Effective Actions in Background Fields

The previous page introduced functional determinants and proper time. We now put that technology to work in the cleanest possible laboratory: a charged quantum field moving in a fixed electromagnetic background. The background is not dynamical at first. It is a probe. By integrating out the charged matter, we ask what local and nonlocal terms are induced in the effective action for the background field.

The payoff is large. A constant magnetic field diagonalizes the one-loop problem into Landau levels, so the determinant can be computed without drawing a single momentum-space diagram. The same answer knows about vacuum polarization, charge renormalization, magnetic susceptibility, and the finite nonlinear photon interactions that become the Euler–Heisenberg effective action. It also reveals a physical distinction that will be crucial for Yang–Mills theory: orbital motion tends to screen, while spin magnetic moments can produce the opposite sign.

Required background. Proper Time, Determinants, and Thermal Traces supplies the determinant signs, heat-kernel normalization, and the open-to-closed worldline construction. Running Charge, Screening, and Antiscreening fixes the interpretation of positive matter contributions to $1/e^2$. The signs below refer first to the coefficient of $F^2$ in the effective action, $\Gamma[A]\supset(4e^2)^{-1}\int F^2$, not directly to the beta function. A positive logarithmic contribution from modes between $m$ and $\Lambda$ raises $1/e^2$; differentiating with respect to the renormalization scale then gives the familiar positive QED beta function.

Background fields and one-loop determinants

Determinant and heat-kernel normalization

We use the Euclidean rescaled gauge-field normalization from the previous pages,

$$\Gamma[A]\supset {1\over4}\int d^4x\,{1\over e^2}F_{\mu\nu}F_{\mu\nu},$$

and charged matter has unit charge,

$$D_\mu=\partial_\mu-iA_\mu, \qquad [D_\mu,D_\nu]=-iF_{\mu\nu}.$$

For a pure magnetic background,

$$F_{12}=B, \qquad F_{\mu\nu}F_{\mu\nu}=2B^2.$$

A complex scalar contributes

$$\Gamma_s^{(1)}[A]=\operatorname{Tr}\log(-D^2+m^2),$$

while a Dirac fermion contributes

$$\Gamma_f^{(1)}[A]=- \operatorname{Tr}\log(\gamma_\mu D_\mu+m).$$

The signs come from ordinary Gaussian integration: bosonic determinants sit in the denominator of the path integral, Grassmann determinants sit in the numerator.

Let $\chi$ denote a charged quantum field and let $A_\mu$ be a fixed external field. The effective action for $A_\mu$ is defined by

$$e^{-\Gamma[A]} = \int \mathcal D\chi\,e^{-S_E[\chi,A]}.$$

If $\chi$ is a complex scalar with

$$S_E[\phi,A] = \int d^4x\,\phi^*(-D^2+m^2)\phi,$$

then the matter integral is Gaussian and gives

$$e^{-\Gamma_s^{(1)}[A]} \propto \left[\operatorname{Det}(-D^2+m^2)\right]^{-1}.$$

Therefore

$$\Gamma_s^{(1)}[A] = \operatorname{Tr}\log(-D^2+m^2).$$

For a Dirac fermion,

$$S_E[\psi,A] = \int d^4x\,\overline\psi(\gamma_\mu D_\mu+m)\psi,$$

and Grassmann integration gives

$$e^{-\Gamma_f^{(1)}[A]} \propto \operatorname{Det}(\gamma_\mu D_\mu+m),$$

so

$$\Gamma_f^{(1)}[A] =-\operatorname{Tr}\log(\gamma_\mu D_\mu+m).$$

The trace includes spacetime, spinor indices when present, and any internal indices. The determinant is a compact way to sum all one-loop diagrams with external $A_\mu$ insertions. This is also why background fields are so efficient pedagogically: one calculation of a spectral trace contains the two-point vacuum polarization, four-photon scattering, and all higher one-loop background vertices.

This is also the cleanest practical definition of “integrating out a heavy charged field.” If the external momenta satisfy $q^2\ll m^2$, the resulting functional can be expanded in local gauge-invariant operators. If $q^2$ is comparable to or larger than $m^2$, the determinant remains meaningful, but its expansion is nonlocal and threshold-dependent.

