Spin & helicity states

Helicity stopped being a bookkeeping choice and became physics the moment the weak interaction was found to care about it. Parity violation means nature distinguishes left from right: the charged-current weak force couples only to left-handed particles and right-handed antiparticles, so the helicity/chirality label of a fermion decides whether it feels the $W$ boson at all. The 1958 Goldhaber experiment pinned the neutrino's helicity at $-1$, and to the precision we can measure, every neutrino ever seen is left-handed — the cleanest handed object in physics. In modern collider calculations the practical payoff is the spinor-helicity formalism: rather than square Dirac traces, theorists compute amplitudes for definite external helicities directly, which collapses page-long QCD expressions into compact "MHV" formulas (Parke–Taylor) and powers the automated matrix-element generators (MadGraph, and the on-shell/BCFW methods behind precision LHC predictions). Experimentally, polarized electron beams at SLAC's SLC measured the left-right asymmetry $A_{LR}$ to determine the weak mixing angle, and a future linear collider (ILC/CLIC) plans beam polarization precisely to project out chosen helicity combinations. And the massless-limit identity helicity $=$ chirality is the conceptual hinge of the whole electroweak theory, where the left-handed doublets and right-handed singlets are assigned by chirality — the frame-independent cousin of the helicity this article builds.

Picture a spin-half particle flying to the right and ask a single yes/no question: is its spin pointing along the flight direction or against it? That projection — spin onto momentum — is the helicity, and it splits every moving fermion into just two states. A right-handed particle carries spin parallel to $\mathbf p$; corkscrew your right hand along the motion and the fingers curl the way the spin points. A left-handed particle carries spin antiparallel. Unlike "spin-up along the lab $z$-axis", this label refers to nothing outside the particle itself, which is exactly why it is robust: rotate your head, tilt the apparatus, and the helicity does not care. The first figure fixes the two handedness states and the eigenvalue arithmetic behind them.

Helicity earns its place because it is conserved for a free particle where $\hat S_z$ is not, but it hides a subtlety that $S_z$ does not: it is not the same for every observer. Since a massive particle travels slower than light, you can always board a faster frame that overtakes it. In that frame the momentum reverses direction while the spin, being unaffected by a boost along its own axis, does not — so the projection of spin on momentum flips sign, and a right-handed electron becomes left-handed. The threshold is the particle's own speed $\beta_p=p/E$: any observer moving faster than that, along the flight line, sees the opposite helicity. The second figure works this out for a clean numerical case.

The starting point is the set of four plane-wave spinors that solve the free Dirac equation for momentum $\mathbf p$. We work in natural units $\hbar=c=1$, in the Dirac (standard) representation, with Thomson's normalization $u^\dagger u=2E$ so that $N=\sqrt{E+m}$. The upper two components are the "large" (particle) spinor; the lower two are fixed by $\boldsymbol\sigma\cdot\mathbf p/(E+m)$ acting on them.

where: $E=\sqrt{p^2+m^2}$ — particle energy; $\mathbf p=(p_x,p_y,p_z)$ — three-momentum; $p=|\mathbf p|$; the lower block is $\tfrac{\boldsymbol\sigma\cdot\mathbf p}{E+m}$ times the upper block. At rest ($\mathbf p=0$) these collapse to $(1,0,0,0)$ and $(0,1,0,0)$ — pure spin-up and spin-down. The antiparticle spinors $v_1,v_2$ have the same structure with the large block in the lower two components. Spin along the lab $z$-axis is measured by the four-by-four operator $\hat S_z=\tfrac12\Sigma_z$. Testing whether the spinors of (1) are its eigenstates is the whole question of this section, and the answer depends entirely on the direction of $\mathbf p$.

where: $\Sigma_z$ — block-diagonal $4\times4$ spin matrix; its eigenvalues are $\pm\tfrac12$. Applying $\hat S_z$ to $u_1$ multiplies the fourth component by $-1$, so $\hat S_z u_1=\tfrac12 u_1$ holds only if that component vanishes, i.e. $p_x+ip_y=0$ — motion along $\pm\hat z$. For general $\mathbf p$, $u_1$ is a superposition of $S_z=+\tfrac12$ and $S_z=-\tfrac12$. The remedy is helicity: spin projected onto the direction of motion. Its operator is built from the same $\boldsymbol\Sigma$ but contracted with the unit momentum, and — crucially — it commutes with the free Dirac Hamiltonian, so helicity and energy can be simultaneously definite.

