
Advanced Quantum Mechanics course sequence · Interactive lesson
Charge conjugation and the positron wavefunction
Topic 8.04 read the positron off the filled sea; here it gets a wavefunction. One antilinear step, complex conjugation followed by iγ², turns every solution for charge q into one for −q, and on free spinors it is the rule Unit 9 uses to build v from u. The skill is knowing exactly what that step reverses and what it leaves alone.
UAQM-08.05 · Lesson 5 of 9 in this unit. Position is not a completion score.Build the model
Complex-conjugate the Dirac equation and γ² hands back a solution of opposite charge.
The key idea
Charge conjugation is the antilinear map ψ → ψc = iγ²ψ*, complex conjugation followed by the matrix iγ², which turns every solution of the Dirac equation for charge q into a solution for charge −q with the same mass.
Example. In the Dirac representation ψc = (ψ₄*, −ψ₃*, −ψ₂*, ψ₁*). The negative-energy rest solution (0, 0, 0, 1)e(+imc²t/ħ), energy −0.511 MeV and spin down, becomes (1, 0, 0, 0)e(−imc²t/ħ): energy +0.511 MeV, spin up, charge +e.
The complete physical model and assumptions
Complex-conjugate the minimally coupled Dirac equation, [γμ(iħ∂μ − qAμ) − mc]ψ = 0, and the charge flips at once: iħ∂μ changes sign while the real potential qAμ does not, and the charge lives only in their relative sign. What spoils the conjugated equation is that the matrices changed too, because in the Dirac representation γ⁰, γ¹ and γ³ are real and γ² alone is imaginary. One matrix repairs it. γ² anticommutes with the three real gammas and commutes with itself, so γ²(γμ)*(γ²)⁻¹ = −γμ for every μ, and ψc = iγ²ψ* solves the equation with q replaced by −q and the same m. All of this is derived here, with (ψc)c = ψ and the equivalent form C ψ̄T with C = iγ²γ⁰. Fed a negative-energy electron solution with momentum p and spin s, conjugation returns a positive-energy solution with −p, −s and charge +e: topic 8.04's hole, now with a wavefunction, and on free spinors the rule v(p, s) = iγ²u(p, s)* that Unit 9 adopts. The model stops at the charge itself. The map is antilinear and leaves ψ†ψ and the probability current untouched, so the one-particle theory reverses the charge only because the equation is told to; ψ̄ γμ ψ itself changes sign under conjugation only for the anticommuting fields of Unit 9. Units are SI with ħ and c explicit, the metric is (+, −, −, −) with xμ = (ct, x), and the electron has q = −e.
1Complex conjugation already flips the charge
Start from Unit 4's minimally coupled equation, [γμ(iħ∂μ − qAμ) − mc]ψ = 0, with Aμ = (φ/c, A) real and q = −e for the electron. The charge lives only in the relative sign of iħ∂μ and qAμ, both in kg m s⁻¹. Complex conjugation reverses the first and not the second: [(γμ)*(−iħ∂μ − qAμ) − mc]ψ* = 0, which is −[(γμ)*(iħ∂μ + qAμ) + mc]ψ* = 0. Inside the bracket the charge now reads −q. Two things are still wrong: the matrices are (γμ)* rather than γμ themselves, and the mass term has the wrong sign against the derivative term. A constant matrix M with M(γμ)*M⁻¹ = −γμ for every μ would repair both at once, since multiplying by M turns the bracket into −[γμ(iħ∂μ + qAμ) − mc]Mψ*. Notice what the conjugation does to a stationary state before any matrix acts: w e(−iEt/ħ) becomes w* e(+iEt/ħ), so the energy reverses as well. Charge and energy flip together, and that pairing is the physics of this topic.
