Antimatter

#1. Introduction to Antimatter

Antimatter consists of antiparticles — particles that have the same mass and spin as their corresponding matter particles, but with certain quantum numbers reversed (charge, lepton number, baryon number, magnetic moment direction). When a particle and its antiparticle meet, they annihilate, converting their combined rest-mass energy into photons (or other particle–antiparticle pairs).

Key historical note: The existence of antimatter was first predicted by Paul Dirac in 1928 through his relativistic wave equation, and the positron (antielectron) was experimentally discovered by Carl Anderson in 1932.

Anti Hydrogen modelAnti Hydrogen model

#2. Properties of Antiparticles

For every particle, its antiparticle satisfies:

PropertyParticleAntiparticle
Massmmmm (same)
Electric Chargeqqq-q (opposite)
Spinssss (same)
Magnetic Momentμ\muμ-\mu (opposite)
Lepton Number LLLLL-L (opposite)
Baryon Number BBBBB-B (opposite)
Linear Momentump\vec{p}p\vec{p} (conserved in annihilation/creation)
EnergyEEEE (same, both positive)
Angular MomentumJ\vec{J}J\vec{J} (same magnitude)

AnnihilationAnnihilation

Conservation Laws in Particle–Antiparticle Processes

In any interaction involving particles and antiparticles, the following quantities are strictly conserved:

Charge: Qi=Qf\text{Charge: } \sum Q_i = \sum Q_f

Linear Momentum: pi=pf\text{Linear Momentum: } \sum \vec{p}_i = \sum \vec{p}_f

Energy: Ei=EfE=(pc)2+(mc2)2\text{Energy: } \sum E_i = \sum E_f \quad \Rightarrow \quad E = \sqrt{(pc)^2 + (mc^2)^2}

Angular Momentum: Ji=Jf\text{Angular Momentum: } \sum \vec{J}_i = \sum \vec{J}_f

Lepton Number: Li=Lf\text{Lepton Number: } \sum L_i = \sum L_f

Baryon Number: Bi=Bf\text{Baryon Number: } \sum B_i = \sum B_f

#3. Particle–Antiparticle Pairs

3.1 Electron ↔ Positron

ee+e^- \longleftrightarrow e^+

PropertyElectron ee^-Positron e+e^+
Massme=9.109×1031 kgm_e = 9.109 \times 10^{-31}\ \text{kg}same
Chargee=1.602×1019 C-e = -1.602 \times 10^{-19}\ \text{C}+e+e
Spin12\frac{1}{2}12\frac{1}{2}
Lepton Number LeL_e+1+11-1

Annihilation reaction:

e+e+2γe^- + e^+ \longrightarrow 2\gamma

Each photon carries energy E=mec2=0.511 MeVE = m_e c^2 = 0.511\ \text{MeV} (in the centre-of-mass frame), emitted back-to-back to conserve momentum.

Pair production (inverse process):

γe+e+\gamma \longrightarrow e^- + e^+

Requires photon energy Eγ2mec2=1.022 MeVE_\gamma \geq 2m_e c^2 = 1.022\ \text{MeV} (in the presence of a nucleus for momentum conservation).

3.2 Proton ↔ Antiproton

ppˉp \longleftrightarrow \bar{p}

PropertyProton ppAntiproton pˉ\bar{p}
Massmp=1.673×1027 kgm_p = 1.673 \times 10^{-27}\ \text{kg}same
Charge+e+ee-e
Spin12\frac{1}{2}12\frac{1}{2}
Baryon Number BB+1+11-1
Quark contentuuduuduˉuˉdˉ\bar{u}\bar{u}\bar{d}

Annihilation:

p+pˉmesons (e.g., pions)orγ-raysp + \bar{p} \longrightarrow \text{mesons (e.g., pions)} \quad \text{or} \quad \gamma\text{-rays}

p+pˉπ++π+π0+p + \bar{p} \longrightarrow \pi^+ + \pi^- + \pi^0 + \cdots

Threshold energy for antiproton production:

Ethreshold=4mpc23.76 GeVE_{\text{threshold}} = 4m_p c^2 \approx 3.76\ \text{GeV}

The antiproton was discovered at the Bevatron accelerator (Berkeley) in 1955 by Segrè and Chamberlain.

