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For a free field in the annihilation-operator vacuum, the coincident two-point function is its mode-power integral. Unsmeared vacuum products generally have an ultraviolet divergence. A finite band or spatial smearing defines a regulated variance without altering the meaning of a mode-by-mode dimensionless power spectrum.

A counterterm is a local term added to a regulated Lagrangian density to cancel ultraviolet divergences and impose chosen renormalization conditions. Field-strength, mass and coupling counterterms respectively adjust propagator normalization, pole position and interaction strength.

The local Lagrangian density whose coefficients absorb the regulated ultraviolet divergences and implement chosen renormalization conditions. In a renormalizable four-dimensional theory, power counting permits operators of mass dimension at most four, subject to the actual symmetries and up to total derivatives. Operators absent from the initial displayed interaction can still be required for counterterm closure.

Cutoff regularization restricts loop momenta to $|\ell |<\Lambda$. The ultraviolet cutoff $\Lambda$ makes individual integrals finite while displaying power and logarithmic ultraviolet divergences explicitly.

The scattering phase shift $\delta (k)=2\arctan (m/k)$ of the Sine-Gordon kink fluctuation operator gives a continuum frequency sum. In mode-number regularization of soliton masses, the zero-frequency translation mode replaces the vacuum oscillator at $k=0$. The bare difference is $-m/2-(2\pi {)}^{-1}{\int }_0^{\Lambda }\delta (k)k/\sqrt{k^2+m^2}dk$. Integration by parts cancels the lower endpoint against $-m/2$ and produces the displayed upper surface term. Although $\delta (\Lambda )$ vanishes, its product with $\omega (\Lambda )$ tends to $2m$. Vacuum normal ordering cancels the logarithmic bulk ultraviolet divergence, leaving this finite one-loop soliton mass correction. Parameters and finite counterterms must be fixed by a stated renormalization condition.

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.

The two-point function Feynman integral with two interaction vertices joined by five parallel scalar propagators has four independent loop momenta:This formula uses Euclidean momenta. In three dimensions the convolution theorem and $G(x)=e^{-m|x|}/(4\pi |x|)$ givewith cutoff regularization required at $\rho =0$. With a lower radial cutoff $a$, the coefficient of $p^2$ has ultraviolet divergence $-\log (1/(ma))/(1536{\pi }^4)$. It therefore generates a divergent wave-function renormalization in three-dimensional sextic scalar field theory. By contrast the three-line sunset diagram has only a momentum-independent logarithmic ultraviolet divergence in three dimensions. This illustrates why a superficial degree of divergence of two allows both a mass counterterm and a kinetic counterterm, while degree zero allows only the former in a two-point function.

An infrared regulator suppresses the small-momentum region responsible for an infrared divergence. Examples are a small positive mass or a finite volume. It must be distinguished from an ultraviolet cutoff: setting external momenta and masses to zero to extract an ultraviolet divergence can introduce a spurious infrared divergence. Keep a positive auxiliary mass until the divergent local coefficients have been separated, then take the desired limit in infrared-safe quantities.

For the QED proper Feynman vertex with both external four-momenta zero, a massive internal fermion and a massless internal photon in Feynman gauge, the even numerator reduces under Lorentz covariance to ${\gamma }^{\mu }[\alpha k^2+\beta m^2]$. The gamma matrices identity ${\gamma }^{\nu }{\gamma }^{\mu }{\gamma }_{\nu }=(2-D){\gamma }^{\mu }$ and the integrated replacement $k_{\rho }{k}_{\sigma }\mapsto g_{\rho \sigma }{k}^2/D$ prove the coefficients. The term $m^2/[k^2(k^2-m^2+i0{)}^2]$ has neither an infrared divergence nor an ultraviolet divergence for $m>0$; its integral in four dimensions is $-i/(16{\pi }^2)$. Only the remaining squared massive denominator supplies the ultraviolet pole.

For finitely many free physical degrees of freedom with nonnegative masses, assign positive weights to bosons and negative weights to fermions. With a common spherical ultraviolet cutoff, the unsubtracted vacuum energy density is finite precisely when the three displayed supertraces vanish. Expanding $\sqrt{p^2+m^2}=p+m^2/(2p)-m^4/(8p^3)+O(p^{-5})$ proves that they cancel the quartic, quadratic and logarithmic ultraviolet divergences. The remaining radial integral converges as ${\int }^{\infty }{p}^{-3}dp$. Equal-mass boson–fermion pairing is sufficient but stronger than the three conditions; a renormalized vacuum energy with a free counterterm need not satisfy them.

An integral over momentum space. For example, state counting in $d$ spatial dimensions uses a measure proportional to $d^{d}p/(2\pi \hslash {)}^d$, whereas a relativistic Feynman integral commonly uses $d^{d}p/(2\pi {)}^d$ in natural units. The convention and any regulator must be specified. Independent loop momenta supply separate integration variables in a loop Feynman diagram; ultraviolet divergences and infrared divergences can require different regularizations.

