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 Incoming links: Fermionic sign

Label

e^{-} (p_{1}, s_{1}) + e^{+} (p_{2}, s_{2}) ⟶ e^{-} (p_{3}, s_{3}) + e^{+} (p_{4}, s_{4}) .

This is Bhabha scattering. There are two diagrams:

the $s$ -channel annihilation topology $e^{-} e^{+} \to γ^{*} (p_{1} + p_{2}) \to e^{-} e^{+}$ ;
the $t$ -channel exchange topology in which the electron and positron lines exchange $γ^{*} (p_{1} - p_{3})$ .

With

s = (p_{1} + p_{2})^{2}

and

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

, a consistent external-fermion ordering gives

i M_{s} i M_{t} i M = [\overline{v} (p_{2}) (- i e γ^{μ}) u (p_{1})] \frac{- i η _{μν}}{s} [\overline{u} (p_{3}) (- i e γ^{ν}) v (p_{4})], = - [\overline{u} (p_{3}) (- i e γ^{μ}) u (p_{1})] \frac{- i η _{μν}}{t} [\overline{v} (p_{2}) (- i e γ^{ν}) v (p_{4})], = i M_{s} + i M_{t} .

The displayed relative minus sign is the fermionic sign from putting the external fermion operators into the common chosen order.

- g

t

and

u

exchanges; antisymmetry of the final identical fermions requires their relative minus. Fermion-antifermion scattering has

t

exchange and

s

annihilation, again with a relative minus. With initial and final states ordered as

b_{1}^{†} d_{2}^{†} ∣0 ⟩

and

b_{3}^{†} d_{4}^{†} ∣0 ⟩

, the normal-ordered current contributes

- d^{†} d \overset{v}{ˉ} γ^{5} v

in the exchange channel but

d b

and

b^{†} d^{†}

in the annihilation channel. These fermionic signs fix the overall sign as well as their interference; changing an external-state phase changes both terms together.