Proper Time, Determinants, and Thermal Traces

The previous page used vacuum polarization to motivate the running charge and introduced the physical image of closed charged worldlines. This page makes that image into a calculational tool. The basic object is no longer a single Feynman diagram written in momentum space, but a functional determinant written as a trace of an evolution operator.

The bridge is Schwinger’s proper-time representation:

$$\operatorname{Tr}\log \mathcal L \quad\longleftrightarrow\quad -\int_0^\infty {ds\over s}\,\operatorname{Tr} e^{-s\mathcal L}.$$

Here $s$ is not physical time. It is a Laplace-transform parameter that probes the spectrum of the Euclidean operator $\mathcal L$. Small $s$ sees large eigenvalues and therefore short distances; large $s$ sees low eigenvalues and therefore infrared physics. This is why proper time is so useful for renormalization: ultraviolet divergences become the small-$s$ asymptotics of a heat kernel.

The same trace language also explains finite-temperature path integrals. A thermal partition function is

$$Z(\beta)=\operatorname{Tr}e^{-\beta H},$$

and the trace turns Euclidean time into a circle of circumference $\beta$. Bosonic variables are periodic, fermionic variables are antiperiodic, and Matsubara frequencies are simply the Fourier modes on that circle. The two themes — determinant traces and thermal traces — are really the same technology used in two different spectral problems.

Required background. Vacuum Polarization and Gauge-Invariant Counterterms fixes the Euclidean gauge-field normalization and scalar-QED determinant sign. Running Charge, Screening, and Antiscreening supplies the closed-worldline interpretation that proper time makes precise below.

A useful reader’s dictionary for this page is:

$$\begin{array}{c|c|c} \text{object} & \text{spectral meaning} & \text{QFT use} \\ \hline \operatorname{Tr}\log\mathcal L & \sum_n\log\lambda_n & \text{one-loop effective action} \\ \operatorname{Tr}e^{-s\mathcal L} & \sum_n e^{-s\lambda_n} & \text{proper-time heat trace} \\ \operatorname{Tr}e^{-\beta H} & \sum_n e^{-\beta E_n} & \text{thermal partition function} \end{array}$$

The notation is similar enough to be dangerous. The parameter $s$ regulates a determinant; the parameter $\beta$ is the physical circumference of the Euclidean-time circle.

Gaussian integrals and one-loop determinants

Trace and determinant conventions. Most formulas on this page are Euclidean. We write a positive Laplace-type operator as

$$\mathcal L=-D^2+m^2+U(x), \qquad D_\mu=\partial_\mu-iA_\mu$$

for a unit-charge complex scalar in a background Abelian field. The operator dimension inside a logarithm is fixed by a reference scale $\mu$, but field-independent constants are suppressed unless they matter.

A real bosonic Gaussian contributes

$$\Gamma^{(1)}_{\rm real}={1\over2}\operatorname{Tr}\log\mathcal L,$$

a complex bosonic Gaussian contributes

$$\Gamma^{(1)}_{\rm complex}=\operatorname{Tr}\log\mathcal L,$$

and a Grassmann Gaussian contributes with the opposite sign,

$$\Gamma^{(1)}_{\rm fermion}=-\operatorname{Tr}\log\mathcal D.$$

These signs are the most common source of mistakes in this subject.

The sign convention can be checked by remembering that $Z=e^{-\Gamma}$. A complex boson gives $Z\propto(\det\mathcal L)^{-1}$, hence $\Gamma=+\operatorname{Tr}\log\mathcal L$. A Grassmann field gives $Z\propto\det\mathcal D$, hence $\Gamma=-\operatorname{Tr}\log\mathcal D$.

Start with an ordinary finite-dimensional Gaussian integral. If $A$ is a real, symmetric, positive $N\times N$ matrix, then

$$\int d^Nx\,\exp\left(-{1\over2}x_iA_{ij}x_j\right) \propto (\det A)^{-1/2}.$$

For complex variables,

$$\int d^Nz^*d^Nz\, \exp\left(-z_i^*A_{ij}z_j\right) \propto (\det A)^{-1}.$$

For Grassmann variables,

$$\int d^N\overline\psi\,d^N\psi\, \exp\left(-\overline\psi_iA_{ij}\psi_j\right) \propto \det A.$$

The infinite-dimensional version is formal until regularized, but its algebra is the same. Consider a complex scalar field in a fixed Euclidean background $A_\mu$,

$$S_E[\phi,A]=\int d^dx\,\phi^*(x)\mathcal L_A\phi(x), \qquad \mathcal L_A=-D_\mu D_\mu+m^2.$$