To see the diagrammatic expansion explicitly, write

$$L_A=L_0+V[A], \qquad G_0=L_0^{-1}.$$

Then

$$\operatorname{Tr}\log L_A = \operatorname{Tr}\log L_0+\operatorname{Tr}\log(1+G_0V),$$

and therefore

$$\operatorname{Tr}\log L_A = \operatorname{Tr}\log L_0 +\operatorname{Tr}(G_0V) -{1\over2}\operatorname{Tr}(G_0VG_0V) +{1\over3}\operatorname{Tr}(G_0VG_0VG_0V)-\cdots.$$

The quadratic term is the vacuum-polarization diagram. The quartic term is one-loop light-by-light scattering. The determinant is not a new approximation; it is the same one-loop expansion written in a form where gauge covariance is much easier to preserve.

Integrating out a charged field gives an effective action for the background field. Expanding $\operatorname{Tr}\log(L_0+V[A])$ reproduces the one-loop diagrams with any number of external background insertions.

A useful identity for variations is

$$\delta\operatorname{Tr}\log L = \operatorname{Tr}(L^{-1}\delta L).$$

It follows immediately from the finite-dimensional formula $\delta\log\det L=\operatorname{tr}(L^{-1}\delta L)$, and it remains valid for regularized functional traces. In the Euclidean functional integral define the source current by $J_\mu(x)\equiv\delta S_E/\delta A_\mu(x)$. Since $\Gamma[A]=-\log Z[A]$, varying the effective action then gives

$${\delta\Gamma[A]\over\delta A_\mu(x)}=\langle J_\mu(x)\rangle_A.$$

Thus the determinant is not merely a formal object: it is the generating functional for the response of the quantum vacuum to the background field.

The worldline representation from the previous page makes the same response geometric. For a complex scalar with $D_\mu=\partial_\mu-iA_\mu$,

$$\boxed{ \Gamma_s^{(1)}[A]-\Gamma_s^{(1)}[0] =-\int_\epsilon^\infty {ds\over s}\,e^{-m^2s} \int_{x(s)=x(0)}\!\mathcal D x\, e^{-\int_0^s d\tau\,\dot x^2/4} \left(e^{+i\oint A_\mu dx^\mu}-1\right). }$$

The subtraction removes the field-independent closed loops. Reversing a loop’s orientation complex-conjugates its Wilson phase, so odd powers of an Abelian background cancel in the unoriented loop average. The first local response is therefore quadratic in $F_{\mu\nu}$, exactly as required by gauge invariance and as found from vacuum polarization.

Proper time in a magnetic field

The proper-time representation turns the determinant into a spectral trace. For a complex scalar,

$$\Gamma_s^{(1)}[A] = -\int_\epsilon^\infty {ds\over s}\, \operatorname{Tr}\,e^{-s(-D^2+m^2)} +\text{counterterms},$$

where $\epsilon\sim \Lambda^{-2}$ is a gauge-covariant ultraviolet cutoff. Subtracting the field-independent vacuum term gives

$$\Gamma_s^{(1)}[A]-\Gamma_s^{(1)}[0] = -\int_\epsilon^\infty {ds\over s}\,e^{-m^2s} \operatorname{Tr}\left(e^{sD^2}-e^{s\partial^2}\right).$$

Now choose a constant magnetic field in the $x^3$ direction,

$$F_{12}=B,$$

and take the gauge

$$A_2=Bx_1, \qquad A_1=A_3=A_4=0.$$

The transverse operator is

$$-D_1^2-D_2^2 = -\partial_1^2-(\partial_2-iBx_1)^2.$$

At fixed momentum $p_2$, this is a harmonic oscillator in $x_1$ with frequency $B$. Its eigenvalues are

$$(2n+1)B, \qquad n=0,1,2,\ldots.$$

The two remaining Euclidean directions are free, so the full scalar spectrum is

$$\lambda_{n,p_3,p_4}=p_3^2+p_4^2+(2n+1)B.$$

The degeneracy per unit area in the $x^1x^2$ plane is

$${B\over2\pi}.$$

Therefore the scalar heat kernel per four-volume is

$${1\over V_4}\operatorname{Tr}e^{-s(-D^2)} = {B\over2\pi}\sum_{n=0}^\infty e^{-s(2n+1)B} \int {d^2p_\parallel\over(2\pi)^2}e^{-sp_\parallel^2},$$

where $p_\parallel=(p_3,p_4)$. Since

$$\int {d^2p_\parallel\over(2\pi)^2}e^{-sp_\parallel^2}={1\over4\pi s}$$

and

$$\sum_{n=0}^\infty e^{-s(2n+1)B} ={1\over2\sinh Bs},$$

we obtain

$$\boxed{ {1\over V_4}\operatorname{Tr}e^{-s(-D^2)} ={1\over(4\pi s)^2}{Bs\over\sinh Bs}. }$$

For a scalar particle, the transverse operator in a constant magnetic field has eigenvalues $(2n+1)B$. For a spinor, the spin magnetic moment shifts them by $\mp B$, producing the four-component spin trace $4\cosh Bs$.

