Quantum Field Theory

5.6 Yukawa TheoryThe interaction between a Dirac fermion of mass m and a real scalar field of mass µ isgoverned by the Yukawa theory,L = 1 2 ∂ µφ∂ µ φ − 1 2 µ2 φ 2 + ¯ψ(iγ µ ∂ µ − m)ψ − λφ ¯ψψ (5.37)which is the proper version of the baby scalar Yukawa theory we looked at in Section 3.Couplings of this type appear in the standard model, between fermions and the Higgsboson. In that context, the fermions can be leptons (such as the electron) or quarks.Yukawa originally proposed an interaction of this type as an effective theory of nuclearforces. With an eye to this, we will again refer to the φ particles as mesons, and theψ particles as nucleons. Except, this time, the nucleons have spin. (This is still nota particularly realistic theory of nucleon interactions, not least because we’re omittingisospin. Moreover, in Nature the relevant mesons are pions which are pseudoscalars, soa coupling of the form φ ¯ψγ 5 ψ would be more appropriate. We’ll turn to this briefly inSection 5.7.3).Note the dimensions of the various fields. We still have [φ] = 1, but the kineticterms require that [ψ] = 3/2. Thus, unlike in the case with only scalars, the couplingis dimensionless: [λ] = 0.We’ll proceed as we did in Section 3, firstly computing the amplitude of a particularscattering process then, with that calculation as a guide, writing down the Feynmanrules for the theory. We start with:5.6.1 An Example: Putting Spin on Nucleon ScatteringLet’s study ψψ → ψψ scattering. This is the same calculation we performed in Section(3.3.3) except now the fermions have spin. Our initial and final states are|i〉 = √ 4E ⃗p E ⃗q b s †⃗p br †⃗q|0〉 ≡ |⃗p, s;⃗q, r〉|f〉 = √ 4E ⃗p ′E ⃗q ′ b s′ †⃗p ′ b r′ †⃗q ′ |0〉 ≡ |⃗p ′ , s ′ ;⃗q ′ , r ′ 〉 (5.38)We need to be a little cautious about minus signs, because the b † ’s now anti-commute.In particular, we should be careful when we take the adjoint. We have〈f| = √ 4E ⃗p ′E ⃗q ′ 〈0| b r′⃗q ′ bs′ ⃗p ′ (5.39)We want to calculate the order λ 2 terms from the S-matrix element 〈f|S − 1 |i〉.(−iλ) 2 ∫d 4 x 1 d 4 x 2 T ( ¯ψ(x1 )ψ(x 1 )φ(x 1 )2¯ψ(x 2 )ψ(x 2 )φ(x 2 ) ) (5.40)– 115 –