The Fermion Propagator

The free fermion propagator is a vacuum-ordered distribution, not merely the formal matrix inverse of $p\!\!\!/-m$. Its two time domains come from particle and antiparticle vacuum contractions; fermionic time ordering fixes their relative minus sign; the spin sums supply $p\!\!\!/\pm m$ on shell; and the Feynman pole prescription combines both residues into $i(p\!\!\!/+m)/(p^2-m^2+i0)$. This page derives that package, including its contact equation and massless limit, for the free Minkowski vacuum. Loop calculations and finite-density propagators are outside its scope.

Required background. Canonical Quantization of the Free Dirac Field supplies the normalized $u/v$ mode expansion, CAR, vacuum, and particle/antiparticle contractions. Scalar Propagators, Ordered Correlators, and Sources supplies the distinction between correlators and Green inverses, the scalar Feynman boundary value, and the energy-contour meaning of $i0$.

Time ordering fixes the reverse-order sign

Work with a free Dirac field of mass $m>0$ in the selected four-dimensional Minkowski vacuum $\lvert\Omega\rangle$. Every point-field expression below is an operator-valued or scalar distribution and is interpreted after smearing. Set $z=x-y$ and define the raw time-ordered vacuum correlator by

$$S_F(z)_{\alpha\beta} \equiv \langle\Omega| \mathrm T\! \left[ \widehat\psi_\alpha(x) \widehat{\bar\psi}_\beta(y) \right] |\Omega\rangle.$$

This definition includes no additional factor of $i$. For two odd operators, time ordering is graded:

$$\begin{aligned} \mathrm T\! \left[ \widehat\psi_\alpha(x) \widehat{\bar\psi}_\beta(y) \right] &= \theta(z^0) \widehat\psi_\alpha(x) \widehat{\bar\psi}_\beta(y)\\ &\quad- \theta(-z^0) \widehat{\bar\psi}_\beta(y) \widehat\psi_\alpha(x). \end{aligned}$$

The minus sign records one exchange of odd operators. It is not optional and does not come from the negative value of $\bar vv$. The convention for $\theta(0)$ does not determine the contact term; the equal-time CAR fixes that distributional jump below. The graded definition and its reverse-order sign are derived in Schwartz 2014, § 12.4, p. 212.

Introduce the two vacuum contractions

$$\begin{aligned} S^+(z)_{\alpha\beta} &\equiv \langle\Omega| \widehat\psi_\alpha(x) \widehat{\bar\psi}_\beta(y) |\Omega\rangle,\\ S^-(z)_{\alpha\beta} &\equiv \langle\Omega| \widehat{\bar\psi}_\beta(y) \widehat\psi_\alpha(x) |\Omega\rangle. \end{aligned}$$

Then

$$S_F(z) = \theta(z^0)S^+(z) - \theta(-z^0)S^-(z).$$

The ordered correlator depends on the selected vacuum. By contrast, the free-field anticommutator $S^++S^-$ is fixed by the CAR algebra once its normalization is fixed.

Spin sums determine the two frequency sectors

In $S^+$, only the $b\,b^\dagger$ contraction survives. In $S^-$, only the $d\,d^\dagger$ contraction survives. Using the invariant measure and the complete spin sums gives, with

$$E_{\mathbf p}=\sqrt{\mathbf p^2+m^2}, \qquad d\Pi_p = \frac{d^3\mathbf p}{(2\pi)^3\,2E_{\mathbf p}},$$

the two contractions

$$\begin{aligned} S^+(z) &= \int d\Pi_p\, (p\!\!\!/+m)e^{-ip\cdot z},\\ S^-(z) &= \int d\Pi_p\, (p\!\!\!/-m)e^{+ip\cdot z}, \qquad p^0=E_{\mathbf p}. \end{aligned}$$

Consequently,

$$\begin{aligned} S_F(z) &= \theta(z^0) \int d\Pi_p\, (p\!\!\!/+m)e^{-ip\cdot z}\\ &\quad- \theta(-z^0) \int d\Pi_p\, (p\!\!\!/-m)e^{+ip\cdot z}. \end{aligned}$$

Both integrands solve the homogeneous Dirac equation in their respective frequency sectors. The second line contains three linked signs: the negative-frequency exponential, the $v$ spin sum $p\!\!\!/-m$, and the minus from exchanging the two fields under time ordering. Changing any one without translating the other two breaks the inverse equation. The canonical contractions, spin sums, and their assembly into the ordered two-point function are worked through in Schwartz 2014, § 12.4.2, pp. 213–215.

