Chirality & Weyl spinors

A four-component Dirac spinor looks like an atom of the theory — indivisible, the smallest object carrying a relativistic electron. It is not. In the right basis the Dirac field falls cleanly into two halves the free equation never mixes: two two-component Weyl spinors of opposite chirality. Chirality is the eigenvalue of the matrix $\gamma^5$ , the deepest handedness in physics — a Lorentz-invariant label no boost or rotation can change, unlike helicity (the projection of spin on momentum), which flips the moment you overtake a massive particle. The two are often confused because they coincide exactly in one limit — massless — and nearly coincide for anything ultrarelativistic. Why does this bookkeeping matter? Because Nature reads the label. The electromagnetic and strong forces are chirality-blind, coupling left and right identically. The weak force is not: its charged current carries the projector $\tfrac12(1-\gamma^5)$ and couples only to the left-chiral half of every fermion — the single interaction that can tell a Weyl spinor from its mirror image. That one fact is the origin of parity violation, of the $V\!-\!A$ structure of beta decay, of why neutrinos come almost only left-handed, and of the helicity suppression that makes the pion prefer to decay to a muon over an electron by four orders of magnitude. This article builds the machinery — $\gamma^5$ , the projectors, the chiral basis, the Weyl equations — and shows the physics it unlocks. A mass term is the one thing that stitches the two halves back together, which is exactly why chirality stops being conserved the instant a particle has mass.

Key takeaways

The Dirac spinor is reducible: in the chiral basis it splits into two two-component Weyl spinors $\psi_L,\psi_R$ that the free, massless equation never couples.
Chirality is the eigenvalue of $\gamma^5$ (with $(\gamma^5)^2=1$ ), projected out by $P_{L,R}=\tfrac12(1\mp\gamma^5)$ ; it is Lorentz-invariant, unlike helicity.
A Dirac mass term $-m\bar\psi\psi=-m(\bar\psi_L\psi_R+\bar\psi_R\psi_L)$ mixes the two chiralities — so chirality is conserved only for $m=0$ .
The weak charged-current vertex $\dfrac{g_w}{2\sqrt2}\gamma^\mu(1-\gamma^5)$ couples only to $\psi_L$ : maximal parity violation, the $V\!-\!A$ law.
For a massive fermion, chirality equals helicity only up to a wrong-chirality probability $\tfrac12(1-\beta)\approx(m/2E)^2$ — vanishing as $\beta\to1$ .
Key idea: a four-component Dirac field is really two glued-together two-component Weyl fields of opposite chirality, and the weak force is the one interaction that can tell them apart.

The core insight

A four-component Dirac field is really two glued-together two-component Weyl fields of opposite chirality; a mass term is the glue, and the weak force is the one interaction that can tell them apart.

Where this lives today

Chirality is the load-bearing wall of the Standard Model, a chiral gauge theory — left-handed fermions sit in weak-isospin doublets, right-handed ones in singlets, so the two chiralities carry different quantum numbers and the theory violates parity. This is why the $W$ couples only to $\psi_L$ , why the Higgs mechanism is needed to give fermions mass without a gauge-violating bare Dirac term, and why the cancellation of gauge anomalies — a purely chiral, $\gamma^5$ -driven effect — forces quarks and leptons into complete generations with just the right charges. The frontier question is the neutrino. If strictly massless it would be a pure Weyl fermion, only ever left-handed; but oscillations prove neutrinos have mass, so a right-handed component must exist — and whether its partner is a distinct Dirac state or the neutrino's own antiparticle (a Majorana mass, pairing a left Weyl spinor with itself) is unknown. Neutrinoless double-beta-decay experiments — KamLAND-Zen, LEGEND, nEXO — hunt for the Majorana signature, a lepton-number-violating decay possible only if the neutrino's two chiral halves are the same field. Meanwhile Weyl's spinors have a second life in condensed matter: in Weyl semimetals such as tantalum arsenide the low-energy excitations obey the massless Weyl equation, with a chiral anomaly read off as a negative magnetoresistance — the 1929 equation, once rejected, now measured on a voltmeter.