The magnetic field has converted the problem into a product of two familiar ingredients: free heat flow in the directions parallel to the field and harmonic-oscillator heat flow in the transverse plane.

Scalar QED: orbital response and the local F² term

The scalar effective action in a constant magnetic field is

$${\Gamma_s^{(1)}[B]-\Gamma_s^{(1)}[0]\over V_4} = -\int_\epsilon^\infty {ds\over s}\,e^{-m^2s} {1\over(4\pi s)^2} \left({Bs\over\sinh Bs}-1\right).$$

The small-$s$ region controls the ultraviolet divergence. For $x=Bs$,

$${x\over\sinh x}=1-{x^2\over6}+{7x^4\over360}+O(x^6).$$

Thus the first field-dependent term in the effective action is

$${\Gamma_s^{(1)}[B]\over V_4} \supset {B^2\over6(4\pi)^2} \int_\epsilon^\infty {ds\over s}\,e^{-m^2s}.$$

The logarithmic part is

$$\int_\epsilon^\infty {ds\over s}\,e^{-m^2s} = \log {1\over m^2\epsilon}+\text{finite} = \log {\Lambda^2\over m^2}+\text{finite}.$$

Therefore

$$\Gamma_s^{(1)}[B] \supset {V_4B^2\over96\pi^2}\log {\Lambda^2\over m^2}.$$

Because $F_{\mu\nu}F_{\mu\nu}=2B^2$, this is

$$\Gamma_s^{(1)}[A] \supset {1\over192\pi^2}\log {\Lambda^2\over m^2} \int d^4x\,F_{\mu\nu}F_{\mu\nu}.$$

Matching to

$${1\over4}\int d^4x\,{1\over e^2}F_{\mu\nu}F_{\mu\nu}$$

gives the scalar contribution. The factor of $1/4$ in the Maxwell term is easy to miss: the coefficient of $\int F^2$ in $\Gamma$ must be multiplied by $4$ to obtain the shift of $1/e^2$.

$$\boxed{ \Delta\left({1\over e^2}\right)_s = {1\over48\pi^2}\log {\Lambda^2\over m^2} = {1\over16\pi^2}{1\over3}\log {\Lambda^2\over m^2}. }$$

This is the same coefficient found from the momentum-space vacuum polarization, now derived from a constant-background heat kernel.

The factor $x/\sinh x$ is less than one for real nonzero $x$. In the thermodynamic language of a charged particle, the orbital motion in a magnetic field reduces the density of low-lying states compared with the naive classical phase-space result. This is the origin of the word diamagnetic in this discussion. In field theory the observable statement is the induced local $F^2$ term; the separation into “magnetic susceptibility” language is an interpretation of the same coefficient.

Spinor QED: spin trace and paramagnetism

For a Dirac fermion, it is useful to square the Dirac operator. Up to an $A$-independent normalization and a choice of determinant branch, the parity-even part may be written as

$$\Gamma_f^{(1)}[A] =-{1\over2}\operatorname{Tr}\log\left[-(\gamma_\mu D_\mu)^2+m^2\right].$$

This rewriting is a calculational device for the even-in-$F$ part of the determinant. It is not a harmless replacement when one studies phases, spectral asymmetry, or anomalies.

Using

$$(\gamma_\mu D_\mu)^2 =D^2-{i\over4}[\gamma_\mu,\gamma_\nu]F_{\mu\nu},$$

the positive Laplace-type operator is

$$-(\gamma_\mu D_\mu)^2+m^2 = -D^2+m^2+{i\over4}[\gamma_\mu,\gamma_\nu]F_{\mu\nu}.$$

The last term is the spin magnetic moment. The determinant branch can matter for parity-odd terms in odd dimensions, but it does not affect the parity-even $F^2$ coefficient we are extracting here. For a pure magnetic field $F_{12}=B$, the spin matrix has eigenvalues $+B$ and $-B$, each twice degenerate in four-component Dirac notation. Hence the spinor heat kernel per four-volume is

$$\boxed{ {1\over V_4}\operatorname{Tr}e^{-s[-(\gamma_\mu D_\mu)^2]} ={4\over(4\pi s)^2}\,Bs\coth Bs. }$$

