
Advanced Quantum Mechanics course sequence · Interactive lesson
Complex scalar field: antiparticles and the charge operator
Here antimatter stops being a filled sea and becomes a second operator. Let the field be complex, carry the extra operator through the energy and the Noether charge, and the π⁺ and π⁻ come out of one Lagrangian with one mass, opposite charges and positive energies, while the Klein-Gordon equation Unit 3 gave up on goes back to work.
UAQM-09.03 · Lesson 3 of 9 in this unit. Position is not a completion score.Build the model
One complex field, two kinds of quanta: H adds them, Q subtracts them.
The key idea
The complex scalar field is a spin-0 field with φ ≠ φ†, whose quanta come in two kinds, particles created by aₖ† and antiparticles created by bₖ†, with the same mass and positive energy and opposite values of the conserved charge Q.
Example. π⁺ and π⁻ are the two quanta of one complex field: each has rest energy 139.57 MeV, their charges are +e and −e, and the pair state a₀†b₀†|0⟩, both at rest, has energy 279.14 MeV and Q = 0.
The complete physical model and assumptions
A real field cannot carry charge. Its reality condition ties the operator on each negative frequency to the one on the positive frequency, aₖ† to aₖ, so every quantum is its own antiparticle. Drop the condition and the Lagrangian ∂μ φ† ∂μ φ − m²φ†φ, which is two real fields of one mass, gains a symmetry rotating φ₁ into φ₂, φ → e(−iα)φ. Canonical quantisation then needs two operators per mode, aₖ on the positive frequencies and bₖ† on the negative, and carrying them through the Legendre transform and Noether's theorem derives the rest: H = Σₖ ωₖ (aₖ†aₖ + bₖ†bₖ) counts both kinds of quantum with positive energy, Q = q Σₖ (aₖ†aₖ − bₖ†bₖ) counts them with opposite signs, and charge conjugation, which swaps aₖ with bₖ and φ with φ†, leaves H alone and so gives both one mass. Unit 3's indefinite Klein-Gordon density was this charge density all along, as Pauli and Weisskopf proposed in 1934: it never had to be positive because it was never a probability, and the negative frequencies need no sea, which bosons could not fill anyway. The model stops in three places. It is free, so a π⁺ and a π⁻ neither attract nor annihilate until the field meets the photon in Unit 10. It treats the pion as a structureless quantum, though the real one is a quark-antiquark bound state. And it assumed commutators; whether that choice is forced is the business of topics 9.04 and 9.08.
1A complex field needs a second operator
Units throughout: ħ = c = 1, so energy, momentum and mass are in MeV and length in MeV⁻¹, converted with ħc = 197.327 MeV fm; the metric is (+, −, −, −). Topic 9.02 is used, not repeated: a real field puts aₖ on e(−ik⋅x) and aₖ† on e(ik⋅x), the second forced by φ = φ†. A complex field obeys (∂μ∂μ + m²)φ = 0 with no reality condition, so the coefficients of the two frequencies are independent; call them aₖ and bₖ†. The conjugate momenta are π = ∂ℒ/∂(∂₀φ) = ∂₀φ† and π† = ∂₀φ, and [φ(r, t), π(r′, t)] = [φ†(r, t), π†(r′, t)] = iδ³(r − r′), with every other equal-time pair commuting, give two independent oscillator algebras per mode. Nothing new is assumed: with φ = (φ₁ + iφ₂)/√2 for two real fields of one mass, aₖ = (a₁ₖ + ia₂ₖ)/√2 and bₖ = (a₁ₖ − ia₂ₖ)/√2, and every commutator follows from topic 9.02's. Dimension check: φ has the dimension of 1/√(ωₖ V), MeV in these units.