The physical picture

Take a Dirac electron and stop thinking of it as one object. In the chiral (Weyl) basis the $\gamma$ -matrices are block off-diagonal and $\gamma^5$ is block diagonal — reading $-1$ on the top two components and $+1$ on the bottom two. Those blocks are the left- and right-chiral Weyl spinors $\psi_L,\psi_R$ , and the projectors $P_L=\tfrac12(1-\gamma^5)$ , $P_R=\tfrac12(1+\gamma^5)$ are the scalpels that separate them. The first figure is that dissection, with the one interaction that acts on only one of the pieces.

Figure

The Dirac field ψ splits under the projectors P_L = ½(1 − γ⁵) and P_R = ½(1 + γ⁵) into two Weyl spinors with γ⁵ eigenvalues −1 and +1; the weak charged current ½(1 − γ⁵) couples only to the left-chiral half ψ_L, while a mass term m is the one operator that bridges ψ_L and ψ_R

The second figure separates the two ideas people most often merge. Helicity is the projection of spin onto momentum — a measurable orientation. Chirality is the $\gamma^5$ label. For a massless particle these coincide: the left Weyl equation forces spin exactly anti-parallel to momentum. But give the particle mass and you can run faster than it; from the overtaking frame its momentum reverses while its spin does not, so its helicity flips — yet its chirality is unchanged, because a boost can never rotate a $(\tfrac12,0)$ Weyl spinor into a $(0,\tfrac12)$ one. Quantitatively, a definite-helicity massive fermion is a mixture of both chiralities, the "wrong" component having probability $\tfrac12(1-\beta)$ , which the figure plots.

Figure

A negative-helicity fermion of velocity β is a mix of chiralities: left-chiral probability ½(1 + β) and right-chiral (wrong) probability ½(1 − β) ≈ (m/2E)². The curve falls to zero as β → 1, where chirality equals helicity; a 1 GeV electron has wrong-chirality probability 6.5×10⁻⁸

The lesson of the two figures together: chirality is the algebraic label the theory is built from, helicity is what a detector measures, and mass is the dial connecting them. At zero mass the two coincide and the weak force's preference for $\psi_L$ becomes a preference for a fixed helicity.

The core equations

Everything begins with $\gamma^5$ , the fifth gamma matrix — the unique (up to sign) matrix built from the four Dirac matrices that anticommutes with all of them. That is what makes it a good "handedness" operator: it commutes with the Lorentz generators (so chirality is invariant) but anticommutes with $\gamma^\mu$ (so it flips under the mass term's axial structure).

\gamma^5 \equiv i\,\gamma^0\gamma^1\gamma^2\gamma^3, \qquad (\gamma^5)^2 = \mathbb{1}, \qquad \{\gamma^5,\gamma^\mu\}=0, \qquad (\gamma^5)^\dagger=\gamma^5 \qquad (1)

where: $\gamma^\mu$ — the four Dirac matrices obeying $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$ ; $\{A,B\}=AB+BA$ is the anticommutator. Reading: $\gamma^5$ is Hermitian with eigenvalues $\pm1$ (since it squares to the identity), and those eigenvalues are the chirality quantum number. Because $\gamma^5$ squares to one, the operators $\tfrac12(1\pm\gamma^5)$ are orthogonal projectors. They slice any Dirac spinor into its two chiral pieces.

P_L=\tfrac12(1-\gamma^5), \quad P_R=\tfrac12(1+\gamma^5), \qquad P_L^2=P_L,\; P_R^2=P_R,\; P_LP_R=0,\; P_L+P_R=1 \qquad (2)

where: $\psi_L\equiv P_L\psi$ and $\psi_R\equiv P_R\psi$ are the left- and right-chiral fields, with $\gamma^5\psi_L=-\psi_L$ and $\gamma^5\psi_R=+\psi_R$ . Reading: the projectors are complete ( $\psi=\psi_L+\psi_R$ always) and mutually exclusive ( $P_LP_R=0$ , so a purely left field has no right part). To see the split, go to the chiral basis, where $\gamma^5$ is diagonal and the Dirac spinor is literally stacked from two Weyl spinors.