Equivalently,

$$4Bs\coth Bs =4{Bs\over\sinh Bs}\cosh Bs.$$

The one-loop spinor effective action is therefore

$${\Gamma_f^{(1)}[B]-\Gamma_f^{(1)}[0]\over V_4} = {1\over2}\int_\epsilon^\infty {ds\over s}\,e^{-m^2s} {4\over(4\pi s)^2}\left(Bs\coth Bs-1\right).$$

For small $x$,

$$x\coth x=1+{x^2\over3}-{x^4\over45}+O(x^6).$$

Thus

$${\Gamma_f^{(1)}[B]\over V_4} \supset {2B^2\over3(4\pi)^2} \int_\epsilon^\infty {ds\over s}\,e^{-m^2s},$$

and hence

$$\Gamma_f^{(1)}[A] \supset {1\over48\pi^2}\log {\Lambda^2\over m^2} \int d^4x\,F_{\mu\nu}F_{\mu\nu}.$$

Matching again to the Maxwell term gives

$$\boxed{ \Delta\left({1\over e^2}\right)_f = {1\over12\pi^2}\log {\Lambda^2\over m^2} = {1\over16\pi^2}{4\over3}\log {\Lambda^2\over m^2}. }$$

The factor $4/3$ has a useful physical decomposition. Write

$$4{x\over\sinh x}\cosh x = 4\left(1-{x^2\over6}+\cdots\right) \left(1+{x^2\over2}+\cdots\right).$$

The coefficient of $x^2$ is

$$4\left(-{1\over6}+{1\over2}\right) =-{2\over3}+2={4\over3}.$$

The $-2/3$ term is the orbital contribution of four fermion components. The $+2$ term is the spin contribution. The spin term dominates. This is the cleanest elementary version of the paramagnetic mechanism that later reappears for spin-one gauge bosons.

The orbital Landau-level factor $x/\sinh x$ begins with a negative $x^2$ correction, while the spin factor $\cosh x$ begins with a positive one. For a Dirac fermion the spin contribution is larger, giving the total coefficient $4/3$.

Renormalized constant-field actions

The proper-time integrals above still display the vacuum-energy and charge-renormalization terms. Once the zero-field vacuum term and the $B^2$ term have been fixed by renormalization conditions, the remaining constant-field action is finite. For one complex scalar,

$$\boxed{ {\Gamma_{s,\mathrm{ren}}^{(1)}(B)\over V_4} =-{1\over16\pi^2}\int_0^\infty {ds\over s^3}\,e^{-m^2s} \left[{Bs\over\sinh Bs}-1+{(Bs)^2\over6}\right]. }$$

For one Dirac fermion, the parity-even part is

$$\boxed{ {\Gamma_{f,\mathrm{ren}}^{(1)}(B)\over V_4} ={1\over8\pi^2}\int_0^\infty {ds\over s^3}\,e^{-m^2s} \left[Bs\coth Bs-1-{(Bs)^2\over3}\right]. }$$

The subtraction signs are fixed by the small-$s$ expansions, not chosen by magnetic intuition. They remove the $B^0$ and $B^2$ terms already assigned to the cosmological constant and the renormalized Maxwell coupling. The first surviving terms are finite:

$${\Gamma_{s,\mathrm{ren}}^{(1)}(B)\over V_4} =-{7B^4\over5760\pi^2m^4}+O(B^6/m^8),$$

and

$${\Gamma_{f,\mathrm{ren}}^{(1)}(B)\over V_4} =-{B^4\over360\pi^2m^4}+O(B^6/m^8).$$

These Euclidean signs are consistent with the familiar positive nonlinear magnetic terms in the Minkowski Lagrangian because a static effective action and a Minkowski Lagrangian differ by the Wick-rotation sign. More importantly, the coefficients are regulator-independent low-energy matching data once the $F^2$ term has been renormalized.

Gauge-covariant cutoffs

A sharp cutoff $|p|<\Lambda$ is natural in ordinary momentum integrals, but it is not gauge invariant in a general background field. Momentum is not a gauge-covariant label once $A_\mu(x)$ is present. A condition such as

$$|p-A(x)|<\Lambda$$

is not a clean gauge-invariant regulator either; it depends on a local gauge choice and does not define a spectral cutoff on a gauge-covariant operator.

