For $g=0$, the invariant scattering amplitude from the tree-level Feynman diagram in i has a pole at $s=M^2$. The intermediate heavy antiparticle can go on shell and live for a time of order its mean lifetime. Near this scattering resonance, inserting a single self-energy is no longer a uniformly small correction. Dyson resummation gives, to leading accuracy in a narrow decay width,Here $M$ is understood as the renormalized resonance mass; the real self-energy shifts the original pole position. The imaginary term removes the unphysical divergence of the fixed-order invariant scattering amplitude.

For equal incoming and outgoing light masses, the two-body momentum ratio cancels in the cross section. The angle-independent invariant scattering amplitude, integrated over the full solid angle with the identical final-state symmetry factor $1/2!$, gives $\sigma =|\mathcal{M}{|}^2/(32\pi s)$. Thus the Breit-Wigner resonance cross section isAt $s\gg M^2$ it reproduces $h^4/(32\pi s^3)$. At the resonance peak,This is also $8\pi /k_{\ast }^2$ for $k_{\ast }=M\beta /2$, the elastic identical-boson S wave partial-wave unitarity limit when this is the only open decay channel. The approximation assumes ${\Gamma }_{\mathrm{phys}}\ll M$ and a resonance sufficiently far from threshold for the width to vary slowly across it.

If one mechanically uses the unsymmetrized ${\Gamma }_{\mathrm{printed}}=2{\Gamma }_{\mathrm{phys}}$ from d, the same displayed Breit-Wigner resonance cross section would have that doubled width and predict a peak $8\pi /(M^2{\beta }^2)$. This is the literal-width substitution, rather than the consistent physical result: the given high-energy cross section already includes the identical-final-particle factor that the printed decay width integral omits.

For an allowed decay channel in iii, $|\mathcal{M}{|}^2=h^2$. Integrating the spatial Dirac delta distribution in the printed Lorentz-invariant phase-space measure sets ${\mathbf{p}}_2=-{\mathbf{p}}_1$. With $p=|{\mathbf{p}}_1|$ and $E_p=\sqrt{p^2+m^2}$, its remaining value isThe positive root is $p_{\ast }=(M/2)\beta$, where $\beta =\sqrt{1-4m^2/M^2}$, and the Dirac delta distribution contributes the Jacobian determinant $E_{\ast }/(2p_{\ast })$. HenceIt follows that the literal printed integral evaluates toThe printed integration ranges count each final configuration twice, once as $({\mathbf{p}}_1,{\mathbf{p}}_2)$ and once in the opposite order. For two identical final bosons, the physical decay width therefore requires an additional identical final-state symmetry factor $1/2!$:There is one allowed decay channel per heavy particle, not an extra factor of two from summing the two charge-conjugate heavy species. Both expressions vanish below the two-body decay threshold $M=2m$, and the limiting value at threshold is zero. For the literally forbidden labels $\Phi \to \varphi \varphi$, the decay width is instead zero because its decay amplitude vanishes, as explained in iii.

The nonzero free scalar propagators join a field to its adjoint:Contractions $⟨T\varphi \varphi ⟩$, $⟨T\Phi \Phi ⟩$ and mixed-species contractions vanish. With all momenta incoming, the Feynman vertices areThe two identical light legs cancel the $1/2!$ in each cubic interaction vertex; there is no remaining factor $1/2$ in its Feynman rule. Multiply each Feynman vertex by $(2\pi {)}^4{\delta }^{(4)}(\sum p)$, and integrate an internal unconstrained four-momentum with $d^{4}p/(2\pi {)}^4$.

The arrows in the figure follow positive Noether charge: a light arrow carries one unit and a heavy arrow two. Consequently a heavy particle $\Phi$ travels opposite its charge arrow, whereas its antiparticle $\bar{\Phi }$ travels along it. At a ${\varphi }^2\Phi$ Feynman vertex, two unit-charge arrows enter and a two-unit arrow leaves; the conjugate Feynman vertex reverses every arrow. The quartic Feynman vertex has one incoming and one outgoing arrow of each weight. These rules prevent attaching a propagator between two unconjugated fields.