The matter path integral is

$$Z_s[A]=\int\mathcal D\phi^*\mathcal D\phi\,e^{-S_E[\phi,A]} \propto \left(\operatorname{Det}\mathcal L_A\right)^{-1}.$$

The one-loop effective action is defined by

$$Z_s[A]=e^{-\Gamma_s^{(1)}[A]},$$

so

$$\boxed{ \Gamma_s^{(1)}[A]=\operatorname{Tr}\log\mathcal L_A. }$$

For a real scalar the answer is half as large. For a Dirac fermion,

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

and the Grassmann integral gives

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

The trace includes the integral over spacetime, the sum over internal indices, and, for fermions, the trace over spinor indices.

For fermions, this first-order determinant may later be squared into a Laplace-type determinant, but that step must be done with care. Squaring gives useful heat-kernel formulas for parity-even terms such as $F^2$, while possible phases of the determinant carry anomaly information. In these notes, whenever we square a Dirac operator we state explicitly which parity-even part is being extracted.

Two comments are worth making immediately. First, a determinant is only meaningful after specifying a regulator and a normalization. Ratios such as

$$\operatorname{Tr}\log\mathcal L_A- \operatorname{Tr}\log\mathcal L_0$$

are better behaved than either determinant separately, because field-independent vacuum-volume terms cancel. Second, the determinant knows about all one-loop diagrams with any number of external background-field insertions. Expanding

$$\mathcal L_A=\mathcal L_0+V[A]$$

gives

$$\operatorname{Tr}\log\mathcal L_A = \operatorname{Tr}\log\mathcal L_0 +\operatorname{Tr}\log(1+\mathcal L_0^{-1}V)$$

and therefore

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

This is the compact determinant form of the one-loop expansion. The quadratic term gives vacuum polarization. The cubic and quartic terms give higher-point background-field vertices.

Proper time and the heat kernel

Suppose $\mathcal L$ has positive eigenvalues $\lambda_n$. Then

$$\operatorname{Tr}\log\mathcal L=\sum_n\log\lambda_n.$$

The logarithm can be represented by a proper-time integral. A useful subtracted identity is

$$\log{\lambda\over\lambda_0} =-\int_0^\infty {ds\over s}\left(e^{-s\lambda}-e^{-s\lambda_0}\right),$$

which follows by differentiating with respect to $\lambda$ and fixing the value at $\lambda=\lambda_0$. In field theory we usually write the regulated version as

$$\operatorname{Tr}\log\mathcal L = -\int_\epsilon^\infty {ds\over s}\,\operatorname{Tr}e^{-s\mathcal L} +\text{constant and counterterms},$$

where

$$\epsilon\sim {1\over\Lambda^2}$$

is a short-proper-time cutoff. The one-loop scalar effective action, after subtracting the $A=0$ vacuum term, is therefore

$$\boxed{ \Gamma_s^{(1)}[A]-\Gamma_s^{(1)}[0] =-\int_\epsilon^\infty {ds\over s}\, \operatorname{Tr}\left(e^{-s\mathcal L_A}-e^{-s\mathcal L_0}\right). }$$

The kernel of $e^{-s\mathcal L}$ is the heat kernel,

$$K(s;x,y)=\langle x|e^{-s\mathcal L}|y\rangle.$$

It obeys

$$\left(\partial_s+\mathcal L_x\right)K(s;x,y)=0, \qquad K(0;x,y)=\delta^{(d)}(x-y).$$

The trace is obtained by closing the endpoints:

$$\operatorname{Tr}e^{-s\mathcal L} = \int d^dx\,\operatorname{tr}K(s;x,x).$$

The lower-case trace is over finite-dimensional indices such as spin, flavor, or gauge representation indices.

Proper time turns the determinant into an open heat kernel. Taking the trace identifies the endpoints and produces a closed charged worldline. Short loops of size $\sqrt{s}$ control ultraviolet terms; long loops probe infrared physics.