3.3 Neutron ↔ Antineutron

nnˉn \longleftrightarrow \bar{n}

PropertyNeutron nnAntineutron nˉ\bar{n}
Massmn=1.675×1027 kgm_n = 1.675 \times 10^{-27}\ \text{kg}same
Electric Charge0000
Spin12\frac{1}{2}12\frac{1}{2}
Magnetic Momentμn=1.913 μN\mu_n = -1.913\ \mu_N+1.913 μN+1.913\ \mu_N (opposite)
Baryon Number BB+1+11-1
Quark contentuddudduˉdˉdˉ\bar{u}\bar{d}\bar{d}

Although the neutron is electrically neutral, its magnetic moment is opposite for the antineutron, making them distinguishable. The antineutron was discovered in 1956.

Annihilation:

n+nˉπ++π+π0+n + \bar{n} \longrightarrow \pi^+ + \pi^- + \pi^0 + \cdots

AntiProton AntiNeutronAntiProton AntiNeutron

3.4 Photon ↔ Itself (Self-Conjugate)

γγ\gamma \longleftrightarrow \gamma

The photon is its own antiparticle. It has:

  • Zero charge Q=0Q = 0
  • Zero mass
  • Spin s=1s = 1
  • Zero lepton number, zero baryon number

Since all additive quantum numbers are zero, the photon satisfies γˉ=γ\bar{\gamma} = \gamma.

Cγ=γ(charge conjugation eigenvalue=1)C|\gamma\rangle = -|\gamma\rangle \quad (\text{charge conjugation eigenvalue} = -1)

3.5 Muon and Antimuon

μμ+\mu^- \longleftrightarrow \mu^+

PropertyMuon μ\mu^-Antimuon μ+\mu^+
Mass105.66 MeV/c2105.66\ \text{MeV}/c^2same
Chargee-e+e+e
Spin12\frac{1}{2}12\frac{1}{2}
Muon Lepton Number LμL_\mu+1+11-1
Lifetimeτ2.197 μs\tau \approx 2.197\ \mu\text{s}same

Decay:

μe+νˉe+νμ\mu^- \longrightarrow e^- + \bar{\nu}_e + \nu_\mu

μ+e++νe+νˉμ\mu^+ \longrightarrow e^+ + \nu_e + \bar{\nu}_\mu

Each decay conserves both electron and muon lepton numbers separately.

3.6 Pion (Pi Meson) ↔ Antipion

Pions are the lightest mesons, composed of quark–antiquark pairs:

PionQuark ContentChargeMass
π+\pi^+udˉu\bar{d}+e+e139.57 MeV/c2139.57\ \text{MeV}/c^2
π\pi^- (antiparticle of π+\pi^+)duˉd\bar{u}e-e139.57 MeV/c2139.57\ \text{MeV}/c^2
π0\pi^012(uuˉddˉ)\frac{1}{\sqrt{2}}(u\bar{u} - d\bar{d})00134.98 MeV/c2134.98\ \text{MeV}/c^2

π+π\pi^+ \longleftrightarrow \pi^-

The π0\pi^0 is self-conjugate (its own antiparticle), similar to the photon.

Pion decay:

π+μ++νμ(τ26 ns)\pi^+ \longrightarrow \mu^+ + \nu_\mu \quad (\tau \approx 26\ \text{ns})

π02γ(τ8.5×1017 s)\pi^0 \longrightarrow 2\gamma \quad (\tau \approx 8.5 \times 10^{-17}\ \text{s})

#4. Quark–Antiquark Annihilation

At the fundamental level, particle–antiparticle annihilation is quark–antiquark annihilation mediated by gauge bosons. For example, in proton–antiproton annihilation:

u+uˉgq+qˉ(via gluon g)u + \bar{u} \longrightarrow g \longrightarrow q + \bar{q} \quad \text{(via gluon } g\text{)}

u+uˉγ or Z0(electroweak)u + \bar{u} \longrightarrow \gamma \text{ or } Z^0 \quad \text{(electroweak)}

The general quark annihilation vertex (QCD):

qi+qˉigawith amplitudegsTijaq_i + \bar{q}_i \longrightarrow g^a \quad \text{with amplitude} \propto g_s\, T^a_{ij}

where gsg_s is the strong coupling constant and TijaT^a_{ij} are the SU(3)SU(3) colour generators.

