Vacuum Polarization, Running Charge, and Screening

Vacuum polarization inserts a charged-fermion loop into the photon propagator. Current conservation makes the insertion transverse; renormalization turns its scalar coefficient into either a momentum-dependent effective charge or a running scheme parameter. This page derives the one-loop contribution of massive Dirac fermions. Both descriptions encode screening, but they are not numerically identical until their subtraction conventions are matched.

Required background. Electron and photon renormalization supplies $Z_3$ and the subtraction conditions. Beta functions and anomalous dimensions supplies the general renormalization-group definition.

Helpful background. Subtracted dispersion relations explains how the timelike spectral cut determines the spacelike function.

The transverse photon self-energy

For an electron of mass $m$, signed charge $q_e=-e$, and $e>0$, the vertex $+iq_e\gamma^\mu$ is $-ie\gamma^\mu$. The one-loop proper photon two-point function is

$$i\Pi^{\mu\nu}(q) =-(-ie)^2\mu^{2\epsilon} \int\frac{\mathrm d^d\ell}{(2\pi)^d} \operatorname{tr}\left[ \gamma^\mu\frac{i(\ell\!\!\!/+m)}{\ell^2-m^2+i0} \gamma^\nu\frac{i(\ell\!\!\!/+q\!\!\!/+m)}{(\ell+q)^2-m^2+i0} \right].$$

The initial minus sign is the closed-fermion-loop sign. Contracting a vertex uses $q\!\!\!/=(\ell+q)\!\!\!/-m-(\ell\!\!\!/-m)$, cancelling one adjacent fermion propagator in each of two terms. Cyclicity of the trace then makes the two remaining integrals equal after a loop-momentum shift. A symmetry-preserving regulator that permits this shift, such as dimensional regularization, therefore gives

$$q_\mu\Pi^{\mu\nu}(q)=0,$$

so Lorentz invariance fixes a single scalar function. Choose the positive-spectral current-correlator convention

$$\Pi^{\mu\nu}(q) =(q^\mu q^\nu-q^2\eta^{\mu\nu})\Pi_M(q^2).$$

The renormalization page instead factors out $q^2\eta^{\mu\nu}-q^\mu q^\nu$ and calls its coefficient $\Pi$. For the same tensor, $\Pi=-\Pi_M$. This tensor-basis sign is separate from the signed fermion charge; reversing a species’ charge leaves this two-vertex loop unchanged.

Any residual $\eta^{\mu\nu}$ term not accompanied by the required $q^2$ is a photon-mass artifact and violates the Ward identity. The spinor loop and its transverse tensor are developed in Schwartz 2014, § 16.2.2, pp. 308–309, Eqs. (16.41)–(16.47).

A subtracted Euclidean effective charge

Keep $m>0$ so that zero-momentum subtraction is infrared finite, and let $\alpha=e^2/(4\pi)$ denote the Thomson-limit coupling. Combining the two denominators, shifting the loop momentum and subtracting the value at $q^2=0$ removes the momentum-independent UV pole. The remaining scalar is

$$\Pi_M(q^2)-\Pi_M(0) =\frac{2\alpha}{\pi}\int_0^1\!\mathrm dx\,x(1-x) \ln\frac{m^2}{m^2-q^2x(1-x)-i0}.$$

The logarithm is continued from spacelike momentum with the displayed Feynman prescription. This is the finite subtraction in Schwartz 2014, § 16.3, pp. 309–310, Eqs. (16.48)–(16.56); his scalar $\Pi_2$ has the factor $e^2$ removed, whereas $\Pi_M$ here includes it.

Let $Q^2=-q^2>0$ be spacelike. To make the sign convention explicit, define the positive Euclidean screening function

$$\Pi_E(Q^2) \equiv-\bigl[\Pi_M(-Q^2)-\Pi_M(0)\bigr].$$

Zero-momentum subtraction then gives the finite one-loop result

$$\boxed{ \Pi_E(Q^2)=\frac{2\alpha}{\pi} \int_0^1\mathrm dx\,x(1-x) \ln\left[1+\frac{Q^2}{m^2}x(1-x)\right]}$$

Dyson summation of transverse insertions motivates the zero-momentum-subtracted Euclidean effective charge

$$\alpha_{\rm eff}(Q^2) =\frac{\alpha}{1-\Pi_E(Q^2)}.$$

This definition is tied to the chosen subtraction and spacelike kinematics. With the one-loop $\Pi_E$ above, the denominator sums repeated one-loop insertions, not every higher-loop correction. It should not be silently identified with an $\overline{\mathrm{MS}}$ coupling.

