The weak neutral current

For twenty-five years the weak interaction had exactly one face: a charged current that always changed the identity of the fermion it touched — an electron became a neutrino, a down quark became an up. Then in 1973 a bubble chamber at CERN photographed a neutrino that struck an electron, kicked it forward, and left as a neutrino, changing nothing. That is a neutral current, and it forced a rewrite of the whole theory. The weak neutral current is the interaction in which a fermion couples to itself — no change of charge or flavour — by exchanging a neutral $Z$ boson, and its very existence is the fingerprint of electroweak unification. The tempting guess is wrong in an instructive way. The $SU(2)_L$ gauge symmetry already contains a neutral member, $W^{(3)}$ , whose current $j^\mu_3=I^{(3)}_W\,\bar f\gamma^\mu\tfrac12(1-\gamma^5)f$ couples only to left-handed states. If the physical $Z$ were simply $W^{(3)}$ , the $Z$ would ignore right-handed fermions entirely — and experiment says it does not. Glashow, Salam and Weinberg resolved this by promoting electromagnetism's $U(1)$ to a new $U(1)_Y$ of weak hypercharge with its own field $B_\mu$ ; the two neutral fields then mix, and the physical photon and $Z$ are two orthogonal rotations of $B_\mu$ and $W^{(3)}_\mu$ through a single angle $\theta_W$ . That one angle fixes everything. Once $\sin^2\theta_W$ is measured, the $Z$ 's coupling to every quark and lepton is predicted with no further freedom: a left-handed piece inherited from $W^{(3)}$ and a right-handed piece inherited from $B_\mu$ , unequal, so the neutral current is part vector and part axial and violates parity. This article builds the mixing, extracts the couplings $c_V=I^{(3)}_W-2Q\sin^2\theta_W$ and $c_A=I^{(3)}_W$ , and runs the machine on the observables that made the $Z$ the most precisely studied particle in physics.

Key takeaways

The weak neutral current couples a fermion to itself through a neutral $Z$ boson — no change of charge or flavour — and its 1973 discovery was the first confirmation of electroweak unification.
The physical $Z$ is not the $SU(2)_L$ field $W^{(3)}$ : it is the orthogonal mixture $Z_\mu=-B_\mu\sin\theta_W+W^{(3)}_\mu\cos\theta_W$ of $W^{(3)}$ with the $U(1)_Y$ field $B_\mu$ , and the photon is the other mixture.
Demanding the photon couple to electric charge fixes $e=g_W\sin\theta_W=g'\cos\theta_W$ and the neutral coupling $g_Z=g_W/\cos\theta_W$ .
The $Z$ couples to both chiralities but unequally: $c_L=I^{(3)}_W-Q\sin^2\theta_W$ , $c_R=-Q\sin^2\theta_W$ , equivalently $c_V=I^{(3)}_W-2Q\sin^2\theta_W$ and $c_A=I^{(3)}_W$ .
Because $c_L\neq c_R$ the current carries a vector and an axial part, so the weak neutral current violates parity — but not maximally, unlike the charged current.
Key idea: the $Z$ boson is a quantum mixture of the $SU(2)_L$ field $W^{(3)}$ (which sees only left-handed states) and the $U(1)_Y$ field $B$ (which sees both equally); the single mixing angle $\theta_W$ then predicts the $Z$ coupling to every fermion, leaving it coupled to both chiralities but unequally.

The core insight

The $Z$ boson is a quantum mixture of the $SU(2)_L$ field $W^{(3)}$ (which sees only left-handed states) and the $U(1)_Y$ field $B$ (which sees both equally); the single mixing angle $\theta_W$ then predicts the $Z$ coupling to every fermion, leaving it coupled to both chiralities but unequally, so the neutral current violates parity.

Where this lives today

The weak neutral current is now the most precisely tested corner of the Standard Model. Between 1989 and 2000 the LEP collider at CERN produced some seventeen million $Z$ bosons, pinning the mass to $M_Z=91.1876\pm0.0021$ GeV and the total width to $\Gamma_Z=2.4952\pm0.0023$ GeV, and reading the number of light neutrino species off the invisible width as $N_\nu=2.984\pm0.008$ — three, and no room for a fourth. The near-vanishing charged-lepton $c_V$ turned the forward-backward and left-right asymmetries into the sharpest determination of $\sin^2\theta_{\text{eff}}^{\text{lept}}=0.23146\pm0.00012$ , a number that in turn feeds a global electroweak fit precise enough to have anticipated the top-quark and Higgs masses before either was found. Away from the $Z$ pole, the same current shows up as parity violation in atoms — the tiny nuclear "weak charge" $Q_W$ of cesium, measured to better than $1\%$ , tests $\sin^2\theta_W$ at almost zero momentum transfer — and in fixed-target experiments like NuTeV, Qweak, and the ongoing MOLLER and P2 programs, which look for any drift of $\sin^2\theta_W$ with energy that would betray new neutral gauge bosons. At the high-energy frontier the LHC produces $Z$ bosons by the hundred-million through Drell–Yan, using them as standard candles for luminosity and as the recoil against which the Higgs and possible heavy $Z'$ resonances are hunted. Fifty years on, the current that changed nothing about the fermion it touched changed almost everything about how we test the electroweak theory.

