Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 301 / 3

Table of contents
- Solution 3

Solution

For metric signature

(+ - - -)

, the momentum-space rules for the Yukawa interaction are:

an internal scalar line contributes $i / (k^{2} - m_{ϕ}^{2} + i ϵ)$ ;
an internal fermion contributes $i (\neq k + m) / (k^{2} - m^{2} + i ϵ)$ ;
each $ϕ \overset{ˉ}{ψ} ψ$ or $ϕ \overset{χ}{ˉ} χ$ vertex contributes $- i g$ ;
incoming and outgoing fermions contribute $u_{r} (p)$ and $\overset{u}{ˉ}_{r} (p)$ , while incoming and outgoing antifermions contribute $\overset{v}{ˉ}_{s} (q)$ and $v_{s} (q)$ ;
every closed fermion loop contributes an additional minus sign.

At leading order the process has one

s

-channel scalar propagator. With

s = (p + q)^{2}

,

i M = [\overset{v}{ˉ}_{s} (q) (- i g) u_{r} (p)] \frac{i}{s - m _{ϕ}^{2} + i ϵ} [\overset{u}{ˉ}_{r^{'}} (p^{'}) (- i g) v_{s^{'}} (q^{'})] .

Thus, up to an irrelevant overall sign,

M = - \frac{g ^{2}}{s - m _{ϕ}^{2} + i ϵ} [\overset{v}{ˉ}_{s} (q) u_{r} (p)] [\overset{u}{ˉ}_{r^{'}} (p^{'}) v_{s^{'}} (q^{'})] .

The fermion spin sums and

tr (\neq a \neq b) = 4 a \cdot b

give

\sum_{r, s} ∣ \overset{v}{ˉ}_{s} (q) u_{r} (p) ∣^{2} = 4 (p \cdot q - m_{ψ}^{2}),

and the analogous final sum is

4 (p^{'} \cdot q^{'} - m_{χ}^{2})

. Therefore

X = \frac{4 g ^{4} ( p \cdot q - m _{ψ}^{2} ) ( p ^{'} \cdot q ^{'} - m _{χ}^{2} )}{( s - m _{ϕ}^{2} ) ^{2}} .

Using

p \cdot q = (s - 2 m_{ψ}^{2}) /2

and its final-state analogue yields

X = \frac{g ^{4} ( s - 4 m _{ψ}^{2} ) ( s - 4 m _{χ}^{2} )}{( s - m _{ϕ}^{2} ) ^{2}} .

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