Weak Interactions

Derivation. Derives Phenomenon 101.1. Suppose the decay had only two products, a daughter nucleus of mass \(M'\) and an electron of mass \(m\), and let the parent of mass \(M\) be at rest. Momentum conservation makes the two momenta equal and opposite, \(\vect{p}'=-\vect{p}\), and energy conservation then reads

The right-hand side is a strictly increasing function of \(p\), so it attains the value \(Mc^{2}\) for exactly one \(p\): a two-body decay of a state of definite mass yields a line, and the electron energy \(E=\sqrt{p^{2}c^{2}+m^{2}c^{4}}\) is fixed. The observed continuum is therefore, by itself, a proof that at least three bodies share the released energy. Since the daughter and the electron are both detected and account between them for less than the release, the third is not detected: it must be neutral, must interact far more weakly than the electron, and must be light enough to leave the endpoint where the two-body kinematics puts it. That is precisely the particle Pauli proposed in 1930 [Pauli:1930] and Cowan and Reines detected twenty-six years later [Cowan:1956]. The argument also shows what would have followed had the spectrum been a line with a lower endpoint — energy conservation itself would have been at stake, which is why Bohr was prepared to give it up.

Derives Proposition 101.3. The electron takes its greatest energy when the daughter and the neutrino move together, so that the pair may be treated as a single body of mass \(M'+m_{\nu}\) recoiling against the electron. That is a two-body decay, and Equation (101.1) with \(M'\to M'+m_{\nu}\) fixes the electron energy uniquely:

Subtract \(m_{e}c^{2}\) and use \(Mc^{2}-m_{e}c^{2}=M'c^{2}+Q\) to factor the numerator as \(\left(Q-m_{\nu}c^{2}\right)\left(2M'c^{2}+Q+m_{\nu}c^{2}\right)\); writing the denominator as \(2(M'c^{2}+m_{e}c^{2}+Q)\) gives the exact form in Equation (101.2), and expanding in \(1/M'c^{2}\) gives the second. The recoil correction is of relative order \(\left(2m_{e}c^{2}+Q\right)/2M'c^{2}\) — dominated by the electron mass, not by \(Q\), which is the trap in this expansion — and for tritium (\(Q=18.6\,\mathrm{keV}\), \(M'c^{2}=2808\,\mathrm{MeV}\)) it is \(1.85\times 10^{-4}\), that is \(3.4\,\mathrm{eV}\) of recoil energy taken by the daughter. That is four times the whole mass sensitivity of Equation (101.23), so it is computed and subtracted rather than neglected. The leading statement is the one that matters here: the endpoint sits a distance \(m_{\nu}c^{2}\) below the release, so a measured endpoint bounds the neutrino mass without any assumption about the interaction that produced it.

Derives Proposition 101.4. The invariant \(s=(p_{\bar{\nu}}+p_{p})^{2}\) is conserved, and the final state exists only if \(\sqrt{s}\,c\geq(m_{n}+m_{e})c^{2}\), the least energy the pair can have. With the proton at rest and the antineutrino massless, \(s c^{2}=m_{p}^{2}c^{4}+2m_{p}c^{2}E_{\bar{\nu}}\), and the inequality rearranges to Equation (101.4). Inserting \(m_{n}c^{2}=939.565\,\mathrm{MeV}\), \(m_{p}c^{2}=938.272\,\mathrm{MeV}\) and \(m_{e}c^{2}=0.511\,\mathrm{MeV}\) [Navas:2024] gives \(1.806\,\mathrm{MeV}\). A fission reactor emits antineutrinos with a spectrum extending to about \(8\,\mathrm{MeV}\), so a useful fraction of them are above threshold; this is why the experiment was done at a reactor and not with a natural source.