\gamma^0=\begin{pmatrix}0&\mathbb{1}\\\mathbb{1}&0\end{pmatrix},\;\; \gamma^i=\begin{pmatrix}0&\sigma^i\\-\sigma^i&0\end{pmatrix},\;\; \gamma^5=\begin{pmatrix}-\mathbb{1}&0\\0&\mathbb{1}\end{pmatrix},\;\; \psi=\begin{pmatrix}\psi_L\\ \psi_R\end{pmatrix} \qquad (3)

where: $\sigma^i$ — the three Pauli matrices; each block is $2\times2$ . Reading: $\gamma^5=\mathrm{diag}(-\mathbb{1},+\mathbb{1})$ makes the upper two-component object $\psi_L$ (eigenvalue $-1$ ) and the lower $\psi_R$ (eigenvalue $+1$ ); the off-diagonal $\gamma^\mu$ link each Weyl spinor only to itself, while a mass term links them to each other. Splitting the Dirac equation in this basis gives the two Weyl equations, exposing the mass term as the sole coupling between chiralities.

i\bar\sigma^\mu\partial_\mu\psi_L=m\,\psi_R,\qquad i\sigma^\mu\partial_\mu\psi_R=m\,\psi_L,\qquad \sigma^\mu=(\mathbb{1},\vec\sigma),\;\bar\sigma^\mu=(\mathbb{1},-\vec\sigma) \qquad (4)

where: at $m=0$ the two lines decouple into independent massless Weyl equations $i\bar\sigma^\mu\partial_\mu\psi_L=0$ and $i\sigma^\mu\partial_\mu\psi_R=0$ . Reading: the right-hand sides carry the only factor of $m$ ; set $m\to0$ and left and right propagate as separate particles — which is why a lone Weyl fermion must be massless. Finally, the one place the label is physical: the weak charged current. The $W$ boson couples through a vector-minus-axial-vector structure that is exactly the left projector.

\mathcal{M}\propto \bar u\,\frac{g_w}{2\sqrt2}\,\gamma^\mu(1-\gamma^5)\,u = \bar u\,\frac{g_w}{\sqrt2}\,\gamma^\mu P_L\,u \qquad (5)

where: $g_w$ — the weak coupling; $u$ — a fermion spinor; $\gamma^\mu(1-\gamma^5)=2\gamma^\mu P_L$ is the $V\!-\!A$ current. Reading: the projector slides onto the spinor, so only the left-chiral part of $u$ survives — the weak force is chirally blind to $\psi_R$ .

Derivation ASplitting the Dirac spinor with the chiral projectors

The goal is to prove that the projectors $P_{L,R}=\tfrac12(1\mp\gamma^5)$ genuinely decompose an arbitrary Dirac spinor into two complete, non-overlapping chiral pieces that are eigenstates of $\gamma^5$ — the algebraic content behind "the Dirac field is reducible into two Weyl spinors." We work entirely from the two defining facts $(\gamma^5)^2=1$ and $\{\gamma^5,\gamma^\mu\}=0$ , so nothing depends on a basis choice; the block form of equation (3) is just a convenient picture of a basis-independent truth.

Step 1. Verify $P_L$ and $P_R$ are idempotent. Using $(\gamma^5)^2=1$ , compute $P_L^2=\tfrac14(1-\gamma^5)(1-\gamma^5)=\tfrac14(1-2\gamma^5+(\gamma^5)^2)=\tfrac14(2-2\gamma^5)=\tfrac12(1-\gamma^5)=P_L$ , and identically $P_R^2=P_R$ . Why: an operator equal to its own square is a true projector — applying it twice changes nothing further.

Step 2. Verify completeness and orthogonality. Adding, $P_L+P_R=\tfrac12(1-\gamma^5)+\tfrac12(1+\gamma^5)=1$ , so $\psi=P_L\psi+P_R\psi=\psi_L+\psi_R$ for every $\psi$ . Multiplying, $P_LP_R=\tfrac14(1-\gamma^5)(1+\gamma^5)=\tfrac14(1-(\gamma^5)^2)=0$ . Why: completeness means no part of the spinor is lost, orthogonality means the two parts share nothing — a genuine direct sum.