In a common finite-volume regularization in quantum field theory, the oscillator contribution is $\Delta M_{\mathrm{osc}}=\frac{1}{2}{\sum }_n({\Omega }_n-{\Omega }_n^{(0)})$, including every discrete and continuum-derived mode and treating translation as a collective coordinate. Physical frequencies are ${\Omega }_n=m{\omega }_n$ when coordinates are measured in units of $m^{-1}$. A mass counterterm is added only afterwards. The vacuum subtraction removes the extensive vacuum energy but can leave an ultraviolet divergence.

Continue the loop momentum to $D=4-\epsilon$ dimensions, keeping $\mathrm{tr}1=4$ and the external physical polarization vectors four dimensional. Introduce Feynman parameters $x,y\ge 0$, $x+y\le 1$, denoting this integration triangle by $\mathcal{T}$. Completing the square givesOdd terms in $k$ integrate to zero. Lorentz invariance of dimensional regularization gives $k_{\mu }{k}_{\nu }\mapsto g_{\mu \nu }{k}^2/D$. The general projected numerator therefore becomes, under the Feynman integral,Only the $k^2$ term can give an ultraviolet divergence. Define, with the conventional scale factor implicit,The gamma function formula givesThus the pole of the numerator integral, including $2{\int }_{\mathcal{T}}dxdy=1$, isThe coefficients $(4,1,1)$ satisfy this condition. A nonzero local divergence would have to be removed by a counterterm. The Ward identity forbids a local $h{A}_{\mu }{A}^{\mu }$ term, and the first electromagnetic gauge-invariant coupling with one Higgs boson and two photons has the form $h{F}_{\mu \nu }{F}^{\mu \nu }$, of engineering dimension five. It is not an independent counterterm of the Standard Model, a renormalizable quantum field theory. The top quark contribution is separately electromagnetic gauge-invariant; no additional W boson loop Feynman diagram is needed to cancel this pole.

The finite remainder must still be retained. For $A=4C$,Putting $D=4$ before evaluating the divergent Feynman integral would incorrectly set this finite contribution to zero.

Route the photon momentum as $l$ and the internal scalar momentum as $l+k$. The transverse projector obeys $P_{\alpha \beta }{l}^{\beta }=0$, so the two cubic vertices giveMultiplying the two vertices and propagators therefore givesFor the seagull, $g^{\alpha \beta }{P}_{\alpha \beta }=d-1$. Including the photon-loop symmetry factor gives insteadThe printed seagull coefficient is twice this result. The vertex's factor two differentiates the two photon fields; closing them into one line requires the compensating symmetry factor. Direct expansion of the action gives the same check: the interaction density $-e^2{A}^2{\varphi }^{†}\varphi$ has one photon contraction, yielding coefficient $-e^2$, not $-2e^2$. The identical-photon origin of the vertex factor is also explicit in the standard scalar-electrodynamics rules.^[ref] This discrepancy does not affect the massless-photon limit in part d or the ultraviolet pole in part f.

After a Wick rotation, the term in question has radial behaviorThe radial integrand is proportional to $k_E$ at the origin and to $k_E^{-3}$ at infinity. Thus it has neither an infrared divergence nor an ultraviolet divergence. In fact the area of the unit three-sphere givesThis also follows from the Feynman parameter identitywith $A=k^2+i0$, $B=k^2-m^2+i0$. Although the parameterized denominator becomes massless at $x=1$, its numerator $1-x$ cancels the apparent endpoint singularity. Since $\beta \to -2$, multiplication by $\beta$ preserves finiteness. No artificial photon mass is needed.

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.

The Ward identity implies $q^{\mu }{\Pi }_{\mu \nu }=0$. Lorentz invariance then gives ${\Pi }_{\mu \nu }=(q^2{g}_{\mu \nu }-q_{\mu }{q}_{\nu })\Pi$, so $g^{\mu \nu }{\Pi }_{\mu \nu }=(d-1)q^2\Pi$. The gamma matrices satisfy ${\gamma }_{\mu }/k{\gamma }^{\mu }=(2-d)/k$. Odd traces vanish, and thereforeSubstituting gives the requested scalar coefficient:To extract the ultraviolet divergence without introducing a Feynman parameter, set $A=(k+q{)}^2-m^2$, $B=k^2-m^2$ and useTranslation invariance of dimensional regularization gives ${\int }_k{A}^{-1}={\int }_k{B}^{-1}={\mathcal{I}}_1$. ConsequentlyIn the pole coefficient one may set $d=4$: the bracket becomesThus the ultraviolet pole of QED vacuum polarization isThe cancellation of the apparent $m^2/q^2$ divergence is essential: gauge invariance forbids a photon mass counterterm. The scalar result at $q^2=0$ is understood by continuation after this cancellation.