At low momentum, expand the logarithm:

$$\begin{aligned} \Pi_E(Q^2) &=\frac{2\alpha}{\pi}\frac{Q^2}{m^2} \int_0^1\mathrm dx\,x^2(1-x)^2 +O(Q^4/m^4)\\ &=\frac{\alpha}{15\pi}\frac{Q^2}{m^2} +O(Q^4/m^4). \end{aligned}$$

The $Q^2/m^2$ suppression is decoupling: a massive charged species cannot produce logarithmic running far below its pair threshold. At the independent checkpoint $Q^2=m^2$,

$$\Pi_E(m^2) =\frac{\alpha}{9\pi} \left[7+\sqrt5\ln(9-4\sqrt5)\right].$$

At $Q^2\gg m^2$,

$$\Pi_E(Q^2) =\frac{\alpha}{3\pi} \left[\ln\frac{Q^2}{m^2}-\frac53\right] +O\!\left(\frac{m^2}{Q^2}\ln\frac{Q^2}{m^2}\right).$$

The positive logarithm makes $\alpha_{\rm eff}$ increase as shorter distances are resolved.

Timelike threshold and analyticity

The Euclidean integral has no imaginary part. For timelike $s=q^2>0$, the logarithm’s denominator becomes negative when $sx(1-x)>m^2$. This interval first appears at

$$s=4m^2,$$

where a timelike virtual photon has enough invariant mass to create a real fermion–antifermion pair. This is the cut of the one-loop massive-fermion contribution, not a classification of every threshold in exact QED. For $s>4m^2$, its endpoints are

$$x_\pm=\frac{1\pm v}{2}, \qquad v=\sqrt{1-\frac{4m^2}{s}}.$$

For a positive real $a$, $\ln(-a-i0)=\ln a-i\pi$. The reciprocal logarithm in $\Pi_M$ therefore has imaginary part $+\pi$ on this interval. Its discontinuity gives

$$\frac{1}{\pi}\operatorname{Im}\Pi_M(s+i0) =\frac{2\alpha}{\pi}\int_{x_-}^{x_+}\!\mathrm dx\,x(1-x) =\frac{\alpha}{6\pi}v(3-v^2).$$

Since $3-v^2=2(1+2m^2/s)$, the spectral density is

$$\rho(s)=\frac{1}{\pi}\operatorname{Im}\Pi_M(s+i0) =\frac{\alpha}{3\pi} \left(1+\frac{2m^2}{s}\right) \sqrt{1-\frac{4m^2}{s}}\, \theta(s-4m^2).$$

The scalar grows only logarithmically at large momentum at this order, so one subtraction suffices. With the subtraction value fixed at zero momentum, the subtracted dispersion relation reconstructs the spacelike function:

$$\Pi_E(Q^2) =Q^2\int_{4m^2}^{\infty} \frac{\mathrm ds}{s(s+Q^2)}\rho(s).$$

This formula supplies an independent sign check: $\rho(s)\ge0$ implies $\Pi_E(Q^2)\ge0$ for $Q^2>0$. Using the Euclidean logarithm directly above timelike threshold would miss both the imaginary part and the correct sheet.

The renormalization-group coupling

In $\overline{\mathrm{MS}}$, the one-loop photon counterterm for $N_f$ active unit-charge Dirac fermions is

$$Z_3=1-\frac{N_f\alpha}{3\pi\bar\epsilon}+O(\alpha^2).$$

Together with $e_0=\mu^\epsilon Z_3^{-1/2}e$, holding the bare coupling fixed gives

$$\boxed{ \frac{\mathrm d\alpha}{\mathrm d\ln\mu} =\frac{2N_f}{3\pi}\alpha^2+O(\alpha^3)}.$$

For charges $Q_f$ in units of $e$, replace $N_f$ by $\sum_f Q_f^2$. At fixed $N_f$, integration gives the one-loop invariant

$$\alpha^{-1}(\mu) +\frac{2N_f}{3\pi}\ln\frac{\mu}{\mu_{\rm ref}} =\text{constant},$$

or

$$\alpha^{-1}(\mu) =\alpha^{-1}(\mu_0) -\frac{2N_f}{3\pi}\ln\frac{\mu}{\mu_0}.$$

The factor is written for $\ln\mu$. Differentiating with respect to $\ln\mu^2$ halves it; mixing the two conventions is a common error. The fixed-bare counterterm derivation appears in Schwartz 2014, § 23.2, pp. 423–425, Eqs. (23.26)–(23.32). His regulator uses $4-d=\epsilon_{\rm S}=2\epsilon$ in the convention of the renormalization prerequisite.