Derivation. Derives Phenomenon 101.5. The cross-section may be predicted from the coupling that Section 101.2.1 extracts from beta decay itself, which is what makes the number a test rather than a fit. At energies far below the mass of the mediating boson the interaction is a contact one, the amplitude is a product of two currents at a point, and by the derivation of Proposition 101.54 — which is performed there in full and is common to every charged-current process of this kind — the cross-section for the nucleon reaction is

with \(E_{e}=E_{\bar{\nu}}-(m_{n}-m_{p})c^{2}\) the positron energy, \(p_{e}c=\sqrt{E_{e}^{2}-m_{e}^{2}c^{4}}\) and \(\lambda=g_{A}/g_{V}=1.2754\) the ratio of the axial and vector nucleon couplings measured in neutron decay [Navas:2024]; the factor \(1+3\lambda^{2}\) is the sum of one Fermi and three Gamow–Teller spin channels, counted in Section 101.2.2. The factor \(\abs{V_{ud}}^{2}=\cos^{2}\theta_{C}=0.949\) is there because this is a hadronic charged current and not a leptonic one: the reaction turns a \(d\) quark into a \(u\), and by Equation (101.63) that vertex carries \(\cos\theta_{C}\). Omitting it overstates the rate by five per cent. At \(E_{\bar{\nu}}=3\,\mathrm{MeV}\) this gives \(E_{e}=1.707\,\mathrm{MeV}\), \(p_{e}c=1.628\,\mathrm{MeV}\) and

which is the observed order of magnitude. For the penetrating power, lead has \(11.34\times 10^{3}\,\mathrm{kg}/\mathrm{m}^{3}\) and molar mass \(0.2072\,\mathrm{kg}/\mathrm{mol}\), hence \(3.30\times10^{28}\) nuclei and \(2.70\times10^{30}\) protons per cubic metre; the mean free path is

which is \(1.5\,\mathrm{yr}\) of travel at the speed of light. The idealization is explicit and is worth stating: a proton bound in a lead nucleus is not a free proton and the reaction on it is not Equation (101.3), so the figure counts protons, not targets. As an order of magnitude for the transparency of matter to neutrinos it is nevertheless the right one, and it is the reason Pauli thought his particle undetectable.

Derives Proposition 101.7. By Definition 100.2 the bilinear \(\bar{\psi}\gamma^{\mu}\psi\) has dimension \(/\mathrm{m}^{3}\), so the product of two currents carries \(\mathrm{m}^{-6}\). An energy density is \(\mathrm{J}/\mathrm{m}^{3}\). Hence \([G]=\mathrm{J}/\mathrm{m}^{3}\times\mathrm{m}^{6} =\mathrm{J}\,\mathrm{m}^{3}\). Since \(\hbar c\) has dimension \(\mathrm{J}\,\mathrm{m}\), dividing by \((\hbar c)^{3}\) leaves \(\mathrm{J}^{-2}\).

Derivation of the conversion. Derives Definition 101.8. \(\hbar c=3.161527\times 10^{-26}\,\mathrm{J}\,\mathrm{m}\) from the defined values of \(\hbar\) and \(c\), so \((\hbar c)^{3}=3.16003\times 10^{-77}\,\mathrm{J}^{3}\,\mathrm{m}^{3}\). One gigaelectronvolt is \(1.602176634\times 10^{-10}\,\mathrm{J}\) exactly, since the elementary charge is fixed by definition since 2019, so \(/\mathrm{GeV}^{2}=3.89566\times 10^{19}\,/\mathrm{J}^{2}\). Then

Two derived numbers used repeatedly below follow at once: the energy at which the dimensionless combination \(G_{F}E^{2}/(\hbar c)^{3}\) reaches unity is

and the combination \((\sqrt{2}G_{F}/(\hbar c)^{3})^{-1/2} =246.2\,\mathrm{GeV}\) is the electroweak scale \(v\) that Electroweak Unification and the Higgs Boson identifies with the vacuum expectation value of the scalar field.

Derives Proposition 101.10. To first order in \(V\) the amplitude for finding the system in \(\ket{f}\) at time \(t\) is

with \(\hbar\omega_{fi}=E_{f}-E_{i}\). Hence

The kernel \(4\sin^{2}(\omega t/2)/(\hbar\omega)^{2}\) has total integral \(\int\dd\omega\,4\sin^{2}(\omega t/2)/(\hbar\omega)^{2} =2\pi t/\hbar^{2}\) and a width in \(\omega\) of order \(1/t\), so for \(t\) large compared with the inverse spread of the continuum it acts as \((2\pi t/\hbar^{2})\delta(\omega)=(2\pi t/\hbar)\delta(E_{f}-E_{i})\). Summing over \(f\) and dividing by \(t\) gives Equation (101.10); the probability grows linearly in \(t\), which is what makes the rate well defined and the decay law exponential once the depletion of \(\ket{i}\) is fed back in. The result is Dirac's [Dirac:1927]; the name is Fermi's [Fermi:1950], and it is the tool with which he wrote [Fermi:1934].