The middle panel in part a has an internal photon of momentum $k$ joining two points on a fermion line; the internal fermion has momentum $p-k$. It is a one-particle-irreducible Feynman diagram with no closed fermion loop sign. In Feynman gauge,orThe decomposition in the source gives $\mathrm{tr}(/p\Sigma )=4p^{2}S(p^2)$. In particular,For a direct ultraviolet divergence extraction, use a Feynman parameter $x$ multiplying the massive denominator, shift $r=k-xp$, and define $\Delta =x{m}^2-x(1-x)p^2$. Then $/p-/k=(1-x)/p-/r$ andOdd $r$ terms integrate to zero. Taking the trace explicitly supplies the weight $(2-d)(1-x)$, whose integral tends to $-1$. HenceAs a check on the same fermion self-energy calculation,This is the ultraviolet pole of QED fermion self-energy. It is safe to extract it at generic external momentum; an on shell or massless limit must distinguish any infrared divergence from this ultraviolet divergence.

Unitary gauge fixes the scalar phase to zero. Locally choose $\alpha =-\pi /f$ in part b; then ${\pi }^{\prime }=0$ and $A_{\mu }^{\prime }=A_{\mu }+{\partial }_{\mu }\pi /(ef)$. The remaining fields are the radial Higgs mode and a massive vector. Its quadratic Lagrangian density is $-F^2/4+m_A^2{A}^2/2$ and its gauge-boson propagator isFor $k^2\ne 0$, introduce the transverse and longitudinal orthogonal projectors $P_{\mu \nu }^T=g_{\mu \nu }-k_{\mu }{k}_{\nu }/k^2$ and $P_{\mu \nu }^L=k_{\mu }{k}_{\nu }/k^2$. Omitting the infinitesimal in this algebraic decomposition,At large momentum in Euclidean space, the transverse coefficient falls as $k^{-2}$ but the longitudinal coefficient approaches a constant of order $m_A^{-2}$. Thus power counting applied directly to unitary gauge Feynman diagrams suggests increasingly severe ultraviolet divergences. This hides the renormalizability of physical scattering amplitudes; it does not disprove it. A finite-$\xi$ R-xi gauge exposes the better propagator scaling while preserving the fields needed to track the cancellations.

Put ${ϵ}_k={\hslash }^2{k}^2/(2m)>0$ for $\mathbf{k}\ne 0$ and $g=u\rho$, using $\rho =N/L^3$. Since the zero mode has zero kinetic energy, the exact kinetic energy in this trial state isTogether with part (d), the fixed-mean-number Bogoliubov variational energy isFor repulsive contact interaction, $u>0$, each ${ϵ}_k+g>g>0$. Using the double-angle hyperbolic-function identities,soThe stationary point therefore satisfiesThis is a minimum, not just a vanishing derivative. Define the positive Bogoliubov quasiparticle dispersionAt the displayed solution ${\theta }_k^{\ast }$, $\cosh 2{\theta }_k^{\ast }=({ϵ}_k+g)/E_k$ and $\sinh 2{\theta }_k^{\ast }=g/E_k$. HenceThe hyperbolic cosine has its unique minimum at zero, proving global minimization of each retained term. Treating equal-magnitude wavevectors together simply multiplies the same function by their multiplicity. At a fixed ultraviolet cutoff, the resulting energy change from $\theta =0$ isIndeed $E_k<{ϵ}_k+g$. More directly, for small positive $\theta$,The pair vacuum expectation value lowers the interaction energy at first order in $\theta$, whereas the nonzero occupation number and its kinetic energy cost start at second order. The two-mode squeezed vacuum admixes opposite-momentum pairs into the condensate, introducing spatial correlations that reduce the chance of two bosons coinciding. These correlations lower repulsive contact interaction enough to outweigh the added kinetic energy. A small quantum depletion of a Bose condensate can therefore lower the energy below the uncorrelated condensate value.

Two qualifications delimit this result. First, the paper does not state the sign of $u$. More generally a finite stable minimum exists only when ${ϵ}_k+g>|g|$ for every retained nonzero mode; repulsion ensures this. For example, with $g=-{ϵ}_k$ in some mode, $f_k(\theta )={ϵ}_k\sinh (2\theta )/2$ tends to $-\infty$ as $\theta \to -\infty$, so an unconditional minimum claim for arbitrary $u$ is false. Also the formula minimizes the truncated Bogoliubov theory energy, not the exact quartic Hamiltonian including its discarded terms, and it must remain in the regime of small quantum depletion of a Bose condensate.

Second, a three-dimensional unregulated contact interaction cannot give a finite absolute beyond-leading-order ground-state energy by this bare sum. For large $k$, $E_k-{ϵ}_k-g=-g^2/(2{ϵ}_k)+O({ϵ}_k^{-2})$, whose three-dimensional sum has an ultraviolet divergence. A finite ultraviolet cutoff, or renormalization of a three-dimensional dilute-gas contact coupling matched to a physical scattering length, is needed. This does not alter the modewise minimization or the finite-cutoff lowering mechanism just derived.