Step 3. Show the pieces are $\gamma^5$ eigenstates. Act with $\gamma^5$ on $\psi_L=\tfrac12(1-\gamma^5)\psi$ : $\gamma^5\psi_L=\tfrac12(\gamma^5-(\gamma^5)^2)\psi=\tfrac12(\gamma^5-1)\psi=-\psi_L$ , and likewise $\gamma^5\psi_R=+\psi_R$ . So $\psi_L$ has chirality $-1$ and $\psi_R$ chirality $+1$ . Why: chirality is defined as the $\gamma^5$ eigenvalue, so this makes the two pieces "left-chiral" and "right-chiral" by definition.

Step 4. Confirm Lorentz invariance of the split. The Lorentz generators of the spinor representation are $S^{\mu\nu}=\tfrac{i}{4}[\gamma^\mu,\gamma^\nu]$ , and since $\gamma^5$ anticommutes with each $\gamma^\mu$ it commutes with every commutator of two of them: $[\gamma^5,S^{\mu\nu}]=0$ . Therefore $[\gamma^5,P_{L,R}]=0$ and a boost or rotation cannot change a spinor's chirality. Why: because $P_{L,R}$ commute with Lorentz transformations, "left-chiral" is a frame-independent statement, which helicity is not.

Go deeper · Derivation B · 1 minWhy a mass term needs both chiralities and the massless limit gives Weyl fermions

Derivation BWhy a mass term needs both chiralities and the massless limit gives Weyl fermions

The aim is to show that the Dirac Lagrangian's two pieces behave oppositely under the chiral split: the kinetic term keeps each chirality to itself while the mass term couples them, so setting $m=0$ leaves two independent Weyl fields. This is the structural reason chirality is conserved only for massless particles, and the origin of the Weyl equations (4). We use the projector identities of the previous derivation with the anticommutation $\{\gamma^5,\gamma^\mu\}=0$ .

Step 1. Establish the key sliding rule for the adjoint. Since $\bar\psi=\psi^\dagger\gamma^0$ and $\gamma^5$ anticommutes with $\gamma^0$ , one finds $\overline{\psi_L}=\overline{P_L\psi}=\psi^\dagger P_L^\dagger\gamma^0=\psi^\dagger P_L\gamma^0=\psi^\dagger\gamma^0 P_R=\bar\psi P_R$ . So the adjoint of a left field carries a right projector. Why: this crossing of the label under conjugation decides which bilinears survive — the engine of the result.

Step 2. Kill the cross terms in the kinetic bilinear. The kinetic term is $i\bar\psi\gamma^\mu\partial_\mu\psi$ . Split $\psi=\psi_L+\psi_R$ and use Step 1 plus $P_L\gamma^\mu=\gamma^\mu P_R$ : a term like $\bar\psi_L\gamma^\mu\psi_R=\bar\psi P_R\gamma^\mu P_R\psi'=\bar\psi\gamma^\mu P_L P_R\psi'=0$ . Only $\bar\psi_L\gamma^\mu\partial_\mu\psi_L$ and $\bar\psi_R\gamma^\mu\partial_\mu\psi_R$ survive. Why: the kinetic term is chirality-diagonal — it never turns a left field into a right one, so by itself it conserves chirality.

Step 3. Show the mass term is purely off-diagonal. For $-m\bar\psi\psi$ , the same crossing gives $\bar\psi_L\psi_L=\bar\psi P_R P_L\psi=0$ , while $\bar\psi_L\psi_R=\bar\psi P_R P_R\psi\neq0$ . Hence $-m\bar\psi\psi=-m(\bar\psi_L\psi_R+\bar\psi_R\psi_L)$ . Why: the mass term does the opposite — it exists only by pairing opposite chiralities, so a nonzero mass forces left and right to mix.

Step 4. Take the massless limit. With $m=0$ the Lagrangian is $i\bar\psi_L\gamma^\mu\partial_\mu\psi_L+i\bar\psi_R\gamma^\mu\partial_\mu\psi_R$ , two decoupled pieces. Their equations of motion, written in two-component form in the chiral basis, are the Weyl equations $i\bar\sigma^\mu\partial_\mu\psi_L=0$ and $i\sigma^\mu\partial_\mu\psi_R=0$ . Why: with nothing connecting them, $\psi_L$ and $\psi_R$ are different particles — precisely Weyl's 1929 massless fermion.