The proper-time cutoff is better because it regulates the spectrum of a covariant operator:

$$\operatorname{Tr}\log L \longrightarrow -\int_{1/\Lambda^2}^\infty {ds\over s}\,\operatorname{Tr}e^{-sL}.$$

If $L$ transforms by conjugation under a gauge transformation,

$$L\mapsto g^{-1}Lg,$$

then

$$\operatorname{Tr}e^{-sL}$$

is gauge invariant. This is why the heat-kernel expansion organizes ultraviolet divergences directly into gauge-invariant local operators,

$$\int d^4x\,1, \qquad \int d^4x\,F_{\mu\nu}F_{\mu\nu}, \qquad \int d^4x\,(D_\rho F_{\mu\nu})(D_\rho F_{\mu\nu}), \qquad\ldots.$$

The first term is vacuum energy. The second renormalizes the gauge coupling. The later terms are higher-derivative effective interactions suppressed by powers of the mass or cutoff.

A related point is that the $F^2$ term extracted above is local. Its finite part depends on the renormalization convention. By contrast, after the $F^2$ coefficient is fixed at a reference scale, the higher-order low-energy terms such as $F^4/m^4$ are genuine predictions of the one-loop theory.

Preview: background-field gauge

The same background-field logic applies to Yang–Mills theory, but now the field being integrated out is partly the gauge field itself. The Abelian sections above use a Hermitian field $A_\mu$ in $D_\mu=\partial_\mu-iA_\mu$. For the non-Abelian preview, it is convenient to absorb the factor $-i$ into an anti-Hermitian Lie-algebra-valued connection,

$$\mathcal A_\mu\equiv -iA_\mu^aT^a, \qquad \mathcal D_\mu=\partial_\mu+\mathcal A_\mu.$$

This is a notation change, not a change of physical convention. Split

$$\mathcal A_\mu=\overline{\mathcal A}_\mu+a_\mu,$$

where $\overline{\mathcal A}_\mu$ is the background and $a_\mu$ is the quantum fluctuation. A background gauge transformation acts as

$$\overline{\mathcal A}_\mu \mapsto g^{-1}\overline{\mathcal A}_\mu g+g^{-1}\partial_\mu g, \qquad a_\mu\mapsto g^{-1}a_\mu g.$$

Thus the fluctuation transforms homogeneously, like matter in the adjoint representation. The background covariant derivative

$$\overline{\mathcal D}_\mu a_\nu = \partial_\mu a_\nu+[\overline{\mathcal A}_\mu,a_\nu]$$

also transforms homogeneously. This makes the gauge-fixing condition

$$\overline{\mathcal D}_\mu a_\mu=0$$

natural: it fixes the quantum gauge redundancy while preserving manifest gauge invariance with respect to the background.

In background Feynman gauge, the quadratic operator for gauge fluctuations has the schematic form

$$\mathcal L_{\mu\nu}^{\rm vec} = -\overline{\mathcal D}^{2}\delta_{\mu\nu} -2\operatorname{ad}(\overline{\mathcal F}_{\mu\nu}),$$

where

$$\operatorname{ad}(\overline{\mathcal F}_{\mu\nu})a_\nu =[\overline{\mathcal F}_{\mu\nu},a_\nu].$$

The associated ghosts contribute the scalar adjoint operator

$$\mathcal L_{\rm gh}=-\overline{\mathcal D}^{2}.$$

The term proportional to $\overline{\mathcal F}_{\mu\nu}$ is the spin-one magnetic-moment coupling of the vector fluctuation. It is the non-Abelian analog of the spin term in the squared Dirac operator, but with a larger spin response. After subtracting unphysical ghost modes, this paramagnetic spin-one term dominates the orbital screening part. That dominance is the physical seed of Yang–Mills antiscreening.

In the background-field method, $\mathcal A_\mu=\overline{\mathcal A}_\mu+a_\mu$. The background transforms as a connection, while the fluctuation transforms homogeneously. The quadratic vector operator contains a spin-one coupling to $\overline{\mathcal F}_{\mu\nu}$, and ghosts remove the unphysical scalar-like components.

The next page turns this preview into the beta-function calculation and compares QED, scalar QED, and Yang–Mills theory in one language.

