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For a complex Klein-Gordon field, $j^{\mu }=i({\varphi }^{\ast }{\partial }^{\mu }\varphi -\varphi {\partial }^{\mu }{\varphi }^{\ast })$ is conserved. A mode $\varphi =N{e}^{-iEt+i\mathbf{p}\cdot \mathbf{x}}$ has $j^0=2E|N{|}^2$, so unrestricted positive- and negative-frequency solutions do not have a positive probability norm under this current. In the quantized complex field, the associated conserved quantity is charge: particles and antiparticles contribute with opposite signs while both have positive excitation energy and positive Fock space norm.

In the scalar transformation $\hat{T}a(p){\hat{T}}^{-1}={\eta }_{T}a(p_P)$, preservation of canonical commutators or the involution condition gives $|{\eta }_T|=1$. Antiunitarity conjugates the phase on a second application, so the condition does not restrict it to plus or minus one. Redefine $a^{\prime }=e^{i\chi }a$; then $\hat{T}{a}^{\prime }{\hat{T}}^{-1}=e^{-2i\chi }{\eta }_T{a}^{\prime }(p_P)$. Choosing $\chi =\arg ({\eta }_T)/2$ makes the intrinsic phase one. Oppositely rephasing the antiparticle annihilator implements the same change as a phase convention for the whole scalar field.

In a sufficiently strong electrostatic step, the positive-energy incident channel overlaps a negative-kinetic-energy channel on the far side. For a charged scalar field the outgoing antiparticle branch has negative charge current, allowing reflected particle current larger than the incident current while signed charge is conserved. The external source supplies energy for particle-antiparticle creation. This requires a multiparticle interpretation; negative transmitted charge current is not a negative probability.

A lepton is a Standard Model fermion without color charge. The charged leptons have electric charge $-1$ and their corresponding neutrinos have charge zero. There are three generations; their antiparticles carry opposite electric charge and lepton number. Charged leptons and neutrinos have lepton number $+1$, with the corresponding antiparticles carrying $-1$.

A Majorana neutrino is its own antiparticle. A corresponding mass term can violate conventional lepton number by two. Observing a wrong-helicity charged-current process is nevertheless suppressed by the very small neutrino mass relative to its energy, rather than having the ordinary charge-conjugate branching fraction.

The muon is the negatively charged second-generation lepton, with electric charge $-1$ and lepton number $+1$. Its antiparticle is the antimuon.

Transitions between a neutral meson carrying flavour and its antiparticle. Decays are described on this two-state subspace by an effective Hamiltonian operator $M-i\Gamma /2$ with Hermitian matrices $M,\Gamma$.

This first-order-in-time field theory has an annihilation-field expansion $\Psi =\int a_{\mathbf{p}}{e}^{-i{\mathbf{p}}^{2}t/(2m)+i\mathbf{p}\cdot \mathbf{x}}{d}^{3}p/(2\pi {)}^3$. Its adjoint creates particles; an antiparticle creation term is not required by a relativistic mass shell or Lorentz-invariant microcausality. It is often an effective low-energy sector of a relativistic charged theory after the antiparticle sector is omitted. This does not mean that antimatter cannot be treated nonrelativistically: one may introduce a separate nonrelativistic field for that species.

Spatial integration of the free Hamiltonian density cancels particle-antiparticle pair terms through $E_p^2-{\mathbf{p}}^2-m^2=0$. The raw Hamiltonian is $\int d{\Pi }_p{E}_p(a^{†}a+b{b}^{†})$. Normal ordering removes the field-independent vacuum energy and gives $H=\int d{\Pi }_p{E}_p(a^{†}a+b^{†}b)$. Both creation operators increase the energy by $E_p$, so a negative-frequency field mode represents a positive-energy antiparticle, not a negative-energy state.

Pair production creates a particle and its antiparticle while conserving energy, momentum and the relevant charges. The available invariant mass must reach the sum of their rest masses.

For a pair of distinct bosonic particle and antiparticle modes, the normalized state obeying $(a+q{b}^{†})|\psi ⟩=(b+q{a}^{†})|\psi ⟩=0$ is $\sqrt{1-|q{|}^2}{\sum }_{n\ge 0}(-q{)}^n|n,n⟩$, for $|q|<1$. It contains equal occupations and is generated by a pair-creation exponential. Infinite products require a separate implementability or regulator check.

