Canonical Quantization of the Free Dirac Field

Canonical quantization of the free Dirac field combines two distinct choices: the canonical anticommutation relations define a graded operator algebra, and the positive-frequency split selects the standard Minkowski vacuum representation. In that representation the negative-frequency wavefunction multiplies an antiparticle creation operator. The CAR identity $dd^\dagger=1-d^\dagger d$ then converts the apparent negative-energy term into a positive antiparticle number operator, while the same algebra gives positive state norms and Pauli exclusion. This page constructs that result for the free field, including its Hamiltonian, charge, and vacuum-relative ordering. It does not prove the spin–statistics theorem or treat interacting fermion loops.

Required background. Plane Waves, Spin Sums, and Bilinears supplies the $u/v$ modes, their covariant normalization, mixed orthogonality, and completeness. Canonical Quantization: Algebra, Representation, and State distinguishes canonical relations from the representation and state chosen below.

Helpful background. Multiparticle States, Statistics, and Fock Organization supplies antisymmetric tensor powers and Fermi Fock space. Exterior and Graded Algebra, Grassmann Variables, and Berezin Integration supplies the graded sign rule; its Grassmann numbers are coefficients, not the operators used here.

The first-order Dirac phase space

Work first with $m>0$ on four-dimensional Minkowski spacetime. At a fixed time, fields are operator-valued distributions on $\mathbb R^3$: pointwise relations below are shorthand for smeared relations. A periodic box and a finite momentum cutoff will be introduced before products at coincident momenta are reordered. Finite-particle vectors provide a common invariant dense domain for the free smeared fields, Hamiltonian, and charge used here.

For the canonical calculation, use the Dirac density in the form

$$\mathcal L = i\psi^\dagger\dot\psi - \psi^\dagger h_D\psi, \qquad h_D = -i\boldsymbol{\alpha}\cdot\nabla+\beta m,$$

where $\alpha^i=\gamma^0\gamma^i$ and $\beta=\gamma^0$. This density differs from the symmetrized density on The Dirac Field by a total derivative under the same boundary conditions. In the canonical variation, $\psi$ and $\psi^\dagger$ are independent Grassmann-odd coordinates. Define the momentum by placing the velocity variation on the right,

$$\left.\delta\mathcal L\right|_{\delta\dot\psi} = \sum_\alpha \pi_{\psi,\alpha}\,\delta\dot\psi_\alpha.$$

This right-derivative convention makes the first-order time derivative unambiguous:

$$\pi_{\psi,\alpha}=i\psi_\alpha^\dagger, \qquad \pi_{\psi^\dagger,\alpha}=0.$$

These equations are second-class constraints, not an invertible Legendre map. With the displayed variable order and graded derivative convention, the kinetic term has reduced kernel $i\delta_{\alpha\beta}\delta^{(3)}(\mathbf x-\mathbf y)$; its inverse is $-i\delta_{\alpha\beta}\delta^{(3)}(\mathbf x-\mathbf y)$. Equivalently, eliminating the two constraints with the graded Dirac bracket gives the reduced equal-time relation

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

Promoting $i$ times this reduced bracket to the graded operator bracket produces the field CAR

$$\begin{aligned} \{\widehat\psi_\alpha(t,\mathbf x), \widehat\psi_\beta^\dagger(t,\mathbf y)\} &= \delta_{\alpha\beta}\delta^{(3)}(\mathbf x-\mathbf y),\\ \{\widehat\psi_\alpha(t,\mathbf x), \widehat\psi_\beta(t,\mathbf y)\} &=0,\\ \{\widehat\psi_\alpha^\dagger(t,\mathbf x), \widehat\psi_\beta^\dagger(t,\mathbf y)\} &=0. \end{aligned}$$

The canonical momentum and equivalent $\{\widehat\psi_\alpha,\widehat{\bar\psi}_\beta\} =(\gamma^0)_{\alpha\beta}\delta^{(3)}$ form are developed in Srednicki 2007, § 37, pp. 236–237.