2Why γ², and only γ², repairs the matrices
In the Dirac representation γ⁰ = diag(1, 1, −1, −1), and γk has off-diagonal blocks σₖ and −σₖ. σₓ and σz are real and σy is imaginary, so γ⁰, γ¹ and γ³ are real while (γ²)* = −γ². M must therefore anticommute with the three real matrices and commute with γ². The Clifford algebra supplies one: γ² anticommutes with every other γμ and commutes with itself, so γ²(γμ)*(γ²)⁻¹ = −γμ for all four μ, derived. Two such matrices differ by one that commutes with every γμ; topic 5.02 showed that such a matrix is a multiple of the identity, so M = ηγ² with η a constant. The convention η = i makes iγ² real and symmetric with (iγ²)² = 1, and gives the component rule ψc = (ψ₄*, −ψ₃*, −ψ₂*, ψ₁*). Apply it twice: (ψc)c = iγ²(iγ²)*ψ = ψ, derived, and true for any η of modulus 1. Dimensions check because iγ² is a pure number: ψc carries the m(−3/2) of ψ, and every term of its equation is a momentum times ψ.
3The same map written as C ψ̄T
Many texts define ψc = Cψ̄T with C = iγ²γ⁰. The two agree, derived: ψ̄T = (ψ†γ⁰)T = (γ⁰)T ψ* = γ⁰ψ* because γ⁰ is real and diagonal, so Cψ̄T = iγ²γ⁰γ⁰ψ* = iγ²ψ*. Written out, C has the entries −1, +1, −1, +1 down its anti-diagonal from the top row; it is real, CT = −C and C² = −1, so C⁻¹ = CT = −C. Its defining property C(γμ)T C⁻¹ = −γμ follows from (γμ)T = γ⁰(γμ)*γ⁰ and the γ² identity. That is the form to carry into other representations. In the Weyl representation γ² is again the only imaginary matrix and the recipe survives; in the Majorana representation all four γ's are imaginary, and there ψc is simply ψ*, up to a phase. The C form also shows that ψc is built from ψ̄, and a two-line check with Unit 5's generators, iγ²S*(iγ²)⁻¹ = S, confirms that ψc transforms under rotations and boosts exactly as ψ does.
4What conjugation does to a negative-energy electron
Write a negative-energy electron solution as ψ = w e(+i|E|t/ħ) e(+ip⋅x/ħ): energy −|E|, momentum p, spin read from Σ. The star inverts both exponentials, so ψc = iγ²w* e(−i|E|t/ħ) e(−ip⋅x/ħ) has energy +|E| and momentum −p. For spin, iγ²Σ*iγ² = −Σ, from σyσₖ*σy = −σₖ, so (ψc)†Σψc = −ψ†Σψ, while helicity, a product of two reversed quantities, survives. With the reversed charge this is topic 8.04's hole: the vacancy left by an electron with −|E|, p, s and −e is a positron with +|E|, −p, −s and +e, and ψc is its wavefunction. In the non-relativistic limit the large components move from the lower pair to the upper, so the positron obeys Pauli's equation with charge +e and the same g = 2 (Unit 6's reduction with q → −q). For a static field the same steps give a spectral rule, derived: H(q)w = Ew implies H(−q)(iγ²w*) = −E(iγ²w*). Hydrogen's ground level at mc² − 13.6 eV becomes a negative-energy level of the +e equation at −mc² + 13.6 eV, and no level sits just below +mc²: the proton binds no positron. Measured: e⁺ and e⁻ g factors agree to (−0.5 ± 2.1) × 10⁻¹².
5Free spinors: v(p, s) = iγ²u(p, s)*
Unit 4 wrote v(p, s) = √((E + mc²)/c)(cσ⋅p ηₛ/(E + mc²); ηₛ) with ηₛ = −iσyχₛ* and called the choice a label convention. Conjugation derives it. Apply iγ², whose blocks are iσy above and −iσy below, to u(p, s)* = √((E + mc²)/c)(χₛ*; cσ*⋅p χₛ*/(E + mc²)). The lower pair becomes −iσyχₛ* = ηₛ, and the upper pair becomes iσy(σ*⋅p)χₛ* = (σ⋅p)(−iσyχₛ*), using σyσₖ*σy = −σₖ again, so v(p, s) = iγ²u(p, s)* exactly. Take cp = 1.000 MeV along z and s = ↑: E = 1.123 MeV and cp/(E + mc²) = 0.612, so u = 1.278(1, 0, 0.612, 0) and v = 1.278(0, −0.612, 0, 1) in √(MeV/c). Checks: v†v = 2.246 MeV/c = 2E/c, and v̄v = −1.022 MeV/c = −2mc. The free limit settles the labels Unit 4 borrowed. v(p, s) exp(ipμxμ/ħ) has energy −E, momentum −p and spin −s, and its conjugate is u(p, s) exp(−ipμxμ/ħ), a positron with energy E, momentum p and spin s.