At equal time, the sum rather than the difference of the contractions is fixed:

$$S^+(0,\mathbf r)+S^-(0,\mathbf r) = \gamma^0\delta^{(3)}(\mathbf r), \qquad \mathbf r=\mathbf x-\mathbf y.$$

This is the canonical field anticommutator written with $\widehat{\bar\psi}=\widehat\psi^\dagger\gamma^0$. It will become the four-dimensional contact term when the time derivative acts on the step functions.

The scalar Feynman kernel lifts to the Dirac inverse

Let the raw scalar Feynman correlator from the prerequisite page be

$$D_F(z) = \int\frac{d^4p}{(2\pi)^4} \frac{i\,e^{-ip\cdot z}} {p^2-m^2+i0}, \qquad (\Box+m^2)D_F=-i\delta^{(4)}.$$

Here $i0$ is the distributional boundary value

$$\frac{1}{p^2-m^2+i0} \equiv \lim_{\epsilon\downarrow0} \frac{1}{p^2-m^2+i\epsilon}, \qquad \epsilon>0,$$

not an ordinary denominator evaluated at $\epsilon=0$.

The Dirac correlator is its first-order lift:

$$\boxed{ S_F(z) = \left( i\gamma^\mu\partial_{z^\mu}+m \right) D_F(z). }$$

For $z^0>0$, the derivative turns $e^{-ip\cdot z}$ into $p\!\!\!/$ and produces $S^+$. For $z^0<0$, it turns $e^{+ip\cdot z}$ into $-p\!\!\!/$, so

$$\left( i\gamma^\mu\partial_\mu+m \right)e^{+ip\cdot z} = -(p\!\!\!/-m)e^{+ip\cdot z},$$

which produces the fermionic reverse-order minus sign. Differentiating the scalar step functions adds no separate contact term here because their equal-time scalar Wightman values agree.

Fourier transformation gives the familiar momentum boundary value

$$\boxed{ S_F(z) = \int\frac{d^4p}{(2\pi)^4} \frac{i(p\!\!\!/+m)} {p^2-m^2+i0} e^{-ip\cdot z}. }$$

The numerator follows from Clifford factorization,

$$(p\!\!\!/-m)(p\!\!\!/+m)=p^2-m^2.$$

Away from the mass shell this is matrix inversion. On shell, $p\!\!\!/+m$ becomes the complete $u$ spin sum at the positive-energy residue. Calling the off-shell numerator itself a spin sum would confuse a matrix polynomial with its on-shell completeness interpretation.

The correlator and the delta-normalized mathematical inverse differ by a factor:

$$G_D(z)\equiv-iS_F(z), \qquad \left( i\gamma^\mu\partial_\mu-m \right)G_D(z) = \delta^{(4)}(z)\mathbf1_4.$$

In four dimensions $[\widehat\psi]=3/2$, so $S_F(z)$ has mass dimension $3$. Its momentum kernel has dimension $-1$: the numerator has dimension $1$ and the denominator dimension $2$. This check is independent of all pole and ordering signs. An independent operator derivation, including the left- and right-acting contact equations, appears in Srednicki 2007, § 42, pp. 268–269. That source uses the $(-+++)$ metric, $e^{+ip\cdot x}$, and $S=i\langle\mathrm T\Psi\bar\Psi\rangle$; translating all three choices gives the $(+---)$ raw-correlator normalization displayed here.

The pole prescription packages both spin sums

The Feynman boundary value places the positive-energy pole below and the negative-energy pole above the real $p^0$ axis:

$$p^0=+E_{\mathbf p}-i0, \qquad p^0=-E_{\mathbf p}+i0.$$

For $z^0>0$, the energy contour closes below and the residue numerator is

$$p\!\!\!/+m = \sum_r u_r(p)\bar u_r(p), \qquad p^0=+E_{\mathbf p}.$$

Including the denominator derivative, the matrix residue is

$$\operatorname*{Res}_{p^0=+E_{\mathbf p}}\widetilde S_F(p) = \frac{i}{2E_{\mathbf p}} \sum_r u_r(p)\bar u_r(p).$$

For $z^0<0$, the contour closes above. At the negative-energy pole, write

$$p^\mu=(-E_{\mathbf p},\mathbf p), \qquad q^\mu=(E_{\mathbf p},-\mathbf p)=-p^\mu.$$