The ultraviolet structure is local because it comes from the short-$s$ expansion of the heat kernel. This is one of the cleanest ways to see why ultraviolet divergences are removable by local counterterms: at very small $s$, the heat kernel samples only an infinitesimal neighborhood of the point $x$. For a Laplace-type operator in flat space,

$$\operatorname{tr}K(s;x,x) \sim {e^{-m^2s}\over(4\pi s)^{d/2}} \operatorname{tr}\left[a_0(x)+s a_1(x)+s^2a_2(x)+\cdots\right].$$

The coefficients $a_n(x)$ are local functions of the background fields and their derivatives. For readers using proper time as a regulator, the power counting is especially transparent. In $d=4$, the terms $a_0$, $a_1$, and $a_2$ multiply respectively $s^{-3}$, $s^{-2}$, and $s^{-1}$ in $ds\,s^{-1}(4\pi s)^{-2}(1+s a_1+s^2a_2+\cdots)$. Thus $a_0$ gives a quartic divergence, $a_1$ a quadratic divergence, and $a_2$ a logarithmic divergence. Gauge coupling renormalization lives in the $a_2$ coefficient.

For the minimal Abelian scalar operator

$$\mathcal L_A=-D^2+m^2, \qquad [D_\mu,D_\nu]=-iF_{\mu\nu},$$

the first gauge-field term is

$$\operatorname{tr}K_A(s;x,x) = {e^{-m^2s}\over(4\pi s)^{d/2}} \left[1-{s^2\over12}F_{\mu\nu}F_{\mu\nu}+O(s^3\partial^2F^2,s^3F^3)\right].$$

The sign is important. The heat trace itself decreases in a weak magnetic field; this is the orbital diamagnetic sign. But the scalar effective action has an overall minus sign in the proper-time integral, so the induced Maxwell term is positive. The distinction between the sign inside the heat kernel and the sign in the effective action is a small bookkeeping point that becomes physically important on the next two pages.

In $d=4$ this gives

$$\Gamma_s^{(1)}[A] \supset -\int_\epsilon^\infty {ds\over s} \int d^4x\,{1\over(4\pi s)^2} \left[-{s^2\over12}F_{\mu\nu}F_{\mu\nu}\right]e^{-m^2s}.$$

Keeping only the logarithmic divergence,

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

If the Maxwell action is written as

$$\Gamma[A]\supset {1\over4e^2}\int d^4x\,F_{\mu\nu}F_{\mu\nu},$$

then matching requires multiplying the coefficient of $\int F^2$ by $4$. Thus one complex scalar shifts

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

which is the scalar QED coefficient used earlier.

Heat kernels as charged worldlines

Proper time converts the spectral operator into a particle path integral. This is the precise bridge between the vacuum-polarization loop of the previous pages and the closed-worldline picture in the manuscript. Begin with the background-field propagator

$$G_A(x,y) =\langle x|\mathcal L_A^{-1}|y\rangle, \qquad \mathcal L_A=-D^2+m^2+U(x).$$

For a positive Euclidean operator,

$$\mathcal L_A^{-1}=\int_0^\infty ds\,e^{-s\mathcal L_A},$$

so

$$G_A(x,y)=\int_0^\infty ds\,K_A(s;x,y).$$

The open heat kernel has the configuration-space representation

$$\boxed{ K_A(s;x,y) =e^{-m^2s} \int_{x(0)=y}^{x(s)=x}\!\mathcal D x(\tau)\, \exp\left[ -\int_0^s d\tau\left({\dot x^2\over4}+U(x)\right) +i\int_y^x A_\mu\,dx^\mu \right]. }$$

The sign of the Wilson phase follows directly from $D_\mu=\partial_\mu-iA_\mu$. Indeed,

$$-D^2=(p-A)^2, \qquad p_\mu=-i\partial_\mu,$$

and integrating the first-order momentum path integral produces $+i\int A_\mu dx^\mu$. Under

$$A_\mu\mapsto A_\mu+\partial_\mu\alpha, \qquad \phi\mapsto e^{+i\alpha}\phi,$$

the open Wilson phase supplies exactly the endpoint phases required for $K_A(s;x,y)$ to transform covariantly.

Taking the trace identifies the endpoints and turns the open trajectory into a closed loop:

$$\boxed{ \operatorname{Tr}e^{-s\mathcal L_A} =e^{-m^2s} \int_{x(s)=x(0)}\!\mathcal D x(\tau)\, \exp\left[ -\int_0^s d\tau\left({\dot x^2\over4}+U(x)\right) +i\oint A_\mu dx^\mu \right]. }$$

The endpoint phases now cancel, so each closed path is weighted by a gauge-invariant Wilson loop. Substituting this trace into the proper-time determinant sums closed charged trajectories of every proper duration $s$. Typical loops have size $|\Delta x|\sim\sqrt{s}$: short loops generate the local ultraviolet expansion, while long loops probe infrared propagation.