Go deeper · Derivation C · 1 minChirality equals helicity only in the massless limit

Derivation CChirality equals helicity only in the massless limit

The goal is to compute, for a massive Dirac particle in a helicity eigenstate, exactly how much of each chirality it contains — and to show the "wrong" chirality has probability $\tfrac12(1-\beta)$ , vanishing only as the speed approaches $c$ . This makes precise the refrain that chirality and helicity "coincide in the massless limit," and it is the derivation behind helicity suppression in weak decays. We use the explicit chiral-basis positive-energy spinor and a helicity eigenspinor $\xi_\mp$ with $\vec\sigma\cdot\hat p\,\xi_\mp=\mp\xi_\mp$ .

Step 1. Write the spinor. In the chiral basis the positive-energy solution is $u(p)=\big(\sqrt{p\cdot\sigma}\,\xi,\ \sqrt{p\cdot\bar\sigma}\,\xi\big)^{\!\top}$ , whose upper block is $\psi_L$ and lower block is $\psi_R$ . Here $p\cdot\sigma=E-\vec p\cdot\vec\sigma$ and $p\cdot\bar\sigma=E+\vec p\cdot\vec\sigma$ . Why: the two halves of this standard on-shell spinor are literally the two chiralities, so their sizes are the chiral content.

Step 2. Choose negative helicity. For $\xi_-$ with $\vec\sigma\cdot\hat p\,\xi_-=-\xi_-$ , we get $\vec p\cdot\vec\sigma\,\xi_-=-|\vec p|\,\xi_-$ , so $p\cdot\sigma\,\xi_-=(E+|\vec p|)\xi_-$ and $p\cdot\bar\sigma\,\xi_-=(E-|\vec p|)\xi_-$ . Therefore $u_-(p)=\big(\sqrt{E+|\vec p|}\,\xi_-,\ \sqrt{E-|\vec p|}\,\xi_-\big)^{\!\top}$ . Why: fixing helicity fixes how the momentum operator acts, turning $\sqrt{p\cdot\sigma}$ into concrete $\sqrt{E\pm|\vec p|}$ .

Step 3. Read off the chiral probabilities. The left-chiral (upper) weight is $E+|\vec p|$ and the right-chiral (lower) is $E-|\vec p|$ ; normalizing by their sum $2E$ , $P_L=\tfrac12(1+\beta)$ and $P_R=\tfrac12(1-\beta)$ with $\beta=|\vec p|/E$ . A negative-helicity particle is mostly left-chiral, with a right-chiral admixture $\tfrac12(1-\beta)$ . Why: this quantifies "chirality is nearly helicity for fast particles" — the mismatch is set by how far $\beta$ is from 1.

Step 4. Take the ultrarelativistic limit. Using $E-|\vec p|=\dfrac{m^2}{E+|\vec p|}\approx\dfrac{m^2}{2E}$ , the wrong-chirality probability is $P_R\approx\Big(\dfrac{m}{2E}\Big)^2$ . As $m\to0$ (or $E\to\infty$ ) this vanishes and the negative-helicity state becomes purely left-chiral — chirality equals helicity. Why: the factor $(m/2E)^2$ is exactly what makes the pion decay to a muon rather than an electron, the worked example below. *As a consistency check: the compact form $(m/2E)^2=(0.511/2000)^2=6.5\times10^{-8}$ agrees with $\tfrac12(1-\beta)$ evaluated directly from $\beta=0.99999987$ , confirming that both expressions for the wrong-chirality probability give the same result.*

Limiting cases

Case	Result	Meaning
Massless, $m=0$	chirality $=$ helicity exactly	$\psi_L,\psi_R$ decouple into independent Weyl fermions; the label is physical spin orientation
Ultrarelativistic, $E\gg m$	helicity $=$ chirality up to $(m/2E)^2$	wrong-chirality probability $\tfrac12(1-\beta)\approx(m/2E)^2$ is tiny but nonzero
Rest frame, $\vec p\to0$	helicity undefined; chirality still defined	with no momentum there is no axis to project spin on, but $\gamma^5$ still labels the state
Dirac mass on	$-m(\bar\psi_L\psi_R+\bar\psi_R\psi_L)$ mixes $L\leftrightarrow R$	chirality is not conserved; the propagator flips chirality once per mass insertion
Pure $V\!-\!A$ weak vertex	only $P_L$ couples ( $P_R$ decoupled)	maximal parity violation; right-handed neutrinos are sterile under the weak force