Cross-section for qqˉq\bar{q} annihilation into lepton pair (Drell–Yan):

σ^(qqˉ+)=4πα23s^eq2Nc1\hat{\sigma}(q\bar{q} \to \ell^+\ell^-) = \frac{4\pi\alpha^2}{3\hat{s}}\, e_q^2 \cdot N_c^{-1}

where s^\hat{s} is the partonic centre-of-mass energy squared, eqe_q is the quark charge, and Nc=3N_c = 3 is the colour factor.

#5. Baryons and Antibaryons

What is a Baryon?

A baryon is a hadron composed of three quarks (qqq), bound together by the strong force (QCD). Baryons carry baryon number B=+1B = +1.

B=13(nqnqˉ)B = \frac{1}{3}(n_q - n_{\bar{q}})

Each quark contributes B=+13B = +\frac{1}{3}; each antiquark contributes B=13B = -\frac{1}{3}.

The antidbaryon is the corresponding antiparticle with three antiquarks (qˉqˉqˉ)(\bar{q}\bar{q}\bar{q}) and B=1B = -1.

Baryon AntiBaryonBaryon AntiBaryon

Associated Baryon Particles

BaryonSymbolQuark ContentChargeMass (MeV/c2c^2)BB
Protonppuuduud+1+1938.3938.3+1+1
Neutronnnuddudd00939.6939.6+1+1
LambdaΛ0\Lambda^0udsuds001115.71115.7+1+1
Sigma-plusΣ+\Sigma^+uusuus+1+11189.41189.4+1+1
Sigma-zeroΣ0\Sigma^0udsuds001192.61192.6+1+1
Sigma-minusΣ\Sigma^-ddsdds1-11197.41197.4+1+1
Xi (Cascade)Ξ0\Xi^0ussuss001314.91314.9+1+1
Xi-minusΞ\Xi^-dssdss1-11321.71321.7+1+1
Omega-minusΩ\Omega^-ssssss1-11672.51672.5+1+1
Delta baryonsΔ++,+,0,\Delta^{++,+,0,-}various+2+2 to 1-11232\sim 1232+1+1

Each of these has a corresponding antibaryon with all quarks replaced by antiquarks and B=1B = -1, e.g., pˉ\bar{p}, nˉ\bar{n}, Λˉ0\bar{\Lambda}^0, Ωˉ+\bar{\Omega}^+, etc.

Baryon number conservation:

ΔB=0in all known interactions\Delta B = 0 \quad \text{in all known interactions}

(Possible violation only in hypothetical proton decay or baryogenesis scenarios.)


Dirac Hole Theory

The Dirac Equation

In 1928, Paul Dirac formulated a relativistic quantum mechanical equation for the electron:

(iγμμmc)ψ=0\left(i\hbar\gamma^\mu \partial_\mu - mc\right)\psi = 0

where γμ\gamma^\mu are the 4×4 Dirac gamma matrices satisfying the Clifford algebra:

{γμ,γν}=γμγν+γνγμ=2gμνI4×4\{\gamma^\mu, \gamma^\nu\} = \gamma^\mu\gamma^\nu + \gamma^\nu\gamma^\mu = 2g^{\mu\nu}\mathbf{I}_{4\times 4}

The energy solutions of the Dirac equation are:

E=±(pc)2+(mc2)2E = \pm\sqrt{(pc)^2 + (mc^2)^2}

This yields both positive and negative energy eigenvalues. The negative-energy solutions presented a conceptual crisis: classically, an electron could cascade down to -\infty energy.

The Dirac Sea

Dirac resolved this using the Pauli Exclusion Principle (valid for fermions like electrons):

Dirac's Hole Hypothesis: The vacuum (ground state) consists of an infinite "sea" of negative-energy states, all completely filled. Since the Pauli principle forbids two fermions in the same state, real electrons cannot fall into these occupied negative-energy states.