6What conjugation cannot reverse
The map is antilinear, (λψ)c = λ*ψc; it also leaves the density alone: (ψc)†ψc = ψT(iγ²)(iγ²)ψ* = ψ†ψ. The current survives too. (ψc)†αₖψc = ψT iγ²αₖ iγ² ψ* is a single number, so transposing it is free for commuting components and gives ψ†(iγ²αₖ*iγ²)ψ = ψ†αₖψ: the probability current of a c-number ψc equals that of ψ, derived. At cp = 2.00 MeV along +z, ψ and ψc both move at −0.969c. The electric current qj therefore reverses only because q was switched by hand; nothing in the one-particle spinor carries the sign. Unit 9's field operators anticommute, ψₐψb† = −ψb†ψₐ + δab δ³(0), so the same transposition costs a minus sign and ψ̄c γμ ψc = −ψ̄ γμ ψ after normal ordering (quoted). The limit q = 0 allows a spinor equal to its own conjugate, ψc = ψ, which in components reads (a, b, −b*, a*): four real parameters, not eight. These Majorana spinors are only named here. Whether neutrinos are Majorana is open: searches for neutrinoless double-β decay have set only lower limits on its half-life, above 10²⁶ years.
Key equations
Read the units, assumptions and limits before substituting.
Conjugating the γ matrices (derived)
γ² (γμ)* (γ²)⁻¹ = −γμ for μ = 0, 1, 2, 3
γ² anticommutes with the three real matrices and commutes with itself, giving each the minus sign it needs. That is the whole reason the operation uses γ² and no other matrix.
Dirac representation: γ⁰, γ¹, γ³ real and γ² imaginary; (γ²)² = −1, so (γ²)⁻¹ = −γ²; all dimensionless 4 × 4 matricesThe conjugate obeys the opposite-charge equation
[γμ(iħ∂μ − qAμ) − mc]ψ = 0 ⇒ [γμ(iħ∂μ + qAμ) − mc]ψc = 0, where ψc = iγ²ψ*
Derived. Only the sign of q relative to iħ∂μ changes; m and the γ's are untouched, so the partner has the electron's mass. Measured: |m₊ − m₋|/m is below 8 × 10⁻⁹.
q in C; Aμ = (φ/c, A) with φ in V and A in T m; iħ∂μ and qAμ both in kg m s⁻¹; ψ and ψc both in m(−3/2)Component form in the Dirac representation
ψc = iγ²ψ* = (ψ₄*, −ψ₃*, −ψ₂*, ψ₁*)
Upper and lower pairs swap, and so do spin-up and spin-down slots. Apply it twice and every component returns, (ψc)c = ψ; it is antilinear, (λψ)c = λ*ψc.
iγ² is real and symmetric with (iγ²)² = 1; the factor i is a phase convention, and any unimodular factor would serveRepresentation-free form (shown equivalent)
ψc = C ψ̄T with C = iγ²γ⁰ and ψ̄ = ψ†γ⁰
Equal to iγ²ψ* because γ⁰ is real and squares to 1. Defining C through (γμ)T lets the map survive a change of representation, where no γ need be the imaginary one.
T is the transpose; C is real with CT = C⁻¹ = −C; the relation C (γμ)T C⁻¹ = −γμ defines C in any representationFree spinors: the convention Unit 9 uses
v(p, s) = iγ² u(p, s)*
Derived. Since iγ²[v(p, s) exp(ipμxμ/ħ)]* = u(p, s) exp(−ipμxμ/ħ), the p and s written on v are the positron's momentum and spin.
u and v in √(kg m s⁻¹) with u†u = v†v = 2E/c; reproduces Unit 4's lower pair ηₛ = −iσyχₛ*What conjugation leaves alone
ψ̄c γμ ψc = ψ̄ γμ ψ, so (ψc)†ψc = ψ†ψ
Derived: density and probability current do not reverse, so qjμ flips only because q was changed by hand. Unit 9's anticommuting fields flip jμ itself (quoted).
jμ = c ψ̄ γμ ψ in m⁻² s⁻¹; holds when the four components are ordinary commuting numbersInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. At θ = 0 the large bar sits in slot 3 of ψ and moves to slot 2 of ψc: lower pair to upper pair, spin up to spin down. Raise p and the small components grow while the two arrows stay exactly opposite; at θ = 90° both shorten to mc²/E.
