Fermions, spin, and anticommutation

A free Dirac field is the smallest relativistic model in which spinor covariance, fermionic statistics, antiparticles, and sign-sensitive correlators all meet. This lesson constructs that model from its action to its Feynman propagator. The decisive result is not merely a familiar formula: it is a chain of checks showing that the same normalization produces the equal-time anticommutator, a positive excitation spectrum, the correct spin-sum residues, and the inverse Dirac kernel.

Required background. Canonical quantization and the free scalar supplies mode expansions, Fock space, and the relation between a vacuum and a propagator. Functional integrals and correlators supplies finite Gaussian integration, sources, and the meaning of an inverse kernel. Helpful background. Lorentz field representations and Poincaré particle representations separates the spinor transformation law from one-particle spin, while spinors, conjugations, bilinears, chirality, and Fierz identities develops the algebra used here.

A Dirac field connects two representation problems

Work in four-dimensional Minkowski spacetime with the inherited metric $\eta_{\mu\nu}=\operatorname{diag}(+,-,-,-)$ and

$$\{\gamma^\mu,\gamma^\nu\}=2\eta^{\mu\nu}\mathbf 1_4, \qquad \overline\psi=\psi^\dagger\gamma^0, \qquad p\!\!\!/\equiv\gamma^\mu p_\mu.$$

No explicit gamma-matrix basis is needed. Under a proper orthochronous Lorentz transformation, a Dirac field transforms in the finite-dimensional spinor representation,

$$\psi'(x)=S(\Lambda)\psi(\Lambda^{-1}x), \qquad S(\Lambda)^{-1}\gamma^\mu S(\Lambda) =\Lambda^\mu{}_{\nu}\gamma^\nu.$$

The Dirac adjoint transforms with $S^{-1}$ because $S^\dagger\gamma^0=\gamma^0S^{-1}$. Consequently $\overline\psi\psi$ is a Lorentz scalar and $\overline\psi\gamma^\mu\psi$ is a Lorentz vector. The matrices $S(\Lambda)$ are not required to be unitary in the ordinary four-component Euclidean inner product.

This field representation is not the same object as the unitary Poincaré representation carried by one-particle states. The plane-wave spinors $u_s(p)$ and $v_s(p)$ connect the two: the index on $\psi_\alpha(x)$ transforms with $S(\Lambda)$, while the label $s$ identifies the two spin states at a massive on-shell momentum. This distinction, including the role of Wigner rotations, is developed in the mathematical preparation above and in Schwartz 2014, §§ 10.2–10.3, pp. 168–172.

The discussion below uses a complex Dirac field of mass $m>0$. The $m=0$ limit exists, but chirality then becomes the more useful organization and the normalizations proportional to $m$ become degenerate. Weyl and Majorana fields require additional dimension-, signature-, and reality-dependent statements; they are not alternative names for the construction on this page.

The action fixes the equation and the two frequency branches

For independent classical variables $\psi$ and $\overline\psi$, with their Grassmann order held fixed when appropriate, take

$$S_0=\int_M\mathrm d^4x\, \overline\psi\left(i\gamma^\mu\partial_\mu-m\right)\psi.$$

Varying $\overline\psi$ gives the Dirac equation; varying $\psi$ and integrating by parts gives its adjoint:

$$\begin{aligned} \left(i\gamma^\mu\partial_\mu-m\right)\psi&=0,\\ i(\partial_\mu\overline\psi)\gamma^\mu+m\overline\psi&=0. \end{aligned}$$

The integration by parts also produces

$$i\int_{\partial M}\mathrm d\Sigma_\mu\, \overline\psi\gamma^\mu\delta\psi.$$

Thus the bulk equation does not by itself specify a boundary-value problem. Compactly supported variations, suitable falloff, or coherent first-order endpoint data are additional assumptions. The symmetrized Dirac density differs from the displayed density by a total derivative and has the same bulk equations but a different explicit boundary term.

Multiplying the equation by the complementary first-order operator gives

$$\left(i\gamma^\mu\partial_\mu+m\right) \left(i\gamma^\nu\partial_\nu-m\right) =-\left(\Box+m^2\right).$$

Every Dirac solution therefore obeys the Klein–Gordon equation component by component, but the converse is false: the first-order equation restricts the spinor amplitude. With $p^0=E_{\mathbf p}=\sqrt{\mathbf p^2+m^2}>0$,

$$\begin{aligned} \psi(x)=u_s(p)e^{-ip\cdot x} &\quad\Longrightarrow\quad (p\!\!\!/-m)u_s(p)=0,\\ \psi(x)=v_s(p)e^{+ip\cdot x} &\quad\Longrightarrow\quad (p\!\!\!/+m)v_s(p)=0. \end{aligned}$$

The second line is a negative-frequency solution, not a state of negative energy. Its coefficient will become an antiparticle creation operator.