2The energy counts both quanta as positive
The Legendre transform gives ℋ = ∂₀φ† ∂₀φ + ∇φ†⋅∇φ + m²φ†φ, a sum of squares, each written with the daggered field on the left. Substituting the modes, each oscillating cross term carries −ωₖ² + k² + m² = 0 and cancels, leaving H = Σₖ ωₖ (aₖ†aₖ + bₖ bₖ†) = Σₖ ωₖ (aₖ†aₖ + bₖ†bₖ) + Σₖ ωₖ. Normal ordering drops the last sum, ωₖ per mode, which is ½ωₖ for each of the two real fields: topic 9.02's zero-point energy, counted twice. What is left gives every quantum positive energy, and the operator on the negative frequency is a creator: [H, bₖ†] = ωₖ bₖ†. The momentum, derived the same way, is P = Σₖ k (aₖ†aₖ + bₖ†bₖ); its zero-point sum Σₖ k vanishes because k and −k pair off. So bₖ† adds energy ωₖ and momentum +k exactly as aₖ† does: a π⁻ of momentum 200 MeV has energy √(200² + 139.57²) = 243.9 MeV, the same as a π⁺.
3The Noether charge, and the sign of the antiparticle
ℒ is unchanged by φ → e(−iα)φ with φ† → e(iα)φ†, and Noether's theorem from topic 9.01 gives jμ = i(φ† ∂μ φ − (∂μ φ†) φ), conserved because ∂μ jμ = i(φ† ∂μ∂μ φ − (∂μ∂μ φ†) φ) = i(−m²φ†φ + m²φ†φ) = 0. In modes, with each daggered field written on the left, Q = q∫d³r j⁰ becomes q Σₖ (aₖ†aₖ − bₖ bₖ†), its oscillating terms cancelling as a conserved charge's must. Two different things sit in that line. The minus sign is structural: it gives [Q, aₖ†] = q aₖ† and [Q, bₖ†] = −q bₖ†, so π⁺ and π⁻ carry +e and −e, and equally [Q, φ] = −qφ, so both terms of φ lower the charge by q, one by removing a particle, the other by adding an antiparticle. The constant is not structural: the two orderings give −qΣₖ 1 and +qΣₖ 1, and their average, the one that makes Q odd under charge conjugation, gives zero, so Q = q Σₖ (aₖ†aₖ − bₖ†bₖ) holds exactly.
4Charge is angular momentum in field space
In the real components the same operator reads Q = iq Σₖ (a₁ₖ†a₂ₖ − a₂ₖ†a₁ₖ), the generator of rotations in the (φ₁, φ₂) plane, and the phase symmetry is that rotation. Each mode is therefore the two-dimensional isotropic oscillator of Quantum Mechanics I: H = ωₖ(nₐ + nb) is its energy, Q/q = nₐ − nb its angular momentum, and a and b its circular quanta. A level holding N quanta in one mode is (N + 1)-fold degenerate, with Q = Ne, (N − 2)e, …, −Ne. This settles what the real-field quanta are: a₁ₖ† = (aₖ† + bₖ†)/√2 makes an equal superposition of π⁺ and π⁻, of definite energy and ⟨Q⟩ = 0, whose charge is found to be +e or −e, never 0. Such states are legitimate in the free field. Once the field couples to electromagnetism, the charge superselection rule, quoted here, admits only states of definite Q, which is why laboratories see π⁺ and π⁻ and never a φ₁ quantum.
5Pauli and Weisskopf's reading, carried out
Unit 3 refused j⁰ as a probability density because it turns negative on negative-frequency solutions. Here it is an operator, and its sign is the charge. In a one-particle packet |f⟩, ⟨f|:j⁰(x):|f⟩ is the bilinear i(F*∂ₜF − F∂ₜF*) of the positive-frequency wavefunction F(x) = ⟨0|φ(x)|f⟩, Unit 3's density up to its 1/2m; in an antiparticle packet |g⟩ it is minus the same bilinear of G(x) = ⟨0|φ†(x)|g⟩. Nothing had to be positive, because nothing was a probability: Pauli and Weisskopf's 1934 proposal, now derived. The Klein-Gordon equation survives as the equation of motion of an operator field, and its negative frequencies need no sea. A bosonic sea could not even be filled, since without exclusion every level takes more quanta, so spin-0 antiparticles sat on Unit 8's list of what hole theory could not do; this field does it. Non-relativistic check: a slow packet has F ≈ e(−imt)ψ/√(2m), which turns the bilinear into |ψ|², Schrödinger's density, and one-particle quantum mechanics serves a slow pion because a pair costs 2m = 279.14 MeV.