The physical picture

Two neutral gauge fields exist before the physics is fixed: $W^{(3)}_\mu$ , the neutral member of the $SU(2)_L$ triplet, which couples only to left-handed weak-isospin; and $B_\mu$ , the single field of $U(1)_Y$ weak hypercharge, which couples to both chiralities. Neither is what an experiment sees. What propagates in the lab is a massless photon and a heavy $Z$ , and each is a fixed rotation of the two originals through the weak mixing angle $\theta_W$ . The first figure is that rotation: it turns the abstract $(B,W^{(3)})$ basis into the physical $(A,Z)$ basis and reads off, from the single demand that the photon couple to electric charge, the coupling relations that tie the weak and electromagnetic charges together.

Figure

The two neutral gauge fields B_μ of U(1)_Y and W³_μ of SU(2)_L rotate through the weak mixing angle θ_W into the physical photon A_μ and the massive Z_μ. With sin²θ_W = 0.23146 the angle is θ_W = 28.7°, and demanding that the photon couple to electric charge fixes e = g_W sin θ_W = g′ cos θ_W and g_Z = g_W / cos θ_W.

The mixing is not a cosmetic relabelling: it changes who the neutral boson talks to. The photon, built to couple to electric charge $Q$ , treats left- and right-handed fermions identically. The $Z$ , forced to be orthogonal to it, inherits a left-handed part from $W^{(3)}$ (weighted by weak isospin $I^{(3)}_W$ ) and a right-handed part from $B_\mu$ (weighted by hypercharge), and those two weights are generically different. The upshot is a coupling with an unequal left/right split — a vector piece $c_V=I^{(3)}_W-2Q\sin^2\theta_W$ and an axial piece $c_A=I^{(3)}_W$ — that differs for every fermion. The second figure is the whole of Thomson's Table 15.1 in one plot: the neutral-current couplings of all twelve fermions, and the near-miracle at the heart of $Z$ physics — the charged lepton whose vector coupling almost exactly vanishes.

Figure

The neutral-current couplings c_V = I₃ − 2Q sin²θ_W and c_A = I₃ of every Standard-Model fermion, in the c_A–c_V plane for sin²θ_W = 0.231. The axial coupling is always ±1/2, but the vector coupling of the charged leptons nearly vanishes, c_V = −0.04, because −1/2 + 2 sin²θ_W ≈ 0 — the near-cancellation that makes the Z almost purely axial to electrons.

The core equations

Everything begins with the rotation that defines the physical neutral bosons. The neutral fields $B_\mu$ and $W^{(3)}_\mu$ are not mass eigenstates; the electroweak symmetry breaking that gives the $Z$ its mass leaves the photon massless, and the eigenstates are a fixed orthogonal rotation of the two through the single weak mixing angle.

A_\mu=B_\mu\cos\theta_W+W^{(3)}_\mu\sin\theta_W,\qquad Z_\mu=-B_\mu\sin\theta_W+W^{(3)}_\mu\cos\theta_W \qquad (1)

where: $A_\mu$ — the photon; $Z_\mu$ — the physical $Z$ boson; $B_\mu$ — the $U(1)_Y$ hypercharge field, coupling $g'$ ; $W^{(3)}_\mu$ — the neutral $SU(2)_L$ field, coupling $g_W$ ; $\theta_W$ — the weak (Weinberg) mixing angle. Read it as: the photon is the particular combination that couples to electric charge, and the $Z$ is simply the orthogonal one — nothing else is available. Demanding that $A_\mu$ reproduce ordinary QED — that it couple to every fermion in proportion to its electric charge $Q$ — is one equation, and it fixes both the electric charge in terms of the two gauge couplings and the strength $g_Z$ of the neutral current.

e=g_W\sin\theta_W=g'\cos\theta_W,\qquad g_Z=\frac{g_W}{\cos\theta_W}=\sqrt{g_W^2+g'^2} \qquad (2)

where: $e$ — the positron charge (the QED coupling); $g_W$ — the $SU(2)_L$ coupling; $g'$ — the $U(1)_Y$ coupling; $g_Z$ — the overall $Z$ coupling. Read it as: the mixing angle is not free once the two couplings are known, $\tan\theta_W=g'/g_W$ , and the $Z$ couples slightly more strongly than the $W$ because $\cos\theta_W<1$ . With the photon pinned to $Q$ , the leftover neutral current — the part that rides the $Z$ — is completely determined. It is the $SU(2)_L$ current minus a $\sin^2\theta_W$ slice of the electromagnetic current, the single most useful line in the subject.