Threshold matching

A mass-independent scheme does not automatically decouple a heavy field. Below a mass $M$, match the full theory onto an EFT without that active species and run with the lower $N_f$. At leading order the coupling is continuous at $\mu=M$; higher orders add a finite decoupling coefficient.

For the exact benchmark

$$N_f=1\quad(\mu<M), \qquad N_f=2\quad(\mu>M), \qquad \alpha(M)=\frac{1}{10},$$

one-loop running with leading matching, one unit of $\ln\mu$ below and above the threshold, gives

$$\alpha^{-1}(M\exp(-1))=10+\frac{2}{3\pi}, \qquad \alpha^{-1}(M\exp(1))=10-\frac{4}{3\pi}.$$

Running through $M$ with one unchanged beta coefficient fails the threshold check.

To compare $\alpha_{\overline{\rm MS}}(\mu)$ with $\alpha_{\rm eff}(Q^2)$, choose a matching point, compute the finite conversion there, and only then compare slopes. Their raw values belong to different definitions.

Screening and the Landau-pole caution

Virtual charged pairs polarize the vacuum so that a distant probe sees a more screened charge than a short-distance probe. Equivalently, the beta function is positive and the renormalized coupling grows toward the ultraviolet. This interpretation is gauge invariant when formulated through the current–current amplitude or a physical effective charge, not through a gauge-dependent photon field normalization alone. The subtracted scattering amplitude and its screened-charge interpretation are developed in Weinberg 1995, § 11.2, pp. 480–482, Eqs. (11.2.22)–(11.2.34).

Formally extending the one-loop solution produces a scale where its denominator vanishes. That “Landau pole” lies outside the regime in which the one-loop approximation can justify itself. It signals that perturbative QED does not supply a controlled ultraviolet completion; it is not a prediction of a physical pole at a trustworthy energy.

Common pitfalls

An $\overline{\mathrm{MS}}$ coupling is not a measured potential by definition. Match it to a physical effective charge before comparing numbers.

Euclidean and timelike polarization are not the same real function. Above $4m^2$, the $+i0$ prescription and branch cut are essential.

Mass-independent running does not implement decoupling automatically. Change the active theory and match at each threshold.

The one-loop Landau pole is not controlled evidence of a state. It occurs where the approximation used to derive it has already failed.

Exercises

Recover decoupling from the timelike cut. Without expanding the Feynman-parameter logarithm, use the dispersion relation to find the coefficient of $Q^2$ in $\Pi_E(Q^2)$ as $Q^2\to0$. Why does its sign not depend on guessing a screening picture?

Solution

At small $Q^2$, the coefficient is the convergent positive spectral moment $\int_{4m^2}^{\infty}\rho(s)\,\mathrm ds/s^2$. Put $v=\sqrt{1-4m^2/s}$, so $s=4m^2/(1-v^2)$ and $\mathrm ds/s^2=v\,\mathrm dv/(2m^2)$. Then

$$\begin{aligned} \int_{4m^2}^{\infty}\frac{\rho(s)}{s^2}\,\mathrm ds &=\frac{\alpha}{12\pi m^2} \int_0^1\!\mathrm dv\,v^2(3-v^2)\\ &=\frac{\alpha}{15\pi m^2}. \end{aligned}$$

Multiplication by $Q^2$ reproduces the low-momentum result. The positive sign follows from the spectral weight and the positive spacelike dispersion kernel; it is already fixed before assigning a screening interpretation.

References

• Matthew D. Schwartz, Quantum Field Theory and the Standard Model, Cambridge University Press (2014), §§ 16.2.2–16.3 and 23.2, doi:10.1017/9781139540940.
• Steven Weinberg, The Quantum Theory of Fields, Volume I: Foundations, Cambridge University Press (1995), § 11.2, doi:10.1017/CBO9781139644167.