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Use dimensionless physical helicity $h=\mathbf{S}\cdot \hat{\mathbf{p}}$, with spin eigenvalues $\pm 1/2$ in natural units. In the Dirac representation,The entries $1$ in the supplied transformation denote two-by-two identities. Since its matrix acts only on the upper/lower spinor blocks, it commutes with the repeated Pauli block. Therefore $u^{-1}=u$ andThe eigenvalues remain the physical spin projection along the direction of motion, $+1/2$ or $-1/2$. A convention omitting the factor one half instead labels them $\pm 1$; this changes no measurement probabilities. A change of gamma-matrix representation cannot change the spin observable.

With the gamma matrices derived in a, the momentum-space Dirac equation readsFor a positive-energy particle at rest, $E=m>0$ and $\mathbf{p}=0$, so $\xi =\eta$. The Dirac adjoint in this basis is $\bar{u}=({\eta }^{†},{\xi }^{†})$, and the specified normalization givesTherefore, for any normalized two-component spinor $\chi$,Equivalently, choosing ${\chi }_{+}=(1,0{)}^{\mathsf{T}}$ and ${\chi }_{-}=(0,1{)}^{\mathsf{T}}$ gives the two independent rest-frame spin states. A strictly massless particle has no rest frame; this step requires $m>0$.

An incoming fermion contributes a Dirac spinor $u_s(p)$ and an outgoing fermion contributes its Dirac adjoint ${\bar{u}}_s(p)=u_s^{†}(p){\gamma }^0$. The label $s$ specifies a spin state; $p$ is the on-shell four-momentum. Incoming and outgoing antifermions instead contribute ${\bar{v}}_s(p)$ and $v_s(p)$.

The exchanged boson is a virtual photon, represented in the Feynman rules by the photon propagator $-i{g}_{\mu \nu }/(q^2+iϵ)$ in Feynman gauge. Its momentum transfer is $q=p_1-p_3=p_4-p_2$; unlike an external real photon, it need not satisfy $q^2=0$. Each Feynman vertex contracts a gamma matrix with this propagator and contributes $-ie{\gamma }^{\mu }$ in a conventional QED sign convention. The coupling constant $e$ sets the interaction strength, with fine-structure constant $\alpha =e^2/(4\pi )$. One may reverse the sign of the photon field and both vertex signs without changing the result. The gamma matrices obey $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2g^{\mu \nu }I$, with Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$.

With $\eta =\mathrm{diag}(1,-1)$, the Clifford algebra gives $({\gamma }^0{)}^2=1$, $({\gamma }^1{)}^2=-1$, and ${\gamma }^0{\gamma }^1=-{\gamma }^1{\gamma }^0$. Henceso $\{{\gamma }^5,{\gamma }^0\}=\{{\gamma }^5,{\gamma }^1\}=0$. Its square isChoose the following gamma matrices, expressed using the Pauli matrices:These satisfy $({\gamma }^0{)}^{†}={\gamma }^0$, $({\gamma }^1{)}^{†}=-{\gamma }^1$ and $({\gamma }^5{)}^{†}={\gamma }^5$. The chirality eigenvalues are therefore $\boxed{+1,-1}$, and we order the spinor as $\psi =({\psi }_R,{\psi }_L{)}^{\mathsf{T}}$ throughout. Its Dirac adjoint is $\bar{\psi }=({\psi }_L^{†},{\psi }_R^{†})$. No additional factor of $i$ multiplies ${\gamma }^0{\gamma }^1$ in this spacetime dimension. This realizes a Dirac field in two spacetime dimensions.