0Dirac=E<0aE,s0bare|0\rangle_{\text{Dirac}} = \prod_{E < 0} a^\dagger_{E,s}|0\rangle_{\text{bare}}

Holes as Antiparticles

If a negative-energy electron (energy E-|E|, charge e-e, momentum p-\vec{p}) is excited out of the Dirac sea into a positive-energy state, it leaves behind a hole.

The hole behaves as a particle with:

  • Energy: +E+|E| (absence of E-|E| energy raises the sea's energy by +E+|E|)
  • Charge: +e+e (absence of e-e charge gives net +e+e)
  • Momentum: +p+\vec{p}
  • Spin: +12+\frac{1}{2} (same as electron)

HolePositron e+\text{Hole} \equiv \text{Positron } e^+

Pair creation in hole theory:

γe (excited from sea)+hole (e+)\gamma \longrightarrow e^- \text{ (excited from sea)} + \text{hole } (e^+)

Pair annihilation:

e falls into holeγ+γe^- \text{ falls into hole} \longrightarrow \gamma + \gamma

Limitation of Hole Theory: It only works for fermions (Pauli exclusion). For bosons (integer spin), there is no exclusion principle to fill the sea. This inadequacy led to the development of Quantum Field Theory.

Dirac Hole ModelDirac Hole Model

Modern Theory of Antiparticles — Quantum Field Theory

In Quantum Field Theory (QFT), particles and antiparticles are both excitations of the same underlying quantum field. There is no Dirac sea; instead, the field is quantised with creation and annihilation operators.

Quantum FieldQuantum Field

Electron Proton FieldElectron Proton Field

The Electron Field

The electron is described by a Dirac spinor field ψ(x)\psi(x), which upon canonical quantisation is expanded as:

ψ(x)=sd3p(2π)312Ep[bpsus(p)eipx+dpsvs(p)e+ipx]\psi(x) = \sum_{s} \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2E_p}} \left[ b^s_{\vec{p}}\, u^s(p)\, e^{-ip\cdot x} + d^{s\dagger}_{\vec{p}}\, v^s(p)\, e^{+ip\cdot x} \right]

where:

  • bpsb^s_{\vec{p}} = annihilation operator for an electron (spin ss, momentum p\vec{p})
  • bpsb^{s\dagger}_{\vec{p}} = creation operator for an electron
  • dpsd^{s\dagger}_{\vec{p}} = creation operator for a positron
  • dpsd^s_{\vec{p}} = annihilation operator for a positron
  • us(p)u^s(p), vs(p)v^s(p) = positive and negative energy Dirac spinors

The anticommutation relations (ensuring Fermi–Dirac statistics):

{bpr,bqs}=(2π)3δ(3)(pq)δrs\{b^r_{\vec{p}},\, b^{s\dagger}_{\vec{q}}\} = (2\pi)^3\,\delta^{(3)}(\vec{p} - \vec{q})\,\delta^{rs}

{dpr,dqs}=(2π)3δ(3)(pq)δrs\{d^r_{\vec{p}},\, d^{s\dagger}_{\vec{q}}\} = (2\pi)^3\,\delta^{(3)}(\vec{p} - \vec{q})\,\delta^{rs}

The conjugate field ψˉ=ψγ0\bar{\psi} = \psi^\dagger \gamma^0 creates electrons and annihilates positrons.

The Proton Field

The proton is a composite particle (three quarks), but at an effective field theory level, it is also described by a Dirac spinor field Ψp(x)\Psi_p(x):

Ψp(x)=sd3p(2π)32Ep[bpsus(p)eipx+dpsvs(p)eipx]\Psi_p(x) = \sum_s \int \frac{d^3p}{(2\pi)^3\sqrt{2E_p}} \left[b^s_{\vec{p}}\, u^s(p)\, e^{-ip\cdot x} + d^{s\dagger}_{\vec{p}}\, v^s(p)\, e^{ip\cdot x}\right]

Here:

  • bpsb^{s\dagger}_{\vec{p}} creates a proton; dpsd^{s\dagger}_{\vec{p}} creates an antiproton
  • The proton and antiproton are excitations of the same proton quantum field Ψp\Psi_p