Choose the mutually consistent normalization package

$$\begin{aligned} \overline u_s(p)u_{s'}(p)&=2m\,\delta_{ss'},& \sum_su_s(p)\overline u_s(p)&=p\!\!\!/+m,\\ \overline v_s(p)v_{s'}(p)&=-2m\,\delta_{ss'},& \sum_sv_s(p)\overline v_s(p)&=p\!\!\!/-m. \end{aligned}$$

It follows that $u_s^\dagger u_{s'}=v_s^\dagger v_{s'} =2E_{\mathbf p}\delta_{ss'}$. The signs in the $v$ norm and completeness relation are linked; changing a spinor normalization requires changing the mode measure, oscillator algebra, state normalization, and propagator residue together. A basis-independent derivation of these identities is given in Schwartz 2014, § 11.2, pp. 188–191.

Anticommutation produces a positive particle spectrum

Define the invariant on-shell measure

$$\mathrm d\Pi_p =\frac{\mathrm d^3\mathbf p}{(2\pi)^3\,2E_{\mathbf p}}$$

and expand the operator-valued field as

$$\begin{aligned} \psi(x) &=\sum_s\int\mathrm d\Pi_p\, \left[ b_s(\mathbf p)u_s(p)e^{-ip\cdot x} +d_s^\dagger(\mathbf p)v_s(p)e^{+ip\cdot x} \right],\\ \overline\psi(x) &=\sum_s\int\mathrm d\Pi_p\, \left[ b_s^\dagger(\mathbf p)\overline u_s(p)e^{+ip\cdot x} +d_s(\mathbf p)\overline v_s(p)e^{-ip\cdot x} \right]. \end{aligned}$$

Impose the mode anticommutation relations

$$\begin{aligned} \{b_s(\mathbf p),b_{s'}^\dagger(\mathbf q)\} &=(2\pi)^3 2E_{\mathbf p}\, \delta_{ss'}\delta^{(3)}(\mathbf p-\mathbf q),\\ \{d_s(\mathbf p),d_{s'}^\dagger(\mathbf q)\} &=(2\pi)^3 2E_{\mathbf p}\, \delta_{ss'}\delta^{(3)}(\mathbf p-\mathbf q), \end{aligned}$$

with every other mode anticommutator zero. The spin sums then give the canonical equal-time relation

$$\{\psi_\alpha(t,\mathbf x), \psi_\beta^\dagger(t,\mathbf y)\} =\delta_{\alpha\beta}\delta^{(3)}(\mathbf x-\mathbf y).$$

This is a normalization test, not an independent convention that can be adjusted afterward. It also implies Pauli exclusion for each regulated mode: $\{b^\dagger,b^\dagger\}=0$ gives $(b^\dagger)^2=0$, and similarly for $d^\dagger$.

Interpret coincident mode products first in a finite spatial box with a finite momentum cutoff; the continuum formulas below are the corresponding compact notation. The canonical Hamiltonian initially contains

$$H_0 =\sum_s\int\mathrm d\Pi_p\,E_{\mathbf p} \left(b_s^\dagger b_s-d_s d_s^\dagger\right).$$

Using $d_sd_s^\dagger=\{d_s,d_s^\dagger\}-d_s^\dagger d_s$ separates a vacuum c-number. Measuring energy relative to the Minkowski vacuum gives

$$H_{\mathrm{exc}}\equiv:H_0: =\sum_s\int\mathrm d\Pi_p\,E_{\mathbf p} \left(b_s^\dagger b_s+d_s^\dagger d_s\right).$$

Both kinds of creation operator therefore raise the energy:

$$[H_{\mathrm{exc}},b_s^\dagger(\mathbf p)] =E_{\mathbf p}b_s^\dagger(\mathbf p), \qquad [H_{\mathrm{exc}},d_s^\dagger(\mathbf p)] =E_{\mathbf p}d_s^\dagger(\mathbf p).$$

The vector phase symmetry has the normal-ordered charge

$$Q\equiv:Q_0: =\sum_s\int\mathrm d\Pi_p\, \left(b_s^\dagger b_s-d_s^\dagger d_s\right).$$

Thus $d_s^\dagger|0\rangle$ has positive energy and charge opposite to $b_s^\dagger|0\rangle$: it is an antiparticle. The minus sign needed for a positive Hamiltonian is the same CAR reordering sign that gives the opposite charge. The complete canonical derivation, including the translation between common oscillator normalizations, appears in Srednicki 2006 draft, § 39, pp. 246–250, PDF and Weinberg 1995, § 7.5, pp. 323–325.