Derives Proposition 101.11. Apply the golden rule of Proposition 101.10 with two light particles in the final state and the recoiling daughter treated as infinitely heavy, so that it absorbs momentum but no energy. With the matrix element independent of the lepton momenta, the entire energy dependence is the density of final states,

the angular integrations giving \(4\pi\) each because an \(s\)-wave matrix element is isotropic. Change variables with \(p^{2}\,\dd p/\dd E=pE/c^{2}\), valid for either particle, and use the delta function to set \(E_{\nu}=E_{0}-E_{e}\):

where the last step used \(p_{\nu}c=\sqrt{E_{\nu}^{2}-m_{\nu}^{2}c^{4}}\). That is Equation (101.11). The Coulomb factor is a separate effect: the emitted electron is not a plane wave but a continuum Coulomb wave in the field of the daughter's charge \(Z\), and the rate is proportional to the squared modulus of that wave at the nucleus. Its non-relativistic form is

with the upper sign for \(\beta^{-}\) emission and the lower for \(\beta^{+}\), \(Z\) being the charge number of the daughter. The sign is the entire physical content of the factor, and the two limits fix it. For \(\beta^{-}\) the electron is attracted, so \(\eta>0\), and \(F\to2\pi\eta>1\) as \(v_{e}\) falls: the Coulomb field draws the outgoing wave onto the nucleus and enhances the low-energy end of the spectrum. For \(\beta^{+}\) the positron is repelled, so \(\eta<0\), and \(F\to2\pi\abs{\eta}\ee^{-2\pi\abs{\eta}}\ll1\), which is the Gamow barrier factor and suppresses the same end. Here \(\alpha\) is the fine-structure constant of Equation (100.3) and \(v_{e}\) the electron speed; \(\eta\) is dimensionless, as it must be.

Derives Proposition 101.12. Integrate Equation (101.12) with \(F=1\) and \(p_{e}c\simeq E_{e}\):

The fifth power is pure phase space — two particles sharing a release, each contributing a \(p^{2}\dd p\) — and it is what Sargent had found empirically in 1933 by plotting the decay constants of the naturally occurring beta emitters against their endpoints on logarithmic axes and obtaining two straight lines of slope five [Sargent:1933]. The two lines, rather than one, are the first appearance in this chapter of a distinction that Section 101.2.2 identifies: allowed and first-forbidden transitions have different matrix elements and therefore different intercepts, but the same slope, because the slope is kinematics.

Derives Proposition 101.14. Linear independence: with \(\acomm{\gamma^{\mu}}{\gamma^{\nu}}=2\eta^{\mu\nu}\identity\), every product of gamma matrices reduces to one of Equation (101.15) up to a sign, and the trace relations of Lemma 100.23 give \(\tr(\Gamma_{i}\Gamma_{j}^{-1}) =4\delta_{ij}\) for the sixteen listed matrices, from which \(\sum_{i}c_{i}\Gamma_{i}=0\) forces every \(c_{i}=0\). Sixteen independent matrices in a sixteen-dimensional space are a basis. The transformation laws follow from the defining property of the spinor representation, \(S^{-1}(\Lambda)\gamma^{\mu}S(\Lambda) =\Lambda^{\mu}{}_{\nu}\gamma^{\nu}\), established in The Dirac Equation: \(\bar{\psi}\chi\) is invariant, \(\bar{\psi}\gamma^{\mu}\chi\) carries one vector index, and so on. That the last two are pseudo-quantities is the statement that \(\gamma^{5}\) anticommutes with every \(\gamma^{\mu}\) and therefore picks up a sign under the improper transformation \(P\), which is proved in Proposition 101.23.