Assume $\lambda >0$ so that the Higgs potential is bounded below, and work at tree level in four spacetime dimensions with metric $(+---)$. For ${\mu }^2>0$, the minimum is $\Phi =0$. The quadratic terms describe three massless gauge bosons and two complex scalars of mass $\sqrt{{\mu }^2}$, equivalently four real scalar degrees of freedom. These are the perturbative particle masses about this vacuum, without a claim about the nonperturbative asymptotic spectrum of an interacting non-Abelian theory.

Expanding the gauge covariant derivative with $T_j={\tau }_j/2$ gives the interaction termsIn the second term, the antisymmetric part of ${\tau }_j{\tau }_k$ vanishes against the symmetric gauge-field product. The corresponding Feynman diagram vertices have one vector and a scalar–antiscalar pair, two vectors and a scalar–antiscalar pair, or four scalar legs with two particles and two antiparticles. The first vertex carries a momentum difference from its derivative. The Pauli matrix in it determines which doublet components couple; the two-vector scalar vertex is proportional to ${\delta }_{jk}$. There is no cubic scalar interaction about the zero-field vacuum.

The gauge fields require their own Yang-Mills Lagrangian, $-W_{\mu \nu }^a{W}^{a\mu \nu }/4$. The non-Abelian field strength contains a term quadratic in the gauge potentials, so squaring it produces three-vector and four-vector self-interaction vertices, with derivative momentum factors in the cubic vertex and SU(2) structure constants in both. They exist in either scalar phase and must be added to the scalar-generated vertices. A gauge-fixed quantum calculation also has gauge-fixing and, where applicable, ghost terms; those are not extra physical boson species.

For ${\mu }^2>0$, the global phase transformation $\Phi \mapsto e^{i\alpha }\Phi$ leaves both the scalar potential and the covariant kinetic term unchanged while leaving $W_j$ fixed. Its Noether charge counts scalar particles minus antiparticles. Thus ${\varphi }_a{\varphi }_b\to {\varphi }_a{\varphi }_b$ and ${\varphi }_a{\varphi }_a^{\ast }\to W_3{W}_3$ are compatible with this charge, whereas ${\varphi }_a{\varphi }_b\to W_j{W}_k$ is forbidden: its initial charge is two and final charge zero. Charge conservation is necessary, not sufficient, for a particular amplitude to be nonzero; other quantum numbers and kinematics still apply.

After adding the hypercharge field, the neutral-vector mass term becomesIts mass matrix has zero determinant. With $G=\sqrt{g^2+g^{\prime 2}}$, the orthonormal combinations areThe unbroken generator is electric charge, $Q=T_3+Y/2$, because the lower doublet component has $T_3=-1/2$ and $Y=1$. The charged combinations $W^{\pm }=(W_1\mp i{W}_2)/\sqrt{2}$ keep mass $gv/2$.

The observed charged weak interaction couples left-chiral fermions within SU(2) doublets. In the Standard Model, $Q_L=(u_L,d_L)$ and $L_L=({\nu }_L,e_L)$ are doublets, while $u_R,d_R,e_R$ are singlets. A vectorlike assignment would also give right-handed charged weak currents, contrary to this observed chiral structure. Once the measured electric charges are specified, $Y=2(Q-T_3)$ fixesAn optional right-handed neutrino is a singlet with $Y=0$. Conjugate left-chiral fields have the opposite hypercharges to these right-chiral fields. These assignments permit gauge-invariant Yukawa interactions; they are not obtained by assigning one common hypercharge to every chirality.

A relativistic quantum field describes both creation and annihilation of excitations; a one-particle wavefunction alone is insufficient when interactions can change particle number. For a free complex scalar field, one conventional mode expansion of a free field isThe apparently negative-frequency term creates a positive-energy antiparticle; it is not a physical particle of unbounded negative energy. The normal-ordered Hamiltonian and conserved global phase charge have the formsParticles and antiparticles have equal mass and opposite additive internal charges. Including both frequency sectors is also essential to the local field commutator: the positive- and negative-frequency terms combine so that spacelike-separated scalar observables commute, respecting microcausality.

A Dirac field similarly contains particle annihilation operators multiplying $u$ spinors and antiparticle creation operators multiplying $v$ spinors. Its canonical anticommutation relations implement Fermi-Dirac statistics, whereas scalar modes use commutators and Bose-Einstein statistics. These assignments are tied to the Spin-statistics theorem under the usual relativistic locality and positivity assumptions. Conserved charge constrains particle creation and annihilation, but particle and antiparticle numbers need not be separately conserved in an interacting theory.