where: $\hat h$ — helicity operator; $\lambda$ — its eigenvalue, $+\tfrac12$ (right-handed) or $-\tfrac12$ (left-handed). The commutator vanishes because $[\hat H_D,\boldsymbol\Sigma]=2i\,\boldsymbol\alpha\times\hat{\mathbf p}$ (inherited from aq-pp-spin-from-dirac) is orthogonal to $\hat{\mathbf p}$. Only two eigenvalues exist because $(\boldsymbol\sigma\cdot\hat{\mathbf p})^2=\hat{\mathbf p}\cdot\hat{\mathbf p}\,\mathbb 1=\mathbb 1$. Solving $\hat h u=\lambda u$ in spherical polar coordinates, with $\hat{\mathbf p}=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$, gives the explicit helicity eigenspinors. Both the upper and lower two-component blocks are proportional to the same two-spinor, because $\boldsymbol\sigma\cdot\hat{\mathbf p}$ has already been diagonalized.

where: $c=\cos(\theta/2)$, $s=\sin(\theta/2)$; $u_\uparrow$ has $\lambda=+\tfrac12$, $u_\downarrow$ has $\lambda=-\tfrac12$. The upper block $\chi_+=(c,\,s\,e^{i\phi})$ satisfies $\boldsymbol\sigma\cdot\hat{\mathbf p}\,\chi_+=+\chi_+$. For motion along $\hat z$ ($\theta=0$) these reduce to the $S_z$ spinors of (1), confirming that helicity and $S_z$ agree only in that direction. Antiparticles carry one extra sign. Because a $v$-spinor describes a negative-energy solution reinterpreted as an antiparticle, its physical spin is measured by the opposite operator, so the naive spinor label must be flipped to read off real spin.

where: $\hat S_z^{(v)}$ — the antiparticle spin operator; the minus sign means $v_1$ is the spin-up antiparticle and $v_2$ the spin-down one, and a $v$-spinor that is a $+\tfrac12$ eigenstate of the naive $\hat h$ is physically a left-handed antiparticle. This bookkeeping is why weak-interaction handedness statements for antiparticles look inverted relative to particles.

1 · A massive particle whose helicity flips under a boost (fully worked)

An electron has mass $m=3$, momentum $p=4$, and energy $E=5$ (units with $E^2=p^2+m^2$), moving along $+\hat z$ prepared with positive helicity ($\lambda=+\tfrac12$). Find its speed, the spinor's lower-to-upper component ratio, and show that an observer boosting at $\beta'=0.9$ along $+\hat z$ sees the helicity flip.

Step 1. Speed and consistency. The velocity is $\beta=p/E=4/5=\mathbf{0.8}$, and $E^2-p^2=25-16=9=m^2$ checks the on-shell condition. Because the electron moves at $0.8c$, a faster observer can overtake it — the door through which helicity dependence enters. The particle is sub-luminal, so its helicity cannot be frame-invariant.

Step 2. The spinor structure. From equation (4) with $\theta=0$, the lower block is $\tfrac{p}{E+m}$ times the upper block: $\tfrac{p}{E+m}=\tfrac{4}{5+3}=\tfrac{4}{8}=\mathbf{0.5}$. The lower two components are half the upper two — the electron is neither at rest (ratio $0$) nor massless (ratio $1$), but partway to becoming a pure chirality state. The "wrong-chirality" admixture is set by this $0.5$; it would vanish only as $m/E\to0$.

Step 3. Boost and reverse the momentum. In a frame moving at $\beta'=0.9$ along $+\hat z$, with $\gamma'=1/\sqrt{1-0.81}=2.294$, the longitudinal momentum transforms as $p_z'=\gamma'(p_z-\beta'E)=2.294\,(4-0.9\cdot5)=2.294\,(-0.5)=\mathbf{-1.147}$. The momentum has reversed sign because the observer now outruns the electron. The energy is $E'=\gamma'(E-\beta'p)=2.294\,(5-3.6)=\mathbf{3.212}$, and $E'^2-p_z'^2=10.32-1.32=9.0=m^2$ ✓. The particle now travels backward in the new frame.

Step 4. Read off the flipped helicity. A boost along $\hat z$ leaves the spin projection $S_z$ unchanged (boosts along the spin axis do not rotate spin), but the momentum reversed, so $h=\mathbf S\cdot\hat{\mathbf p}$ changes sign: the state that was $\lambda=+\tfrac12$ is now measured as $\lambda=-\tfrac12$. The right-handed electron has become left-handed simply by changing observer — the concrete meaning of "helicity is not Lorentz invariant for $m\neq0$." Had the electron been massless, $\beta=1$ and no observer could overtake it, so the flip would be impossible and helicity would be absolute.