Common trapChirality is the same thing as helicity

It is not — this is the single most common confusion in the subject. Helicity is $\hat h=\vec S\cdot\hat p/|\vec S|$ , the projection of spin onto momentum: a physical, measurable orientation, but frame-dependent for any massive particle, because you can overtake the particle and reverse $\hat p$ while its spin stays put, flipping the helicity. Chirality is the eigenvalue of $\gamma^5$ : an algebraic, Lorentz-invariant label that no boost or rotation can change, because $\gamma^5$ commutes with all the Lorentz generators. The two agree exactly only for massless particles, where the Weyl equation locks spin anti-parallel (left) or parallel (right) to momentum. For a massive fermion a helicity eigenstate is a mixture of chiralities, with the wrong-chirality fraction $\tfrac12(1-\beta)\approx(m/2E)^2$ . The weak force couples to definite chirality ( $P_L$ ), and it is only because most weak-decay fermions are ultrarelativistic that this looks like a coupling to definite helicity. Say "left-handed" carefully: for the interaction it means chirality; for a detector it means helicity.

Worked examples

1 · Helicity suppression of pion decay (fully worked)

The charged pion decays by the weak interaction, $\pi^-\to\ell^-\bar\nu_\ell$ , with $\ell=e$ or $\mu$ . Naive phase space favours the electron mode (more energy released), yet experiment finds the electron mode enormously suppressed. Predict the ratio $\Gamma(\pi\to e\nu)/\Gamma(\pi\to\mu\nu)$ from chirality and compare to data. Use $m_e=0.511$ MeV, $m_\mu=105.66$ MeV, $m_\pi=139.57$ MeV.

Step 1. Set up the helicity constraint. The pion is spin-0, so in its rest frame the $\ell^-$ and $\bar\nu$ fly back-to-back with spins summing to zero — they must have the same helicity. The antineutrino is essentially purely right-handed, forcing the charged lepton into its right-helicity state. But the weak vertex $\tfrac12(1-\gamma^5)$ demands left chirality, so the decay proceeds only through the lepton's wrong-chirality admixture.

Step 2. Insert the suppression factor. The wrong-chirality probability (Derivation 3) is $\tfrac12(1-\beta_\ell)$ , so the amplitude carries $\sqrt{\tfrac12(1-\beta_\ell)}$ . Working out the full matrix element, the rate is proportional to $m_\ell^2$ times a phase-space factor: $\Gamma\propto m_\ell^2\,(m_\pi^2-m_\ell^2)^2$ . The rate vanishes as $m_\ell\to0$ — a massless lepton could not decay this way at all.

Step 3. Form the ratio. Dividing the two modes,

\frac{\Gamma(\pi\to e\nu)}{\Gamma(\pi\to\mu\nu)}=\frac{m_e^2}{m_\mu^2}\left(\frac{m_\pi^2-m_e^2}{m_\pi^2-m_\mu^2}\right)^{\!2}.

The mass ratio $(m_e/m_\mu)^2=(0.511/105.66)^2=2.34\times10^{-5}$ hugely favours the muon, opposite to the naive guess.

Step 4. Put in numbers. With $m_\pi^2=1.948\times10^4$ , $m_\mu^2=1.116\times10^4$ , $m_e^2=0.261$ (MeV $^2$ ), the phase-space factor is $(19479.5/8315.8)^2=(2.343)^2=5.49$ . Multiplying: $2.34\times10^{-5}\times5.49=\mathbf{1.28\times10^{-4}}$ .

Step 5. Compare. The measured value is $\Gamma(\pi\to e\nu)/\Gamma(\pi\to\mu\nu)=1.230\times10^{-4}$ ; the tiny gap from our $1.28\times10^{-4}$ is radiative corrections (theory with them: $1.2352\times10^{-4}$ ). A four-order-of-magnitude suppression predicted purely by chirality — one of the sharpest confirmations of the $V\!-\!A$ structure.

2 · Wrong-chirality content of a fast electron

A 1.0 GeV electron is produced in a definite negative-helicity state. What fraction of it is right-chiral (the "wrong" chirality for the weak force)?

