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The free Dirac propagator is $S_F(p)=i(/p+m)/(p^2-m^2+iϵ)$ and inverts the Dirac kinetic operator with Feynman boundary conditions.

The leading scattering amplitude is the sum of two tree-level Feynman diagrams, with the outgoing photons attached in either order to the fermion line. The Dirac propagator carries momentum $p-k_1$ or $p-k_2$ for incoming electron momentum $p$. Their sum obeys the two-photon fermion Ward identity, although individual diagrams need not. Crossing symmetry relates this amplitude to Compton scattering.

Take Hermitian generators with $[T^a,T^b]=i{f}^{abc}{T}^c$, $(T_{\mathrm{ad}}^b{)}_{ac}=-i{f}^{bac}$, and $D_{\mu }={\partial }_{\mu }-ig{A}_{\mu }^b{T}_{\mathrm{ad}}^b$. For $S_{\psi }=\int \bar{\psi }({\gamma }_{\mu }{D}_{\mu }+m)\psi$ and $\{{\gamma }_{\mu },{\gamma }_{\nu }\}=2{\delta }_{\mu \nu }$, the Dirac propagator is ${\delta }^{ac}(-i/p+m)/(p^2+m^2)$. Expansion of $e^{-S_{\psi }}$ gives the fermion-gluon vertex $+ig{\gamma }_{\mu }(T_{\mathrm{ad}}^b{)}_{ac}$. There is no two-gluon seagull vertex in this first-order kinetic action, and a closed fermion loop has a minus sign.

For two odd fermionic operators, time ordering contributes a minus sign whenever the chronological rearrangement exchanges them. For several fields the sign is the sign of a permutation restricted to the odd fields. This convention makes the free Dirac propagator a Green function of the Dirac equation with its contact term fixed by the canonical anticommutation relations.

A closed chain of Dirac propagators is a fermion loop. Its closed fermion loop sign distinguishes it from a bosonic loop. Tracing spinor indices produces gamma matrix traces, and the high-momentum powers of the propagators determine possible ultraviolet divergences.

For a free Dirac field with action $\int \bar{\psi }K\psi$ and independent odd sources coupled as $\bar{\psi }J+\bar{J}\psi$, translation invariance of the Berezin integral and completion of the square give $Z[J,\bar{J}]=Z[0,0]\exp (-i\bar{J}{K}_F^{-1}J)$. Here $K_F^{-1}$ has Feynman i-epsilon prescription boundary conditions, and each bilinear includes its spacetime integrations. A left Grassmann derivative with respect to $\bar{J}$, followed by a right Grassmann derivative with respect to $J$, has normalized value $-i{K}_F^{-1}$ at zero sources. Multiplying by $-1$ removes the two insertion factors $i^2$ and yields the Dirac propagator $S_F=i{K}_F^{-1}$. Keeping source order fixed prevents a spurious fermionic sign.

For action $-\int \bar{\psi }(\gamma \cdot \partial +m)\psi$, choose gamma matrices with $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$ and signature $(-,+,\dots ,+)$. The Fourier kernel of $D=\gamma \cdot \partial +m$ is $i\gamma \cdot p+m$. Its algebraic inverse is $(-i\gamma \cdot p+m)/(p^2+m^2)$. The Grassmann Gaussian integral for $e^{-i\bar{\psi }D\psi }$ has covariance $-i{D}_F^{-1}$, giving the displayed Dirac propagator. Its denominator has positive-energy poles below and negative-energy poles above the real energy axis by the Feynman i-epsilon prescription. Both the metric and the factor of $i$ in the derivative kernel must be specified when comparing propagator formulas.

A Feynman diagram with one independent closed cycle and hence one independent loop momentum. An $n$-vertex polygon of Dirac propagators in four dimensions has superficial degree of divergence $4-n$ before symmetry cancellations. For a fermion loop, the closed fermion loop sign is separate from any symmetry factor or pairwise charge conjugation cancellation.