Then

$$p\!\!\!/+m = -(q\!\!\!/-m) = -\sum_r v_r(q)\bar v_r(q).$$

The derivative of the denominator is now $-2E_{\mathbf p}$, so its minus sign cancels the numerator minus:

$$\operatorname*{Res}_{p^0=-E_{\mathbf p}}\widetilde S_F(p) = \frac{i}{2E_{\mathbf p}} \sum_r v_r(q)\bar v_r(q).$$

Thus the single off-shell numerator yields the $u$ spin sum at one pole and minus the $v$ spin sum at the other. Closing below for $z^0>0$ is clockwise, whereas closing above for $z^0<0$ is counterclockwise. Those orientations then reproduce exactly $\theta(z^0)S^+-\theta(-z^0)S^-$. The $i0$ is therefore vacuum boundary data, not a decorative convergence symbol or a shift that may be dropped before the distribution acts on a test function.

Feynman pole placement is different from causal support. A retarded inverse places both energy poles below the real axis; an advanced inverse places both above. The Feynman distribution places one on each side because it follows frequency under time ordering rather than response under future or past support.

The contact equation is the equal-time CAR

Apply the Dirac operator to the momentum representation:

$$\begin{aligned} &\left( i\gamma^\mu\partial_{z^\mu}-m \right)S_F(z)\\ &\quad= \int\frac{d^4p}{(2\pi)^4} \frac{i(p\!\!\!/-m)(p\!\!\!/+m)} {p^2-m^2+i0} e^{-ip\cdot z}\\ &\quad= i\delta^{(4)}(z)\mathbf1_4. \end{aligned}$$

The same result follows directly from the ordered form. Away from $z^0=0$, the two contractions solve the homogeneous equation. Differentiating the step functions gives

$$\begin{aligned} &i\gamma^0\delta(z^0) \left[ S^+(0,\mathbf r)+S^-(0,\mathbf r) \right]\\ &\quad= i\gamma^0\delta(z^0) \gamma^0\delta^{(3)}(\mathbf r) = i\delta^{(4)}(z)\mathbf1_4. \end{aligned}$$

Equivalently, the ordered correlator has the equal-time jump

$$S_F(0^+,\mathbf r)-S_F(0^-,\mathbf r) = \gamma^0\delta^{(3)}(\mathbf r).$$

This second derivation checks the time-ordering minus, the spin-sum normalization, the overall factor of $i$, and the gamma-matrix placement all at once.

Feynman ordering is not causal support

For the free field, the operator anticommutator is the c-number matrix distribution

$$\{\widehat\psi_\alpha(x), \widehat{\bar\psi}_\beta(y)\} = \left[S^+(z)+S^-(z)\right]_{\alpha\beta}.$$

Indeed, if the scalar commutator distribution from the prerequisite page is

$$C(z) = \int d\Pi_p \left(e^{-ip\cdot z}-e^{+ip\cdot z}\right),$$

then

$$S^+(z)+S^-(z) = \left(i\gamma^\mu\partial_{z^\mu}+m\right)C(z).$$

Because $C$ has causal support and differentiation does not enlarge support, the anticommutator vanishes at spacelike separation. The Wightman functions and $S_F$, however, are not cone-supported; vacuum correlations may be nonzero at spacelike separation. Microcausality concerns the graded operator anticommutator, not the time-ordered expectation value. The spinor anticommutator and its light-cone support are derived in Schwartz 2014, § 12.6.1, pp. 221–222. Independently, the explicit nonzero spacelike positive-frequency function in Weinberg 1995, § 5.2, pp. 201–203 shows why causal cancellation of the anticommutator does not imply spacelike vanishing of a Wightman or Feynman function.

The distinction also separates three questions:

• $S^\pm$ are homogeneous, state-dependent vacuum contractions;
• $S_F$ is the vacuum time-ordered correlator selected by one pole on each side of the energy axis;
• retarded and advanced Dirac inverses are selected by future or past support.

They share the same differential operator but not the same state, ordering, support, or momentum boundary value. Finite-density and thermal propagators change the state-dependent contractions and contour data, so they cannot be obtained by silently reusing the vacuum formula.