Electric fields and the origin of imaginary parts

A constant magnetic field is Euclidean-friendly. Its proper-time factors contain hyperbolic functions such as

$${Bs\over\sinh Bs}, \qquad Bs\coth Bs.$$

A constant electric field in Minkowski space is reached by analytic continuation. In the simplest pure-field case, this continuation morally sends

$$B\mapsto iE,$$

so hyperbolic functions turn into trigonometric functions:

$${Bs\over\sinh Bs} \mapsto {Es\over\sin Es}.$$

The magnetic-field heat kernel has no poles on the positive real proper-time axis. After analytic continuation to an electric field, factors such as $Es/\sin Es$ have poles at $s=n\pi/E$. The prescription $m^2\to m^2-i0$ tells how to pass them and produces an imaginary part.

The poles at

$$s={n\pi\over E}, \qquad n=1,2,\ldots,$$

are not a mathematical nuisance. They signal that the vacuum in a background electric field is unstable to pair creation. In the language of the effective action, the vacuum persistence amplitude has the form

$$\langle0_{\rm out}|0_{\rm in}\rangle_A =e^{i\Gamma_M[A]},$$

and an imaginary part of the Minkowski effective action gives a decay probability. We will return to this real-time interpretation later, when in/out vacua and pair creation are treated directly. For now, the lesson is simply this: the same proper-time determinant that renormalizes $F^2$ in a magnetic background also knows about vacuum instability in an electric background.

For weak slowly varying fields, after the $F^2$ term is renormalized, the remaining local effective action begins with fourth-order invariants. In Minkowski notation these are built from

$$\mathcal F={1\over2}(\mathbf B^2-\mathbf E^2), \qquad \mathcal G=\mathbf E\cdot\mathbf B.$$

The precise coefficients depend on the spin and charge of the particle in the loop. Their existence is the important structural point: integrating out massive charged matter produces local nonlinear photon interactions at energies small compared with the mass.

Summary

A one-loop effective action in a background field is a determinant. Proper time turns that determinant into a heat-kernel trace. For a constant magnetic field, the heat kernel is exactly computable because the transverse motion is a Landau-level problem.

For a complex scalar in four dimensions,

$${1\over V_4}\operatorname{Tr}e^{-s(-D^2)} ={1\over(4\pi s)^2}{Bs\over\sinh Bs},$$

and the logarithmic $B^2$ term gives

$$\Delta\left({1\over e^2}\right)_s ={1\over16\pi^2}{1\over3}\log {\Lambda^2\over m^2}.$$

For a Dirac fermion,

$${1\over V_4}\operatorname{Tr}e^{-s[-(\gamma_\mu D_\mu)^2]} ={4\over(4\pi s)^2}Bs\coth Bs,$$

and

$$\Delta\left({1\over e^2}\right)_f ={1\over16\pi^2}{4\over3}\log {\Lambda^2\over m^2}.$$

The scalar result comes from orbital Landau motion. The spinor result is larger because the spin magnetic moment contributes a paramagnetic term. These positive shifts of $1/e^2$ are cutoff matching statements; when translated into the running of $e(\mu)$, they give the familiar positive QED beta function. The background-field method generalizes this logic to Yang–Mills theory, where spin-one gauge fluctuations produce the antiscreening sign.

After subtracting the vacuum and $F^2$ terms, the constant-field proper-time integrals are finite. Their expansion begins at $B^4/m^4$ and gives the low-energy nonlinear photon interactions encoded by the Euler–Heisenberg action.

Common pitfalls

Proper time is not inverse temperature. Both $s$ and $\beta$ appear in traces and can produce hyperbolic functions, but $s$ is a spectral parameter while $\beta$ is the circumference of the physical Euclidean-time circle.

An ordinary momentum cutoff is not gauge covariant in a general background. A proper-time cutoff regulates the spectrum of a covariant operator and organizes divergences into gauge-invariant local terms.

The unrenormalized $F^2$ coefficient is not a finite prediction. It renormalizes the gauge coupling. The higher-dimension terms left after the vacuum and $F^2$ subtractions are finite low-energy matching data.

Gauge-field normalization changes displayed coefficients. Some references keep $e$ in $D_\mu=\partial_\mu-ieA_\mu$ and write $F^2/4$. Here the background is rescaled so the kinetic term contains $1/e^2$ and positively charged unit matter couples through $D_\mu=\partial_\mu-iA_\mu$.