At the fundamental quark level, the proton is built from quark fields qf(x)q_f(x) (where f=u,d,s,f = u, d, s, \ldots):

u-quark field: qu(x)=d3p(2π)32Ep[bpu(p)eipx+dpv(p)eipx]u\text{-quark field: } q_u(x) = \int \frac{d^3p}{(2\pi)^3\sqrt{2E_p}} \left[b_{\vec{p}}\, u(p)\, e^{-ip\cdot x} + d^\dagger_{\vec{p}}\, v(p)\, e^{ip\cdot x}\right]

The dd^\dagger operator here creates an anti-up quark uˉ\bar{u}.


Crossing Symmetry

Crossing symmetry is a fundamental property of scattering amplitudes in QFT. It states that the amplitude for a process involving a particle in the initial state is related to the amplitude for the same process with the corresponding antiparticle in the final state (with reversed 4-momentum), and vice versa.

Formally, for an S-matrix element:

M(A+BC+D)=M(B+CˉAˉ+D)\mathcal{M}\bigl(A + B \to C + D\bigr) = \mathcal{M}\bigl(B + \bar{C} \to \bar{A} + D\bigr)

by analytically continuing the external momenta.

Consider the generic 222 \to 2 scattering. Crossing symmetry connects three "channels":

ChannelProcessMandelstam Variable
s-channelA+BC+DA + B \to C + Ds=(pA+pB)2s = (p_A + p_B)^2
t-channelA+CˉBˉ+DA + \bar{C} \to \bar{B} + Dt=(pApC)2t = (p_A - p_C)^2
u-channelA+DˉBˉ+CA + \bar{D} \to \bar{B} + Cu=(pApD)2u = (p_A - p_D)^2

The three Mandelstam variables satisfy:

s+t+u=imi2s + t + u = \sum_i m_i^2

The same Feynman amplitude M(s,t,u)\mathcal{M}(s, t, u), when analytically continued, describes all three channels.

Example: Compton Scattering vs. Pair Annihilation

s-channel (Compton scattering):

e+γe+γe^- + \gamma \longrightarrow e^- + \gamma

t-channel (related by crossing):

e+e+γ+γ(pair annihilation)e^- + e^+ \longrightarrow \gamma + \gamma \quad \text{(pair annihilation)}

These two processes share the same Feynman diagrams — the difference is only which legs are "crossed" (moved from initial to final state with reversed momentum):

Meγeγ(s,t)=Mee+γγ(t,s)\mathcal{M}_{e^-\gamma \to e^-\gamma}(s,t) = \mathcal{M}_{e^-e^+ \to \gamma\gamma}(t,s)

Mathematical Statement

If a particle AA with momentum pp is in the initial state, crossing symmetry allows us to replace it with its antiparticle Aˉ\bar{A} with momentum p-p in the final state:

Initial state particle A(p)Final state antiparticle Aˉ(p)\text{Initial state particle } A(p) \quad \longleftrightarrow \quad \text{Final state antiparticle } \bar{A}(-p)

For spinor fields, this replacement is:

us(p) (initial electron)vˉs(p) (final positron)u^s(p) \text{ (initial electron)} \quad \longleftrightarrow \quad \bar{v}^s(-p) \text{ (final positron)}

uˉs(p) (final electron)vs(p) (initial positron)\bar{u}^s(p) \text{ (final electron)} \quad \longleftrightarrow \quad v^s(-p) \text{ (initial positron)}

#Summary Table

ConceptKey Equation / Result
Antiparticle rest massmAˉ=mAm_{\bar{A}} = m_A
Annihilation energyE=2mAc2E = 2m_A c^2 (at rest)
Pair production thresholdEγ2mAc2E_\gamma \geq 2m_A c^2
Dirac equation(iγμμmc)ψ=0(i\hbar\gamma^\mu\partial_\mu - mc)\psi = 0
Mandelstam constraints+t+u=mi2s + t + u = \sum m_i^2
Baryon number conservationΔB=0\Delta B = 0
Lepton number conservationΔL=0\Delta L = 0

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Last updated on 4/7/2026