For a slowly varying field, the Wilson loop measures the flux through the trajectory. Expanding that flux gives local powers of $F_{\mu\nu}$; averaging over loop orientation removes the term linear in $F$ and leaves $F^2$ as the first Abelian contribution. This is the worldline explanation of why the same determinant reproduces the transverse vacuum-polarization term.

Proper-time closure and thermal closure must nevertheless be distinguished. In a determinant, $s$ is integrated from zero to infinity. In a thermal trace, the Euclidean-time circumference $\beta$ is fixed by the temperature. The paths are closed in both cases, but the parameters play different physical roles.

Time ordering inside an operator trace

The notation $\operatorname{Tr}e^{-s\mathcal L}$ looks innocent, but when $\mathcal L$ is built from noncommuting operators the trace is not a classical phase-space integral. The distinction is already visible in a finite-dimensional Hilbert space.

For two noncommuting operators $A$ and $B$, one cannot write

$$e^{A+B}=e^Ae^B.$$

Instead, introduce

$$B(t)=e^{-tA}Be^{tA}, \qquad 0\le t\le1.$$

Duhamel’s formula gives

$$e^{A+B} =e^A\,T\exp\left(\int_0^1dt\,B(t)\right),$$

where $T$ orders larger $t$ to the left. Expanding to first order reproduces

$$e^{A+B}=e^A+\int_0^1dt\,e^{(1-t)A}Be^{tA}+O(B^2).$$

This is the operator origin of time-ordered perturbation theory. In a trace, cyclicity identifies the beginning and end of the interval, so insertions live on a circle.

Time ordering also produces contact terms. If $q(t)$ and $p(t)$ are Heisenberg operators with

$$[q(t),p(t)]=i,$$

then

$$T\,q(t)p(t') =\theta(t-t')q(t)p(t')+\theta(t'-t)p(t')q(t).$$

Differentiating gives

$$\partial_t\langle Tq(t)p(t')\rangle =\langle T\dot q(t)p(t')\rangle +\delta(t-t')\langle[q(t),p(t')]\rangle.$$

At equal time the commutator is $i$, so

$$\boxed{ \partial_t\langle Tq(t)p(t')\rangle =\langle T\dot q(t)p(t')\rangle +i\delta(t-t') }$$

in a state normalized so that $\langle1\rangle=1$. This delta function is the tiny but crucial trace of noncommutativity. Any path-integral derivation that replaces operators by ordinary functions must reproduce these contact terms through its discretization prescription.

First-order path integrals and closed trajectories

The thermal trace of a quantum-mechanical Hamiltonian is

$$Z_{\rm qm}(\beta)=\operatorname{Tr}e^{-\beta H(\hat p,\hat q)}.$$

Insert complete sets of $q$ and $p$ eigenstates at many intermediate Euclidean times. Using

$$\langle q|p\rangle=e^{ipq}$$

up to the standard normalization, one obtains the first-order phase-space path integral

$$\boxed{ Z_{\rm qm}(\beta) = \int_{q(\beta)=q(0)}\mathcal Dp\,\mathcal Dq\, \exp\left[\int_0^\beta d\tau\, \left(i p\dot q-H(p,q)\right) \right]. }$$

The trace is responsible for the periodic boundary condition. In several dimensions this becomes

$$\int_0^\beta d\tau\, \left(i p_i\dot q^i-H(p,q)\right).$$

The term $i p_i\dot q^i$ is the Euclidean version of the symplectic one-form. It is why the phase-space path integral remembers the canonical commutator.

The operator trace $\operatorname{Tr}e^{-\beta H}$ becomes a path integral over closed Euclidean trajectories. In first-order form the weight contains $i\int p_i dq^i$, which encodes the canonical commutation relations and fixes the ordering prescription.

For a positively charged unit particle with $D_i=\partial_i-iA_i$, the corresponding Hamiltonian is

$$H(\mathbf p,\mathbf x)= {1\over2M}\left(\mathbf p-\mathbf A(\mathbf x)\right)^2+V(\mathbf x),$$

the momentum integral is Gaussian. Completing the square gives the second-order path integral

$$Z_{\rm qm}(\beta) = \int_{\mathbf x(\beta)=\mathbf x(0)}\mathcal D\mathbf x(\tau) \exp\left[ -\int_0^\beta d\tau\left({M\over2}\dot{\mathbf x}^2+V(\mathbf x)\right) +i\oint \mathbf A\cdot d\mathbf x \right].$$

The vector potential appears through the same Wilson phase as in the relativistic heat kernel. A particle of charge $q$ is obtained by replacing $A_\mu$ with $qA_\mu$; changing the sign of $q$ complex-conjugates the phase but leaves the magnetic spectrum unchanged.