Live results
- ENERGY OF ψc = −E(ψ)
- 1.562 mc²
- DENSITY ψ†ψ, BOTH ROWS
- 3.124 mc
- VELOCITY vz, BOTH ROWS
- -0.768 c
- ⟨Σz⟩ OF ψc
- -0.500
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. Electron momentum pz: 1.2 mc; Rest-frame spin angle θ from +z: 60 °
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The tempting reading is that ψc is the positron version of whatever electron state is fed in: conjugate a 1 MeV electron moving along +z with spin up, and out comes a 1 MeV positron moving along +z with spin up. It is wrong twice. Conjugating a positive-energy electron solution gives a negative-energy solution of the q = +e equation, which is not a positron state at all; read through hole theory, its vacancy is the electron you started with. The positron wavefunction comes from conjugating the negative-energy electron solution whose absence the positron is, and it carries the opposite momentum and opposite spin of that missing state. The second slip is to think the conjugated spinor knows its charge. Its density and probability current are those of the original; the +e comes from the equation it obeys, and becomes a property of the state only for Unit 9's anticommuting fields.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A free electron solution at rest is ψ = N(0, 0, 1, 1)e(+imc²t/ħ), with N a real constant. Find ψc = iγ²ψ*, its energy and spin direction, compare the two densities, and check that conjugating twice returns ψ.
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.
- Read off ψ: only the lower pair is filled, and at rest H = βmc² gives −mc² there, so iħ∂ₜ ψ = −mc²ψ, an energy of −0.511 MeV. The lower pair (1, 1)/√2 is the σₓ = +1 eigenvector, so the spin points along +x.
- Apply the component rule (ψ₄*, −ψ₃*, −ψ₂*, ψ₁*). The spatial entries are real, so the star acts only on the phase: ψc = N(1, −1, 0, 0)e(−imc²t/ħ).
- Energy: iħ∂ₜ now returns +mc² = +0.511 MeV. Both phases turn at mc²/ħ = (8.187 × 10⁻¹⁴ J)/(1.0546 × 10⁻³⁴ J s) = 7.76 × 10²⁰ rad s⁻¹, in opposite senses.
- Spin: the filled pair is now the upper one, (1, −1)/√2, the σₓ = −1 eigenvector, so ⟨Σₓ⟩ goes from +1 to −1, as (ψc)†Σψc = −ψ†Σψ requires.
- Densities: ψ†ψ = N²(1 + 1) = 2N² and (ψc)†ψc = N²(1 + 1) = 2N², equal. Conjugating N(1, −1, 0, 0)e(−imc²t/ħ) again gives N(0, 0, 1, 1)e(+imc²t/ħ), so (ψc)c = ψ.
Answer. ψc = N(1, −1, 0, 0)e(−imc²t/ħ): energy +0.511 MeV (from −0.511 MeV), spin along −x (from +x), the same density 2N², and (ψc)c = ψ.
MediumExample 2
A positron is to move with momentum 0.800 MeV/c along +y, spin up along z. Build u(p, ↑) for the electron mass, form iγ²u*, and show it is Unit 4's v(p, ↑). Then show that iγ²u, without the complex conjugate, fails the v equation, and check v†v = 2E/c and v̄v = −2mc.
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.
- Kinematics: E = √(0.800² + 0.511²) MeV = 0.9493 MeV, E + mc² = 1.4603 MeV, a = cp/(E + mc²) = 0.5478, and the prefactor √((E + mc²)/c) = 1.2084 in √(MeV/c).
- With σ⋅p = pσy and σy(1, 0) = (0, i): u(p, ↑) = 1.2084(1, 0, 0, 0.5478i) = (1.2084, 0, 0, 0.6620i).