An antiparticle need not be a distinct species. A neutral real scalar field is self-conjugate; the photon is also its own antiparticle. A Majorana fermion obeys a self-conjugacy condition and cannot carry an unbroken additive charge that distinguishes it from its antiparticle. Electric neutrality alone does not force self-conjugacy when other conserved quantum numbers distinguish the states.

Use natural units and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. Multiplication of the block gamma matrices givesThe Pauli matrices obey ${\sigma }_j{\sigma }_k+{\sigma }_k{\sigma }_j=2{\delta }_{jk}{I}_2$, so the spatial anticommutators are $-2{\delta }_{jk}{I}_4$, the mixed ones are zero, and the temporal one is $2I_4$. ThusBecause partial derivatives commute, only the symmetric part of ${\gamma }^{\mu }{\gamma }^{\nu }$ survives in their contraction. Multiplying the Dirac equation on the left by its opposite-mass factor therefore gives the flat-space Dirac factorizationHence every component satisfies $\boxed{(\square +m^2)\psi =0}$, the Klein-Gordon equation. This implication does not remove the first-order spinor constraints imposed by the Dirac equation.

For Lorentz covariance, transform coordinates by $x^{\prime }=Lx$ and the Dirac spinor by ${\psi }^{\prime }(x^{\prime })=S(L)\psi (x)$. The chain rule gives ${\partial }_{\mu }^{\prime }=(L^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$. ConsequentlyThe cancellation uses $L^{\mu }{}_{\rho }(L^{-1}{)}^{\nu }{}_{\mu }={\delta }^{\nu }{}_{\rho }$. Thus the Lorentz transformation maps every solution to another, establishing the Lorentz covariance of the Dirac operator.

The infinitesimal identity follows directly from the Clifford algebra. Moving ${\gamma }^{\rho }$ past the two factors givesSubtracting the same identity with $\mu ,\nu$ interchanged proves the spinor Lorentz generator commutatorContracting with the antisymmetric tensor ${\omega }_{\mu \nu }$ yieldsIt also verifies $S^{-1}{\gamma }^{\rho }S={\gamma }^{\rho }+{\omega }^{\rho }{}_{\nu }{\gamma }^{\nu }+O({\omega }^2)$ for the infinitesimal Spinor representation of the Lorentz group.

For a positive-energy four-momentum, put $\psi =u(p)e^{-ip\cdot x}$, $E=\sqrt{{\mathbf{p}}^2+m^2}>0$. The Dirac equation becomes $(/p-m)u=0$, where the Feynman slash notation uses $/p={\gamma }^{\mu }{p}_{\mu }=E{\gamma }^0-\mathbf{\gamma }\cdot \mathbf{p}$. Splitting $u=(\xi ,\eta {)}^T$ givesThe second equation sets $\eta =(\mathbf{\sigma }\cdot \mathbf{p})\xi /(E+m)$, and the first is then the mass shell, since $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2={\mathbf{p}}^2{I}_2$. Choosing two orthonormal two-spinors ${\chi }_s$ gives the Dirac plane waves in the standard representation:The negative-frequency solutions are independentlywith orthonormal ${\eta }_s$. They are the antiparticle modes after field quantization. The normalization gives $u_r^{†}{u}_s=v_r^{†}{v}_s=2E{\delta }_{rs}$ and ${\overline{u}}_r{u}_s=2m{\delta }_{rs}$, ${\overline{v}}_r{v}_s=-2m{\delta }_{rs}$, using the Dirac adjoint $\overline{u}=u^{†}{\gamma }^0$.

For a spatial rotation, ${\sigma }^{ij}={ϵ}_{ijk}\mathrm{diag}({\sigma }_k,{\sigma }_k)$. Hence the intrinsic rotation generators areFor a massive particle in its rest frame, the two independent spinors therefore carry the two-dimensional spin one-half representation, with $J_3$ eigenvalues $\pm 1/2$ and Casimir $s(s+1)=3/4$. A $2\pi$ rotation acts as $-I$, the spinor sign under a full spatial rotation. Thus the particle has spin $\boxed{s=1/2}$; four spinor components encode particle and antiparticle sectors, not four spin states of one particle. In the massless limit the corresponding physical labels are helicities $\pm 1/2$ rather than a rest-frame spin basis.