j^\mu_Z=j^\mu_3-\sin^2\theta_W\,j^\mu_{\text{EM}},\qquad j^\mu_3=\bar f\gamma^\mu I^{(3)}_W\tfrac12(1-\gamma^5)f,\qquad j^\mu_{\text{EM}}=Q\,\bar f\gamma^\mu f \qquad (3)

where: $j^\mu_3$ — the third-component weak-isospin current (left-handed only); $j^\mu_{\text{EM}}$ — the electromagnetic current (both chiralities); $I^{(3)}_W$ — weak isospin, $+\tfrac12$ for up-type and neutrinos, $-\tfrac12$ for down-type and charged leptons. Read it as: the $Z$ inherits its left-handedness from $j^\mu_3$ and its right-handed contamination from the $\sin^2\theta_W$ dose of $j^\mu_{\text{EM}}$ . Rewriting equation (3) as a Dirac bilinear gives the $Z$ vertex in the vector/axial form used for every calculation. The Feynman rule is $\dfrac{-ig_Z}{2}\gamma^\mu(c_V^f-c_A^f\gamma^5)$ , and the two coefficients follow by matching $\gamma^\mu$ and $\gamma^\mu\gamma^5$ .

c_V^f=I^{(3)}_W-2Q_f\sin^2\theta_W,\qquad c_A^f=I^{(3)}_W \qquad (4)

where: $c_V^f$ — the vector coupling of fermion $f$ ; $c_A^f$ — the axial coupling; $Q_f$ — its electric charge in units of $e$ . Read it as: the axial coupling is pure weak isospin (always $\pm\tfrac12$ ), while the vector coupling carries the whole $\sin^2\theta_W$ dependence — and can be tuned to nearly zero for the charged leptons. The same content in the chiral basis makes the left/right asymmetry manifest, which is the cleanest way to see parity violation. The left-handed coupling picks up the isospin, the right-handed one does not.

c_L^f=I^{(3)}_W-Q_f\sin^2\theta_W,\qquad c_R^f=-Q_f\sin^2\theta_W,\qquad c_V=c_L+c_R,\ c_A=c_L-c_R \qquad (5)

where: $c_L^f$ — coupling to the left-handed chiral component; $c_R^f$ — coupling to the right-handed one. Read it as: right-handed fermions couple only through their charge times $\sin^2\theta_W$ (the $B_\mu$ inheritance), so a right-handed neutrino ( $Q=0$ ) decouples entirely — the $Z$ is blind to it. Finally, the couplings become an observable through the partial decay width. Each open channel $Z\to f\bar f$ contributes a width set by the sum of squared couplings and a colour factor, and summing over channels gives the total $Z$ width.

\Gamma(Z\to f\bar f)=\frac{g_Z^2 M_Z}{48\pi}\big(c_V^2+c_A^2\big)N_c=\frac{G_F M_Z^3}{6\sqrt2\,\pi}\big(c_V^2+c_A^2\big)N_c \qquad (6)

where: $M_Z=91.19\ \mathrm{GeV}/c^2$ ; $N_c$ — colour factor ( $1$ for leptons, $3$ for quarks); $G_F=1.166\times10^{-5}\ \mathrm{GeV}^{-2}$ . Read it as: measure the partial widths and you measure $c_V^2+c_A^2$ for each fermion — which, run backwards through equation (4), is a measurement of $\sin^2\theta_W$ .

Derivation AHow the photon and the Z emerge from the neutral fields

The goal is to show, from the neutral-sector couplings alone, that the physical photon and $Z$ are forced rotations of $B_\mu$ and $W^{(3)}_\mu$ , and that the single requirement "the photon couples to electric charge" simultaneously fixes the hypercharge assignments, the relation $e=g_W\sin\theta_W=g'\cos\theta_W$ , and the neutral current $j^\mu_Z=j^\mu_3-\sin^2\theta_W\,j^\mu_{\text{EM}}$ with coupling $g_Z=g_W/\cos\theta_W$ . Nothing is put in by hand except the demand that ordinary electromagnetism survive; everything else is bookkeeping on that demand.

Step 1. Write the two neutral interactions before mixing. The $SU(2)_L$ field couples to weak isospin, $g_W W^{(3)}_\mu j^\mu_3$ with $j^\mu_3=\bar f\gamma^\mu I^{(3)}_W P_L f$ , and the $U(1)_Y$ field couples to hypercharge, $g' B_\mu j^\mu_Y$ with $j^\mu_Y=\bar f\gamma^\mu\tfrac{Y}{2}f$ and the Gell-Mann–Nishijima relation $Q=I^{(3)}_W+Y/2$ . Why: these are the only two neutral currents the gauge symmetry allows, and the charge formula is what lets us trade hypercharge for the observable electric charge in the next step.

Step 2. Rotate to the physical basis using equation (1) inverted, $W^{(3)}_\mu=A_\mu\sin\theta_W+Z_\mu\cos\theta_W$ and $B_\mu=A_\mu\cos\theta_W-Z_\mu\sin\theta_W$ , and collect the coefficient of $A_\mu$ . It is $g_W\sin\theta_W\,j^\mu_3+g'\cos\theta_W\,j^\mu_Y$ . Why: the photon must couple to the electromagnetic current $j^\mu_{\text{EM}}=j^\mu_3+j^\mu_Y$ (that is what $Q=I^{(3)}_W+Y/2$ means), so this coefficient has to collapse to $e\,j^\mu_{\text{EM}}$ .

Step 3. Match the photon coupling. For $g_W\sin\theta_W\,j^\mu_3+g'\cos\theta_W\,j^\mu_Y$ to equal $e(j^\mu_3+j^\mu_Y)$ , the two prefactors must be equal and both equal to $e$ : $g_W\sin\theta_W=g'\cos\theta_W=e$ . Why: this is the single equation that ties the weak and electromagnetic charges together — it says the mixing angle is exactly the angle whose tangent is $g'/g_W$ , and it is why measuring $e$ and $\theta_W$ measures the weak coupling.

