From Spinors to Fermion Fields

A free Dirac field combines numerical spinor modes with fermionic annihilation and creation operators. The spinors carry Lorentz and spin information; the operators carry occupation and exchange statistics. Their combination has the canonical equal-time anticommutator and vanishing spacelike anticommutators. The spin sums make this locality check explicit and also explain why ordinary numerical $u$ and $v$ spinors appear on external particle legs.

Required background. Free Dirac Spinors fixes normalization and spin sums; Fermionic Anticommutation Relations supplies the mode algebra; From Fock Space to Quantum Fields establishes the scalar construction and field-matrix-element viewpoint.

Helpful background. Dirac Negative-Energy Solutions derives the positive free particle and antiparticle energies; Spin-Statistics Preview states the broader theorem’s assumptions.

Numerical spinors and operator coefficients

Use $\hbar=c=1$, metric $(+---)$, and $m>0$. All labels $p=(E_{\mathbf p},\mathbf p)$ are future-directed and on shell. Adopt

$$\begin{aligned} d\Pi_p&=\frac{d^3p}{(2\pi)^3\,2E_{\mathbf p}},\\ u_s^\dagger(p)u_r(p)&=v_s^\dagger(p)v_r(p) =2E_{\mathbf p}\delta_{sr}. \end{aligned}$$

The covariant mode anticommutators are

$$\{a_s(\mathbf p),a_r^\dagger(\mathbf q)\} =2E_{\mathbf p}(2\pi)^3 \delta_{sr}\delta^3(\mathbf p-\mathbf q),$$

with the identical formula for $b,b^\dagger$. All remaining anticommutators vanish, including those between the particle and antiparticle species. The vacuum is annihilated by both $a$ and $b$.

The free field is

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

Every component of $\widehat\psi$ is an operator-valued distribution. The columns $u_s$ and $v_s$ are ordinary numerical solutions of the Dirac mode equation. They commute with the ladder operators and do not themselves satisfy canonical anticommutators.

The negative-frequency factor multiplying $b^\dagger$ does not create a negative-energy physical particle. The normal-ordering calculation on Dirac Negative-Energy Solutions gives positive energies for both species and opposite additive charges. That Hamiltonian and charge derivation is separate from the locality calculation performed here.

The field anticommutator from the spin sums

Let $z=x-y$ and define

$$\mathcal A_{\alpha\beta}(z) =\{\widehat\psi_\alpha(x), \widehat{\bar\psi}_\beta(y)\}.$$

The mode anticommutators remove one momentum integral. The established spin sums

$$\begin{aligned} \sum_su_s(p)\bar u_s(p)&=p\!\!\!/+m,\\ \sum_sv_s(p)\bar v_s(p)&=p\!\!\!/-m. \end{aligned}$$

then give the numerical matrix distribution

$$\mathcal A(z)=\int d\Pi_p\, \left[(p\!\!\!/+m)e^{-ip\cdot z} +(p\!\!\!/-m)e^{ip\cdot z}\right].$$

The second term has a plus because $\{b^\dagger,b\}=\{b,b^\dagger\}$. This is the sign needed for the spacelike cancellation below.

Use the scalar commutator distribution, with no additional factor of $i$ in its definition,

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

Acting with $i\gamma^\mu\partial_\mu+m$ yields

$$\mathcal A(z) =(i\gamma^\mu\partial_\mu+m)C(z).$$

For the positive exponential, the derivative contributes $-p\!\!\!/$ and the minus in $C$ reverses it. Together with the mass term, this gives exactly $+(p\!\!\!/-m)e^{ip\cdot z}$. The anticommutator is state-independent in this free theory, although individual Wightman expectation values need not be.

The other free field anticommutators $\{\widehat\psi_\alpha(x),\widehat\psi_\beta(y)\}$ and the corresponding barred pair vanish by the mode algebra.

Equal-time normalization and spacelike locality

The scalar distribution has equal-time data

$$C(0,\mathbf z)=0,\qquad \partial_{z^0}C(0,\mathbf z)=-i\delta^3(\mathbf z).$$

Its spatial derivatives at that time vanish distributionally. Therefore

$$\mathcal A(0,\mathbf z)=\gamma^0\delta^3(\mathbf z).$$

Using $\bar\psi=\psi^\dagger\gamma^0$ gives the canonical component relation

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

The scalar causality calculation establishes $C(z)=0$ for spacelike $z$. A local differential operator cannot enlarge its distributional support. Hence $\mathcal A(z)=0$ at spacelike separation. The two frequency sectors in the field are both needed for this result.

This is graded locality for odd fermionic fields. Even local observables constructed from them commute when their supports are spacelike separated. For example, interchanging two fermion bilinears requires four odd-field interchanges, whose signs multiply to $+1$. Coincident-point products still need a suitable definition or renormalization prescription; the spacelike algebra does not remove that ultraviolet issue.