- Conjugate, then apply (ψ₄*, −ψ₃*, −ψ₂*, ψ₁*): iγ²u* = (−0.6620i, 0, 0, 1.2084). Unit 4's v(p, ↑) has η = −iσy(1, 0) = (0, 1) below and aσy(0, 1) = (−0.5478i, 0) above, times 1.2084: the same spinor.
- Without the star, iγ²u = (+0.6620i, 0, 0, 1.2084). The first row of (γμ pμ + mc)v = 0 demands (E + mc²)vA = cσ⋅p vB. In MeV √(MeV/c) the left side is 1.4603 × 0.6620i = 0.9667i and the right side is −0.9667i, leaving 1.933i: this is v(−p, ↑), the spinor of the wrong momentum.
- Norms: v†v = 0.6620² + 1.2084² = 0.4383 + 1.4603 = 1.8985 MeV/c = 2E/c, and v̄v = 0.4383 − 1.4603 = −1.0220 MeV/c = −2mc. Conjugating back, iγ²[v exp(ipμxμ/ħ)]* = u exp(−ipμxμ/ħ): energy +E, momentum along +y, spin up, the positron asked for.
Answer. iγ²u* = (−0.662i, 0, 0, 1.208) √(MeV/c) = v(p, ↑). Without the star the first row leaves 1.93i MeV √(MeV/c): that is v(−p, ↑). v†v = 1.899 MeV/c = 2E/c and v̄v = −1.022 MeV/c = −2mc.
HardExample 3
A negative-energy electron solution has momentum 2.00 MeV/c along +z and Sz = +ħ/2: ψ = N(a, 0, −1, 0)e(+i|E|t/ħ)e(+ipz/ħ). Find |E| and a, show that ψc is u(−pẑ, ↓) times its plane wave, compute the velocity jz/ρ of both, and decide what electric current each carries, comparing with topic 8.04's sea with one vacancy.
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.
- |E| = √(2.00² + 0.511²) MeV = 2.064 MeV, and a = cp/(|E| + mc²) = 2.00/2.575 = 0.777. Unit 4's v(−pẑ, ↓) has η = −iσy(0, 1) = (−1, 0) below and −aσz(−1, 0) = (a, 0) above: the given ψ, with N = √(2.575 MeV/c) = 1.605 √(MeV/c).
- Conjugate: (ψ₄*, −ψ₃*, −ψ₂*, ψ₁*) = N(0, 1, 0, a), and the plane wave becomes e(−i|E|t/ħ)e(−ipz/ħ). Upper pair (0, 1), lower pair −aσz(0, 1) = (0, a): this is u(−pẑ, ↓), with energy +2.064 MeV, pz = −2.00 MeV/c and Sz = −ħ/2.
- Velocity: with ψ = (φ; χ), jz/ρ = 2c Re(φ†σzχ)/ψ†ψ. For ψ, 2a(−1)/(1 + a²) = −1.553/1.603 = −0.969; for ψc the same expression gives 2(1)(−a)/(1 + a²) = −0.969 again. Both move at −0.969c, and pc²/|E| = 2.00/2.064 = 0.969 confirms the size.
- So the probability current is unchanged, as derived for c-number spinors. The missing electron moves along −z although its momentum is +z, topic 4.04's reversed current, and the positron moves along −z because its momentum is −z. Helicity is positive for both: +p with +ħ/2, and −p with −ħ/2.
- Charge: the missing electron carries (−e)(−0.969cρ) = +0.969ecρ along z, so the sea with that vacancy carries −0.969ecρ. Reading ψc with q = +e gives (+e)(−0.969cρ) = −0.969ecρ, the same; reading it with the electron's −e would give the wrong sign. The spinor alone cannot tell which.
Answer. |E| = 2.064 MeV, a = 0.777; ψc = N(0, 1, 0, 0.777)e(−i|E|t/ħ)e(−ipz/ħ) = u(−pẑ, ↓): +2.064 MeV, −2.00 MeV/c, Sz = −ħ/2. Both move at −0.969c; the current −0.969ecρ along z follows only once q = +e is assigned.
Quick check
Select an answer and explain it before checking.
Review & continue
Charge conjugation is the antilinear map ψ → ψc = iγ²ψ*, complex conjugation followed by the matrix iγ², which turns every solution of the Dirac equation for charge q into a solution for charge −q with the same mass.
- 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.