Each squared term has

K_{ss} = K_{tt} = (N_{c}^{2} - 1) / (4 N_{c}^{2})

. The cross term instead has the color average

K_{s t} = \frac{1}{N _{c}^{2}} \sum_{a, b} tr (T^{a} T^{b} T^{a} T^{b}) .

To evaluate it, use the SU(N) fundamental completeness relation

\sum_{a} (T^{a})_{ij} (T^{a})_{k l} = \frac{1}{2} (δ_{i l} δ_{jk} - \frac{1}{N _{c}} δ_{ij} δ_{k l}) .

This follows by completing the orthogonal basis of traceless Hermitian matrices with the identity. Since

tr T^{b} = 0

, it gives

\sum_{a} T^{a} T^{b} T^{a} = - T^{b} / (2 N_{c})

, and hence

K_{s t} = - \frac{N _{c}^{2} - 1}{4 N _{c}^{3}} = - \frac{1}{N _{c}} K_{ss} .

Multiplying the two squared Bhabha scattering terms by

K_{ss}

and its cross term by

K_{s t}

gives

\overline{∣ M ∣^{2}} = g^{4} \frac{N _{c}^{2} - 1}{2 N _{c}^{2}} [\frac{t ^{2} + u ^{2}}{s ^{2}} + \frac{s ^{2} + u ^{2}}{t ^{2}} - \frac{2}{N _{c}} \frac{u ^{2}}{s t}] .

For the SU(3) gauge group, this is

\overline{∣ M ∣^{2}} = \frac{4 g ^{4}}{9} [\frac{t ^{2} + u ^{2}}{s ^{2}} + \frac{s ^{2} + u ^{2}}{t ^{2}}] - \frac{8 g ^{4}}{27} \frac{u ^{2}}{s t} .

The negative color charge overlap reverses the sign of the QED interference. Since

s t < 0

, this cross contribution is positive in the physical region. No additional spinor calculation is needed; all the new information is in the SU(N) fundamental completeness relation.

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Past exam of the advanced quantum field theory course of the University of Oxford / 2022 / 2 / b / ii / Solution 2026-10-25

The two Bhabha scattering tree-level Feynman diagrams are the two panels of Solution with a photon mediator:

s

-channel annihilation with internal momentum

p_{1} + p_{2}

, and

t

-channel exchange with internal momentum

p_{1} - p_{3}

. Each interaction vertex preserves the electron species. There is no independent

u

-channel Feynman diagram, since the outgoing electron and Positron are distinguishable.

Removing the color charge factor from Solution and replacing

g

by

e

gives

\overline{∣ M_{s} ∣^{2}} = 2 e^{4} \frac{t ^{2} + u ^{2}}{s ^{2}} .

Crossing symmetry interchanges

s

and

t

for the exchange Feynman diagram, giving

\overline{∣ M_{t} ∣^{2}} = 2 e^{4} \frac{s ^{2} + u ^{2}}{t ^{2}} .

Squaring the sum of the two scattering amplitudes also produces the interference term

2 Re \overline{M_{s} M_{t}^{*}}

. The given third term must be this cross term: it has both propagator denominators

s t

, whereas a squared individual Feynman diagram has

s^{2}

or

t^{2}

. With the correct relative fermionic sign, its value is

4 e^{4} u^{2} / (s t)

. Thus

\overline{∣ M ∣^{2}} = 2 e^{4} [\frac{t ^{2} + u ^{2}}{s ^{2}} + \frac{s ^{2} + u ^{2}}{t ^{2}} + \frac{2 u ^{2}}{s t}] .

The coefficient of the last displayed fraction is positive, but

s > 0

and

t < 0

in the physical scattering region, so this interference is destructive. The relative sign originates in the two possible external fermion contractions; changing an external-state phase changes both scattering amplitudes together.^[ref]

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