Independent global phase symmetries would rotate $\varphi \mapsto e^{-i\alpha }\varphi$ and $\Phi \mapsto e^{-i\beta }\Phi$. The cubic interaction vertex preserves only $2\alpha +\beta =0$ modulo $2\pi$. The continuous internal symmetry of a classical field theory is therefore the circle groupNormalize the Noether charge so that $q_{\varphi }=1$. Then $q_{\Phi }=-2$, $q_{\bar{\varphi }}=-1$ and $q_{\bar{\Phi }}=2$. The Noether current iswhere $A\overleftrightarrow{\partial }B=A\partial B-(\partial A)B$. For example, the Euler-Lagrange field equations givewhich cancel with the relative coefficient $-2$. The quartic interaction vertex separately preserves both original phase rotations.

There is also a discrete charge conjugation symmetry $(\varphi ,\Phi )\mapsto ({\varphi }^{†},{\Phi }^{†})$ for these real couplings. Including it, the internal symmetry is $U(1)⋊{\mathbb{Z}}_2\cong O(2)$, with charge conjugation reversing the Noether charge. The connected symmetry and its conserved current are the $U(1)$ above.

Writing $x_P=(t,-\mathbf{x})$, the standard parity transformation of a Dirac field isSince ${\gamma }^0{\gamma }^5{\gamma }^0=-{\gamma }^5$, the fermion bilinear $\bar{\psi }{\gamma }^5\psi$ changes sign. Invariance of the pseudoscalar Yukawa interaction therefore requiresThus $\varphi$ is a pseudoscalar, despite having the usual scalar-form kinetic term and mass term. In this convention $({\gamma }^5{)}^{†}=-{\gamma }^5$, but $\bar{\psi }{\gamma }^5\psi$ is Hermitian because of the additional ${\gamma }^0$ in the Dirac adjoint. A real $g$ is consequently consistent with a Hermitian Lagrangian density.

In perturbation theory about the parity-invariant vacuum state, renormalization can preserve this non-anomalous parity symmetry in quantum field theory. Every local counterterm must then be parity even, at every loop order. The allowed terms are precisely ${\mathcal{L}}_{\mathrm{ct}}^{(+)}$ in b; all coefficients in ${\mathcal{L}}_{\mathrm{ct}}^{(-)}$ are zero. In particular, no $\varphi$ or ${\varphi }^3$ counterterm is generated, whereas ${\varphi }^4$ is allowed and required. The one-point and three-point pure-boson vertices also vanish to all orders: a parity-odd Lorentz scalar cannot be formed from their at most two independent external four-momenta. This last reasoning is specific to these low-point vertices; it does not assert that parity symmetry in quantum field theory is an independent $\varphi \mapsto -\varphi$ internal symmetry with the fermions held fixed.

Expanding $e^{i{S}_{\mathrm{int}}}$ gives the Feynman vertexThere is one scalar leg and a continuous oriented fermion line; no extra factor $i$ is absorbed into ${\gamma }^5$. The Dirac propagator is $S_F(k)=i(/k+M)/(k^2-M^2+i0)$, where $/k={\gamma }^{\mu }{k}_{\mu }$. Each closed fermion loop supplies a minus sign.

The tadpole diagram is proportional tobecause both numerator traces vanish by a. This is zero before the loop momentum integration.

For either orientation of the triangle Feynman diagram, let $a,b,c$ be its three internal four-momenta. Its numerator trace isAfter using cyclic property of the trace, each expanded term is a trace of ${\gamma }^5$ times at most three ordinary gamma matrices. Such traces vanish in four dimensions. For an odd number of ordinary gamma matrices, conjugation by ${\gamma }^5$ changes the sign. For two ordinary gamma matrices, the contracted Clifford algebra identity ${\gamma }_{\rho }{\gamma }^{\mu }{\gamma }^{\nu }{\gamma }^{\rho }=4{\eta }^{\mu \nu }I$ givesThe zero-matrix case is a. Hence both the one-point and three-point one-loop bosonic vertices vanish, independently of their external momenta. No cancellation between different loop orientations or delicate shift of a divergent integral is needed.

The mass dimensions are $[\varphi ]=1$, $[\psi ]=3/2$, $[g]=0$. Up to total derivatives, power counting allows local counterterms of mass dimension at most four. A convenient basis for the counterterm Lagrangian compatible with the actual parity symmetry in quantum field theory and Dirac fermion number conservation isThe optional constant $\delta \Lambda$ renormalizes the vacuum energy density. The wavefunction renormalization, mass counterterms and Yukawa coupling renormalization correct the terms already present. The quartic interaction must also be included: the fermion loop with four external bosonic legs is logarithmically divergent, even if the renormalized quartic coupling is set to zero at one chosen renormalization scale. The displayed original interactions alone do not form a closed set of counterterms.