Step 4. Read off the $Z$ current. The coefficient of $Z_\mu$ is $g_W\cos\theta_W\,j^\mu_3-g'\sin\theta_W\,j^\mu_Y$ ; substitute $j^\mu_Y=j^\mu_{\text{EM}}-j^\mu_3$ and $g'=g_W\tan\theta_W$ to get $\tfrac{g_W}{\cos\theta_W}\big(j^\mu_3-\sin^2\theta_W\,j^\mu_{\text{EM}}\big)$ . Hence $g_Z=g_W/\cos\theta_W$ and $j^\mu_Z=j^\mu_3-\sin^2\theta_W\,j^\mu_{\text{EM}}$ . Why: the $Z$ is whatever is left after the photon has claimed the pure-charge combination, and that leftover is unavoidably a mix of the left-handed isospin current and a $\sin^2\theta_W$ slice of the vector-like charge current — the origin of parity violation in the neutral sector. *Why the algebra closes: with $g'=g_W\tan\theta_W$ , the $Z$ coupling reduces to $g_W/\cos\theta_W$ ; the photon coefficient reduces to $e\,j^\mu_{\text{EM}}$ with $e=g_W\sin\theta_W$ ; and the neutral current is exactly $j^\mu_Z=j^\mu_3-\sin^2\theta_W j^\mu_{\text{EM}}$ .*

Go deeper · Derivation B · 2 minThe vector and axial couplings of the Z to any fermion

Derivation BThe vector and axial couplings of the Z to any fermion

The aim is to convert the compact current $j^\mu_Z=j^\mu_3-\sin^2\theta_W\,j^\mu_{\text{EM}}$ into the explicit couplings $c_V=I^{(3)}_W-2Q\sin^2\theta_W$ and $c_A=I^{(3)}_W$ that fill Thomson's Table 15.1, and to expose why the axial coupling is pure weak isospin while the vector coupling carries all of the mixing-angle dependence. This is the step that makes the neutral current predictive: a single measured number, $\sin^2\theta_W$ , then determines the $Z$ coupling to every quark and lepton in the Standard Model.

Step 1. Insert the explicit currents. Using $j^\mu_3=I^{(3)}_W\,\bar f\gamma^\mu\tfrac12(1-\gamma^5)f$ (left-handed projector $P_L=\tfrac12(1-\gamma^5)$ ) and $j^\mu_{\text{EM}}=Q\,\bar f\gamma^\mu f$ , the neutral current is $j^\mu_Z=\bar f\gamma^\mu\big[I^{(3)}_W\tfrac12(1-\gamma^5)-Q\sin^2\theta_W\big]f$ . Why: everything now sits inside one Dirac bilinear, so the vector ( $\gamma^\mu$ ) and axial ( $\gamma^\mu\gamma^5$ ) parts can simply be read off by grouping terms.

Step 2. Separate vector from axial. Expand the bracket: $\big[\tfrac12 I^{(3)}_W-Q\sin^2\theta_W\big]-\tfrac12 I^{(3)}_W\gamma^5$ . Writing the standard vertex as $\tfrac12(c_V-c_A\gamma^5)$ and matching coefficients gives $c_V=I^{(3)}_W-2Q\sin^2\theta_W$ and $c_A=I^{(3)}_W$ . Why: the axial part comes purely from the $\gamma^5$ inside $j^\mu_3$ — the electromagnetic current has none — so $c_A$ knows nothing about the mixing angle and is always $\pm\tfrac12$ , whereas $c_V$ is where the $B_\mu$ admixture lives.

Step 3. Cross-check in the chiral basis. Since $c_L=\tfrac12(c_V+c_A)$ and $c_R=\tfrac12(c_V-c_A)$ , substitute to find $c_L=I^{(3)}_W-Q\sin^2\theta_W$ and $c_R=-Q\sin^2\theta_W$ . A right-handed fermion couples only through $-Q\sin^2\theta_W$ ; set $Q=0$ and it decouples. Why: this is the clean statement that the $Z$ 's right-handed coupling is entirely the hypercharge inheritance — a right-handed neutrino is sterile because it carries neither isospin nor charge.

Step 4. Tabulate and spot the cancellation. For the charged leptons $I^{(3)}_W=-\tfrac12$ , $Q=-1$ , so $c_V=-\tfrac12+2\sin^2\theta_W=-\tfrac12+0.462=-0.038$ , while $c_A=-\tfrac12$ . For up-type quarks $c_V=+\tfrac12-\tfrac43\sin^2\theta_W=+0.19$ ; for down-type $c_V=-\tfrac12+\tfrac23\sin^2\theta_W=-0.35$ ; for neutrinos $c_V=c_A=+\tfrac12$ . Why: the accidental near-vanishing of the charged-lepton $c_V$ — because $2\sin^2\theta_W\approx\tfrac12$ — makes the $Z$ almost purely axial to electrons, and it is the reason the leptonic forward-backward asymmetry is small and exquisitely sensitive to the precise value of $\sin^2\theta_W$ . *Why the table works: $c_V=I^{(3)}_W-2Q\sin^2\theta_W$ and $c_A=I^{(3)}_W$ reproduce the full set of couplings — $\nu$ : $(0.5,\,0.5)$ ; $\ell$ : $(-0.038,\,-0.5)$ ; $u$ : $(0.19,\,0.5)$ ; $d$ : $(-0.35,\,-0.5)$ — at $\sin^2\theta_W=0.231$ .*