Use Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$ and the convention $S_{fi}={\delta }_{fi}+i(2\pi {)}^4{\delta }^{(4)}(p_f-p_i){\mathcal{M}}_{fi}$. Put $m=m_t$, $y={\lambda }_t/\sqrt{2}$, $Q=e_t=2/3$ andThe Feynman i-epsilon prescription is understood in subsequent denominators. Contracting the spinor indices around the fermion loop gives the traceHere $/p={\gamma }^{\rho }{p}_{\rho }$ is Feynman slash notation. The closed fermion loop sign is $-1$, the three interaction vertices contribute $(-iy)(-ieQ{)}^2$, and the three Dirac propagators contribute $i^3$. Their product, including the color charge trace, is $-N_{c}y{e}^2{Q}^2$. Thus one ordering hasMultiplying by the incoming polarization vectors and adding the exchanged Feynman diagram givesThe displayed four-dimensional Feynman integral is formal until regularized; subsequent manipulations use dimensional regularization, which permits the required loop momentum shifts.

At second order in the interaction term, attach the two external fermion fields to different vertices and contract the remaining two fermion fields with each other and the two vector fields with each other. The factor $1/2!$ in the expansion is cancelled by the two ways to assign the external ends. There is no closed fermion loop, and hence no extra closed-loop fermionic sign.

After removing the two external Dirac propagators, the internal momentum assignment $l$ for the photon and $p+l$ for the fermion gives the one-loop insertionThus the insertion is proportional to exactly the propagator product specified in the source. Define $\mathcal{I}=-i\Sigma$. With Fourier convention $\psi (x)\sim e^{ip\cdot x}$, the reduced free inverse kernel is $K_0(p)=-/p-m$, since the printed Dirac propagator equals $i{K}_0^{-1}$. Repeated insertions giveThe factors of $i$ and the sign of the named self-energy are convention dependent; the cancelling counterterm is fixed once this insertion convention is specified. The internal photon joins two vertices on one directed fermion line, as in the first panel of the figure in 1b.

At first order the Dyson expansion inserts $i{S}_{\mathrm{int}}=\lambda \int d^{4}x{\bar{\psi }}_{\alpha }(x)({\gamma }_5{)}_{\alpha \beta }{\psi }_{\beta }(x)\varphi (x)$. Contract the scalar field at the vertex with the external scalar. The four remaining fermion fields have two admissible pairings: both external fermions attach to the vertex, or the two vertex fermions contract with each other while the external pair contracts freely. Wick's theorem givesThe minus sign in the second term is the fermion sign of the closed contraction: $⟨T{\bar{\psi }}_{\alpha }(x){\psi }_{\beta }(x)⟩=-i{S}_F(0{)}_{\beta \alpha }$. The first term is connected; the second is a scalar-attached fermion tadpole multiplied by a disconnected free external fermion propagator. The full time-ordered expectation includes both topologies, whereas the connected generating functional retains only the first. There is no order-${\lambda }^2$ contribution in this pure Yukawa expansion because the total number of scalar fields would be odd.

For a parity-preserving free propagator, the tadpole coefficient is $\mathrm{tr}[{\gamma }_5{S}_F(0)]$, whose integrand contains only $\mathrm{tr}{\gamma }_5$ and $\mathrm{tr}({\gamma }_5{\gamma }^{\mu })$. Both vanish. Thus the second term is a legitimate formal Wick contraction but is zero with a compatible regulator; normal ordering the interaction removes it from the outset.

The Slavnov-Taylor identity implies transversality of the full perturbative gluon self-energy:The quark loop is separately transverse. This follows without evaluating its Feynman integral: contract its vector Feynman vertex with $k_{\mu }$ and use $/k=(/l+/k-m)-(/l-m)$ between its two Dirac propagators. One obtains a difference of two loop integrals that cancel after a shift of $l$ in dimensional regularization. The colour factor is proportional to $\mathrm{tr}(T^a{T}^b)$ and does not obstruct this argument.

The four-gluon tadpole diagram vanishes after removal of $m_g$. Thus the nontransverse parts of the two-three-gluon-vertex loop and the ghost loop must cancel. Contracting the preceding ghost-loop pole of the gluon self-energy givesFor QCD, set $C_A=3$. No evaluation of the remaining three-gluon Feynman diagram is required. Its uncontracted transverse part is not fixed by this identity.

