Quantum Field Theory Course

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Part IV, Chapter 6 | Page 2 of 4

Vacuum Polarization

The photon self-energy, charge screening, and the running of $\alpha$

6.6 The One-Loop Photon Self-Energy

Vacuum polarization is the process by which a virtual photon fluctuates into an electron-positron pair. This modifies the photon propagator. The one-loop correction to the photon propagator is given by the fermion loop:

$$i\Pi^{\mu\nu}(q) = (-1)(-ie)^2 \int \frac{d^dk}{(2\pi)^d} \frac{\text{Tr}\left[\gamma^\mu(\not\!k + m)\gamma^\nu(\not\!k - \not\!q + m)\right]}{(k^2 - m^2 + i\epsilon)((k-q)^2 - m^2 + i\epsilon)}$$

The overall factor of $(-1)$ comes from the closed fermion loop (Fermi statistics). The trace is over Dirac indices. By gauge invariance (the Ward identity), the tensor structure must be transverse:

$$\Pi^{\mu\nu}(q) = (q^2 g^{\mu\nu} - q^\mu q^\nu)\,\Pi(q^2)$$

The scalar function $\Pi(q^2)$ is the vacuum polarization function. The $q^\mu q^\nu$ term does not contribute to physical processes (it couples to conserved currents), so all the physics is in $\Pi(q^2)$.

💡Why Transverse?

Gauge invariance requires $q_\mu \Pi^{\mu\nu}(q) = 0$. This is a consequence of current conservation $\partial_\mu J^\mu = 0$ in QED. The transversality condition $\Pi^{\mu\nu} \propto (q^2 g^{\mu\nu} - q^\mu q^\nu)$ is the unique rank-2 tensor satisfying this. Dimensional regularization automatically preserves this structure, whereas a naive momentum cutoff would violate it.

6.7 Evaluating the Loop Integral

Step 1: Feynman parametrization. Combine the two propagators using:

$$\frac{1}{AB} = \int_0^1 dx\;\frac{1}{[xA + (1-x)B]^2}$$

Setting $A = k^2 - m^2$ and $B = (k-q)^2 - m^2$, and shifting$k \to \ell = k - (1-x)q$:

$$xA + (1-x)B = \ell^2 - \Delta, \qquad \Delta = m^2 - x(1-x)q^2$$

Step 2: Dirac trace. The numerator trace evaluates to (in $d$ dimensions):

$$\text{Tr}\left[\gamma^\mu(\not\!k + m)\gamma^\nu(\not\!k - \not\!q + m)\right] = 4\left[k^\mu(k-q)^\nu + k^\nu(k-q)^\mu - g^{\mu\nu}(k \cdot (k-q) - m^2)\right] + O(d-4)$$

After shifting to $\ell$ and using the fact that odd powers of $\ell$ vanish under symmetric integration, the numerator becomes:

$$N^{\mu\nu} = 4\left[\frac{2}{d}\ell^2 g^{\mu\nu} - g^{\mu\nu}(\ell^2 - \Delta + m^2 - x(1-x)q^2) + 2x(1-x)(q^2 g^{\mu\nu} - q^\mu q^\nu)\right]$$

Step 3: The $d$-dimensional integral. Using the master formula for dimensional regularization:

$$\int \frac{d^d\ell}{(2\pi)^d} \frac{\ell^2}{(\ell^2 - \Delta)^2} = \frac{i}{(4\pi)^{d/2}} \frac{d}{2}\frac{\Gamma(1 - d/2)}{\Gamma(2)}\left(\frac{1}{\Delta}\right)^{1-d/2}$$

After careful algebra, the non-transverse pieces cancel (as guaranteed by gauge invariance), and we obtain the transverse result:

$$\Pi(q^2) = -\frac{8e^2}{(4\pi)^{d/2}} \int_0^1 dx\;x(1-x)\;\frac{\Gamma(2-d/2)}{\Delta^{2-d/2}}$$

6.8 The UV Divergence and Renormalization

Expanding around $d = 4$ (setting $d = 4 - \varepsilon$) and using$\Gamma(\varepsilon/2) = 2/\varepsilon - \gamma_E + O(\varepsilon)$:

$$\Pi(q^2) = -\frac{2\alpha}{\pi}\int_0^1 dx\;x(1-x)\left[\frac{2}{\varepsilon} - \gamma_E + \ln(4\pi) - \ln\frac{\Delta}{\mu^2}\right] + O(\varepsilon)$$

Using the $\overline{\text{MS}}$ scheme (absorbing $-\gamma_E + \ln(4\pi)$into the renormalization scale), and evaluating the $x$-integral of $x(1-x)$:

$$\int_0^1 dx\;x(1-x) = \frac{1}{6}$$

The UV-divergent part is:

$$\boxed{\Pi(q^2)\big|_{\text{div}} = -\frac{\alpha}{3\pi}\cdot\frac{2}{\varepsilon}}$$

This divergence is absorbed by the photon field-strength renormalization $Z_3$:

$$Z_3 = 1 - \frac{\alpha}{3\pi}\cdot\frac{2}{\varepsilon} + \text{finite}$$

The renormalized vacuum polarization (subtracting $\Pi(0)$ to enforce the on-shell condition that the photon propagator has unit residue at $q^2 = 0$) is:

$$\boxed{\hat{\Pi}(q^2) \equiv \Pi(q^2) - \Pi(0) = -\frac{2\alpha}{\pi}\int_0^1 dx\;x(1-x)\ln\!\left(\frac{m^2 - x(1-x)q^2}{m^2}\right)}$$

This is manifestly UV-finite: the $1/\varepsilon$ poles cancel in the subtraction. It vanishes at $q^2 = 0$ as required by the renormalization condition.