The massless numerator preserves the prescription

The controlled $m\to0$ limit is

$$S_F^{(0)}(z) = \int\frac{d^4p}{(2\pi)^4} \frac{i\,p\!\!\!/} {p^2+i0} e^{-ip\cdot z},$$

with

$$i\gamma^\mu\partial_\mu S_F^{(0)}(z) = i\delta^{(4)}(z)\mathbf1_4.$$

No factor $1/m$ is present, so the distributional numerator has a regular massless limit even though the massive rest-spin projectors do not. At on-shell residues the basis is rebuilt with helicity spinors. Since

$$P_Lp\!\!\!/=p\!\!\!/P_R, \qquad P_Rp\!\!\!/=p\!\!\!/P_L,$$

the numerator maps between the chiral index carried by $\psi$ and the opposite projector carried by $\bar\psi$. The chiral projectors and massless decoupling are developed in Schwartz 2014, § 11.1, pp. 185–188; the two displayed identities follow directly from $\{\gamma_5,p\!\!\!/\}=0$. Weyl Fields and Chirality develops that decomposition; the present result is only the massless Dirac inverse.

Common pitfalls

Omitting the fermionic time-ordering sign. Reversing two odd fields introduces a minus. Without it, the negative-energy residue disagrees with the $v$ spin sum and the equal-time contact equation fails.

Calling the off-shell numerator a spin sum. $p\!\!\!/+m$ is an off-shell matrix polynomial in the inverse. It becomes a complete spin sum only after a pole puts the momentum on shell.

Dropping the factor that distinguishes a correlator from an inverse. The raw correlator satisfies $(i\gamma\cdot\partial-m)S_F=i\delta^{(4)}$. The delta-normalized inverse is $G_D=-iS_F$.

Calling the Feynman correlator retarded. One Feynman pole lies above and one below the real energy axis, and $S_F$ is not future-supported. Retarded response puts both poles below.

Erasing $i0$ before integration. The rational functions agree away from the mass shell, but their boundary values differ by on-shell distributions. The pole prescription is part of the definition.

Using the massive spin projectors at zero mass. The propagator numerator has a finite limit, but the $1/(2m)$ projectors and rest-frame spin vector do not. Rebuild the residue basis in helicity variables.

Check your understanding

Check 1: identify the negative-energy residue

At the pole $p^0=-E_{\mathbf p}$, introduce $q=(E_{\mathbf p},-\mathbf p)$. Show that the numerator becomes minus the $v$ spin sum with label $q$.

Solution

The pole four-vector is $p=-q$. Therefore

$$p\!\!\!/+m = -q\!\!\!/+m = -(q\!\!\!/-m).$$

On shell, $q\!\!\!/-m=\sum_rv_r(q)\bar v_r(q)$. The residue numerator is thus $-\sum_rv_r(q)\bar v_r(q)$, exactly the sign required by the reverse-time piece of the ordered correlator.

Check 2: derive the contact term without Fourier inversion

Apply $i\gamma^\mu\partial_\mu-m$ to $\theta(z^0)S^+(z)-\theta(-z^0)S^-(z)$.

Solution

The operator annihilates both terms away from $z^0=0$. Since $\partial_0\theta(z^0)=\delta(z^0)$ and $\partial_0[-\theta(-z^0)]=+\delta(z^0)$, the contact part is

$$i\gamma^0\delta(z^0) \left[S^+(0,\mathbf r)+S^-(0,\mathbf r)\right].$$

The equal-time CAR makes the bracket $\gamma^0\delta^{(3)}(\mathbf r)$. Since $(\gamma^0)^2=1$, the result is $i\delta^{(4)}(z)$.

Check 3: translate between correlator conventions

Suppose a source defines the delta-normalized Green matrix $G_D$ by $(i\gamma\cdot\partial-m)G_D=\delta^{(4)}$. Express it in terms of the raw vacuum correlator $S_F$ used here.

Solution

This page found

$$(i\gamma\cdot\partial-m)S_F=i\delta^{(4)}.$$

Multiplication by $1/i=-i$ therefore gives

$$G_D=-iS_F.$$

The two objects have the same pole placement and numerator but different overall normalization. A formula must state which defining equation it uses.

Where the propagator is used next

• Grassmann Functional Integrals for Free Fermions derives the same inverse from a regulated Berezin Gaussian with explicit source order.
• Lorentzian Boundary Conditions and the $i\epsilon$ Prescription develops vacuum boundary data and contour deformation beyond this one-energy calculation.
• Weyl Fields and Chirality projects the massless inverse into chiral two-component sectors.
• Fermion Signs and Closed Loops develops the extra signs from closed fermion contractions in perturbative amplitudes.
• Finite-Density Correlators and Charge Susceptibilities develops state-dependent propagators at nonzero density.

References

• Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge University Press, 2014. DOI.
• Srednicki, Mark. Quantum Field Theory. Cambridge University Press, 2007. DOI.
• Weinberg, Steven. The Quantum Theory of Fields. Volume I: Foundations. Cambridge University Press, 1995. DOI.