6Checks: the real limit, C and the pion masses
Three checks close the derivation. The real limit: set bₖ = aₖ, so that φ = φ†, and Q = q Σₖ (aₖ†aₖ − aₖ†aₖ) vanishes identically. A real field has no current and carries no charge, and its quantum is its own antiparticle, as the π⁰ is. Charge conjugation: the unitary C with C aₖ C⁻¹ = bₖ and C bₖ C⁻¹ = aₖ leaves the vacuum and H unchanged and sends φ to φ† and Q to −Q, so particle and antiparticle have the same ωₖ at every k and therefore one mass. That is derived for this free field; for interacting fields the equality follows from the CPT theorem, quoted. It is tested: the charged pions have m = 139.57039 ± 0.00018 MeV, and (m(π⁺) − m(π⁻))/m = (2 ± 5) × 10⁻⁴, a difference of (0.03 ± 0.07) MeV, zero within its uncertainty. The coupling to electromagnetism, made by replacing ∂μ with a covariant derivative containing qAμ, turns this jμ into the source of the photon field; it is named here and built in Unit 10.
Key equations
Read the units, assumptions and limits before substituting.
Complex Klein-Gordon Lagrangian density
ℒ = ∂μ φ† ∂μ φ − m²φ†φ
Two of topic 9.02's real fields, φ = (φ₁ + iφ₂)/√2, with a phase symmetry φ → e(−iα)φ that rotates φ₁ into φ₂. Varying φ† gives (∂μ∂μ + m²)φ = 0.
ħ = c = 1, metric (+, −, −, −): φ in MeV, ℒ in MeV⁴, m the quantum's mass in MeV. φ and φ† are varied independently.Mode expansion in a box of volume V
φ(x) = Σₖ [aₖ e(−ik⋅x) + bₖ† e(ik⋅x)] / √(2ωₖ V)
Two operators per mode because φ ≠ φ†: aₖ rides the positive frequencies, bₖ† the negative. Setting bₖ = aₖ gives back the real field.
ωₖ = √(k² + m²) in MeV, k⋅x = ωₖ t − k⋅r; [aₖ, aₖ′†] = [bₖ, bₖ′†] = δₖₖ′ and every other pair commutes.Noether current of the phase symmetry
jμ = i(φ† ∂μ φ − (∂μ φ†) φ)
∂μ jμ = 0 for every solution of the field equation. Its indefinite j⁰ is a charge density, so nothing requires it to be positive.
jμ in MeV³, a number current; j⁰ is Unit 3's Klein-Gordon density without its factor 1/2m. Derived from φ → e(−iα)φ.Hamiltonian, normal ordered
H = Σₖ ωₖ (aₖ†aₖ + bₖ†bₖ)
[H, bₖ†] = +ωₖ bₖ†: the operator on the negative frequencies creates positive energy. Nothing lies below the vacuum, so there is no sea to fill.
ωₖ in MeV (ħωₖ in J in SI). Normal ordering drops Σₖ ωₖ, twice the Σₖ ωₖ/2 that one real field drops.Charge operator
Q = q Σₖ (aₖ†aₖ − bₖ†bₖ)
[Q, aₖ†] = q aₖ†, [Q, bₖ†] = −q bₖ† and [Q, H] = 0: a π⁺ adds e, a π⁻ takes it away, and states of different charge never mix.
q is the charge of one a-quantum, +e for the π⁺. Reordering adds −qΣₖ 1 or +qΣₖ 1; their average adds nothing.Charge conjugation
C aₖ C⁻¹ = bₖ, C bₖ C⁻¹ = aₖ, so C φ C⁻¹ = φ†
Particle and antiparticle share ωₖ at every k, hence one mass. Measured: (m(π⁺) − m(π⁻))/m = (2 ± 5) × 10⁻⁴, with m = 139.57039 MeV.