The classical partition function would be

$$Z_{\rm cl}(\beta)=\int {d^nq\,d^np\over(2\pi)^n}\,e^{-\beta H(p,q)}.$$

The quantum trace reduces to this expression only in an appropriate semiclassical or high-temperature limit. In the full quantum trace, the variables are functions of Euclidean time, and nonzero Fourier modes carry the memory of operator ordering.

For a bosonic coordinate on the thermal circle,

$$q(\tau)=\sum_{n\in\mathbb Z}q_n e^{i\omega_n\tau}, \qquad \omega_n={2\pi n\over\beta}.$$

When $\beta$ is small, the nonzero modes have large frequencies. In many problems the zero mode dominates the leading classical thermodynamics, while the nonzero modes produce quantum corrections.

Thermal circles and Matsubara determinants

In quantum field theory the same trace gives the finite-temperature Euclidean path integral,

$$Z(\beta)=\operatorname{Tr}e^{-\beta H} =\int_{\text{thermal b.c.}}\mathcal D\Phi\,e^{-S_E[\Phi]}.$$

Euclidean time is compact:

$$\tau\sim\tau+\beta.$$

Bosons are periodic,

$$\phi(\tau+\beta,\mathbf x)=\phi(\tau, \mathbf x),$$

while fermions are antiperiodic,

$$\psi(\tau+\beta,\mathbf x)=-\psi(\tau,\mathbf x).$$

The minus sign for fermions is required by the trace over fermionic states and the Grassmann nature of fermionic coherent states. It leads to different Matsubara frequencies:

$$\omega_n^{\rm B}={2\pi n\over\beta}, \qquad \omega_n^{\rm F}={(2n+1)\pi\over\beta}.$$

A thermal trace compactifies Euclidean time to a circle of circumference $\beta$. Bosonic fields are periodic and have frequencies $2\pi n/\beta$; fermionic fields are antiperiodic and have frequencies $(2n+1)\pi/\beta$.

The harmonic oscillator is the cleanest check. Its exact thermal trace is

$$Z_{\rm osc} =\operatorname{Tr}e^{-\beta\omega(a^\dagger a+1/2)} =\sum_{n=0}^\infty e^{-\beta\omega(n+1/2)} ={1\over2\sinh(\beta\omega/2)}.$$

The Euclidean path integral gives the same result as a determinant,

$$Z_{\rm osc} \propto \left[\operatorname{Det}_{\rm P}(-\partial_\tau^2+\omega^2)\right]^{-1/2},$$

where P means periodic boundary conditions. The eigenvalues are

$$\lambda_n=\left({2\pi n\over\beta}\right)^2+\omega^2.$$

The infinite product is regulated, but its $\omega$ dependence is fixed by

$$\prod_{n=-\infty}^{\infty} \left[\left({2\pi n\over\beta}\right)^2+\omega^2\right] \propto \sinh^2{\beta\omega\over2}.$$

Thus the determinant knows the oscillator spectrum.

For a free scalar field, each spatial momentum mode is an oscillator with

$$E_{\mathbf p}=\sqrt{\mathbf p^2+m^2}.$$

The one-loop thermal free energy density is therefore

$$f_B(\beta) ={1\over2}\int {d^{d-1}p\over(2\pi)^{d-1}}E_{\mathbf p} +{1\over\beta}\int {d^{d-1}p\over(2\pi)^{d-1}} \log\left(1-e^{-\beta E_{\mathbf p}}\right),$$

after the usual normalization of the zero-point energy. For a free Dirac fermion the antiperiodic determinant gives the Fermi–Dirac factor

$$\log\left(1+e^{-\beta E_{\mathbf p}}\right)$$

with the appropriate spin and particle–antiparticle degeneracies and with the overall fermionic sign in the free energy.