Use units $\hslash =c=1$ and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. The canonical momentum of the real scalar field is $\pi =\partial \mathcal{L}/\partial \dot{\varphi }=\dot{\varphi }$. Canonical quantization replaces the classical fields by operator-valued distributions and imposes the equal-time canonical commutation relationsThese relations are imposed on an initial slice, with smeared fields understood. In the Heisenberg picture, simultaneous unitary time evolution of both operators preserves their commutators. Since the momentum is $\dot{\varphi }$, the requested equal-time commutator is $[\varphi (t,\mathbf{x}),\dot{\varphi }(t,\mathbf{y})]=i{\delta }^{(3)}(\mathbf{x}-\mathbf{y})$.

The Legendre transform in mechanics, applied pointwise, gives the canonical Hamiltonian density of a real scalar field. ThusTo obtain its dynamics rather than assume them, use the Heisenberg equation of motion $\dot{O}=i[H,O]$. The momentum-square term gives $[H,\varphi ]=-i\pi$. The field-square term gives $[H,\pi ]=i{m}^2\varphi$, while differentiating the Dirac delta function in the gradient term and integrating by parts gives $[H,\pi ]=-i{\nabla }^2\varphi +i{m}^2\varphi$. ConsequentlyBoundary terms vanish for suitably decaying smeared fields. This is the Klein-Gordon equation as an operator identity.

A spatial Fourier transform reduces the Klein-Gordon equation to ${\ddot{\varphi }}_{\mathbf{p}}+E_p^2{\varphi }_{\mathbf{p}}=0$, where $E_p=\sqrt{{\mathbf{p}}^2+m^2}$. Its two frequency branches and the Hermitian adjoint condition giveThe invariant-normalized scalar mode extraction can be written particularly simply asSubstituting the mode expansion of a free field proves these inversion formulas: the opposite-frequency coefficient is proportional to $E_p-E_q$ and vanishes after spatial Fourier inversion. Alternatively they are the corresponding Klein-Gordon inner products. Taking their commutators using the equal-time fields gives $(E_p+E_q)(2\pi {)}^3{\delta }^{(3)}(\mathbf{p}-\mathbf{q})$ in the mixed case. In the two equal-type cases the delta function enforces $\mathbf{q}=-\mathbf{p}$, multiplying $E_p-E_q=0$. ThereforeConversely these mode commutators reproduce $[\varphi ,\dot{\varphi }]=i{\delta }^{(3)}$: the two frequency contributions each supply half the spatial Dirac delta function.

For the covariant oscillator Hamiltonian of a real scalar field, insert the mode expansion of a free field into $H$. Spatial integration sets the two momenta equal in the mixed terms and opposite in the two-creator or two-annihilator terms. The latter have coefficient $-E_p^2+{\mathbf{p}}^2+m^2=0$. The mixed terms giveThe constant $E_0$ is the divergent zero-point energy; a finite box and a momentum cutoff make its meaning explicit. Normal ordering sets the free vacuum energy to zero and leaves all commutators unchanged. Directly applying the mode commutators to the mixed expression yieldsChoose the Fock vacuum by $a(p)|0⟩=0$. Each creation operator adds a boson of energy $E_p$ and four-momentum $p$, while each annihilation operator removes one. Products of commuting creation operators give bosonic statistics from commuting creation operators and unrestricted occupation numbers. The real scalar field has one spin-zero species, identical to its antiparticle. Its one-particle states obey $⟨q|p⟩=(2\pi {)}^{3}2E_p{\delta }^{(3)}(\mathbf{p}-\mathbf{q})$, precisely the relativistic normalization of a one-particle state.

For the Feynman propagator, put $z=x-y$ and contract the two fields in the Fock vacuum. Only $a{a}^{†}$ survives, giving the Wightman functionThe stipulated time-ordered product convention is $i{\Delta }_F(z)=\theta (z^0)W(z)+\theta (-z^0)W(-z)$. In a contour integral over $p^0$, the Feynman i-epsilon prescription places the positive-energy pole just below the real axis and the negative-energy pole just above it. For $z^0>0$ close below, clockwise, obtaining $-i{e}^{-i{E}_p{z}^0}/(2E_p)$; for $z^0<0$ close above, obtaining $-i{e}^{i{E}_p{z}^0}/(2E_p)$. Changing $\mathbf{p}\mapsto -\mathbf{p}$ in the second contribution therefore provesActing with $\square +m^2$ multiplies the integrand by $-(p^2-m^2)$. The remaining factor tends to $-1$ as a distribution, so Fourier inversion givesThe sign also follows from the derivative jump of a free scalar time-ordered two-point function: the first time derivative of $i{\Delta }_F$ jumps by $⟨[\dot{\varphi },\varphi ]⟩=-i{\delta }^{(3)}$, so that of ${\Delta }_F$ jumps by $-{\delta }^{(3)}$.