With $q=p^{\prime }-p$ and the routing just described, multiplication of the three QED Feynman rules for the Feynman vertices, the two Dirac propagators and the photon propagator gives $(-i{e}_D{)}^3{i}^2(-i)=-e_D^3$. Hence the unrenormalized amputated Feynman vertex through this order isFor the renormalized Feynman vertex, add $-i{e}_D{\delta }_1{\gamma }^{\mu }$ to this expression. Contributions from replacing parameters inside the one-loop term by their bare counterparts enter at higher loop order.

The massive Dirac propagator isUse the gamma matrix trace identities $\mathrm{tr}{\gamma }^{\mu }=0$ and $\mathrm{tr}({\gamma }^{\mu }{\gamma }^{\rho })=4g^{\mu \rho }$. They giveThe integrand is odd under $k\mapsto -k$, and dimensional regularization preserves that symmetry. The same vanishing holds for $S=i{S}_F$. One must not assert that $S(0)$ itself is zero: its term proportional to $m1$ generally is nonzero and divergent, but its trace against a single gamma matrix is zero. This one-photon fermion loop is also the simplest instance of Furry's theorem.

For commuting real scalar fields, remove the closed fermion loop sign and replace the Dirac propagators by scalar propagators. There are no spinor indices, hence no matrix trace. With two independent real scalar fields ${\chi }_1,{\chi }_2$ and interaction $g\varphi {\chi }_1{\chi }_2$, the same normalization givesThe overall sign has reversed relative to part (i) because $(ig{)}^2$ is now unaccompanied by a closed fermion loop sign. The same formula holds in the source's kernel convention, since there are again four propagators.

There is an additional convention to specify if the two symbols are intended to denote the same real scalar field. For the literal interaction $g\varphi {\chi }^2$, there are two identical connected Wick contractions between ${\chi }^2(x)$ and ${\chi }^2(y)$, so the coefficient becomes $-2g^2$. If the conventional interaction is instead $(g/2)\varphi {\chi }^2$, each vertex is $ig$ and the loop has symmetry factor $1/2$, giving coefficient $-g^2/2$. In both interpretations the closed fermion loop sign and spinor matrix trace disappear; the differing numerical factor comes from the interaction normalization and field identity.

Normalize the generating functional by its value with all sources zero, and take the connected part of the time-ordered correlation function. Source differentiation of the interaction exponential is equivalent to the Dyson series followed by Wick's theorem. Write $D(x-y)=⟨T\varphi (x)\varphi (y){⟩}_0$ and $F_{ij}(x-y)=⟨T{\psi }_i(x){\bar{\psi }}_j(y){⟩}_0$ temporarily, so that every contraction has an unambiguous meaning.

The second-order Dyson series contributesThere are two connected external scalar field attachments:They become equal after exchanging the integration variables $x,y$, cancelling $2!$. The connected fermion contractions have the closed fermion loop sign:The minus sign can also be checked directly: the pairing of the first with the fourth field and the second with the third has positive pairing parity, but $⟨T{\bar{\psi }}_i(x){\psi }_j(y)⟩=-F_{ji}(y-x)$. Multiplying it by $(ig{)}^2=-g^2$ leaves a positive result:A contraction of the two external fields with each other times a vacuum bubble cancels against the normalization. Separate fermion tadpole integrals at $x$ and $y$ belong to a product of one-point functions, removed by taking the connected part.

To match the source's generating functional convention, distinguish its displayed kernels from contractions. The positive $iJ\Delta J/2$ exponent gives $D=-i\Delta$. With the source insertion order in equation (5), the positive $i\bar{\eta }{S}_F\eta$ exponent similarly gives $F=-i{S}_F$. The two scalar propagators and two Dirac propagators supply a total phase $(-i{)}^4=1$. Thus, writing $S=S_F$ as in the requested formula,The matrix trace closes the spinor indices; no extra factor of two or additional symmetry factor remains for a Dirac field with this Yukawa interaction.