This step defines the graded algebra. It does not yet select a Hilbert space, a vacuum, or a particle interpretation. Those enter through the mode split and state condition below. Treating the first-order action as an ordinary nonsingular bosonic Legendre transform would conceal precisely this distinction.

Mode operators matched to the spinor normalization

Keep the future-directed label $p^\mu=(E_{\mathbf p},\mathbf p)$ and the invariant measure

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

The accepted spinor normalization is

$$\begin{aligned} u_r^\dagger(p)u_s(p) &= v_r^\dagger(p)v_s(p) =2E_{\mathbf p}\delta_{rs},\\ \sum_r u_r(p)\bar u_r(p)&=p\!\!\!/+m,\\ \sum_r v_r(p)\bar v_r(p)&=p\!\!\!/-m. \end{aligned}$$

The matched operator expansion is

$$\widehat\psi(x) = \sum_{r=1}^{2} \int d\Pi_p \left[ b_r(\mathbf p)u_r(p)e^{-ip\cdot x} + d_r^\dagger(\mathbf p)v_r(p)e^{+ip\cdot x} \right],$$

with

$$\widehat{\bar\psi}(x) = \sum_{r=1}^{2} \int d\Pi_p \left[ b_r^\dagger(\mathbf p)\bar u_r(p)e^{+ip\cdot x} + d_r(\mathbf p)\bar v_r(p)e^{-ip\cdot x} \right].$$

The mode CAR use the same covariant normalization:

$$\begin{aligned} \{b_r(\mathbf p),b_s^\dagger(\mathbf q)\} &= (2\pi)^3\,2E_{\mathbf p}\, \delta_{rs}\delta^{(3)}(\mathbf p-\mathbf q),\\ \{d_r(\mathbf p),d_s^\dagger(\mathbf q)\} &= (2\pi)^3\,2E_{\mathbf p}\, \delta_{rs}\delta^{(3)}(\mathbf p-\mathbf q). \end{aligned}$$

Every other elementary anticommutator vanishes. The coefficient of the negative-frequency mode is $d^\dagger$, not $d$: after the vacuum is chosen, it creates a positive-energy antiparticle. Negative frequency describes the wavefunction in the classical mode expansion; it is not the energy of the quantum state.

Srednicki derives the mode expansion by inversion and obtains the covariant mode CAR in Srednicki 2007, § 39, pp. 246–248. His mostly-plus slash and positive-frequency phase must be translated together into the site’s convention; the invariant measure and $(2\pi)^3 2E_{\mathbf p}$ CAR factor are unchanged. Schwartz instead places $1/\sqrt{2E_{\mathbf p}}$ in the expansion and uses noncovariant mode CAR in Schwartz 2014, § 12.3, pp. 211–212; rescaling its operators by $\sqrt{2E_{\mathbf p}}$ gives the package above.

Rescaling the spinors, measure, or mode operators is possible, but all three must be translated together. The package above gives covariantly normalized one-particle momentum states and makes the field CAR a direct completeness test.

Equal-time CAR from completeness

Set $\mathbf r=\mathbf x-\mathbf y$. The mode algebra gives

$$\begin{aligned} &\{\widehat\psi_\alpha(t,\mathbf x), \widehat\psi_\beta^\dagger(t,\mathbf y)\}\\ &\quad= \int d\Pi_p \sum_r \left[ u_{r,\alpha}(p)u_{r,\beta}^\dagger(p) e^{+i\mathbf p\cdot\mathbf r} + v_{r,\alpha}(p)v_{r,\beta}^\dagger(p) e^{-i\mathbf p\cdot\mathbf r} \right]. \end{aligned}$$

Change $\mathbf p\mapsto-\mathbf p$ in the second term and define $\widetilde p=(E_{\mathbf p},-\mathbf p)$. The preceding spin sums imply

$$\begin{aligned} &\sum_r \left[ u_r(p)u_r^\dagger(p) + v_r(\widetilde p)v_r^\dagger(\widetilde p) \right]\\ &\quad= \left[ (p\!\!\!/+m) + (\widetilde p\!\!\!/-m) \right]\gamma^0\\ &\quad= 2E_{\mathbf p}\,\mathbf1_4. \end{aligned}$$