Step 1. Get the velocity. With $E=1000$ MeV and $m_e=0.511$ MeV, $|\vec p|=\sqrt{E^2-m^2}\approx E$ to one part in $10^7$ , so $\beta=|\vec p|/E=0.99999987$ .

Step 2. Apply the result. From Derivation 3 the wrong-chirality probability is $P_R=\tfrac12(1-\beta)$ . Compute $1-\beta=1.31\times10^{-7}$ , so $P_R=6.5\times10^{-8}$ .

Step 3. Cross-check with the compact form. $P_R\approx(m/2E)^2=(0.511/2000)^2=6.5\times10^{-8}$ — the two agree. A GeV-scale electron is left-chiral to better than one part in ten million, which is why "helicity" and "chirality" are used interchangeably in high-energy beams despite being conceptually distinct.

Physics of the portrait

Weyl spinors carry the name of the mathematician who wrote them down a year after Dirac, then watched the idea sit dormant until the weak force made it indispensable.

Origins · 1 minHermann Weyl and how this was discovered

Hermann Weyl

1885–1955 · mathematician of symmetry, gauge invariance, and the two-component spinor

Hermann Weyl moved fluently between pure mathematics and fundamental physics and left permanent marks on both. A student of Hilbert at Göttingen, he built much of the representation theory of Lie groups — the "Weyl group," "character formula," and "Weyl chamber" are his — the very machinery that classifies how spinors transform under the Lorentz group. In 1918 he proposed the first gauge theory; the original scale-invariance idea was wrong, but the principle of local invariance, transplanted to the quantum phase, became the organizing law of modern particle physics. In 1929, in "Elektron und Gravitation," he introduced the two-component massless spinor equation now named after him, showing a relativistic spin-½ particle need not have four components if it has no mass. Pauli attacked it for manifestly violating parity, and it was dismissed as unphysical for nearly thirty years. Then in 1956–57 parity violation was discovered in weak decays, and the two-component neutrino leapt from discarded curiosity to the natural description of the weakly interacting world. Weyl, who had left Germany in 1933 for the Institute for Advanced Study alongside Einstein, died in 1955 before the full vindication — but his spinors, gauge principle, and group theory now sit at the foundation of the Standard Model, and "Weyl fermion" reentered the language in 2015 when the massless quasiparticles were observed in a crystal.

History

1928

— Dirac writes his relativistic wave equation, forcing four-component spinors

1929

— Weyl proposes the two-component massless spinor equation

1930

— Pauli postulates the neutrino, later the archetypal chiral fermion

1956

— Lee and Yang question whether parity is conserved in weak decays

1957

— Wu's cobalt-60 experiment confirms parity violation; the two-component neutrino is revived

1958

— Feynman–Gell-Mann and Sudarshan–Marshak formulate the V−A theory

2015

— Weyl fermions observed as quasiparticles in the semimetal tantalum arsenide

Going further

Texts. Peskin & Schroeder, An Introduction to Quantum Field Theory, §3.2 (the chiral basis and Weyl spinors) and §3.5 · Schwartz, Quantum Field Theory and the Standard Model, Ch. 10–11 (chirality, helicity, and Weyl vs Dirac) · Griffiths, Introduction to Elementary Particles, 2nd ed., §10.7 (neutral and charged weak currents) · Thomson, Modern Particle Physics, §11.3 (chirality and the weak interaction) · Halzen & Martin, Quarks and Leptons, Ch. 12 (parity violation and the $V\!-\!A$ current). The key results of this article are: (1) $P_{L,R}^2=P_{L,R}$ , $P_LP_R=0$ , $P_L+P_R=1$ ; (2) $-m\bar\psi\psi=-m(\bar\psi_L\psi_R+\bar\psi_R\psi_L)$ with the kinetic term chirality-diagonal; (3) $P_R=\tfrac12(1-\beta)\approx(m/2E)^2$ ; (4) $\Gamma(\pi\to e\nu)/\Gamma(\pi\to\mu\nu)=(m_e/m_\mu)^2\,[(m_\pi^2-m_e^2)/(m_\pi^2-m_\mu^2)]^2=1.28\times10^{-4}$ , against the measured $1.230\times10^{-4}$ .