C is unitary with C|0⟩ = |0⟩; it leaves H and P unchanged and sends Q to −Q. Derived for the free field.Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Set Nₐ = Nb = 2 and slide d to 0: the solid charge curve flattens to zero everywhere while the dashed energy curve doubles, a neutral cloud of pairs that still holds 558 MeV. No separation changes either total: charge is conserved, and a free field has no interaction energy.
Live results
- NET CHARGE Q
- 1 e
- REST ENERGY (Nₐ + Nb) mc²
- 418.7 MeV
- CHARGE DENSITY AT x = 0
- 0.033 q fm⁻¹
- ENERGY DENSITY AT x = 0
- 13.88 MeV fm⁻¹
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Particles Nₐ in the left packet: 2; Antiparticles Nb in the right packet: 1; Packet separation d: 10 fm
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The commonest error is to read the negative-frequency term as negative energy. Because bₖ† e(ik⋅x) oscillates as e(+iωₖ t), students take it to describe a quantum of energy −ωₖ, and conclude that a spin-0 field, like Dirac's, needs those states filled by a sea. But φ is an operator, not a wavefunction, and the frequency of a term reports what its operator does, not what some state has. In the Heisenberg picture every creator evolves as e(iHt) c† e(−iHt) = c† e(+iωt), aₖ† included, so e(+iωₖ t) in φ is the signature of adding energy ωₖ, and [H, bₖ†] = +ωₖ bₖ† says it outright. The spectrum is bounded below by the vacuum and nothing needs filling; nor could a bosonic sea be filled, since without exclusion every level takes more quanta.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A box of charged-pion field, m = 139.57 MeV, holds |s⟩ = a†(0) a†(k) b†(−k)|0⟩ with k = 150 MeV along x, where a†(k) means aₖ†. Find the energy, momentum and charge of |s⟩, then write C|s⟩ and give its energy, momentum and charge.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Each creator raises H by ωₖ = √(k² + m²), whichever letter it carries: ω(0) = 139.57 MeV and ω(150 MeV) = √(150² + 139.57²) MeV = √(41 980) MeV = 204.89 MeV.
- H is normal ordered, so there is no zero-point term: E = 139.57 + 204.89 + 204.89 = 549.35 MeV. Dimension check: every term is √(MeV²), an energy.
- P = Σₖ k (nₐ + nb) adds the labels with plus signs whatever the letter: P = 0 + k + (−k) = 0. The antiparticle carries the momentum written on its b†, here −k.
- Q = q(nₐ − nb) = e(2 − 1) = +e: the state holds two π⁺ and one π⁻.
- C swaps a and b: C|s⟩ = b†(0) b†(k) a†(−k)|0⟩, two π⁻ and one π⁺. H and P are unchanged by C, so E = 549.35 MeV and P = 0 again, while Q = −e.
Answer. E = 549.35 MeV, P = 0, Q = +e. C|s⟩ = b†(0) b†(k) a†(−k)|0⟩, with E = 549.35 MeV, P = 0 and Q = −e.
MediumExample 2
Two quanta of the real component φ₁ are placed in the k = 0 mode of the charged-pion field: |s⟩ = (a₁†)²|0⟩/√2, with a₁† = (a† + b†)/√2. Find the probability of each value of Q, the mean and spread of Q, and the energy. Is |s⟩ a state of definite charge?
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Expand: (a₁†)² = ½(a†² + 2a†b† + b†²), the cross term doubled because a† and b† commute.