Landau levels as a trace calculation

A constant magnetic field gives a useful test of all these ideas. Consider a positively charged unit spinless particle moving in two spatial dimensions with $M=1$ and magnetic field $B>0$. Its Hamiltonian is

$$H={1\over2}\left(\mathbf p-\mathbf A\right)^2, \qquad \partial_xA_y-\partial_yA_x=B.$$

The kinetic momenta

$$\Pi_x=p_x-A_x, \qquad \Pi_y=p_y-A_y$$

satisfy

$$[\Pi_x,\Pi_y]=+iB.$$

The sign fixes which linear combination is the raising operator; the spectrum depends only on $|B|$. For $B>0$, the energy levels are

$$E_n=B\left(n+{1\over2}\right), \qquad n=0,1,2,\ldots,$$

and the degeneracy in area $A_{\rm area}$ is

$${A_{\rm area}B\over2\pi}.$$

The one-particle thermal trace is

$$\boxed{ Z_B(\beta) ={A_{\rm area}B\over2\pi} \sum_{n=0}^\infty e^{-\beta B(n+1/2)} ={A_{\rm area}B\over4\pi\sinh(\beta B/2)}. }$$

At high temperature,

$$Z_B(\beta) ={A_{\rm area}\over2\pi\beta} \left[1-{\beta^2B^2\over24}+O(\beta^4B^4)\right].$$

The leading term is the classical phase-space answer. The $B^2$ correction is the first orbital quantum correction.

A constant magnetic field turns the transverse kinetic momenta into an oscillator. Each Landau level has degeneracy $A_{\rm area}B/(2\pi)$, so the thermal trace is a geometric series in $e^{-\beta B}$.

The four-dimensional heat kernel in a constant magnetic field is the same calculation with two additional free Euclidean directions. For the scalar operator $-D^2$,

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

Expanding for small $sB$,

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

Since $F_{\mu\nu}F_{\mu\nu}=2B^2$ for a purely magnetic field in Euclidean space, this agrees with the heat-kernel coefficient

$$1-{s^2\over12}F_{\mu\nu}F_{\mu\nu}+\cdots.$$

This is the calculation that the next pages will refine for scalar, spinor, and vector fluctuations in background fields.

A useful diagnostic is that the magnetic factor is dimensionless: $Bs$ is the only possible combination because $[B]=2$ and $[s]=-2$. Expanding in $Bs$ is therefore a derivative/locality expansion for fields that are weak on the scale set by the eigenvalues being integrated out.

Summary

A one-loop path integral is a functional determinant. Bosonic fluctuations give inverse determinants in $Z$ and positive $\operatorname{Tr}\log$ terms in the effective action; fermionic fluctuations give determinants in $Z$ and negative $\operatorname{Tr}\log$ terms in the effective action.

Schwinger proper time rewrites the determinant as a heat trace:

$$\operatorname{Tr}\log\mathcal L =-\int_\epsilon^\infty {ds\over s}\operatorname{Tr}e^{-s\mathcal L} +\text{counterterms}.$$

The heat kernel $K(s;x,y)$ solves a diffusion equation in the auxiliary variable $s$. Its small-$s$ expansion is local and controls ultraviolet divergences. In four-dimensional scalar QED, the $F^2$ term in this expansion reproduces the scalar contribution

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

The open heat kernel is also a charged-particle path integral. With $D_\mu=\partial_\mu-iA_\mu$, each path carries $e^{+i\int A\cdot dx}$. Taking the trace identifies its endpoints, producing a gauge-invariant Wilson loop and the closed-worldline representation of the determinant.

Thermal traces use the same logic with physical Euclidean time compactified to a circle of circumference $\beta$. Periodic bosonic fields have frequencies $2\pi n/\beta$, antiperiodic fermionic fields have frequencies $(2n+1)\pi/\beta$, and functional determinants reduce to products over these modes. Landau levels provide an especially concrete example: a magnetic field discretizes transverse motion, and the trace becomes a weighted sum over oscillator levels.

Common pitfalls

Proper time is not thermal Euclidean time. Proper time $s$ is a spectral parameter used to represent logarithms and inverse operators. Thermal Euclidean time $\tau$ has physical period $\beta=1/T$.

Determinant powers and signs matter. A real scalar, complex scalar, and Dirac fermion differ by factors of $1/2$ and by an overall sign. These differences produce different one-loop coefficients.

A quantum trace is not automatically a classical phase-space integral. Replacing $\operatorname{Tr}e^{-\beta H(\hat p,\hat q)}$ by $\int dpdq\,e^{-\beta H(p,q)}$ requires a justified semiclassical limit. Operator ordering and contact terms otherwise remain.

Thermal boundary conditions determine the spectrum. Periodic versus antiperiodic conditions change bosonic frequencies into fermionic Matsubara frequencies.

The small-$s$ expansion is ultraviolet. Large $s$ is where zero modes, masslessness, and infrared divergences enter.

