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It is convenient to evaluate the one-particle irreducible vertices using the Euclidean action and then translate the counterterms back to the Minkowski spacetime Lagrangian density. For the tadpole integral, analytic continuation of solution givesIndeed $\Gamma (-1+ϵ/2)=\Gamma (ϵ/2)/(ϵ/2-1)=-2/ϵ+O(1)$. The negative pole is a property of analytic continuation, not the value of a convergent positive integral.

The Euclidean space inverse two-point vertex has the one-loop correctionThe factor $1/2$ is the symmetry factor weight of the tadpole diagram. A mass counterterm $\delta m^2={\lambda }_0{m}^2/(16{\pi }^2ϵ)$ cancels this pole. There is no momentum-dependent pole at this order, and hence no one-loop wave-function renormalization.

For the four-point vertex, the three bubble diagrams giveOne can check the relative minus sign independently from the one-loop effective action: its $\frac{1}{2}\mathrm{Tr}\log (1+G{\lambda }_0{\varphi }^2/2)$ term has a positive term linear in ${\lambda }_0{\varphi }^2$ and a negative term quadratic in it. Thus $\delta \lambda =3{\lambda }_0^2/(16{\pi }^2ϵ)$ cancels the quartic pole.

In the Euclidean action, these two counterterms are $+\delta m^2{\varphi }^2/2+\delta \lambda {\varphi }^4/4!$. Returning to the Minkowski spacetime Lagrangian density givesFor a direct sign check with Feynman rules, the two amputated loop pole insertions are $+i{\lambda }_0{m}^2/(16{\pi }^2ϵ)$ and $+i3{\lambda }_0^2/(16{\pi }^2ϵ)$, while the counterterm insertions are $-i\delta m^2$ and $-i\delta \lambda$.

As usual, the displayed pole residues suppress renormalization scale powers: $[{\lambda }_0]=ϵ$ away from four dimensions. The next part restores these powers explicitly with a dimensionless renormalized coupling. Replacing the tree coupling by its renormalized counterpart inside a one-loop residue changes only higher-loop terms.

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.

The momentum space Feynman rules use the scalar propagator and two Dirac propagatorsThe cubic scalar field interaction vertex is $-i\lambda$. Each of the two flavour-changing Yukawa coupling interaction vertices, $\varphi \bar{\Psi }\chi$ and $\varphi \bar{\chi }\Psi$, has factor $-ig{\delta }_{ab}$ in Dirac spinor indices. A fermion flow arrow continues through each interaction vertex, but its species changes. There is no $\varphi \bar{\Psi }\Psi$ or $\varphi \bar{\chi }\chi$ interaction vertex.

Every interaction vertex carries $(2\pi {)}^4{\delta }^{(4)}(\sum p)$ with inward four-momenta. Integrate each independent loop momentum with $d^4\ell /(2\pi {)}^4$, divide by the Feynman-diagram symmetry factor, and include a minus sign and a gamma-matrix trace for each closed fermion loop. The $i0$ is the Feynman i-epsilon prescription.

For an amputated scattering amplitude, incoming fermions supply $u^s(p)$ and outgoing fermions supply ${\bar{u}}^s(p)$; incoming antifermions supply ${\bar{v}}^s(p)$ and outgoing antifermions supply $v^s(p)$. An external real scalar field particle supplies $1$. Use the standard relativistic normalization of a one-particle state; the external factors $1/\sqrt{2E}$ belong to that normalization and the Lorentz-invariant phase-space measure, rather than an additional factor in the invariant scattering amplitude.

To include all momentum-dependent contributions to the two-point function, put one external line at each interaction vertex. Let $r$ be the number of lines connecting the vertices, and $t_A,t_B$ their numbers of tadpole diagrams. For vertex valences $v_A,v_B\in \{4,6\}$,The integer $r$ is odd, and $r=1$ gives a one-particle-reducible Feynman diagram and is excluded. Thus the complete list is

The coefficient of the mixed graph includes the reflected external attachment as well as its reciprocal Feynman-diagram symmetry factor. If both external lines attach to the same interaction vertex, no external momentum enters the connecting internal lines, and that Feynman diagram cannot contribute to wave-function renormalization.

For incoming external momentum $p$, assign the connecting lines momenta $k_1,\dots ,k_{r-1},p-{\sum }_j{k}_j$. DefineThe momentum space Feynman rules give, in the order listed,Each $C$ is a reciprocal Feynman-diagram symmetry factor, including any equivalent external attachments. There are respectively $2,3,4,4$ independent loop momenta.

If “contribute to wavefunction renormalization” means requiring a divergent kinetic counterterm after subtraction of all subdivergences, only the last topology qualifies. The first three still have nontrivial external momentum dependence and give finite contributions to a normalization condition; their raw divergent factors arise from mass counterterms and interaction vertex counterterms. Keeping all four here makes this distinction explicit. The ultraviolet divergence analysis in d establishes it directly.