This free charged field has a global charge symmetry. Dynamical QED adds gauge constraints and further qualifications on physical charged operators; the present calculation is not a construction of a strictly local gauge-invariant charged observable in that interacting theory.

External spinors are field matrix elements

Define covariantly normalized one-particle states $|\mathbf p,s\rangle=a_s^\dagger(\mathbf p)|0\rangle$ and antiparticle states $|\overline{\mathbf p,s}\rangle =b_s^\dagger(\mathbf p)|0\rangle$. Direct contraction gives

$$\langle0|\widehat\psi_\alpha(x)|\mathbf p,s\rangle =u_{s,\alpha}(p)e^{-ip\cdot x},$$ $$\langle\overline{\mathbf p,s}| \widehat\psi_\alpha(x)|0\rangle =v_{s,\alpha}(p)e^{ip\cdot x}.$$

Thus the familiar columns are numerical coefficients obtained between the field vacuum and normalized particle states. Fermionic exchange signs come from the operators and the ordering of multiparticle states. Assigning anticommutation rules to the numerical external spinor entries would confuse those roles.

In a fermionic functional integral, the symbols $\psi$ and $\bar\psi$ instead denote independent Grassmann integration variables. That is a representation of the quantum correlation calculation. It is not a statement that an ordinary Dirac wavefunction has become a column of measurable Grassmann numbers.

A wrong-statistics test and the theorem boundary

An instructive free-field test replaces both species’ mode anticommutators with positive-norm bosonic commutators, retaining the same mode expansion. The resulting field commutator would be

$$\mathcal B(z) =(i\gamma^\mu\partial_\mu+m) \left[W(z)+W(-z)\right],$$

where $W(z)=\int d\Pi_p\,e^{-ip\cdot z}$. The plus inside the bracket replaces the commutator distribution $C$. For spacelike $z=(0,\mathbf r)$, $\mathbf r\ne0$, one has $W(z)=W(-z)$, and the trace is

$$\operatorname{tr}\mathcal B(z)=8m\,W(z).$$

For the massive scalar function this is nonzero. Thus this proposed bosonic spinor construction fails the desired spacelike commutativity despite using the same numerical Dirac solutions. Other attempts to alter the ladder signs must also confront Hilbert-space positivity and energy boundedness.

This calculation is a consistency check for the specified free construction. It is not a proof of the full spin–statistics theorem. The theorem preview states the usual relativistic covariance, locality, positive physical norm, spectrum, vacuum, and domain assumptions. A minus sign under a $2\pi$ spin rotation alone does not derive an exchange law.

The resulting free field is now sufficient to interpret the Dirac Feynman propagator as a specified vacuum time-ordered two-point function, including its fermionic ordering sign. The inverse and pole derivations remain with that owner.

Exercises

Recover the equal-time identity. Starting from $\mathcal A=(i\gamma\cdot\partial+m)C$, identify which term survives at equal time and explain the final factor of $\gamma^0$.

Solution

Only $i\gamma^0\partial_0C$ survives, giving $i\gamma^0(-i\delta^3)=\gamma^0\delta^3$. To replace $\bar\psi$ by $\psi^\dagger$, multiply the anticommutator on the right by $\gamma^0$. Since $(\gamma^0)^2=I$, the result is $I\delta^3$.

Remove the antiparticle term. What is lost if one keeps only $a_su_se^{-ipx}$ in the field while keeping the same CAR for $a$?

Solution

The anticommutator retains only $\int d\Pi_p(p\!\!\!/+m)e^{-ipz}$. It is a positive-frequency distribution, not the derivative of the spacelike-vanishing commutator difference. In general it neither has the full canonical equal-time identity nor vanishes spacelike. A particle-sector projection changes the local field problem.

Spinor coefficients and exchange. Why can commuting numerical external spinors appear in a two-fermion amplitude that changes sign when identical external states are exchanged?

Solution

The external spinors are matrix-element coefficients. The state is built with ordered fermionic creation operators, whose interchange changes its sign. The amplitude inherits that sign from the state and field algebra, without changing the arithmetic of numerical spinor entries.

References

• Pauli, Wolfgang. “The Connection Between Spin and Statistics.” Physical Review 58, 716–722 (1940). doi:10.1103/PhysRev.58.716. Historical relativistic free-field consistency argument.
• Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge University Press (2014). doi:10.1017/9781139540940. Fermionic quantization, external states, and propagators.
• Tong, David. Quantum Field Theory. University of Cambridge lecture notes (2006), sections 5.1–5.5. Quantizing the Dirac field. Mode statistics, field anticommutators, and the free Dirac correlator.