The Wilson-phase sign follows the covariant derivative. With $D_\mu=\partial_\mu-iA_\mu$, the worldline phase is $e^{+i\int A\cdot dx}$. Reversing the charge convention reverses this sign without changing the even-in-$F$ terms studied here.

Exercises

Exercise 1: Track Gaussian determinant powers and signs

Let $A$ be a positive $N\times N$ matrix. Show that real bosons, complex bosons, and Grassmann variables give determinant powers $-1/2$, $-1$, and $+1$, respectively, in the partition function. Translate these powers into contributions to the effective action $\Gamma=-\log Z$.

Solution

Diagonalize $A$ by a unitary or orthogonal transformation. For a real variable,

$$\int_{-\infty}^{\infty}dx\,e^{-\lambda x^2/2} =\sqrt{2\pi\over\lambda},$$

so $N$ real variables give

$$Z_{\rm real}\propto\prod_i\lambda_i^{-1/2} =(\det A)^{-1/2}.$$

For a complex variable $z=x+iy$, the Gaussian has two real components and gives one inverse power of the eigenvalue:

$$Z_{\rm complex}\propto\prod_i\lambda_i^{-1} =(\det A)^{-1}.$$

For Grassmann variables,

$$\int d\overline\psi_i\,d\psi_i\,e^{-\lambda_i\overline\psi_i\psi_i} \propto \lambda_i,$$

up to a convention-dependent sign in the measure. Therefore

$$Z_{\rm Grassmann}\propto\prod_i\lambda_i=\det A.$$

Since $\Gamma=-\log Z$,

$$\Gamma_{\rm real}^{(1)}={1\over2}\operatorname{Tr}\log A, \qquad \Gamma_{\rm complex}^{(1)}=\operatorname{Tr}\log A, \qquad \Gamma_{\rm Grassmann}^{(1)}=-\operatorname{Tr}\log A.$$

Exercise 2: Prove the subtracted proper-time identity

Prove the subtracted proper-time identity

$$\log{\lambda\over\lambda_0} =-\int_0^\infty {ds\over s}\left(e^{-s\lambda}-e^{-s\lambda_0}\right)$$

for $\lambda,\lambda_0>0$.

Solution

Define

$$I(\lambda)=-\int_0^\infty {ds\over s}\left(e^{-s\lambda}-e^{-s\lambda_0}\right).$$

Differentiate with respect to $\lambda$:

$${dI\over d\lambda} =-\int_0^\infty {ds\over s}(-s)e^{-s\lambda} =\int_0^\infty ds\,e^{-s\lambda} ={1\over\lambda}.$$

Thus

$$I(\lambda)=\log\lambda+C.$$

At $\lambda=\lambda_0$, the integrand vanishes and $I(\lambda_0)=0$. Hence

$$0=\log\lambda_0+C,$$

so $C=-\log\lambda_0$. Therefore

$$I(\lambda)=\log{\lambda\over\lambda_0}.$$

The subtraction is essential: each exponential integral separately is divergent at small $s$, but their difference is finite.

Exercise 3: Recover the equal-time contact term

Using

$$Tq(t)p(t')=\theta(t-t')q(t)p(t')+\theta(t'-t)p(t')q(t),$$

show that

$$\partial_t\langle Tq(t)p(t')\rangle =\langle T\dot q(t)p(t')\rangle+i\delta(t-t')$$

when $[q(t),p(t)]=i$.

Solution

Differentiate the time-ordered product:

$$\partial_t\left[\theta(t-t')q(t)p(t')\right] =\delta(t-t')q(t)p(t')+\theta(t-t')\dot q(t)p(t'),$$

and

$$\partial_t\left[\theta(t'-t)p(t')q(t)\right] =-\delta(t'-t)p(t')q(t)+\theta(t'-t)p(t')\dot q(t).$$

Since $\delta(t'-t)=\delta(t-t')$, the delta-function part is

$$\delta(t-t')\left[q(t)p(t')-p(t')q(t)\right].$$

At the support of the delta function, $t=t'$, so this is

$$\delta(t-t')[q(t),p(t)]=i\delta(t-t').$$

The remaining terms are precisely $T\dot q(t)p(t')$. Taking the expectation value gives the result.

Exercise 4: Reconstruct the oscillator determinant

The periodic operator $-\partial_\tau^2+\omega^2$ on a circle of circumference $\beta$ has eigenvalues

$$\lambda_n=\left({2\pi n\over\beta}\right)^2+\omega^2, \qquad n\in\mathbb Z.$$

Use the product identity

$$\sinh x=x\prod_{n=1}^{\infty}\left(1+{x^2\over\pi^2n^2}\right)$$

to show that the determinant reproduces the $\omega$ dependence of the harmonic-oscillator partition function.