If “most general” is interpreted before imposing the parity symmetry in quantum field theory examined in d, Lorentz invariance, a Hermitian action and Dirac fermion number conservation also allowHere all coefficients can be real in the paper's gamma matrix convention; the last term is Hermitian in the action up to a total derivative. It is an axial kinetic normalization, which could alternatively be removed by a suitable perturbative field redefinition. Every term in ${\mathcal{L}}_{\mathrm{ct}}^{(-)}$ is odd under the actual parity symmetry in quantum field theory, so its coefficient vanishes in a symmetry-preserving renormalization scheme. Listing these terms makes explicit which assumption excludes a tadpole counterterm, a cubic scalar term, a scalar Yukawa interaction, a pseudoscalar Dirac mass term and an axial kinetic term.

Distinct gamma matrices anticommute, and $({\gamma }^0{)}^2=I$, $({\gamma }^j{)}^2=-I$ for $j=1,2,3$. Moving a particular ${\gamma }^{\mu }$ through the other three factors in ${\gamma }^5$ gives three minus signs. ThereforePairing equal factors in the square involves $3+2+1=6$ interchanges. The Clifford algebra consequently givesIn particular $({\gamma }^5{)}^{-1}=-{\gamma }^5$; replacing it by the conventional chirality matrix would change this answer.

The trace is unchanged by a similarity transformation. Since $({\gamma }^0{)}^{-1}={\gamma }^0$ and ${\gamma }^0{\gamma }^5{\gamma }^0=-{\gamma }^5$, cyclic property of the trace gives $\mathrm{tr}{\gamma }^5=-\mathrm{tr}{\gamma }^5$. AlsoThus

Set the renormalized $g_3$ to zero. For a six-scalar vertex, b gives $\omega =-n_2$, so a new overall logarithmic divergence can only occur in graphs with $n_2=0$. With $b=6,f=0$, the line counts become $I_f=n_1$ and $I_b=n_1-3$, hence $L=n_1-2$.

At one loop this forces $n_1=3$ and $I_b=0$: the only candidate is a closed triangle Feynman diagram with two external scalar legs at each vertex. Its numerator at large loop momentum is proportional toThis vanishes for the specified two-component gamma matrices, each of which is traceless. Replacing one leading $/k$ by the fermion mass or an external momentum lowers the superficial degree of divergence from zero to at most $-1$. Thus the apparent logarithm cancels. Graphs with a $g_2$ vertex already have negative overall degree; the pure-scalar one-loop triangle, for instance, has $\omega =-3$. There is no one-loop ultraviolet subdivergence inside these proper one-loop vertices. ThereforeThe six-point correlation function need not vanish: finite terms are allowed. Nor is it correct to say that all odd gamma-matrix traces vanish in $2+1$ dimensions. In this representation,The cancellation above follows from repeating the same leading loop momentum, not from a false four-dimensional trace rule.

At two loops, $n_1=4$, $I_f=4$ and $I_b=1$ are possible. A closed fermion box with a scalar line joining two of its vertices leaves six external scalar legs. It has $L=2$, $\omega =0$, and four leading gamma matrices; the one-loop odd-numerator argument no longer applies.

A background-field check demonstrates that its logarithm is actually nonzero. Put $\varphi =v+\eta$ with constant $v$. Define the background fermion mass $\mu =m+g_1{v}^2/2$, the background scalar squared mass ${\mathcal{B}}^2=M^2+g_2{v}^2/2$, and the induced three-leg Yukawa coupling $y=g_{1}v$. The two-loop quantum effective potential contains the sunset diagram with two fermion lines and one scalar line. After a Wick rotation, its relevant integral, apart from a nonzero overall combinatorial factor, isHere the dots and gamma-matrix traces are Euclidean. With $A=k^2+{\mu }^2$, $B=l^2+{\mu }^2$, $C=(k-l{)}^2+{\mathcal{B}}^2$, its numerator is exactlyThe first three terms reduce to products of one-loop tadpole integrals, which have no logarithmic pole near three dimensions in dimensional regularization. The remaining term isFor positive regulator masses, the large-momentum region of $H$ has a positive, nonzero logarithmic divergence. One can see this directly by restricting the relative angles and the ratio $|k|/|l|$ to compact intervals away from zero: the remaining common radial integral is proportional to ${\int }^{\Lambda }dr/r$. Its proper one-loop subgraphs are ultraviolet convergent.