Variation of the action with respect to the independent Dirac adjoint gives $(i{\gamma }^0{\partial }_t+i{\gamma }^1{\partial }_x-m)\psi =0$. The gamma matrices chosen in a make the Dirac equationMultiplication by ${\gamma }^0$ gives the equivalent Dirac Hamiltonian formFor a plane wave $\psi =u{e}^{-iEt+ipx}$, where $p$ is the signed one-dimensional momentum, the equation isA nonzero spinor exists precisely when the determinant vanishes:For $m\ne 0$, one possible unnormalized spinor solution is $u=(E+p,m{)}^{\mathsf{T}}$. Alternatively, $(p{\sigma }_3+m{\sigma }_1{)}^2=(p^2+m^2)1$ follows from the Pauli matrix multiplication law, proving the same relativistic energy-momentum relation. Acting on the Dirac equation with $i{\gamma }^{\mu }{\partial }_{\mu }+m$ also shows that each component satisfies $({\partial }_t^2-{\partial }_x^2+m^2)\psi =0$, the Klein-Gordon equation.

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Use natural units $\hslash =c=1$ and the mostly-minus Minkowski metric. Question 3 has two spatial dimensions: its gamma matrices are the specified two-by-two matrices. The printed ${\sigma }^{\mu }$ in its kinetic term is interpreted as those ${\gamma }^{\mu }$, rather than as three unmodified Pauli matrices. All spin sums in question 2 instead use four-component Dirac spinors in four spacetime dimensions.

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 $A,B$.

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.

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

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.

A nonderivative interaction of two real scalar legs with a fermion pair in $2+1$ dimensions. The engineering dimensions of relativistic fields are $[\varphi ]=1/2$ and $[\psi ]=1$, so $g$ is a marginal coupling by classical power counting. The internal symmetry $\varphi \mapsto -\varphi$ forbids odd-scalar interactions, but allows a sextic scalar interaction. The two-component Dirac field representation must be distinguished from four-component gamma matrices in four spacetime dimensions.

For $g_{ab}={\Omega }^2{\eta }_{ab}$, use the vierbein $e^{\mu }{}_a=\Omega {\delta }^{\mu }{}_a$ and constant flat gamma matrices. The tetrad postulate and ${\Gamma }^c{}_{ab}={\delta }_a^c{\partial }_b\log \Omega +{\delta }_b^c{\partial }_a\log \Omega -{\eta }_{ab}{\eta }^{cd}{\partial }_d\log \Omega$ give the displayed lowered spin connection. With $L={\hat{\gamma }}^c{\partial }_c\log \Omega$, its spinor action is ${\nabla }_a={\partial }_a+[{\hat{\gamma }}_a,L]/4$. The Lorentz index lowering matters in indefinite signature.

For two outgoing massless leptons, ${\ell }_{\mu }=\bar{u}(q_1){\gamma }_{\mu }(1-{\gamma }^5)v(q_2)$ is transverse to $q=q_1+q_2$. This follows from the massless Dirac equation and anticommutation of the chirality matrix with the gamma matrices. Thus the $q^{\mu }{f}_{-}(q^2)$ term in a pseudoscalar-to-pseudoscalar form factor does not contribute to this scattering amplitude. A nonzero charged-lepton mass spoils exact transversality and restores a mass-suppressed scalar-form-factor contribution.

A twistor spinor has vanishing gamma-trace-free part of its spinor covariant derivative. In $n$ dimensions the displayed equation uses gamma matrices satisfying $\{{\gamma }_a,{\gamma }_b\}=2g_{ab}$. It is weaker than being a parallel spinor. In four dimensions it implies $[{\gamma }_a{\nabla }_b+{\gamma }_b{\nabla }_a-g_{ab}{\gamma }^c{\nabla }_c/2]ϵ=0$.

Write the QED vertex as $-ie({\gamma }^{\mu }+{\Lambda }^{\mu })$. In Feynman gauge and $d=4-2ϵ$, the one-loop ultraviolet pole has the local form ${\Lambda }_{\mathrm{UV}}^{\mu }=e^2{\gamma }^{\mu }/(16{\pi }^2ϵ)$. Its coefficient follows from the large-momentum gamma matrices contraction ${\gamma }^{\alpha }/k{\gamma }^{\mu }/k{\gamma }_{\alpha }\mapsto (d-2{)}^2{k}^2{\gamma }^{\mu }/d$. It is independent of external kinematics, whereas the full finite vertex requires additional structures off shell. Its cancellation verifies the Ward identity for QED renormalization constants.