Go deeper · Derivation C · 2 minThe Z partial width and counting the neutrino generations

Derivation CThe Z partial width and counting the neutrino generations

The goal is to turn the couplings into the measured $Z$ decay widths — deriving $\Gamma(Z\to f\bar f)=\tfrac{G_F M_Z^3}{6\sqrt2\,\pi}(c_V^2+c_A^2)N_c$ — and then to show how the invisible width, the part the $Z$ spends decaying into neutrinos it cannot show you, counts the number of light neutrino species and finds exactly three. This is the calculation that let LEP weigh the particle content of the universe from a single resonance curve.

Step 1. Assemble the width from the vertex. For a heavy $Z$ decaying to a massless fermion pair, spin-averaging the squared matrix element of the vertex $\tfrac{-ig_Z}{2}\gamma^\mu(c_V-c_A\gamma^5)$ and folding in the two-body phase space gives $\Gamma=\tfrac{g_Z^2 M_Z}{48\pi}(c_V^2+c_A^2)N_c$ . Why: the vector and axial parts do not interfere for unpolarized, massless fermions, so their squares simply add — the width depends only on $c_V^2+c_A^2$ , never on the sign or the split.

Step 2. Trade $g_Z$ for $G_F$ . Using $g_Z^2=g_W^2/\cos^2\theta_W$ and the charged-current relation $G_F/\sqrt2=g_W^2/8M_W^2$ with $M_W=M_Z\cos\theta_W$ , one finds $g_Z^2=4\sqrt2\,G_F M_Z^2$ , so $\Gamma=\tfrac{G_F M_Z^3}{6\sqrt2\,\pi}(c_V^2+c_A^2)N_c$ . Why: rewriting in terms of the precisely known $G_F$ and $M_Z$ turns the width into an absolute prediction with no adjustable normalization — only $\sin^2\theta_W$ (inside the couplings) remains.

Step 3. Evaluate the width unit. With $G_F=1.166\times10^{-5}\ \mathrm{GeV}^{-2}$ and $M_Z=91.19$ GeV, the prefactor is $\tfrac{G_F M_Z^3}{6\sqrt2\,\pi}=0.332$ GeV. A single neutrino has $c_V^2+c_A^2=\tfrac14+\tfrac14=\tfrac12$ and $N_c=1$ , giving $\Gamma_\nu=0.332\times0.5=0.166$ GeV $=166$ MeV per species. Why: the neutrino channels are invisible — no track, no calorimeter deposit — so their combined width can only be inferred as a deficit in the total, which is exactly what makes the count a clean subtraction.

Step 4. Count the neutrinos. The measured total width $\Gamma_Z=2.495$ GeV, minus the visible (charged-lepton and hadronic) channels of $\approx1.997$ GeV, leaves an invisible width $\Gamma_{\text{inv}}=0.499$ GeV. Dividing by the per-species $166$ MeV gives $N_\nu=0.499/0.166=3.0$ . Why: had there been a fourth light neutrino, the $Z$ resonance would be visibly wider and shorter; LEP's precision fit gives $N_\nu=2.984\pm0.008$ — three, and only three, light neutrino generations couple to the $Z$ . *Why the numbers check out: the prefactor $G_F M_Z^3/6\sqrt2\pi=0.332$ GeV gives $\Gamma_\nu=166$ MeV and $\Gamma_\ell=83.4$ MeV per species; the tree-level total is $\Gamma_Z=2.42$ GeV, with QCD corrections raising the hadronic part to the measured $2.495$ GeV; and $\Gamma_{\text{inv}}/\Gamma_\nu=3$ .*

Limiting cases

Case	Result	Meaning
Right-handed fermion, $Q=0$	$c_R=-Q\sin^2\theta_W=0$	a right-handed neutrino decouples entirely — the $Z$ is blind to it (it is sterile)
Neutrinos	$c_V=c_A=+\tfrac12$ , $c_R=0$	pure left-handed, like the charged current; the $Z$ sees only $\nu_L$
Charged leptons	$c_V=-\tfrac12+2\sin^2\theta_W\approx-0.04$	vector coupling nearly cancels; the $Z$ is almost purely axial to $e,\mu,\tau$
$\sin^2\theta_W\to0$ (no mixing)	$Z\to W^{(3)}$ , $c_V\to c_A=I^{(3)}_W$	the $Z$ would couple to left-handed states only — the wrong, pre-unification limit
$\sin^2\theta_W\to1$	photon becomes $W^{(3)}$ , $B\to$ neutral	unphysical; used only to check the rotation is a genuine $SO(2)$
Low energy $q^2\ll M_Z^2$	contact term $\propto G_F$ , no flavour change	neutral-current neutrino scattering and atomic parity violation live here