Finally, the coefficient multiplying $H$ containsThe two-loop sextic counterterm in quartic Yukawa theory follows because its $v^6$ coefficient is $g_1^4\ne 0$ for a nonzero coupling. Other two-loop vacuum topologies reduce to products of finite-in-dimensional-regularization tadpoles and do not cancel this logarithmic pole. ThusA sextic counterterm is required even if the renormalized sextic coupling is set to zero at a chosen renormalization scale. A theory constrained to $g_3=0$ at every scale lacks counterterm closure when $g_1\ne 0$. The decoupled exception $g_1=0$ is the three-dimensional scalar quartic interaction, which needs no sextic counterterm by power counting.

Use one-particle-irreducible correlation functions to identify independent local counterterms. The exact internal symmetry $\varphi \mapsto -\varphi$ excludes odd $b$, and Dirac fermion number conservation requires equal numbers of $\psi$ and $\bar{\psi }$, so $f$ is even. From b, the complete list of potentially overall divergent vertices with external legs is:

The zero-leg vacuum diagrams, with $\omega =3-n_2$, may also require a vacuum energy density subtraction. Four-fermion vertices have $\omega \le -1$ and do not require a new four-fermion counterterm. These are superficial bounds: a particular allowed diagram can vanish or be convergent.

Locality and Lorentz invariance restrict the divergent parts to a finite basis. The scalar two-point vertex can contain a constant and $p^2$; there is no Lorentz scalar linear in a single momentum. The fermion two-point vertex can contain $I$ and $/p$. The remaining listed vertices have no allowed derivative divergence. A complete counterterm Lagrangian, up to total derivatives, is thereforeThe coefficients supply wavefunction renormalization, mass counterterms and renormalizations of the three interactions already present. Recursive subtraction of ultraviolet subdivergences then leaves only these local overall divergences. This finite counterterm closure is why the complete theory is expected to have perturbative renormalizability. It does not mean every listed coefficient first becomes nonzero at one loop, or that setting an allowed coupling to zero is stable under renormalization.

Let $I_b,I_f$ count internal scalar and fermion lines and $L$ count independent loops in a connected amputated Feynman diagram. At large loop momentum, the scalar propagator falls as $k^{-2}$ and the Dirac propagator falls as $k^{-1}$, including its numerator. Each loop integration is $d^{3}k$. The nonderivative interaction vertices add no powers. ThereforeCount endpoints of each kind of line and use the connected-graph loop order identity:Eliminating $L,I_b,I_f$ givesThe cancellation of $n_1,n_3$ reflects their dimensionless couplings; every insertion of the positive-dimensional $g_2$ improves the superficial degree of divergence by one.

The printed exact homogeneity is shorthand for ultraviolet power counting, not an identity for a massive integrated correlation function. For example, even the free amputated scalar two-point vertex ${\Gamma }^{(2)}(p)=p^2-M^2$ obeys ${\Gamma }^{(2)}(\lambda p)={\lambda }^2{p}^2-M^2$, which differs from ${\lambda }^2{\Gamma }^{(2)}(p)$ when $M\ne 0$. Logarithmic divergences, actual numerator cancellations and ultraviolet subdivergences also require separate treatment.

In $2+1$ spacetime dimensions, the dimensionless action gives $[\mathcal{L}]=3$ and $[{\partial }_{\mu }]=1$. The scalar kinetic term requires $2[\varphi ]+2=3$, whereas the fermion kinetic term requires $2[\psi ]+1=3$. Thus the engineering dimensions of relativistic fields areMatching each interaction to mass dimension three givesHere the brackets mean exponents of mass in natural units. The coefficients $g_1,g_3$ are marginal couplings by classical power counting, and $g_2$ is a relevant coupling.

For a single species both panels of the figure in c contribute, now with $j=1$. They are the $s$-channel annihilation and $t$-channel scalar exchange, where the Mandelstam variables are $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$, $u=(p_1-p_4{)}^2$ and $s+t+u=4m^2$. Fix the external-state order consistently with c. The exchange of identical external fermionic operators gives a relative minus sign:Changing the convention for ordering all external states can change the overall sign, but not this relative fermionic sign.