💡Charge Screening

The vacuum around a charge is filled with virtual $e^+e^-$ pairs. Like a dielectric medium, these pairs partially screen the bare charge. At larger distances (lower $q^2$), more screening occurs and the effective charge is smaller. At shorter distances (higher $q^2$), you "penetrate" the cloud and see more of the bare charge. This is why $\alpha$ increases with energy.

6.9 The Dressed Photon Propagator

The full (dressed) photon propagator is obtained by summing the geometric series of one-particle-irreducible (1PI) insertions:

$$\tilde{D}^{\mu\nu}(q) = \frac{-ig^{\mu\nu}}{q^2} + \frac{-ig^{\mu\alpha}}{q^2}\;i(q^2 g_{\alpha\beta} - q_\alpha q_\beta)\Pi(q^2)\;\frac{-ig^{\beta\nu}}{q^2} + \cdots$$

Summing the geometric series, the transverse part of the dressed propagator becomes:

$$\tilde{D}^{\mu\nu}_T(q) = \frac{-i(g^{\mu\nu} - q^\mu q^\nu/q^2)}{q^2[1 - \Pi(q^2)]}$$

The effect of vacuum polarization is to modify the effective coupling between two charged particles. In a scattering process, the exchanged photon propagator carries a factor:

$$\frac{e^2}{q^2} \to \frac{e^2}{q^2[1 - \hat\Pi(q^2)]} = \frac{e^2_{\text{eff}}(q^2)}{q^2}$$

This defines an effective, momentum-dependent charge.

6.10 The Running Coupling Constant

The effective fine-structure constant at momentum transfer $q^2$ is:

$$\alpha_{\text{eff}}(q^2) = \frac{\alpha}{1 - \hat\Pi(q^2)}$$

For large spacelike momenta $|q^2| \gg m^2$ (where $q^2 < 0$ in our convention), the logarithm in $\hat\Pi$ can be evaluated:

$$\hat\Pi(q^2) \approx -\frac{\alpha}{3\pi}\ln\!\left(\frac{-q^2}{m^2}\right) + O\!\left(\frac{m^2}{q^2}\right) \quad\text{for } |q^2| \gg m^2$$

This gives the famous running coupling:

$$\boxed{\alpha(Q^2) = \frac{\alpha}{1 - \frac{\alpha}{3\pi}\ln\!\left(\frac{Q^2}{m^2}\right)}} \qquad (Q^2 \equiv -q^2 > 0)$$

Numerically, with $\alpha(0) \approx 1/137.036$:

  • At the electron mass scale: $\alpha(m_e^2) \approx 1/137$
  • At the $Z$ boson mass ($M_Z \approx 91$ GeV): $\alpha(M_Z^2) \approx 1/128$
  • Extrapolating to the Planck scale: $\alpha(M_{\text{Pl}}^2)$ would hit a pole (Landau pole)

The $\beta$-function governing the running is:

$$\beta(\alpha) = \mu\frac{\partial\alpha}{\partial\mu} = \frac{2\alpha^2}{3\pi} + O(\alpha^3)$$

The positive sign means QED is not asymptotically free — the coupling grows at high energies. This is in contrast to QCD, where the $\beta$-function is negative.

💡The Landau Pole

The running coupling formula has a pole at $Q^2 = m^2 e^{3\pi/\alpha}$, a fantastically large scale ($\sim 10^{286}$ GeV). This "Landau pole" signals that QED perturbation theory breaks down at extremely high energies. In practice, this is irrelevant because new physics (the electroweak unification) enters long before this scale. But it tells us that QED is an effective field theory, not a fundamental theory valid to all energies.

6.11 Physical Consequences: The Uehling Potential

Vacuum polarization modifies the Coulomb potential at short distances. Fourier transforming the modified photon propagator gives the Uehling potential:

$$V(r) = -\frac{\alpha}{r}\left[1 + \frac{2\alpha}{3\pi}\int_1^\infty du\;\left(1 + \frac{1}{2u^2}\right)\frac{\sqrt{u^2 - 1}}{u^2}\;e^{-2mru}\right]$$

For $r \gg 1/m$ (distances much larger than the electron Compton wavelength), the correction is exponentially suppressed. For $r \ll 1/m$:

$$V(r) \approx -\frac{\alpha}{r}\left[1 + \frac{\alpha}{3\pi}\left(\ln\frac{1}{mr} + \text{const}\right)\right]$$

This correction shifts hydrogen energy levels. It contributes about $-27$ MHz to the Lamb shift of the $2S_{1/2}$ state, which is roughly 3% of the total Lamb shift of $\sim 1058$ MHz.

Key Concepts (Page 2)

  • • Vacuum polarization: a virtual $e^+e^-$ loop dresses the photon propagator
  • • Gauge invariance forces $\Pi^{\mu\nu}$ to be transverse: $\Pi^{\mu\nu} = (q^2 g^{\mu\nu} - q^\mu q^\nu)\Pi(q^2)$
  • • The UV divergence $\sim 2/\varepsilon$ is absorbed by photon field renormalization $Z_3$
  • • The renormalized $\hat\Pi(q^2) = \Pi(q^2) - \Pi(0)$ is UV-finite
  • • Running coupling: $\alpha(Q^2) = \alpha/[1 - (\alpha/3\pi)\ln(Q^2/m^2)]$
  • • QED is not asymptotically free: $\beta > 0$, coupling grows at high energy
  • • The Uehling potential modifies Coulomb's law at short distances and contributes to the Lamb shift
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