Common trapAssuming the Z boson is just the W³ of SU(2)_L

It is not — and the difference is the whole content of electroweak unification. If the physical $Z$ were literally the neutral member $W^{(3)}$ of the weak-isospin triplet, its current would be $j^\mu_3=I^{(3)}_W\bar f\gamma^\mu\tfrac12(1-\gamma^5)f$ , which couples only to left-handed particles (right-handed states carry $I^{(3)}_W=0$ ). Such a $Z$ would give right-handed electrons and right-handed quarks exactly zero neutral-current coupling — flatly contradicted by measured $Z$ decay rates and by the nonzero right-handed couplings $c_R=-Q\sin^2\theta_W$ . The physical $Z$ is the orthogonal mixture of $W^{(3)}$ and the hypercharge field $B$ , the one that is not the photon, and it therefore drags in a right-handed piece proportional to charge. The tell is the vector coupling: a pure $W^{(3)}$ would have $c_V=c_A=I^{(3)}_W$ for everyone, whereas the real $Z$ has $c_V=I^{(3)}_W-2Q\sin^2\theta_W$ , unequal to $c_A$ , which is exactly why the neutral current — unlike a naive $SU(2)_L$ current — couples to both chiralities and violates parity only partially.

Worked examples

1 · The Z couplings of the electron and the up quark (fully worked)

Using $\sin^2\theta_W=0.231$ , compute $c_V$ , $c_A$ , $c_L$ , $c_R$ for the electron ( $I^{(3)}_W=-\tfrac12$ , $Q=-1$ ) and the up quark ( $I^{(3)}_W=+\tfrac12$ , $Q=+\tfrac23$ ), and say what the near-vanishing electron $c_V$ implies.

Step 1. Electron vector and axial couplings from equation (4): $c_V=I^{(3)}_W-2Q\sin^2\theta_W=-\tfrac12-2(-1)(0.231)=-0.5+0.462=\mathbf{-0.038}$ , and $c_A=I^{(3)}_W=\mathbf{-0.5}$ . The vector part is nearly zero because $2\sin^2\theta_W=0.462$ almost exactly cancels the $-\tfrac12$ .

Step 2. Electron chiral couplings from equation (5): $c_L=I^{(3)}_W-Q\sin^2\theta_W=-0.5-(-1)(0.231)=\mathbf{-0.269}$ , and $c_R=-Q\sin^2\theta_W=-(-1)(0.231)=\mathbf{+0.231}$ . Check: $c_V=c_L+c_R=-0.269+0.231=-0.038$ ✓ and $c_A=c_L-c_R=-0.5$ ✓. The right-handed electron couples at nearly the same magnitude as the left — the $Z$ is far from maximally parity-violating here.

Step 3. Up quark: $c_V=+\tfrac12-2(\tfrac23)(0.231)=0.5-0.308=\mathbf{+0.192}$ , $c_A=\mathbf{+0.5}$ ; and $c_L=+\tfrac12-\tfrac23(0.231)=\mathbf{+0.346}$ , $c_R=-\tfrac23(0.231)=\mathbf{-0.154}$ . Here the left/right split is large.

Step 4. Interpret the electron result. Because $c_V\approx0$ , the $Z$ couples to the electron almost purely axially, so the leptonic forward-backward and left-right asymmetries — which scale as $c_Vc_A$ — are small but extraordinarily sensitive to $\sin^2\theta_W$ : a $1\%$ shift in $\sin^2\theta_W$ moves $c_V$ by $\sim20\%$ . This near-cancellation is why the cleanest measurement of $\sin^2\theta_W$ comes from leptonic asymmetries at the $Z$ pole.

2 · The invisible width and the number of neutrinos (fully worked)

Predict the $Z$ partial width to a single neutrino species, then use the measured invisible width to count the light neutrino generations. Take $G_F=1.166\times10^{-5}\ \mathrm{GeV}^{-2}$ , $M_Z=91.19$ GeV, $\Gamma_Z^{\text{meas}}=2.495$ GeV, and a visible width of $1.997$ GeV.

Step 1. Evaluate the width prefactor from equation (6): $\dfrac{G_F M_Z^3}{6\sqrt2\,\pi}=\dfrac{(1.166\times10^{-5})(91.19)^3}{6\sqrt2\,\pi}=\dfrac{(1.166\times10^{-5})(7.58\times10^{5})}{26.66}=\mathbf{0.332\ GeV}$ .

Step 2. A neutrino has $c_V=c_A=+\tfrac12$ , so $c_V^2+c_A^2=\tfrac14+\tfrac14=\tfrac12$ , and $N_c=1$ . Then $\Gamma_\nu=0.332\times0.5=\mathbf{0.166\ GeV}=166$ MeV per species.

Step 3. The invisible width is the total minus the visible: $\Gamma_{\text{inv}}=2.495-1.997=\mathbf{0.498\ GeV}$ . This is the width the $Z$ spends on final states that leave no trace in the detector — neutrinos.