In momentum space, the Feynman rules for the real scalar field areEach interaction vertex supplies $(2\pi {)}^6{\delta }^{(6)}(\sum p)$, with all its momenta counted inward. Impose momentum conservation, integrate each independent loop momentum with $d^{6}p/(2\pi {)}^6$, and divide by the Feynman-diagram symmetry factor. For a connected amputated scattering amplitude, remove the external scalar propagators and retain a single overall momentum conservation factor. These signs follow by expanding $\exp (i\int {\mathcal{L}}_{\mathrm{int}})$; the factorials in the Lagrangian density cancel the corresponding attachments in the Wick theorem. The Feynman i-epsilon prescription specifies the scalar propagator poles.

The auxiliary field $\sigma$ has quadratic kernel $-1$ and no kinetic term. Inverting that kernel, rather than introducing an artificial scalar mass, gives the momentum space Feynman rulesThe auxiliary-field propagator is independent of momentum; in position space it is $-i{\delta }^{(4)}(x-y)$. At each interaction vertex attach $(2\pi {)}^4{\delta }^{(4)}(\sum p)$ for momentum conservation. Integrate independent loop momenta with $d^{4}p/(2\pi {)}^4$, respect fermion flow, include the closed fermion loop sign $-1$, and use the appropriate symmetry factor. Incoming and outgoing fermions carry $u_s(p)$ and ${\bar{u}}_s(p)$; incoming and outgoing antiparticles carry ${\bar{v}}_s(p)$ and $v_s(p)$. Exchanging external fermions introduces the fermionic sign. Here $/p={\gamma }^{\mu }{p}_{\mu }$ is Feynman slash notation.

For a dimensional check, $[\psi ]=3/2$, $[\sigma ]=2$, $[g]=-1$ in mass dimensions. Eliminating $\sigma$ consequently produces a four-fermion interaction of negative coupling mass dimension, not a propagating scalar degree of freedom.

Use $\hslash =c=1$, the Minkowski metric $\eta =\mathrm{diag}(1,-1)$, and $p^2=(p^0{)}^2-(p^1{)}^2$. The momentum-space Feynman rules for the real scalar field areAt each interaction vertex impose two-dimensional momentum conservation; before removing the overall conservation factor this is $(2\pi {)}^2{\delta }^{(2)}({\sum }_j{p}_j)$. Integrate each independent loop momentum with $d^{2}k/(2\pi {)}^2$, and divide a Feynman diagram by its symmetry factor. External lines carry the same scalar propagator in an unamputated two-point correlation function; they are removed in an amputated scattering amplitude.

The quadratic action is inverted with the Feynman i-epsilon prescription to obtain the scalar propagator. Expanding the Dyson series gives $-i{g}_4/4!$ per insertion; the $4!$ attachments in the Wick theorem cancel the factorial. Fourier integration over the insertion point supplies the momentum conservation factor. Automorphisms that leave the labelled external attachments fixed account for the remaining symmetry factor. In two spacetime dimensions the mass dimensions are $[\varphi ]=0$ and $[g_4]=2$.

Interpret the repeated species label in the Yukawa interaction as summed over $i=1,2$: each Feynman vertex is diagonal in species and equals $-ig$. Define the invariant scattering amplitude by $S-1=i(2\pi {)}^4{\delta }^{(4)}(P_f-P_i)\mathcal{M}$. There is one tree-level Feynman diagram, an $s$-channel scalar exchange. In the left panel, $j=2$ for this part; the right panel only occurs in d. Arrows show fermion-number flow, so an antifermion arrow opposes its physical momentum.

Writing $k=p_1+p_2=p_3+p_4$ and $s=k^2$, the momentum-space Feynman rules giveThe scalar carries no spin label. Average over the $2\times 2$ incoming spin states and sum the four outgoing combinations. For the real scalar Yukawa interaction, the conjugate of ${\bar{v}}_2{u}_1$ is ${\bar{u}}_1{v}_2$. The fermion spin sums then yieldThe bar denotes the initial spin average, not a Dirac adjoint. The gamma matrix trace identities giveand the second trace has the same value because $p_3+p_4=p_1+p_2$ and all external masses are $m$. ThereforeThis is the fixed-order expression away from $s=M^2$. If that pole is accessible, a physical scattering resonance requires a scalar self-energy and decay width; the tree formula alone is not finite there.

In unitary gauge, ${\varphi }^{†}\varphi -v^2/2=vH+H^2/2$, so this Higgs potential gives $M_H^2=2\kappa v^2$, a cubic coefficient $\kappa v$ and a quartic coefficient $\kappa /4$. The corresponding Feynman rules are $-3i{M}_H^2/v$ and $-3i{M}_H^2/v^2$. Writing the potential with an additional factor $1/2$ changes the relation between its named coupling and the Higgs boson mass; it does not change these mass-based rules.