The $2E_{\mathbf p}$ cancels the denominator in $d\Pi_p$, leaving

$$\{\widehat\psi_\alpha(t,\mathbf x), \widehat\psi_\beta^\dagger(t,\mathbf y)\} = \delta_{\alpha\beta} \int\frac{d^3\mathbf p}{(2\pi)^3} e^{+i\mathbf p\cdot\mathbf r} = \delta_{\alpha\beta}\delta^{(3)}(\mathbf r).$$

The remaining equal-time field anticommutators vanish because the $b$–$d$ and like-creation or like-annihilation mode anticommutators vanish. This reverse derivation is an independent normalization check: a missing $2E_{\mathbf p}$, a doubled momentum sector, or an inconsistent $v$ label would prevent the spatial delta distribution from appearing.

The selected fermionic Fock representation

Choose a normalized vacuum vector $\lvert\Omega\rangle$ satisfying

$$b_r(\mathbf p)\lvert\Omega\rangle = d_r(\mathbf p)\lvert\Omega\rangle =0$$

in the distributional sense. For square-integrable wave packets, define

$$\begin{aligned} b^\dagger(f) &= \sum_r\int d\Pi_p\, f_r(\mathbf p)b_r^\dagger(\mathbf p),\\ d^\dagger(g) &= \sum_r\int d\Pi_p\, g_r(\mathbf p)d_r^\dagger(\mathbf p). \end{aligned}$$

The mode CAR give

$$\begin{aligned} \|b^\dagger(f)\Omega\|^2 &= \sum_r\int d\Pi_p\,|f_r(\mathbf p)|^2,\\ \|d^\dagger(g)\Omega\|^2 &= \sum_r\int d\Pi_p\,|g_r(\mathbf p)|^2. \end{aligned}$$

Both particle and antiparticle sectors therefore have positive Hilbert norm. The negative Lorentz scalar $\bar vv=-2m$ from the classical spinors is unrelated to this norm.

The representation is the antisymmetric Fock space over the particle and antiparticle one-particle spaces,

$$\mathcal F_D = \mathcal F_-\!\left( \mathcal H_{\mathrm p}\oplus\mathcal H_{\bar{\mathrm p}} \right) = \bigoplus_{n=0}^{\infty} \bigwedge\nolimits^n \left( \mathcal H_{\mathrm p}\oplus\mathcal H_{\bar{\mathrm p}} \right).$$

For a normalized discrete mode $a$, the CAR imply $N_a=a^\dagger a$ and

$$N_a^2=N_a, \qquad (a^\dagger)^2=0.$$

Its occupation is therefore $0$ or $1$. Products of distinct creation operators change sign under exchange, realizing the antisymmetric multiparticle sectors rather than imposing a separate sign rule afterward. The vacuum condition and positive-frequency split are additional state and representation data; they do not follow from the abstract CAR algebra alone.

Hamiltonian and the antiparticle sign

To manipulate coincident mode products honestly, place the system in a periodic box of volume $V$ and retain a finite momentum set $\Lambda$. Use discrete operators with

$$\{b_{\mathbf k r},b_{\mathbf l s}^\dagger\} = \{d_{\mathbf k r},d_{\mathbf l s}^\dagger\} = \delta_{\mathbf k\mathbf l}\delta_{rs}.$$

The one-particle Dirac Hamiltonian gives energy $+E_{\mathbf k}$ on the $u$ mode and $-E_{\mathbf k}$ on the negative-frequency $v$ wave. Spatial integration removes cross terms by the mixed $u$–$v$ orthogonality. The regulated canonical Hamiltonian is therefore

$$\begin{aligned} H_\Lambda &= \sum_{\mathbf k\in\Lambda,r} E_{\mathbf k} \left( b_{\mathbf k r}^\dagger b_{\mathbf k r} - d_{\mathbf k r}d_{\mathbf k r}^\dagger \right)\\ &= \sum_{\mathbf k\in\Lambda,r} E_{\mathbf k} \left( N^b_{\mathbf k r} + N^d_{\mathbf k r} -1 \right). \end{aligned}$$