Step 4. Divide: $N_\nu=\dfrac{\Gamma_{\text{inv}}}{\Gamma_\nu}=\dfrac{0.498}{0.166}=\mathbf{3.0}$ . A fourth light neutrino would have widened the $Z$ resonance by another $166$ MeV — easily seen. The LEP fit gives $N_\nu=2.984\pm0.008$ . The neutral current, through one invisible width, establishes that there are exactly three light neutrino generations.

3 · Parity violation made quantitative: the leptonic asymmetry

The polarization (left-right) asymmetry parameter for a fermion is $A_f=\dfrac{2c_Vc_A}{c_V^2+c_A^2}$ . Evaluate it for the electron and explain why its smallness is a feature, not a bug.

Step 1. Insert the electron couplings from Example 1, $c_V=-0.038$ , $c_A=-0.5$ : the numerator is $2c_Vc_A=2(-0.038)(-0.5)=0.038$ , and the denominator is $c_V^2+c_A^2=0.00144+0.25=0.2514$ .

Step 2. Divide: $A_e=0.038/0.2514=\mathbf{0.151}$ . The measured LEP/SLD value is $A_e=0.1515\pm0.0019$ — a $15\%$ asymmetry between left- and right-handed electron couplings, and a direct, quantitative measure of parity violation in the neutral current.

Step 3. Note the leverage. Since $A_e$ is nearly linear in the small $c_V$ , and $c_V=-\tfrac12+2\sin^2\theta_W$ , a measurement of $A_e$ to $1\%$ pins $\sin^2\theta_W$ to about $\pm0.0002$ . The tiny $c_V$ that made the asymmetry small also made it the single most sensitive probe of the weak mixing angle — turning a near-cancellation into the sharpest ruler in electroweak physics.

Physics of the portrait

The neutral current is Weinberg's prediction made flesh. His 1967 paper wrote the mixing, fixed the couplings, and named a neutral partner to the $W$ — a particle nobody had seen and few believed in — six years before a bubble chamber found its footprint.

Origins · 1 minSteven Weinberg and how this was discovered

Steven Weinberg

1933–2021 · the unified electroweak theory and the prediction of the neutral current

Steven Weinberg's 1967 paper "A Model of Leptons" — barely three pages — is among the most cited in the history of physics, and the weak neutral current is its most direct consequence. Building on Sheldon Glashow's 1961 observation that a unified weak-electromagnetic theory needs a fourth gauge boson, and on Abdus Salam's parallel work, Weinberg combined $SU(2)_L\times U(1)_Y$ with spontaneous symmetry breaking to give the $W$ and $Z$ their masses while keeping the photon massless. The mixing angle that now bears his name rotated the neutral fields into the physical photon and $Z$ , and — crucially — his Lagrangian forced a neutral current: a $Z$ that couples to every fermion with strengths $c_V=I^{(3)}_W-2Q\sin^2\theta_W$ , $c_A=I^{(3)}_W$ , and no freedom left once $\sin^2\theta_W$ was fixed. The prediction was ignored for four years, because no one knew whether the theory even made sense at the quantum level; only when Gerard 't Hooft and Martinus Veltman proved it renormalizable in 1971 did the community take the neutral current seriously. In 1973 the Gargamelle bubble chamber at CERN photographed neutrino events with no outgoing muon — the smoking gun of a neutral current — and in 1983 the $Z$ itself was produced at CERN's proton–antiproton collider. Weinberg, Glashow and Salam shared the 1979 Nobel Prize, two years before the $Z$ was seen directly, on the strength of the neutral current alone. The whole modern edifice of precision electroweak physics — every $Z$ -pole measurement at LEP — is the working-out of the couplings Weinberg wrote down.

History

1961

— Glashow proposes an SU(2)×U(1) theory with a neutral partner to the W

1967

— Weinberg's "A Model of Leptons" fixes the mixing angle and predicts the Z couplings

1971

— 't Hooft and Veltman prove the electroweak theory renormalizable

1973

— Gargamelle at CERN observes muonless neutral-current neutrino events

1983

— UA1 and UA2 discover the Z boson directly at the SppS collider

Going further

Texts. Thomson, Modern Particle Physics, §15.2–15.3 (electroweak unification and the couplings of the $Z$ ), Table 15.1 and Eqs. (15.14), (15.17)–(15.38) — the source this article grounds · Halzen & Martin, Quarks and Leptons, §13 (the standard electroweak model) · Griffiths, Introduction to Elementary Particles, 2nd ed., §9.4–9.5 · Peskin & Schroeder, An Introduction to QFT, §20.2 (the Glashow–Weinberg–Salam theory) · Schwartz, QFT and the Standard Model, Ch. 29. The key results to keep in view: the rotation (1) with the photon matched to $Q$ gives $e=g_W\sin\theta_W=g'\cos\theta_W$ and $j^\mu_Z=j^\mu_3-\sin^2\theta_W j^\mu_{\text{EM}}$ ; the couplings $c_V=I^{(3)}_W-2Q\sin^2\theta_W$ and $c_A=I^{(3)}_W$ reproduce the fermion table at $\sin^2\theta_W=0.231$ ; the width formula $\Gamma=\tfrac{G_F M_Z^3}{6\sqrt2\pi}(c_V^2+c_A^2)N_c$ gives $\Gamma_\nu=166$ MeV and $\Gamma_\ell=83.4$ MeV, with $\Gamma_{\text{inv}}/\Gamma_\nu=3$ and $A_e=2c_Vc_A/(c_V^2+c_A^2)=0.151$ . Constants used throughout: $M_Z=91.19$ GeV, $g_Z=g_W/\cos\theta_W$ , $\theta_W=28.7°$ .