Anticommutation is imposed here as part of the free quantum theory. Its necessity for a local, positive relativistic theory is the content of the spin–statistics theorem, not a consequence of the classical Dirac action alone.

The Feynman propagator is a graded inverse

For fermionic operators, time ordering includes an exchange sign:

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

Using the vacuum annihilation conditions, the mode expansion, and the two spin sums gives

$$\begin{aligned} [S_F(x-y)]_{\alpha\beta} &\equiv \langle0|\mathrm T\psi_\alpha(x)\overline\psi_\beta(y)|0\rangle,\\ S_F(x-y) &=\int\frac{\mathrm d^4p}{(2\pi)^4}\, \frac{i(p\!\!\!/+m)}{p^2-m^2+i0} e^{-ip\cdot(x-y)}. \end{aligned}$$

Three independent facts are visible in this formula:

• the numerator $p\!\!\!/+m$ is fixed by the spin sum and by inversion of the first-order operator;
• the two poles are the particle and antiparticle mass shells; and
• the $+i0$ prescription is vacuum boundary data, not a gamma-matrix identity.

The fastest algebraic check is

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

It implies the distributional contact equation

$$\left(i\gamma^\mu\partial_{x^\mu}-m\right)S_F(x-y) =i\delta^{(4)}(x-y)\mathbf 1_4.$$

The discontinuity generated when the time derivative acts on the step functions reproduces the equal-time CAR. This closes a useful round trip: mode normalization fixes the CAR, the CAR and spin sums fix the time-ordered function, and the time-ordered function inverts the same Dirac operator. The operator derivation and contour prescription are given in Schwartz 2014, § 12.4.2, pp. 213–215 and Srednicki 2006 draft, § 42, Eqs. (42.6)–(42.14), pp. 268–269, PDF. For that draft, $g_{\mu\nu}=-\eta_{\mu\nu}$, so its scalar product and $p\!\!\!/$ have the opposite sign for the same contravariant momentum and gamma matrices. Its $e^{+ipx}$ is therefore our $e^{-ipx}$, and its named propagator is $S_{\rm draft}=iS_F$. Translate all three choices together; the vacuum correlator and its contact equation then agree.

The Feynman propagator is not a probability amplitude for a particle to travel from $y$ to $x$, and it need not vanish at spacelike separation. Microcausality concerns the graded commutator: two odd fermion fields anticommute at spacelike separation, whereas even local observables commute. For the free Dirac field, the particle and antiparticle contractions combine in $\{\psi_\alpha(x),\bar\psi_\beta(y)\}$ and vanish outside the light cone. This does not make either contraction or the Feynman correlator vanish there.

A finite Grassmann Gaussian checks the inverse and the sign

The functional calculation should begin at a finite regulator. Let $\psi_i$ and $\overline\psi_i$ be independent Grassmann generators, let $A$ be an invertible finite matrix, and declare the paired measure

$$\mathcal D_R =\mathrm d\overline\psi_N\,\mathrm d\psi_N\cdots \mathrm d\overline\psi_1\,\mathrm d\psi_1.$$

With the source order shown explicitly,

$$\begin{aligned} \mathcal Z_E[\overline\eta,\eta] &=\int\mathcal D_R\, \exp\!\left( -\overline\psi A\psi +\overline\eta\psi +\overline\psi\eta \right)\\ &=\det A\, \exp\!\left(\overline\eta A^{-1}\eta\right). \end{aligned}$$

After division by the identical zero-source integral, $Z_E=\exp(\overline\eta A^{-1}\eta)$. If every derivative is a left derivative, the written order gives

$$\left. \frac{\partial^L}{\partial\eta_j} \frac{\partial^L}{\partial\overline\eta_i} Z_E \right|_{\eta=\overline\eta=0} =(A^{-1})_{ij}.$$

