Past exam of the quantum field theory course of the University of Oxford / 2020 / 1

For four-momentum $k$ flowing along the fermion flow, the momentum space Feynman rules areThe scalar propagator and Dirac propagator invert their respective quadratic kernels. The interaction vertex follows by expanding $\exp (i\int {\mathcal{L}}_{\mathrm{int}})$, fixing its sign. Species is conserved along each open fermion line; there is no vertex changing species 1 into species 2.

An interaction vertex carries $(2\pi {)}^4{\delta }^{(4)}(\sum k)$ with inward four-momenta. Impose momentum conservation and integrate each independent loop momentum with $d^{4}k/(2\pi {)}^4$. Include the closed fermion loop sign $-1$ for every fermion loop, the relevant fermionic signs when rearranging external states, and the Feynman-diagram symmetry factor. External incoming and outgoing fermions carry $u_s(p)$ and ${\bar{u}}_s(p)$; incoming and outgoing antifermions carry ${\bar{v}}_s(p)$ and $v_s(p)$. An external real scalar field leg has factor 1 in an amputated scattering amplitude. The Feynman i-epsilon prescription specifies the poles.

Label the initial species-1 and species-2 fermions by $(p_1,s_1)$ and $(p_2,s_2)$, and the final species-1 and species-2 fermions by $(p_3,s_3)$ and $(p_4,s_4)$. There is exactly one tree-level Feynman diagram: the $t$-channel scalar propagator joins the continuous species-1 line $1\to 3$ to the continuous species-2 line $2\to 4$. Its exchanged four-momentum $q_t$ determines $t$, in the notation of Mandelstam variables:The top row of the figure displays this Feynman diagram. A crossed attachment $1\to 4$ would change species at an interaction vertex and is forbidden. There is no initial antifermion for an annihilation channel.

With all four external fermions of species 1, both final attachments are possible. The middle row of the figure gives the two tree-level Feynman diagrams:

For a fixed ordering of external fermion states, the fermionic sign from exchanging the final identical fermions givesHere $u_j=u_{s_j}(p_j)$ are Dirac spinors. In particular, the scattering amplitude changes sign when the complete final labels $3$ and $4$ are exchanged, as required for identical fermions.

Let $p_1,p_3$ label the incoming and outgoing fermion, and $p_2,p_4$ the incoming and outgoing antifermion. The bottom row of the figure gives the two tree-level Feynman diagrams:

Thus the Mandelstam variables in the two denominators areThe two scattering amplitude terms have a relative fermionic sign of minus when written with spinor products $({\bar{u}}_3{u}_1)({\bar{v}}_2{v}_4)$ and $({\bar{v}}_2{u}_1)({\bar{u}}_3{v}_4)$. The overall phase depends on the chosen ordering of the external states. No $u$ channel joins an incoming fermion to an outgoing antifermion through a species-preserving Yukawa interaction.

Define $\mathcal{M}$ by removing the factor $i(2\pi {)}^4{\delta }^{(4)}(p_1+p_2-p_3-p_4)$ from the connected transition matrix element. The single tree-level Feynman diagram in i givesThere is no identical-particle exchange term because the species differ. Each incoming fermion has two equally populated spin angular momentum states in an unpolarized beam. Their probabilities multiply to $1/4$, giving a spin average over the four initial choices. The unobserved final spin angular momentum states are distinct outcomes, so their probabilities are added by a spin sum, with no final averaging:Taking the complex conjugate of ${\bar{u}}_3{u}_1$ gives ${\bar{u}}_1{u}_3$. The fermion spin sum therefore converts each squared bilinear into a gamma-matrix trace:Using $\mathrm{tr}1=4$, $\mathrm{tr}{\gamma }^{\mu }=0$, $\mathrm{tr}({\gamma }^{\mu }{\gamma }^{\nu })=4{\eta }^{\mu \nu }$, and the mass shell conditions $p_j^2=m^2$, each gamma-matrix trace isThe second equality follows from $q^2=2m^2-2p_1\cdot p_3$; momentum conservation gives the identical result for $p_2,p_4$. HenceThe printed prefactor $g^2$ is incorrect for the stated coupling. Each of the two Yukawa interaction vertices contributes one $g$ to the scattering amplitude, so its square is proportional to $g^4$. As a concrete counterexample, take $g=2$, $m>0$, and forward elastic kinematics $p_3=p_1$, $p_4=p_2$. Then $q^2=0$ and the spin average is $16g^4=256$, whereas the printed expression gives $16g^2=64$.

For a connected amputated Feynman diagram, write $L$ for the number of independent loop momenta. Each loop momentum integration scales as ${\lambda }^4$, each internal scalar propagator as ${\lambda }^{-2}$, and each internal Dirac propagator as ${\lambda }^{-1}$ when both four-momenta and mass are scaled. The Yukawa coupling has mass dimension zero. ThusCounting the two fermion ends and one scalar end at each interaction vertex, and using connectedness, givesConsequently $n_f=n_v$, $L=n_s+1$, andThe requested exponent requires amputation and removal of the overall momentum delta function. An ordinary unamputated connected Green function has an additional $n$ external scalar propagators, giving mass dimension $4-3n$ after stripping that delta function, or $-3n$ if it is retained. For a concrete check, the free two-point scalar propagator scales as ${\lambda }^{-2}$, not ${\lambda }^2$. The exponent $4-n$ is the engineering result for the amputated object. A fixed renormalization scale introduces logarithms; the displayed homogeneity is a naive power counting statement rather than an exact quantum scaling law at fixed subtraction scale.

More generally, with $E_s$ external scalar ends and $E_f$ external fermion ends, the same counting gives the superficial degree of divergenceThis is an upper bound on the overall ultraviolet divergence. Numerator cancellations and Lorentz invariance may lower it; an apparently convergent whole Feynman diagram can still have ultraviolet subdivergences. Local subtraction of those ultraviolet subdivergences is required first.

For pure scalar ends, $E_s=0,1,2,3,4$ can require counterterms. Lorentz invariance reduces the two-point terms to a mass counterterm and wave-function renormalization; derivatives of a single scalar field are total derivatives, and a one-derivative cubic scalar term cannot be a Lorentz scalar. With two external fermions, $E_s=0$ requires their kinetic term and mass counterterm, while $E_s=1$ requires Yukawa coupling renormalization. Four external fermions give $D=-2$, so no independent four-fermion counterterm is required.

A complete counterterm Lagrangian consistent with the species symmetry can therefore be writtenThe vacuum energy density term matters for an unnormalized vacuum functional and cancels from normalized Green functions. The tadpole counterterm can impose $⟨\varphi ⟩=0$. There is no $\varphi \mapsto -\varphi$ symmetry of the massive Yukawa interaction, so linear, cubic and quartic scalar terms cannot be excluded. For instance, in a constant background the divergent fermion loop contribution contains $(m+g\varphi {)}^4$, exhibiting all these powers. This is counterterm closure of a massive Yukawa theory.

The two species have identical couplings, so their common unitary flavour rotations preserve the Lagrangian density and their counterterms can be chosen species independent. Equality of the scalar and fermion masses, however, is not protected: $\delta m_s^2$ and $\delta m_f$ are independent. Including the displayed finite set of operators yields a renormalizable quantum field theory. Restricting the bare Lagrangian density to the terms printed in the question is not closed under renormalization.