Chirality in QED interactions

Chirality stopped being a curiosity of the Dirac algebra the day the weak interaction turned out to couple only to left-chiral fermions. The modern electroweak theory is chiral to its foundations: left-handed quarks and leptons live in $SU(2)_L$ doublets while their right-handed partners are singlets that the $W$ never touches, and the requirement that the resulting chiral gauge anomalies cancel is what forces quarks and leptons to come in matched generations with just the right charges. The single most direct measurement of chirality is the left–right asymmetry $A_{LR}$ at the SLC: the SLD experiment at SLAC collided longitudinally polarized electrons — beams prepared in a definite chirality — against positrons at the $Z$ pole and measured the difference in $Z$ production rate between left- and right-handed electron beams. Because a 45.6 GeV electron's helicity is its chirality to eleven decimal places (Example 1), that asymmetry reads out the chiral couplings of the $Z$ directly, yielding the electroweak mixing angle $\sin^2\theta_W^{\text{eff}}=0.23098\pm0.00026$ — one of the most precise numbers in the Standard Model. The same chiral bookkeeping runs through neutrino physics (only left-handed neutrinos have ever been seen), through the Higgs mechanism (the Yukawa term $\bar\psi_L\phi\,\psi_R$ is the chirality-flip that turns massless Weyl fermions into massive Dirac ones), and through the long-standing problem of putting chiral fermions on a lattice, where the Nielsen–Ninomiya theorem shows Weyl's decomposition resists naive discretization.

The vertex is where a fermion line meets a photon line, and the algebra of that meeting is the whole story. Decompose the incoming and outgoing electrons into their left- and right-handed chiral parts. The photon carries a $\gamma^\mu$, and $\gamma^\mu$ slides through a projector by flipping it, $P_R\gamma^\mu=\gamma^\mu P_L$. Chase that flip through the current and the mixed pieces collapse to zero: a left-handed electron can only emerge left-handed, a right-handed one right-handed. The first figure is that selection rule drawn as three vertices — two that survive, one that cannot exist.

The selection rule is exact, but it is a statement about chirality — the abstract $\gamma^5$ label — not about the spin a detector reads out. Those coincide only in the massless limit. For a real electron with mass, a state of definite helicity (spin along its motion) is a superposition of both chiralities, and the amount of the "wrong" chirality is controlled entirely by how relativistic the particle is. The second figure makes that quantitative: it plots the exact chiral content of a right-handed-helicity electron against its speed, and the mixing has a strikingly simple form.

Chirality begins with one new matrix. In four spacetime dimensions the Dirac algebra admits a fifth anticommuting matrix, conventionally built from the product of all four $\gamma^\mu$. It squares to the identity, is Hermitian, and — the property that does all the work — anticommutes with every one of the original gamma matrices.

where: $\gamma^\mu$ — the four Dirac matrices; the factor $i$ makes $\gamma^5$ Hermitian and $(\gamma^5)^2=+1$; the anticommutator $\{\gamma^5,\gamma^\mu\}=\gamma^5\gamma^\mu+\gamma^\mu\gamma^5=0$ is equivalent to $\gamma^5\gamma^\mu=-\gamma^\mu\gamma^5$. Read it as: $\gamma^5$ is a genuinely new element of the algebra, not a function of the others, and it clashes with $\gamma^\mu$ by a sign. Because $\gamma^5$ has eigenvalues $\pm1$, we can build projectors onto its two eigenspaces. These are the chiral (left- and right-handed) projection operators, and they obey the standard projector algebra of any pair of complementary projectors.

where: $P_R$ projects onto the right-handed chiral eigenstate ($\gamma^5u_R=+u_R$) and $P_L$ onto the left-handed ($\gamma^5u_L=-u_L$); any spinor decomposes as $u=P_Lu+P_Ru=u_L+u_R$. For antiparticle spinors the roles swap, $v_R=P_Lv$ and $v_L=P_Rv$, because $v$ enters the theory with the opposite $\gamma^5$ assignment. The reading: $P_L$ and $P_R$ cut every Dirac spinor cleanly into two non-overlapping halves. The physics enters through the QED current. The photon couples to $\bar\psi\gamma^\mu\psi$, a pure vector. Sliding a projector past the $\gamma^\mu$ flips it, $P_R\gamma^\mu=\gamma^\mu P_L$, and the vector current splits into two same-chirality pieces with the cross terms gone.

where: $\bar\psi_L=\bar\psi P_R$ and $\bar\psi_R=\bar\psi P_L$ (the bar flips the projector through $\gamma^0$); the vector current is chirality-diagonal. Contrast the scalar bilinear, $\bar\psi\psi=\bar\psi_L\psi_R+\bar\psi_R\psi_L$, which is purely chirality-flipping — this is why a Dirac mass term connects the two halves. Read it as: the photon preserves chirality, mass violates it. The abstract chirality label is tied to the measurable helicity by the particle's speed. Projecting a definite-helicity spinor onto its chiral components and squaring gives an exact probability that involves only $\beta=p/E$ — no reference to the representation survives.

where: $h=+$ denotes a right-handed helicity state (spin along momentum); $\beta=p/E$; the "wrong" chirality is the left-handed component of a right-helicity electron. As $\beta\to1$ the wrong fraction vanishes and helicity equals chirality; at rest ($\beta=0$) the split is an even $\tfrac12/\tfrac12$. The reading: helicity and chirality are the same observable only in the ultra-relativistic limit.