Reversing the two odd derivatives gives the negative matrix element, just as reversing $\psi_i\overline\psi_j$ does. For one pair the complete test is

$$\int\mathrm d\overline\psi\,\mathrm d\psi\, e^{-a\overline\psi\psi}=a, \qquad \langle\psi\overline\psi\rangle=\frac1a, \qquad \langle\overline\psi\psi\rangle=-\frac1a.$$

At finite regulator, taking $A$ to be the Euclidean Dirac kernel therefore reproduces its inverse. With the vacuum boundary condition and analytic continuation stated, that inverse becomes the Feynman kernel above. The Gaussian also gives $\det A$, in contrast with the inverse determinant power of commuting variables. Normalizing by $\mathcal Z_E[0,0]$ removes this determinant only when the kernel, regulator, boundary conditions, and retained modes are exactly the same. Zero modes require explicit saturation or projection; one may not write $A^{-1}$ and divide by a vanishing integral. The ordered finite identity and its field-theory application are developed in Schwartz 2014, § 14.6, pp. 270–272 and Srednicki 2006 draft, § 44, pp. 276–281, PDF. In his Eq. (44.40), set $M=-A$ and reverse each adjacent $\mathrm d\psi_i\,\mathrm d\bar\psi_i$ pair to obtain our declared measure. The resulting factor $(-1)^N$ cancels the same factor in $\det(-A)$; the source exponential becomes $\exp(\bar\eta A^{-1}\eta)$.

This Grassmann calculation checks the inverse kernel and exchange signs. It does not by itself construct a Hilbert space, prove positivity, or select the vacuum prescription; the canonical construction supplies those distinct ingredients.

Keep the signs attached to their causes

When two sources use different conventions, translate a complete package and then rerun a convention-independent check.

Possible sign change	What fixes it	A check that survives translation
Metric or Fourier phase	The declared $\eta_{\mu\nu}$ and transform pair	Pole at $p^2=m^2$ and the Dirac contact equation
Negative-frequency spinor	The phase $e^{+ip\cdot x}$ and $(p\!\!\!/+m)v=0$	Completeness $\sum_s v_s\overline v_s=p\!\!\!/-m$
Antiparticle Hamiltonian	Reordering $d d^\dagger$ with the CAR	$[H,d^\dagger]=E_{\mathbf p}d^\dagger$
Fermionic time ordering	One odd exchange	Equal-time discontinuity reproduces the CAR
Grassmann source result	Measure, source, insertion, and derivative order	The one-pair integral returns $1/a$ and its exchanged negative
Closed fermion loop	Reordering a closed chain of contractions	Derive it in perturbation theory; do not infer it from this free propagator alone

A useful calculation header is therefore short: metric and Fourier phase; gamma and adjoint convention; spinor and oscillator normalization; ordering of every Grassmann measure, source, and derivative; and the boundary prescription selecting the inverse.

Scope and failure boundaries

This construction establishes the free complex Dirac field in the Minkowski vacuum. It does not establish any of the following.

• The spin–statistics theorem. CAR are imposed and checked here; the theorem requires locality, positivity, covariance, and further structural assumptions.
• Chiral or Majorana dynamics. At $m=0$, chirality needs its own two-component organization. A Majorana condition needs a compatible reality structure and changes the particle–antiparticle interpretation.
• An interacting field. Yukawa and gauge couplings introduce renormalization, composite operators, loop signs, and possible anomalies.
• A representation-independent vacuum. The particle split uses time translation and the Minkowski vacuum. Curved spacetime, thermal states, and time-dependent backgrounds can require different representations and propagators.
• A continuum determinant. The finite Grassmann determinant has an exact meaning. Its continuum magnitude and phase require regulator, renormalization, boundary, and zero-mode data.

These are limits on the claim, not defects in the free model. Within its scope, the CAR, Hamiltonian, charge, spin sums, and propagator form one mutually testable construction.

Exercises

Recover the equal-time field anticommutator

Use the mode expansion, covariant mode CAR, and completeness relations to derive $\{\psi_\alpha(t,\mathbf x),\psi_\beta^\dagger(t,\mathbf y)\} =\delta_{\alpha\beta}\delta^{(3)}(\mathbf x-\mathbf y)$.