The second line is exactly where anticommutation changes the conclusion: $dd^\dagger=1-d^\dagger d$. The regulated vacuum energy is

$$E_{\mathrm{vac},\Lambda} = -\sum_{\mathbf k\in\Lambda,r}E_{\mathbf k}.$$

Vacuum-relative normal ordering defines

$$H_{\mathrm{exc}} \equiv :H_\Lambda: = \sum_{\mathbf k\in\Lambda,r} E_{\mathbf k} \left( N^b_{\mathbf k r}+N^d_{\mathbf k r} \right) \geq0.$$

Every finite-particle excitation has energy equal to a sum of positive $E_{\mathbf k}$ values, and

$$\begin{aligned} [H_{\mathrm{exc}},b_{\mathbf k r}^\dagger] &=E_{\mathbf k}b_{\mathbf k r}^\dagger,\\ [H_{\mathrm{exc}},d_{\mathbf k r}^\dagger] &=E_{\mathbf k}d_{\mathbf k r}^\dagger. \end{aligned}$$

Normal ordering is a prescription relative to the selected free vacuum. It sets the reference energy of this nongravitational free model; it does not prove that absolute vacuum energy vanishes, nor does it renormalize arbitrary interacting composite operators.

The negative-frequency mode also gives a sharp one-oscillator test of the statistics choice. Its unreordered contribution is $H_-=-E\,dd^\dagger$. With CAR, it becomes $E\,d^\dagger d-E$, so the excitation gap is positive and a mode cannot be occupied twice. If one instead imposed $[d,d^\dagger]=+1$, then

$$H_-=-E(d^\dagger d+1),$$

which is unbounded below as bosonic occupation increases. Reversing the commutator sign would repair the energy gap only by giving

$$\|d^\dagger\Omega\|^2 = \langle\Omega|dd^\dagger|\Omega\rangle =-1.$$

Thus this free-mode construction cannot have positive norm and energy with commutators. This diagnostic is not the relativistic spin–statistics theorem, whose proof requires locality, Lorentz covariance, and further spectral assumptions. The regulated Hamiltonian, its CAR reordering, and the wrong-statistics instability provide an independent check in Schwartz 2014, § 12.5.2, pp. 217–218; the covariantly normalized derivation is also given in Srednicki 2007, § 39, pp. 248–249. A structural derivation from the canonical conjugate through the equal-time and mode CAR to the reordered positive Hamiltonian appears in Weinberg 1995, § 7.5, pp. 323–325, after translation to the site’s index and normalization conventions.

Charge separates particles from antiparticles

The inherited global phase convention is $\widehat\psi\mapsto e^{-i\alpha}\widehat\psi$, with current $j^\mu=\widehat{\bar\psi}\gamma^\mu\widehat\psi$. At finite cutoff, the raw charge is

$$\begin{aligned} Q_{\mathrm{raw},\Lambda} &= \int_V d^3\mathbf x\, \widehat\psi^\dagger\widehat\psi\\ &= \sum_{\mathbf k\in\Lambda,r} \left( b_{\mathbf k r}^\dagger b_{\mathbf k r} + d_{\mathbf k r}d_{\mathbf k r}^\dagger \right)\\ &= \sum_{\mathbf k\in\Lambda,r} \left( N^b_{\mathbf k r} - N^d_{\mathbf k r} +1 \right). \end{aligned}$$

Choose the selected vacuum to be neutral. Fermionic normal ordering then gives

$$Q = :Q_{\mathrm{raw},\Lambda}: = \sum_{\mathbf k\in\Lambda,r} \left( N^b_{\mathbf k r}-N^d_{\mathbf k r} \right).$$