Solution

Up to an $\omega$-independent constant,

$$\operatorname{Det}_{\rm P}(-\partial_\tau^2+\omega^2) \propto \omega^2\prod_{n=1}^{\infty} \left[\left({2\pi n\over\beta}\right)^2+\omega^2\right]^2.$$

Factor out the $\omega$-independent pieces:

$$\operatorname{Det}_{\rm P} \propto \omega^2\prod_{n=1}^{\infty} \left(1+{\beta^2\omega^2\over4\pi^2n^2}\right)^2.$$

Set

$$x={\beta\omega\over2}.$$

The product identity gives

$$\sinh{\beta\omega\over2} ={\beta\omega\over2} \prod_{n=1}^{\infty} \left(1+{\beta^2\omega^2\over4\pi^2n^2}\right).$$

Therefore

$$\operatorname{Det}_{\rm P}(-\partial_\tau^2+\omega^2) \propto \sinh^2{\beta\omega\over2}.$$

A real oscillator has

$$Z\propto\left(\operatorname{Det}_{\rm P}(-\partial_\tau^2+\omega^2)\right)^{-1/2} \propto {1\over\sinh(\beta\omega/2)}.$$

The normalization fixed by canonical quantization gives

$$Z_{\rm osc}={1\over2\sinh(\beta\omega/2)}.$$

Exercise 5: Sum the Landau-level thermal trace

For a unit-charge spinless particle in two dimensions with $M=1$ and magnetic field $B>0$, use the Landau spectrum

$$E_n=B\left(n+{1\over2}\right)$$

and degeneracy $A_{\rm area}B/(2\pi)$ to derive

$$Z_B(\beta)={A_{\rm area}B\over4\pi\sinh(\beta B/2)}.$$

Then expand the answer for $\beta B\ll1$.

Solution

The trace is degeneracy times the Boltzmann sum:

$$Z_B(\beta) ={A_{\rm area}B\over2\pi} \sum_{n=0}^{\infty}e^{-\beta B(n+1/2)}.$$

The sum is geometric:

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

Thus

$$Z_B(\beta) ={A_{\rm area}B\over4\pi\sinh(\beta B/2)}.$$

For small $x$,

$${1\over\sinh x}={1\over x}-{x\over6}+O(x^3).$$

With $x=\beta B/2$,

$$Z_B(\beta) ={A_{\rm area}B\over4\pi} \left({2\over\beta B}-{\beta B\over12}+O(\beta^3B^3)\right).$$

Therefore

$$\boxed{ Z_B(\beta) ={A_{\rm area}\over2\pi\beta} \left[1-{\beta^2B^2\over24}+O(\beta^4B^4)\right]. }$$

The first term is the classical phase-space result; the $B^2$ term is the leading orbital quantum correction.

Exercise 6: Extract the scalar-QED logarithm from the heat kernel

Use the four-dimensional scalar heat kernel in a constant magnetic field,

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

to recover the logarithmic scalar-QED correction to $1/e^2$.

Solution

Expand the magnetic factor:

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

The $B$-independent term contributes only to the vacuum energy. The $B^2$ term gives, for a complex scalar,

$$\Gamma_s^{(1)}[B]-\Gamma_s^{(1)}[0] =-\int_\epsilon^\infty {ds\over s}\,e^{-m^2s} V_4{1\over(4\pi s)^2}\left(-{B^2s^2\over6}\right).$$

Thus

$$\Gamma_s^{(1)}[B] \supset {V_4B^2\over6(4\pi)^2} \int_\epsilon^\infty {ds\over s}e^{-m^2s}.$$

The logarithmic part is

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

Since $F_{\mu\nu}F_{\mu\nu}=2B^2$ for this background,

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

Therefore

$$\Gamma_s^{(1)}[A] \supset {1\over6(4\pi)^2}\cdot {1\over2} \log{\Lambda^2\over m^2} \int d^4x\,F_{\mu\nu}F_{\mu\nu} ={1\over192\pi^2} \log{\Lambda^2\over m^2} \int d^4x\,F_{\mu\nu}F_{\mu\nu}.$$

Matching to

$${1\over4e^2}\int d^4x\,F_{\mu\nu}F_{\mu\nu}$$

gives

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

References

• 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.