Put $S(p)=i(/p+m)/(p^2-m^2+i0)$ for the Dirac propagator, and ${\int }_{\ell }={\mu }_R^{2ϵ}\int d^{4-2ϵ}\ell /(2\pi {)}^{4-2ϵ}$ for a translation-invariant dimensional regularization of the loop momentum integral. The two oriented scattering amplitudes areThe common minus sign is the closed fermion loop sign. There is no additional symmetry factor for either labelled orientation. Each expression is a trace over spinor indices; all external photon propagators are amputated. If $K(p)=(/p+m)/(p^2-m^2+i0)$, the common factor is $-(-ie{)}^3{i}^3=-e^3$, so, for example, $i{\mathcal{M}}_A=-e^3{\int }_{\ell }\mathrm{tr}[{\gamma }^{\mu }K(\ell +k_1){\gamma }^{\nu }K(\ell -k_3){\gamma }^{\rho }K(\ell )]$.

Use the source's charge-conjugation matrix convention $C{\gamma }^{\mu }{C}^{-1}=-({\gamma }^{\mu }{)}^T$. It impliesA trace equals the trace of its matrix transpose, which reverses the order of all factors. Inserting the displayed identities givesThere is one minus sign from each of the three photon interaction vertices, and none from the transformed Dirac propagators. In the last line, change loop momentum to ${\ell }^{\prime }=-\ell -k_1$. Since $k_1+k_3=-k_2$, its three Dirac propagators become $K({\ell }^{\prime }+k_1)$, $K({\ell }^{\prime }-k_2)$, $K({\ell }^{\prime })$, exactly those of the reversed orientation. HenceThis is Furry's theorem. The regularization in quantum field theory matters: the individual Feynman integrals are superficially divergent, so an unregulated shift of a divergent integral is not a proof. A common dimensional regularization preserves charge conjugation and the vector Ward identity, making the change of variables legitimate. No chirality matrix occurs, so the axial triangle anomaly is not involved.

For an $n$-vertex fermion loop, reversing the fermion flow reverses the ordered product inside its trace. The same charge-conjugation matrix calculation transforms every Dirac propagator by $p\mapsto -p$ and contributes $-1$ for each vector photon interaction vertex. Relabel the loop momentum so that the external labels are fixed. The reversed Feynman diagram therefore obeysFor odd $n\ge 3$, the $(n-1)!$ cyclic orders pair into opposite orientations and cancel. For $n=1$, the single photon tadpole maps into itself with a minus sign and is zero. The common closed fermion loop sign is present on both sides and does not supply the factor $(-1{)}^n$. This proves Furry's theorem for all the requested one-loop Feynman diagrams. The charge conjugation argument also gives vanishing odd-photon correlation functions to all orders in vacuum QED, provided no charge conjugation symmetry breaking background or charged external state is introduced.

At large loop momentum, each Dirac propagator scales as ${\ell }^{-1}$. One four-dimensional loop and four internal Dirac propagators therefore give the superficial degree of divergenceThus a possible ultraviolet divergence in the sum is logarithmic. By locality its divergent coefficient is a degree-zero polynomial in the external four-momenta, namely a constant Lorentz tensor,The photon Ward identities apply to the complete sum for arbitrary external four-momenta, including off shell ones: $k_{1\mu }{\mathcal{M}}^{\mu \nu \rho \sigma }=0$, and likewise for the other legs. A constant tensor with $k_{1\mu }{\mathcal{M}}_{\mathrm{div}}^{\mu \nu \rho \sigma }=0$ for every $k_1$ must be zero. Hence $a=b=c=0$ andEquivalently, a logarithmic divergence would require a local dimension-four $A^4$ counterterm, forbidden by Abelian gauge invariance. A gauge-invariant four-photon local operator starts with four gauge field strengths, for example $(F_{\mu \nu }{F}^{\mu \nu }{)}^2$, of mass dimension eight; it cannot be this power counting divergence. Individual boxes need not be finite; their forbidden divergent parts cancel in the sum. There are no divergent proper loop subgraphs at this loop order. With the massive Dirac field in the source, there is also no soft or collinear infrared divergence at generic external kinematics. The argument establishes ultraviolet finiteness, and does not claim the absence of physical threshold singularities.