The commutators

$$\begin{aligned} [Q,\widehat\psi]&=-\widehat\psi,\\ [Q,b_{\mathbf k r}^\dagger]&=+b_{\mathbf k r}^\dagger,\\ [Q,d_{\mathbf k r}^\dagger]&=-d_{\mathbf k r}^\dagger \end{aligned}$$

show that $b^\dagger$ creates a unit-charge particle and $d^\dagger$ creates a unit-opposite-charge antiparticle. With $U(\alpha)=e^{i\alpha Q}$, the first relation reproduces $U\widehat\psi U^{-1}=e^{-i\alpha}\widehat\psi$. The particle–antiparticle interpretation is therefore fixed simultaneously by positive energy and the conserved charge, not by the frequency exponent alone. The same phase convention and particle-minus-antiparticle charge appear in Schwartz 2014, § 10.4, p. 174 and Srednicki 2007, § 39, pp. 249–250.

Dynamics, locality, and the massless limit

The Hamiltonian commutators reproduce the time dependence in the field expansion, while the mode CAR give a compact spacetime consistency check. Define the scalar Pauli–Jordan distribution

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

The spin sums yield

$$\{\widehat\psi_\alpha(x), \widehat{\bar\psi}_\beta(y)\} = \left[ \left( i\gamma^\mu\partial_{x^\mu}+m \right) D(x-y) \right]_{\alpha\beta}.$$

Because $D$ has causal support, this graded field anticommutator vanishes at spacelike separation. At equal time, $D(0,\mathbf r)=0$ and $\partial_0D(0,\mathbf r)=-i\delta^{(3)}(\mathbf r)$, recovering the canonical $\{\widehat\psi,\widehat\psi^\dagger\}$ relation after multiplying on the right by $\gamma^0$. The spacelike CAR check is also carried out in Schwartz 2014, § 12.6, p. 222. It is a microcausality check for the free field, not a proof that half-integer spin must obey CAR.

The massless theory uses the same operator CAR and Fock construction for nonzero momentum, but the spin basis must be rebuilt in helicity variables. One keeps

$$\begin{aligned} u_h^\dagger(p)u_{h'}(p) &= v_h^\dagger(p)v_{h'}(p) =2E_{\mathbf p}\delta_{hh'},\\ \sum_h u_h(p)\bar u_h(p) &= \sum_h v_h(p)\bar v_h(p) =p\!\!\!/. \end{aligned}$$

with the frequency sector still specified by the exponential. Rest-frame spin projectors and the zero-momentum point are not obtained by substituting $m=0$ into the massive formulas. Weyl Fields and Chirality develops the chiral decomposition.

Finally, this representation uses the time-translation symmetry of flat spacetime to choose positive frequency and a vacuum. The CAR algebra alone does not provide a preferred Fock state on a general time-dependent background, nor does this free construction establish uniqueness among all infinite-system representations. Those are limitations of the claim, not failures of the flat-space calculation.

Common pitfalls

Treating the first-order action as unconstrained. The momenta are constraints and the reduced graded bracket supplies the field CAR. An ordinary nonsingular Legendre transform skips the canonical issue that must be resolved.

Changing one normalization in isolation. The invariant measure, spinor norms, and covariant mode CAR are one package. A mismatch leaves an unwanted factor in the equal-time delta distribution.

Calling negative frequency negative state energy. The $v$ wavefunction multiplies $d^\dagger$. After CAR reordering, that operator creates a positive-energy state with charge opposite to the particle.

Using $\bar vv$ as a Fock norm. The Lorentz scalar is negative under the chosen massive normalization, whereas $\|d^\dagger(g)\Omega\|^2$ is positive. They are different pairings on different objects.

Dropping the regulator before reordering. At coincident continuum momenta, the constant is proportional to $\delta^{(3)}(\mathbf0)$. Derive it in a finite box or after smearing, declare the subtraction, and only then return to continuum notation.