Prerequisites. The Dirac Equation · Gamma Matrices and the Clifford Algebra · Spin from the Dirac Equation · Spin and Helicity States. Leads to. Chiral Structure of the Weak Force · V−A Theory · Parity Violation · Majorana vs Dirac Neutrinos.

Ask the tutor: Ask me to walk through the pion helicity-suppression amplitude line by line, or to show explicitly why $\gamma^5$ commutes with every Lorentz generator but anticommutes with every $\gamma^\mu$ .

What is the difference between chirality and helicity in one sentence?

Helicity is the physical projection of a particle's spin onto its direction of motion — something a detector can measure but that flips if you boost to a frame moving faster than the particle — whereas chirality is the abstract, Lorentz-invariant eigenvalue of the matrix $\gamma^5$ that no change of frame can alter. They agree exactly for massless particles and approximately (to within $(m/2E)^2$ ) for ultrarelativistic ones, which is why the words are often used interchangeably in practice even though they mean different things. The weak force couples to chirality; a detector reports helicity; mass is what makes the two disagree.

Why must a Weyl fermion be massless?

Because a Dirac mass term $-m(\bar\psi_L\psi_R+\bar\psi_R\psi_L)$ requires both a left- and a right-chiral field to pair up, and a Weyl fermion is by definition only one of them. With just $\psi_L$ and no $\psi_R$ , there is no gauge- and Lorentz-invariant Dirac mass to write — the needed term does not exist. There is one loophole: if the particle is its own antiparticle you can build a Majorana mass pairing $\psi_L$ with its own charge conjugate, exactly the possibility being tested for neutrinos. Absent that, a lone Weyl spinor propagates at the speed of light forever.

Is a neutrino a Weyl spinor?

In the original Standard Model, yes — the neutrino was taken as exactly massless and hence a single left-chiral Weyl fermion, explaining why only left-handed neutrinos and right-handed antineutrinos are ever seen. Oscillations changed that: they require neutrino masses, so a right-handed (or Majorana) component must exist and the neutrino is no longer a pure Weyl particle. It remains overwhelmingly left-chiral in any weak process because it is produced ultrarelativistically, so the Weyl picture stays an excellent approximation. Whether the mass is Dirac or Majorana is one of the biggest open questions in particle physics.

Why does the weak force care about chirality but the electromagnetic force does not?

It is in the structure of the currents. The electromagnetic (and strong) current is a pure vector, $\bar\psi\gamma^\mu\psi$ , which splits into $\bar\psi_L\gamma^\mu\psi_L+\bar\psi_R\gamma^\mu\psi_R$ with equal strength — it treats the two chiralities identically and conserves parity. The weak charged current is vector-minus-axial-vector, $\bar\psi\gamma^\mu(1-\gamma^5)\psi$ , and the relative minus sign cancels the right-chiral contribution exactly, leaving $2\bar\psi_L\gamma^\mu\psi_L$ . That is a design feature of the gauge theory — left-handed fermions in weak-isospin doublets, right-handed in singlets — not something derivable from electromagnetism.

Where does the (m/2E)² suppression come from?

From the mismatch between what the weak vertex wants (definite chirality) and what a massive helicity eigenstate can supply. A helicity eigenstate of a massive fermion is not a chirality eigenstate; its wrong-chirality content is $\tfrac12(1-\beta)$ , and since $1-\beta=m^2/[E(E+|\vec p|)]\approx m^2/2E^2$ , that probability is $\approx(m/2E)^2$ . When a decay is forced through the wrong chirality — as pion decay is, by angular momentum conservation — the amplitude carries one power of $m/2E$ and the rate its square, which is why the rate scales as $m_\ell^2$ and the pion favours the heavier muon.

Does chirality being non-conserved for massive particles make it useless?

No — it means chirality is a good quantum number only in the massless limit, but it remains the right variable to build the theory from. The Standard Model is written in chiral fields because that is how the gauge symmetry acts; masses arise from the Higgs connecting the left and right pieces, and the small chirality violation they induce is precisely calculable (the $(m/2E)^2$ factors). So chirality is exactly conserved by the gauge interactions and broken only in the controlled, mass-proportional way the Higgs sector dictates — the bookkeeping that makes the electroweak theory consistent.