For the complex scalar field, the canonical momenta are $\pi ={\dot{\varphi }}^{†}$ and ${\pi }^{†}=\dot{\varphi }$. Split $H=H_0+H_I$. The interaction picture is obtained from the Schrödinger picture by $|{\Psi }_I(t)⟩=e^{i{H}_{0}t}|{\Psi }_S(t)⟩$ and $O_I(t)=e^{i{H}_{0}t}{O}_S{e}^{-i{H}_{0}t}$, choosing a common time origin. The fields therefore obey the free Klein-Gordon equation, while the states evolve through $H_I(t)$. The canonical quantization of a complex scalar field givesThe adjoint expansion contains $a^{†}$ and $b$. Particle and antiparticle modes are independent; their nonzero commutators areAll cross-species commutators, and all pairs of creation operators or of annihilation operators, vanish. These formulas reproduce $[\varphi (\mathbf{x}),\pi (\mathbf{y})]=[{\varphi }^{†}(\mathbf{x}),{\pi }^{†}(\mathbf{y})]=i{\delta }^{(3)}(\mathbf{x}-\mathbf{y})$ and the remaining vanishing equal-time commutators.

There are no derivative interactions, so the interaction Hamiltonian is $H_I(t)=-\int d^{3}x{\mathcal{L}}_I(x)$. Differentiating the interaction picture state gives $i{\partial }_t|{\Psi }_I⟩=H_I(t)|{\Psi }_I⟩$. Its time-evolution operator satisfiesIteration gives ordered integration regions $t_1>\cdots >t_n$. Each is $1/n!$ of the full region with a time-ordered product, yielding the Dyson seriesWith the usual asymptotic scattering prescription, taking $t_0\to -\infty$ and $t\to +\infty$ provesTime ordering is essential because different interaction Hamiltonians need not commute.

At first order, use the connected tree scattering amplitude, equivalently the normal-ordered quartic vertex with the vacuum and one-particle mass contributions removed. This qualification matters if the local product is interpreted literally without normal ordering: single-vertex tadpole diagrams and vacuum diagrams are not the requested connected two-particle amplitude. The complex scalar quartic contact vertex is obtained fromIn a fully connected matrix element, an incoming particle and outgoing antiparticle attach to the two $\varphi$ factors; an outgoing particle and incoming antiparticle attach to the two ${\varphi }^{†}$ factors. Each pair has two assignments, giving $2!2!=4$. The external relativistic normalization of a one-particle state cancels each mode measure: for example $[\varphi (x),a^{†}(p_1)]=e^{-i{p}_{1}x}$ and $[b(q_2),\varphi (x)]=e^{i{q}_{2}x}$. Consequently the connected integrand is $4e^{-i(p_1+q_1-p_2-q_2)\cdot x}$, andThus the invariant connected amplitude isThe Dirac delta function expresses four-momentum conservation.

For the identity contribution, commute $a(p_2)$ through $a^{†}(p_1)$ and independently $b(q_2)$ through $b^{†}(q_1)$; all terms leaving a creation operator against the left Fock vacuum or an annihilation operator against the right Fock vacuum vanish. HenceThere is no exchange term between the two distinct particle and antiparticle species.

The baryon asymmetry is an excess of net baryon number over zero, conveniently measured by ${\eta }_B=(n_B-n_{\bar{B}})/n_{\gamma }$. Ordinary matter contains abundant baryons, whereas no comparable primordial antibaryon population is observed. Large nearby domains of matter and antimatter would produce radiation through particle-antiparticle annihilation at their boundaries. Antiparticles produced in energetic collisions are not evidence for a cosmological antimatter domain. The small nonzero baryon-to-photon ratio inferred from Big Bang nucleosynthesis is independently tested by baryon-density consistency between nucleosynthesis and the microwave background.

Starting with zero net baryon number, the Sakharov conditions require baryon-number violation, C and CP violation, and departure from thermal equilibrium. Conservation of baryon number would keep the initial value zero. C symmetry or CP symmetry would pair baryon production with equal antibaryon production; CP rate cancellation in baryogenesis explains why C violation alone does not suffice. In thermal equilibrium with zero imposed charge chemical potentials, CPT symmetry forbids the required net baryon number. An asymmetric rate in one reaction must not be mistaken for net production after all reverse reactions are included.