Calling normal ordering a general renormalization theorem. It subtracts the selected free-vacuum expectation value. Interacting composite operators require a local renormalization prescription and can mix under scale changes.

Claiming that CAR alone select the vacuum. The algebra, representation, and state are distinct. Positive frequency and $b\Omega=d\Omega=0$ are additional Minkowski-vacuum data.

Check your understanding

Check 1: recover the field delta distribution

Starting from the two spin sums, prove

$$\sum_r \left[ u_r(p)u_r^\dagger(p) + v_r(\widetilde p)v_r^\dagger(\widetilde p) \right] = 2E_{\mathbf p}\mathbf1_4.$$

Solution

Use $\bar u=u^\dagger\gamma^0$ and $\bar v=v^\dagger\gamma^0$:

$$\begin{aligned} \sum_r u_r(p)u_r^\dagger(p) &=(p\!\!\!/+m)\gamma^0,\\ \sum_r v_r(\widetilde p)v_r^\dagger(\widetilde p) &=(\widetilde p\!\!\!/-m)\gamma^0. \end{aligned}$$

The mass terms cancel. The spatial components of $p+\widetilde p$ cancel, while the time components add, so

$$(p\!\!\!/+\widetilde p\!\!\!/)\gamma^0 = 2E_{\mathbf p}\gamma^0\gamma^0 = 2E_{\mathbf p}\mathbf1_4.$$

Insertion into the mode anticommutator cancels the invariant-measure denominator and leaves the Fourier representation of $\delta^{(3)}(\mathbf x-\mathbf y)$.

Check 2: compare CAR and commutators for one antiparticle mode

For $H_-=-Edd^\dagger$, compute the excitation gap and one-particle norm using (a) $\{d,d^\dagger\}=1$, (b) $[d,d^\dagger]=+1$, and (c) $[d,d^\dagger]=-1$.

Solution

For CAR,

$$H_-=E\,d^\dagger d-E.$$

The state $d^\dagger\Omega$ has norm $1$, lies $E$ above the vacuum, and a second application of $d^\dagger$ vanishes. With the positive commutator,

$$H_-=-E\,d^\dagger d-E,$$

so arbitrarily high occupation drives the energy downward. With the negative commutator, the energy gap changes sign as desired, but

$$\|d^\dagger\Omega\|^2 = \langle\Omega|dd^\dagger|\Omega\rangle =-1.$$

Only CAR pass both tests in this free-mode construction.

Check 3: verify the charge assignment

Use $Q=\sum(N^b-N^d)$ to compute its commutator with $b^\dagger$, $d^\dagger$, $b$, and $d^\dagger v e^{+ip\cdot x}$.

Solution

For a fermionic number operator,

$$[N^b,b^\dagger]=b^\dagger, \qquad [N^d,d^\dagger]=d^\dagger.$$

Therefore

$$[Q,b^\dagger]=b^\dagger, \qquad [Q,d^\dagger]=-d^\dagger, \qquad [Q,b]=-b.$$

Both terms in $\widehat\psi$ have charge $-1$ under the adjoint action: $[Q,bu e^{-ipx}]=-bu e^{-ipx}$ and $[Q,d^\dagger v e^{+ipx}]=-d^\dagger v e^{+ipx}$. Hence $[Q,\widehat\psi]=-\widehat\psi$, as required by the declared phase convention.

Where the construction continues

• The Fermion Propagator derives time ordering, the equal-time contact term, and the Feynman pole prescription from this operator field.
• Grassmann Functional Integrals for Free Fermions gives the regulated Berezin-Gaussian representation and declares every source-order sign.
• Weyl Fields and Chirality quantizes the massless chiral sectors with helicity-adapted modes.
• Majorana Fields and Reality Conditions removes the independent particle–antiparticle charge distinction for a self-conjugate field.
• The Spin–Statistics Connection develops the theorem-level relation among spin, locality, positivity, and statistics that this free-mode diagnostic does not prove.

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.