Solution

At equal time, the $b$ contraction contributes

$$\int\frac{\mathrm d^3\mathbf p}{(2\pi)^3\,2E_{\mathbf p}} \sum_su_s(\mathbf p)u_s^\dagger(\mathbf p) e^{+i\mathbf p\cdot(\mathbf x-\mathbf y)}.$$

The $d$ contraction has the opposite spatial phase. Change its integration variable $\mathbf p\mapsto-\mathbf p$ and define $p=(E_{\mathbf p},\mathbf p)$ and $q=(E_{\mathbf p},-\mathbf p)$. Since

$$\begin{aligned} \sum_su_s(p)u_s^\dagger(p)&=(p\!\!\!/+m)\gamma^0,\\ \sum_sv_s(q)v_s^\dagger(q)&=(q\!\!\!/-m)\gamma^0, \end{aligned}$$

their sum is

$$(p\!\!\!/+q\!\!\!/)\gamma^0 =2E_{\mathbf p}\mathbf 1_4.$$

This cancels the $2E_{\mathbf p}$ in the on-shell measure. The remaining Fourier integral is

$$\delta_{\alpha\beta} \int\frac{\mathrm d^3\mathbf p}{(2\pi)^3} e^{+i\mathbf p\cdot(\mathbf x-\mathbf y)} =\delta_{\alpha\beta}\delta^{(3)}(\mathbf x-\mathbf y).$$

The cancellation of the mass terms is why both frequency branches are needed.

Diagnose the antiparticle rather than a negative-energy particle

For one regulated momentum-spin mode, let $\{d,d^\dagger\}=1$ and let $d|0\rangle=0$. Starting from the raw contribution $-Edd^\dagger$, find the vacuum-relative Hamiltonian, the energies of the two states, and their charges if the normal-ordered charge contribution is $-d^\dagger d$.

Solution

The CAR give

$$-Edd^\dagger =-E(1-d^\dagger d) =-E+Ed^\dagger d.$$

Subtracting the vacuum c-number leaves $H=Ed^\dagger d$. The empty state has vacuum-relative energy zero, while

$$H\,d^\dagger|0\rangle =E\,d^\dagger|0\rangle.$$

Because $(d^\dagger)^2=0$, the mode has no second occupied state. With $Q=-d^\dagger d$, the excitation obeys

$$Q\,d^\dagger|0\rangle =-d^\dagger|0\rangle.$$

It therefore has positive energy and charge opposite to the corresponding $b^\dagger$ excitation. Calling $e^{+ip\cdot x}$ a negative-frequency term does not assign negative energy to the state created by its coefficient.

Check the propagator by both operator and Grassmann routes

First multiply the momentum-space Feynman kernel by $p\!\!\!/-m$. Then use the one-pair Grassmann integral with $a\ne0$ to explain why exchanging the two fermionic insertions reverses the inverse-kernel sign.

Solution

Clifford factorization gives

$$(p\!\!\!/-m) \frac{i(p\!\!\!/+m)}{p^2-m^2+i0} =i\frac{p^2-m^2}{p^2-m^2+i0},$$

which equals $i$ as the boundary-value distribution. Fourier transformation therefore gives the contact term $i\delta^{(4)}(x-y)\mathbf 1_4$.

For one Grassmann pair,

$$e^{-a\overline\psi\psi} =1+a\psi\overline\psi.$$

With the declared measure, $\int\mathrm d\overline\psi\,\mathrm d\psi\,\psi\overline\psi=1$, so the zero-source integral is $a$. The normalized insertion is $\langle\psi\overline\psi\rangle=1/a$. Since $\overline\psi\psi=-\psi\overline\psi$, reversing the insertion gives $-1/a$. Both routes therefore identify an inverse kernel, while the Grassmann route also exposes the exchange sign directly.

Continue through both free-field branches

This lesson follows the functional-integral construction and completes the fermion branch. Study vector fields and gauge redundancy alongside it, then follow that branch through symmetry, currents, and Ward identities. Once the free-fermion signs and the symmetry constraints are both secure, continue to perturbative expansion and Feynman rules, where contraction signs, fermion lines, and closed loops are derived from an interacting expansion.

References

• Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge University Press, 2014. DOI.
• Srednicki, Mark. Quantum Field Theory. Prepublication draft, 2006. Author’s draft and published-edition errata. Open PDF.
• Weinberg, Steven. The Quantum Theory of Fields, Volume I: Foundations. Cambridge University Press, 1995. DOI.