Squaring the Dirac operator does not erase spin. Dropping the Pauli term turns a spinor into several scalar-like degrees of freedom and gives the wrong coefficient. The determinant phase must also be retained when parity-odd terms or anomalies are in scope.

Hermitian and anti-Hermitian connection notation must not be mixed silently. The Abelian sections use $D=\partial-iA$ with Hermitian $A$. The background-field-gauge preview writes $\mathcal A=-iA^aT^a$ so that $\mathcal D=\partial+\mathcal A$ is the same derivative in anti-Hermitian notation; its connection transforms with $+g^{-1}\partial g$.

Exercises

Exercise 1: Derive the scalar magnetic heat kernel

Derive the scalar heat kernel in a constant magnetic field,

$${1\over V_4}\operatorname{Tr}e^{-s(-D^2)} ={1\over(4\pi s)^2}{Bs\over\sinh Bs}.$$

Use the Landau-level spectrum and degeneracy.

Solution

For $F_{12}=B$, the transverse eigenvalues are

$$(2n+1)B, \qquad n=0,1,2,\ldots,$$

with degeneracy per transverse area

$${B\over2\pi}.$$

The two directions parallel to the magnetic field are free, so

$${1\over V_4}\operatorname{Tr}e^{-s(-D^2)} = {B\over2\pi}\sum_{n=0}^\infty e^{-s(2n+1)B} \int {d^2p\over(2\pi)^2}e^{-sp^2}.$$

The free Gaussian integral is

$$\int {d^2p\over(2\pi)^2}e^{-sp^2}={1\over4\pi s}.$$

The Landau-level sum is

$$\sum_{n=0}^\infty e^{-s(2n+1)B} ={e^{-Bs}\over1-e^{-2Bs}} ={1\over2\sinh Bs}.$$

Combining these factors gives

$${B\over2\pi}\cdot {1\over2\sinh Bs}\cdot {1\over4\pi s} ={B\over16\pi^2s\sinh Bs} ={1\over(4\pi s)^2}{Bs\over\sinh Bs}.$$

Exercise 2: Match the scalar contribution to the Maxwell term

Using

$${x\over\sinh x}=1-{x^2\over6}+O(x^4),$$

extract the logarithmic scalar contribution to $1/e^2$.

Solution

The scalar effective action is

$${\Gamma_s^{(1)}[B]-\Gamma_s^{(1)}[0]\over V_4} = -\int_\epsilon^\infty {ds\over s}\,e^{-m^2s} {1\over(4\pi s)^2} \left({Bs\over\sinh Bs}-1\right).$$

Using the expansion,

$${Bs\over\sinh Bs}-1=-{B^2s^2\over6}+O(B^4s^4).$$

Thus

$${\Gamma_s^{(1)}[B]\over V_4} \supset {B^2\over6(4\pi)^2} \int_\epsilon^\infty {ds\over s}e^{-m^2s}.$$

The logarithmic part is

$$\int_\epsilon^\infty {ds\over s}e^{-m^2s} = \log {\Lambda^2\over m^2}+\text{finite}.$$

Therefore

$$\Gamma_s^{(1)}[B] \supset {V_4B^2\over96\pi^2}\log {\Lambda^2\over m^2}.$$

Since

$$V_4B^2={1\over2}\int d^4x\,F_{\mu\nu}F_{\mu\nu},$$

we have

$$\Gamma_s^{(1)}[A] \supset {1\over192\pi^2}\log {\Lambda^2\over m^2} \int d^4x\,F_{\mu\nu}F_{\mu\nu}.$$

Matching to

$${1\over4}\Delta\left({1\over e^2}\right) \int d^4x\,F_{\mu\nu}F_{\mu\nu}$$

gives

$$\Delta\left({1\over e^2}\right)_s ={1\over48\pi^2}\log {\Lambda^2\over m^2} ={1\over16\pi^2}{1\over3}\log {\Lambda^2\over m^2}.$$

Exercise 3: Separate orbital and spin contributions in QED

Show that the spinor heat-kernel factor produces the coefficient $4/3$ in spinor QED. More precisely, use

$$4x\coth x=4{x\over\sinh x}\cosh x$$

and expand to order $x^2$.