2 · A helicity eigenstate that is not an $S_z$ eigenstate

Take a particle moving along the $+\hat x$ axis ($\theta=\tfrac{\pi}{2}$, $\phi=0$) in a positive-helicity state. Build its two-spinor and show explicitly that $\langle\hat S_z\rangle=0$, so it is maximally not an $S_z$ eigenstate.

Step 1. Evaluate the half-angles. For $\theta=\tfrac{\pi}{2}$, $c=\cos\tfrac{\pi}{4}=s=\sin\tfrac{\pi}{4}=\tfrac{1}{\sqrt2}$, and $\phi=0$ so $e^{i\phi}=1$. The upper two-spinor is $\chi_+=\tfrac{1}{\sqrt2}(1,\,1)$. This is spin pointing along $+\hat x$ — exactly what "spin parallel to the motion" should mean.

Step 2. Verify it is a helicity eigenstate. With $\hat{\mathbf p}=\hat x$, $\boldsymbol\sigma\cdot\hat{\mathbf p}=\sigma_x=\begin{pmatrix}0&1\\1&0\end{pmatrix}$, and $\sigma_x\chi_+=\tfrac{1}{\sqrt2}(1,\,1)=\chi_+$. So $\hat h\chi_+=+\tfrac12\chi_+$ ✓. It is a clean helicity eigenstate with $\lambda=+\tfrac12$.

Step 3. Now test $\hat S_z$. Compute $\hat S_z\chi_+=\tfrac12\sigma_z\chi_+=\tfrac12\cdot\tfrac{1}{\sqrt2}(1,\,-1)$, which is not proportional to $\chi_+$ — it is $\tfrac{1}{\sqrt2}\chi_-$-like — so $\chi_+$ is not an $S_z$ eigenstate. The expectation value is $\langle\hat S_z\rangle=\chi_+^\dagger\hat S_z\chi_+=\tfrac12\cdot\tfrac12(1\cdot1+1\cdot(-1))=\mathbf{0}$. A particle moving along $\hat x$ with definite helicity has exactly zero average spin along $\hat z$ — it is a 50/50 mixture of $S_z$ up and down, the sharpest possible illustration that helicity and $S_z$ are different questions off-axis.

Helicity is not just a computational convenience; it is the label that a deep theorem says defines what a relativistic particle even is. That theorem is Eugene Wigner's 1939 classification of the unitary representations of the Poincaré group, which showed that every elementary particle is fixed by just two invariants — its mass and its spin (or, for massless particles, its helicity).

Texts. Thomson, Modern Particle Physics, §4.8 (pp. 122–125), Eqs. 4.48–4.65 — the explicit spinors and the helicity eigenstates followed here · Griffiths, Introduction to Elementary Particles, 2nd ed., Ch. 7, §7.1–7.3, Eqs. 7.46–7.51 — the four canonical solutions and the remark that clean spin states need $z$ along the motion · Peskin & Schroeder, An Introduction to Quantum Field Theory, §3.2–3.3 — helicity spinors and the massless limit · Halzen & Martin, Quarks and Leptons, §5.2–5.4 · Schwartz, Quantum Field Theory and the Standard Model, Ch. 11 — the spinor-helicity formalism and modern amplitudes. The key results threading through this article are: $\hat S_z u_1=\tfrac12u_1$ requires $p_x+ip_y=0$; $[\hat H_D,\boldsymbol\Sigma\cdot\hat{\mathbf p}]=0$ follows from the orthogonality of $\boldsymbol\alpha\times\hat{\mathbf p}$ to $\hat{\mathbf p}$; $(\boldsymbol\sigma\cdot\hat{\mathbf p})^2=1\Rightarrow\lambda=\pm\tfrac12$; $\boldsymbol\sigma\cdot\hat{\mathbf p}\,\chi_+=\chi_+$ for $\chi_+=(\cos\tfrac\theta2,\sin\tfrac\theta2 e^{i\phi})$; and the boost example with $m=3,p=4,E=5,\beta=0.8$ yields $p_z'=-1.147,E'=3.212$ at $\beta'=0.9$, flipping the helicity from $+\tfrac12$ to $-\tfrac12$.

Prerequisites. The Dirac Equation · Dirac Equation Solutions · Spin from the Dirac Equation. Leads to. Covariant Form of the Dirac Equation and Chirality · Antiparticles.

Ask the tutor: Ask me to boost a specific helicity spinor into a new frame and watch its handedness change, or to derive the helicity eigenspinors for any direction step by step.