Prerequisites. The Weak Charged Current · Chiral Structure of the Weak Interaction · Weak Isospin and Hypercharge · The Weinberg (Weak Mixing) Angle. Leads to. Z Boson Decays · The Z Resonance at LEP · Electroweak Unification · The Higgs Mechanism.

Ask the tutor: Ask me to rotate the neutral fields into the photon and $Z$ step by step, or to compute the $c_V$ , $c_A$ couplings and $Z$ partial width for any fermion you name.

Why did the discovery of the neutral current matter so much in 1973?

Because it was the first evidence that the electromagnetic and weak forces are two faces of one interaction. Before 1973 every known weak process changed a fermion's charge — beta decay, muon decay, all charged-current. Weinberg's 1967 theory required an additional neutral process in which nothing changes charge, mediated by a new $Z$ boson. When Gargamelle photographed a neutrino elastically scattering off an electron with no outgoing muon, it confirmed that the $Z$ exists and that the whole unified $SU(2)_L\times U(1)_Y$ structure is real. The Nobel Prize went to Glashow, Salam and Weinberg in 1979 largely on this evidence — a full four years before anyone produced a $Z$ directly.

What is the difference between the charged current and the neutral current?

The charged current couples fermions to the $W^\pm$ and always changes electric charge and flavour: an electron becomes a neutrino, a down quark becomes an up. It is purely left-handed — a maximal $(1-\gamma^5)$ V−A structure. The neutral current couples fermions to the $Z$ and changes nothing: the fermion keeps its charge and its flavour. Crucially it couples to both chiralities, with $c_L=I^{(3)}_W-Q\sin^2\theta_W$ and $c_R=-Q\sin^2\theta_W$ . So the neutral current still violates parity (because $c_L\neq c_R$ ) but not maximally, unlike the charged current. The single mixing angle $\theta_W$ ties the two currents together.

Why does the Z couple to both left- and right-handed fermions, but the W does not?

Because they have different parentage. The $W^\pm$ come entirely from the $SU(2)_L$ triplet, and $SU(2)_L$ acts only on left-handed doublets — right-handed fermions are singlets under it, so they never touch a $W$ . The $Z$ , however, is a mixture of the $SU(2)_L$ field $W^{(3)}$ and the $U(1)_Y$ hypercharge field $B$ . The $B$ field couples to hypercharge, which every fermion carries regardless of chirality, so the $Z$ inherits a right-handed coupling proportional to $-Q\sin^2\theta_W$ through its $B$ component. The right-handed part of the $Z$ coupling is, quite literally, its share of the photon-like field $B$ .

What does sin²θ_W actually measure, and why is 0.23 special?

It measures how much the hypercharge field $B$ is mixed into the physical $Z$ (and, equivalently, how much $W^{(3)}$ is mixed into the photon): $\tan\theta_W=g'/g_W$ , the ratio of the two gauge couplings. Its value $\sin^2\theta_W\approx0.231$ is not predicted by the Standard Model — it is an input, measured. What makes $0.23$ special is an accident: it is close to $\tfrac14$ , so $2\sin^2\theta_W\approx\tfrac12$ , and the charged-lepton vector coupling $c_V=-\tfrac12+2\sin^2\theta_W$ nearly vanishes. That near-cancellation makes leptonic asymmetries small but exquisitely sensitive to the exact value, which is why $\sin^2\theta_W$ is now known to five decimal places.

How does the neutral current become Fermi-like at low energy?

Exactly as the charged current does. The $Z$ propagator carries a denominator $q^2-M_Z^2$ , and at momentum transfers far below the $91$ GeV $Z$ mass it freezes to $-M_Z^2$ , collapsing the exchange into a four-fermion contact interaction with strength $\propto G_F$ . This is the regime of neutrino–electron elastic scattering, coherent neutrino–nucleus scattering, and atomic parity violation — all low-energy neutral-current processes that measure $\sin^2\theta_W$ far from the $Z$ pole. The difference from the charged-current Fermi theory is only the couplings: the contact term inherits the vector/axial structure $c_V,c_A$ rather than the pure V−A of the $W$ , so it probes both chiralities.

If the Z violates parity, why is the violation called "not maximal"?

Maximal parity violation means the interaction couples to only one chirality — as the charged current does, with $c_R=0$ for the $W$ . The neutral current, by contrast, has $c_R=-Q\sin^2\theta_W\neq0$ for charged fermions, so both handedness's participate, just unequally. The degree of violation is captured by the asymmetry parameter $A_f=2c_Vc_A/(c_V^2+c_A^2)$ : for a maximally violating coupling it would be $\pm1$ , but for the electron it is only $0.15$ because $c_V$ nearly cancels. Neutrinos are the exception — with $Q=0$ they have $c_R=0$ and couple to the $Z$ purely left-handedly, so for neutrinos the neutral current is maximally parity-violating.