Solution

The two factors have small-$x$ expansions

$${x\over\sinh x}=1-{x^2\over6}+O(x^4),$$

and

$$\cosh x=1+{x^2\over2}+O(x^4).$$

Therefore

$$4{x\over\sinh x}\cosh x =4\left(1-{x^2\over6}+O(x^4)\right) \left(1+{x^2\over2}+O(x^4)\right).$$

Multiplying gives

$$4{x\over\sinh x}\cosh x =4\left(1+\left[-{1\over6}+{1\over2}\right]x^2+O(x^4)\right).$$

Hence

$$4{x\over\sinh x}\cosh x =4\left(1+{x^2\over3}+O(x^4)\right) =4+{4\over3}x^2+O(x^4).$$

The orbital part contributes

$$4\left(-{1\over6}\right)=-{2\over3},$$

while the spin factor contributes

$$4\left({1\over2}\right)=2.$$

Their sum is

$$-{2\over3}+2={4\over3}.$$

This is the coefficient that leads to

$$\Delta\left({1\over e^2}\right)_f ={1\over16\pi^2}{4\over3}\log {\Lambda^2\over m^2}.$$

Exercise 4: Verify background-gauge covariance

Let

$$\mathcal A_\mu=\overline{\mathcal A}_\mu+a_\mu.$$

Under a background gauge transformation,

$$\overline{\mathcal A}_\mu \mapsto g^{-1}\overline{\mathcal A}_\mu g+g^{-1}\partial_\mu g, \qquad a_\mu\mapsto g^{-1}a_\mu g.$$

Show that $\overline{\mathcal D}_\mu a_\nu$ transforms homogeneously.

Solution

The background covariant derivative is

$$\overline{\mathcal D}_\mu a_\nu =\partial_\mu a_\nu+[\overline{\mathcal A}_\mu,a_\nu].$$

A compact way to prove the transformation law is to let it act on an adjoint test field $X$, which transforms as

$$X\mapsto X'=g^{-1}Xg.$$

The transformed background derivative is

$$\overline{\mathcal D}'_\mu X' =\partial_\mu(g^{-1}Xg) +[g^{-1}\overline{\mathcal A}_\mu g+g^{-1}\partial_\mu g,g^{-1}Xg].$$

Using

$$\partial_\mu g^{-1}=-g^{-1}(\partial_\mu g)g^{-1},$$

the terms involving $\partial_\mu g$ cancel, leaving

$$\overline{\mathcal D}'_\mu X' =g^{-1}(\partial_\mu X+[\overline{\mathcal A}_\mu,X])g =g^{-1}(\overline{\mathcal D}_\mu X)g.$$

Taking $X=a_\nu$ gives

$$\boxed{ \overline{\mathcal D}'_\mu a'_\nu =g^{-1}(\overline{\mathcal D}_\mu a_\nu)g. }$$

Thus the background gauge condition $\overline{\mathcal D}_\mu a_\mu=0$ is covariant under background gauge transformations.

Exercise 5: Locate the electric-field proper-time poles

Explain why the analytic continuation $B\mapsto iE$ changes the magnetic heat-kernel factor $Bs/\sinh Bs$ into $Es/\sin Es$, and identify the proper-time pole locations.

Solution

Using

$$\sinh(iEs)=i\sin(Es),$$

we find

$${Bs\over\sinh Bs}\bigg|_{B=iE} ={iEs\over\sinh(iEs)} ={iEs\over i\sin(Es)} ={Es\over\sin(Es)}.$$

The denominator vanishes when

$$\sin(Es)=0,$$

so the positive real proper-time poles are at

$$\boxed{ s_n={n\pi\over E}, \qquad n=1,2,\ldots. }$$

The prescription for passing these poles is tied to the Minkowski $i0$ prescription. Their contribution produces an imaginary part of the effective action, which is the determinant-language signal of pair creation in an electric field.

References

• Heisenberg, Werner, and Hans Euler. “Folgerungen aus der Diracschen Theorie des Positrons.” Zeitschrift für Physik 98 (1936): 714–732.
• Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge University Press, 2014.
• Schwinger, Julian. “On Gauge Invariance and Vacuum Polarization.” Physical Review 82, no. 5 (1951): 664–679.
• Srednicki, Mark. Quantum Field Theory. Cambridge University Press, 2007.
• Weinberg, Steven. The Quantum Theory of Fields, Volume II: Modern Applications. Cambridge University Press, 1996.
• Zee, A. Quantum Field Theory in a Nutshell. 2nd ed. Princeton University Press, 2010.
• Zinn-Justin, Jean. Quantum Field Theory and Critical Phenomena. 4th ed. Oxford University Press, 2002.