A simple example is heavy-boson decay baryogenesis. Take equally many heavy $X$ bosons and their antiparticles, with decay channelsThe symbols in parentheses are branching fractions. The two $X$ channels carry baryon numbers $2/3$ and $-1/3$, so both cannot conserve one assigned baryon number for $X$. Their mean baryon number is $r-1/3$, while the mean for $\bar{X}$ is $-\bar{r}+1/3$. Consequently,where $ϵ$ is the asymmetry per decaying particle. CP violation permits $r\ne \bar{r}$ without violating the equality of total decay widths required by CPT symmetry. Microscopically a direct CP asymmetry requires interference between decay amplitudes with both relative weak phases and absorptive phases; specifying complex couplings alone would not establish unequal partial decay widths.

For radiation domination, a sufficient decay condition is ${\Gamma }_X\lesssim H(T=M_X)=1.66\sqrt{g_{\ast }}{M}_X^2/m_P$. The heavy population can then survive as the equilibrium abundance drops, before decaying while inverse production is suppressed. Its actual abundance differs from its thermal equilibrium value. If subsequent baryon-number washout is negligible and cosmological entropy conservation holds, an initial total abundance $Y_X=(n_X+n_{\bar{X}})/s$ producesHere $s$ is the entropy density. Residual baryon-number washout multiplies this expression by an efficiency factor below one. The example realizes all three Sakharov conditions; it illustrates a mechanism, rather than identifying the actual cosmological source of the baryon asymmetry.

Use signature $(+,-,-,-)$ and the free Dirac equation $(i{\gamma }^{\mu }{\partial }_{\mu }-m)\psi =0$. For positive $p^0=E_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$, substitution of the two frequency signs givesMultiplying by the opposite factor shows $p^2=m^2$. Each equation has two independent solutions for a massive field.

In the Dirac basis, ${\gamma }^0=\mathrm{diag}(I,-I)$ and ${\gamma }^i=\left(\begin{array}{cc}0& {\sigma }_i\\ -{\sigma }_i& 0\end{array}\right)$. For orthonormal two-component spin labels ${\xi }_s,{\eta }_s$, convenient solutions areThe two-block equations verify these expressions using $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2={\mathbf{p}}^{2}I$. At rest the positive-frequency solutions have only upper components and the negative-frequency solutions only lower components. Their normalizations are $u_s^{†}{u}_r=v_s^{†}{v}_r=2E_{\mathbf{p}}{\delta }_{sr}$ and ${\bar{u}}_s{u}_r=2m{\delta }_{sr}$, ${\bar{v}}_s{v}_r=-2m{\delta }_{sr}$. Their completeness relations areThe field expansion combines fermionic annihilation operators multiplying the positive-frequency waves with antiparticle fermionic creation operators multiplying the negative-frequency waves. The latter frequency sign is not the energy of a physical negative-energy antiparticle state.

With a vacuum annihilated by all $a_{\mathbf{p}}^s,b_{\mathbf{p}}^s$, define the particle and antiparticle occupation operators of a Dirac fieldEach counts its corresponding occupied modes. The total positive particle count isCommuting it through either creation operator gives $[N_{\mathrm{total}},a^{†}]=a^{†}$ and $[N_{\mathrm{total}},b^{†}]=b^{†}$, so both kinds of one-particle states have eigenvalue one. In a normalized discrete mode, $(a^{†}a{)}^2=a^{†}a$ by the anticommutation relations, so its occupation eigenvalues are $0,1$.

For a Dirac field there is also a distinct conserved signed fermion number. Inserting the mode expansion into the normal-ordered local density givesSpatial integration makes the same-sector momenta equal. The cross terms vanish by $u^{†}(\mathbf{p})v(-\mathbf{p})=0$; normal ordering of $b{b}^{†}$ supplies the minus sign for the antiparticles. Thus the density integral is not the total positive particle count. It generates the global phase symmetry, and for electron charge $-e$ the electric charge is $Q_{\mathrm{em}}=-e{Q}_D$.

The free normal-ordered Hamiltonian isso $N_e,N_{\bar{e}},N_{\mathrm{total}}$ and $Q_D$ are conserved in the free theory. In an interacting theory, pair creation can change the total count while preserving the signed charge. This distinguishes a positive occupation number operator from the charge behind Dirac fermion number conservation.