1 · The wrong-chirality fraction of an electron (fully worked)

A right-helicity electron has total energy $E=1.00$ MeV ($m_e=0.511$ MeV). What fraction of it is "wrong" (left-) chirality, and how does that compare to a 45.6 GeV LEP beam electron?

Step 1. Momentum and speed. $p=\sqrt{E^2-m^2}=\sqrt{1.000-0.511^2}=\sqrt{0.739}=0.860$ MeV, so $\beta=p/E=0.860$. The electron is already fairly relativistic but nowhere near the massless limit.

Step 2. Chiral content from equation (4). $P(\text{wrong})=(1-\beta)/2=(1-0.860)/2=\mathbf{0.070}$, and $P(\text{right})=(1+\beta)/2=0.930$. As a cross-check via the projector form, $\kappa=p/(E+m)=0.860/1.511=0.569$, and $\tfrac12(1-\kappa)^2/(1+\kappa^2)=\tfrac12(0.431)^2/1.324=0.070$ — identical. A 1 MeV right-helicity electron is 93% right-chiral, 7% wrong.

Step 3. The collider contrast. For $E=45.6$ GeV, $\gamma=E/m=8.9\times10^4$, so $1-\beta\approx1/2\gamma^2=6.3\times10^{-11}$ and $P(\text{wrong})=(1-\beta)/2\approx m^2/4E^2=\mathbf{3.1\times10^{-11}}$. At the Z pole helicity and chirality agree to eleven decimal places — which is exactly why a polarized-beam experiment can treat a measured beam helicity as a pure chirality and read off electroweak couplings directly.

2 · Counting the surviving amplitudes in $e^+e^-\to\mu^+\mu^-$

At $\sqrt{s}=91$ GeV, far above $m_\mu$, how many of the sixteen helicity amplitudes are nonzero, and what does the survivor pattern predict for the angular distribution?

Step 1. Enumerate. Each of the four particles ($e^-,e^+,\mu^-,\mu^+$) has two helicity states, so there are $2^4=\mathbf{16}$ configurations to consider.

Step 2. Apply chirality conservation at each vertex (Derivation C). The annihilation current forces $e^-$ and $e^+$ to have opposite helicities — 2 of 4 initial states survive. The production current forces $\mu^-,\mu^+$ opposite — 2 of 4 final states survive. Nonzero amplitudes: $2\times2=\mathbf{4}$; the remaining twelve vanish.

Step 3. Read off the physics. The four survivors are $(e^-_Re^+_L\to\mu^-_R\mu^+_L)$, $(e^-_Re^+_L\to\mu^-_L\mu^+_R)$, and the two with initial helicities swapped. Because a right-handed initial current defines a spin-1 axis along the beam, the amplitudes carry the rotation matrices $\tfrac12(1+\cos\theta)$ and $\tfrac12(1-\cos\theta)$; squaring and summing gives the classic $\mathrm{d}\sigma/\mathrm{d}\Omega\propto1+\cos^2\theta$ for unpolarized beams, with a $\cos\theta$ interference term once parity-violating $Z$ exchange is added. Four amplitudes, and their angular structure is the $1+\cos^2\theta$ line shape seen at every $e^+e^-$ collider.

Chirality entered physics as a piece of pure mathematics — a two-component way of writing the Dirac equation — decades before anyone knew nature would use it to build the weak force. The man who wrote it down was chasing symmetry, not weak decays.

Texts. Thomson, Modern Particle Physics, §6.4–§6.4.2 and Eqs. (6.27)–(6.38) (the source this article grounds) · Peskin & Schroeder, An Introduction to QFT, §3.2 and §5.2 (chiral spinors, helicity amplitudes) · Schwartz, QFT and the Standard Model, Ch. 11 (spinors and chirality) · Griffiths, Introduction to Elementary Particles, 2nd ed., §7.7 · Halzen & Martin, Quarks and Leptons, §5.3–§6. The key relations used throughout are: (1) $\{\gamma^5,\gamma^\mu\}=0$, $(\gamma^5)^2=1$; (2) $P_R\gamma^\mu=\gamma^\mu P_L$ and $\bar u_R\gamma^\mu u_L=0$; (3) $P(\text{wrong})=\tfrac12(1-\kappa)^2/(1+\kappa^2)=(1-\beta)/2$, evaluated at $E=1$ MeV ($0.070$) and $E=45.6$ GeV ($3.1\times10^{-11}=m^2/4E^2$); (4) the $16\to4$ amplitude count for $e^+e^-\to\mu^+\mu^-$.

Prerequisites. The Dirac Equation · Gamma Matrices and the Clifford Algebra · Spin and Helicity States · Dirac Plane-Wave Solutions. Leads to. The Weak Charged Current · V−A Structure of the Weak Interaction · Parity Violation · $e^+e^-\to\mu^+\mu^-$.

Ask the tutor: Ask me to walk through why $\bar u_L\gamma^\mu u_R=0$ line by line, or to derive the $(1-\beta)/2$ wrong-chirality fraction for any beam energy you name.