The exchange $p_1\leftrightarrow -p_4$ must include crossing symmetry the associated external spinors, as well as continuing the four-momenta. It interchanges $s$ and $t$, leaves $u$ fixed, and interchanges the two spinor pairings. Hence the analytically continued invariant scattering amplitude is antisymmetric:The first relation is understood with consistently chosen spinor crossing phases. It is not a substitution of negative-energy arguments into the original positive-energy $u$ spinors without changing external states.

Without repeating the spin sum calculation, c fixes the square of the first diagram. Crossing symmetry fixes the second by $s\leftrightarrow t$. Consequently the strongest immediate inference isThe interference function is not determined by the separate diagram squares. In particular adding the two squares and dropping interference is unjustified. The Cauchy-Schwarz inequality applied to the arrays of spin amplitudes also gives $|C(s,t)|\le 2|(s-4m^2)(t-4m^2)|$ on the physical scattering region.

For completeness, its value can be checked with one contracted trace, without another diagram calculation. The cross product isThe relative minus in the invariant scattering amplitude gives $C=-2K$. Thus the fully evaluated result, beyond the symmetry inference alone, isEvery term respects the inferred $s,t$ symmetry. In the massless limit the interference numerator reduces to $st$, which is nonzero at a generic scattering angle.

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.

Every positive-energy Dirac spinor obeys $(/p-m)u^s(\mathbf{p})=0$. Therefore the fermion spin sum satisfiesSince ${\bar{u}}^s=u^{s†}{\gamma }^0$ and $({\gamma }^0{)}^2=I$, the trace of each outer product is its contracted scalar. The prescribed normalization and the two spin states yieldNow make exactly the stated polynomial assumption and apply polynomial reduction of a slashed vector. By a, on the mass shell $p^2=m^2$ write $P=A/p+BI$. ThenThe matrices $I$ and $/p$ are linearly independent for a physical nonzero $p$, so $B=mA$. The gamma matrix trace identities, $\mathrm{tr}{\gamma }^0=0$ and $\mathrm{tr}({\gamma }^{\mu }{\gamma }^0)=4{\eta }^{\mu 0}$, giveThus $A=1$ andNo explicit representation of the Dirac spinors has been used. The same reasoning covers $m=0$ provided $p^0>0$.

Using Feynman slash notation, the symmetric tensor $p_{\mu }{p}_{\nu }$ kills the antisymmetric commutator of the gamma matrices. The Clifford algebra givesConsequently $(/p{)}^{2r}=(p^2{)}^{r}I$ and $(/p{)}^{2r+1}=(p^2{)}^r/p$. For a scalar-coefficient polynomial $F(z)={\sum }_n{c}_n{z}^n$, split its even and odd powers:This reduction assumes scalar coefficients; arbitrary noncommuting matrix coefficients would not give the stated scalar functions $A,B$.

Assume $m_{\psi }>0$, as required to have a rest frame. Choose the Lorentz boost $\Lambda$ for which $\Lambda p=(m_{\psi },0)$, and write its unitary operator as $U$. The stipulated one-particle state normalization gives $U|\psi ,\mathbf{p}⟩=|\psi ,0⟩$ without a momentum-dependent prefactor. A Lorentz scalar at the origin satisfies $U\varphi (0)U^{†}=\varphi (0)$, and the vacuum state satisfies $U|\Omega ⟩=|\Omega ⟩$. Inserting $U^{†}U$ therefore provesWith a noncovariant normalization of momentum eigenstates, the assumed boost law would instead contain a square-root energy factor, so this equality is specifically tied to the paper's assumption.

Let $q^{\mu }=(E_{\mathbf{q}}^{\chi },\mathbf{q})$ denote the four-momentum of the other energy eigenstate. Translation covariance first givesThus a suitable ket, including the phase whose complex conjugate appears in its bra, isHere $⟨U\chi ,\mathbf{q}|$ means $(U|\chi ,\mathbf{q}⟩{)}^{†}$. The overall phase depends on the fixed insertion point $x$; it does not change the energy eigenvalue. It cannot simply be dropped if the equality of complex matrix elements is to hold exactly.

The transformed state has four-momentum $q^{\prime }=\Lambda q$. Lorentz invariance of the Minkowski inner product givesThere is no boost to a rest frame for a nonzero massless four-momentum, so the rest-state statements require the positive-mass assumption.