- On the vacuum, writing |nₐ, nb⟩: a†²|0⟩ = √2|2, 0⟩, a†b†|0⟩ = |1, 1⟩ and b†²|0⟩ = √2|0, 2⟩. So |s⟩ = ½|2, 0⟩ + (1/√2)|1, 1⟩ + ½|0, 2⟩, normalised since ¼ + ½ + ¼ = 1.
- Q = e(nₐ − nb): Q = +2e with probability ¼, 0 with probability ½ and −2e with probability ¼. Then ⟨Q⟩ = 0, ⟨Q²⟩ = 4e²(¼ + ¼) = 2e² and ΔQ = √2 e = 1.41e.
- Every component holds two quanta at rest, so H|s⟩ = 2m|s⟩ with 2m = 279.14 MeV: the energy is sharp. The three components are the N = 2 level of the field-space oscillator, Q = +2e, 0, −2e.
- |s⟩ superposes π⁺π⁺, π⁺π⁻ and π⁻π⁻. It is a legitimate state of the free field, but once the field couples to the photon the charge superselection rule, quoted, forbids preparing a superposition of different Q.
Answer. P(+2e) = 0.25, P(0) = 0.50, P(−2e) = 0.25; ⟨Q⟩ = 0 and ΔQ = 1.41e; E = 279.14 MeV exactly. Not a state of definite charge.
HardExample 3
A free charged-pion gas at T = 150 MeV (kB = 1) is given net positive charge: ρ = exp[−(H − μNQ)/T]/Z, with NQ = Q/q and μ = 40 MeV. Find the mean π⁺ and π⁻ occupations of the k = 0 mode and of a mode with |k| = 300 MeV, compare their ratios with e(2μ/T), and show why μ must stay below m = 139.57 MeV.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- H − μNQ = Σₖ [(ωₖ − μ) aₖ†aₖ + (ωₖ + μ) bₖ†bₖ]. Both operators are diagonal in the same occupation numbers, so ρ factorises into independent Bose oscillators: π⁺ of energy ωₖ − μ, π⁻ of energy ωₖ + μ. Hence n± = 1/(exp[(ωₖ ∓ μ)/T] − 1), each exponent MeV over MeV.
- k = 0, ω = 139.57 MeV: n₊ = 1/(e(99.57/150) − 1) = 1/(1.9422 − 1) = 1.061 and n₋ = 1/(e(179.57/150) − 1) = 1/(3.3106 − 1) = 0.433, so the mode carries net charge 0.629e and n₊/n₋ = 2.45.
- |k| = 300 MeV, ω = 330.88 MeV: n₊ = 1/(e(290.88/150) − 1) = 1/(6.953 − 1) = 0.168 and n₋ = 1/(e(370.88/150) − 1) = 1/(11.852 − 1) = 0.0921, so n₊/n₋ = 1.82.
- Boltzmann limit: when both exponentials are large the ratio tends to e(2μ/T) = e(0.533) = 1.70 for every k. The fast mode, at 1.82, is close; the k = 0 mode, at 2.45, is not, because Bose enhancement favours the lower energy ω − μ.
- Limits: at μ = 0 the occupations are equal, 1/(e(139.57/150) − 1) = 0.651 each at k = 0, so a hot neutral gas holds pairs with no net charge. As μ → m, n₊ at k = 0 diverges and charge condenses into that mode; beyond m it would be negative and Tr ρ would diverge.
Answer. k = 0: n₊ = 1.06, n₋ = 0.433, ratio 2.45. |k| = 300 MeV: n₊ = 0.168, n₋ = 0.0921, ratio 1.82, against the Boltzmann value e(2μ/T) = 1.70. The chemical potential must stay below m = 139.57 MeV in size.
Quick check
Select an answer and explain it before checking.
Review & continue
The complex scalar field is a spin-0 field with φ ≠ φ†, whose quanta come in two kinds, particles created by aₖ† and antiparticles created by bₖ†, with the same mass and positive energy and opposite values of the conserved charge Q.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.