Derives Proposition 101.15. Take the non-relativistic limit of the nucleon current, in which the large components of the nucleon spinors dominate. For the scalar and the time component of the vector, \(\bar{\psi}_{p}\psi_{n}\) and \(\bar{\psi}_{p}\gamma^{0}\psi_{n}\) both reduce to \(\phi_{p}^{\dagger}\phi_{n}\) acting on the two-component nucleon spinors: the operator is the identity in spin space, so the nucleon spin is untouched and by conservation the pair carries spin zero, hence \(\Delta J=0\). For the axial spatial components and the space components of the tensor, \(\bar{\psi}_{p}\gamma^{i}\gamma^{5}\psi_{n}\) and \(\bar{\psi}_{p}\sigma^{ij}\psi_{n}\) reduce to \(\phi_{p}^{\dagger}\sigma^{i}\phi_{n}\): the operator is a Pauli matrix, which flips the nucleon spin, so the pair carries spin one and \(\Delta J\in\set{0,\pm1}\). A spin-one pair cannot be emitted in a \(0\to0\) transition, since there is nothing to absorb its unit of angular momentum. The remaining components — the space part of the vector, the time part of the axial vector and the pseudoscalar — carry an explicit factor of the nucleon velocity in the non-relativistic reduction and are suppressed by \(v_{\mathrm{nucleon}}/c\sim10^{-1}\); they are the leading terms of the forbidden transitions.

Parity: in the allowed approximation the lepton wave functions are constants evaluated at the nucleus, so the whole dependence on the nuclear coordinates sits in the two operators just obtained, \(\identity\) and \(\sigma^{i}\), and both are even under spatial reflection — the identity trivially, the Pauli matrix because a spin is an axial vector. A parity-even operator has vanishing matrix element between nuclear states of opposite parity, so the nuclear parity cannot change. The created pair does carry intrinsic parity \(-1\), a fermion and an antifermion having opposite intrinsic parities, but that factor is the same in every beta decay and multiplies every amplitude alike; what distinguishes one transition from another is the orbital factor \((-1)^{\ell}\), which is \(+1\) here and is exactly what changes sign in a first-forbidden transition.

Derives Proposition 101.17. Collect the factors of Proposition 101.11. The rate is the golden-rule expression Equation (101.10) with the squared coupling \(G_{F}^{2}\abs{V_{ud}}^{2}\) in front, the nuclear matrix elements squared and summed over the two spin channels with their multiplicities — one for the singlet, three for the triplet, which is the origin of the factor \(3\) in Equation (101.5) — and the lepton phase space integrated, which is \((m_{e}c^{2})^{5}f(Z,E_{0})\) by Equation (101.16). Half-life and rate are related by \(t_{1/2}=\hbar\ln 2/(\hbar\Gamma)=\ln2/\Gamma\). Assembling the numerical factors of the two-particle phase space, which are \((2\pi)^{-3}\) per particle from the density of states and \(2\pi/\hbar\) from Equation (101.10), leaves \(K=2\pi^{3}\hbar\ln2\,(\hbar c)^{6}/(m_{e}c^{2})^{5}\). Its value follows from \(\hbar=6.582120\times 10^{-25}\,\mathrm{GeV}\,\mathrm{s}\) and \(m_{e}c^{2}=0.51099895\,\mathrm{MeV}\):

and multiplying by \((\hbar c)^{6}=9.98576\times 10^{-154}\,\mathrm{J}^{6}\,\mathrm{m}^{6}\) after converting \(\mathrm{GeV}^{-4}\) to \(\mathrm{J}^{-4}\) gives the SI form in Equation (101.18).

Derivation. Derives Phenomenon 101.18. A \(0^{+}\to0^{+}\) transition is pure Fermi by Proposition 101.15: the Gamow–Teller operator cannot connect two spinless states. The Fermi operator is the isospin raising or lowering operator, whose matrix element between members of an isospin multiplet is fixed by the algebra and not by the nuclear wave function at all: for a triplet, \(\abs{M_{\mathrm{F}}}^{2}=2\). Hence Equation (101.17) loses every trace of nuclear structure and reduces to

where \(\Delta_{R}^{V}\) is the transition-independent radiative correction and \(\mathcal{F}t\) denotes \(ft\) after the small nucleus-dependent corrections have been removed. With \(G_{F}/(\hbar c)^{3}=1.1663787\times 10^{-5}\,/\mathrm{GeV}^{2}\), \(\abs{V_{ud}}=0.97435\) and \(\Delta_{R}^{V}=0.0240\) [Navas:2024],

against the measured \(3072\,\mathrm{s}\). The agreement to a part in \(10^{3}\) is what turns Equation (101.20) around: read backwards it is the most precise determination of \(\abs{V_{ud}}\) available, and hence the first-row test of the unitarity of the mixing matrix in Section 101.5. The constancy of \(\mathcal{F}t\) across the multiplet is separately the evidence for the conserved vector current of Section 101.4.2: if the strong interaction renormalized the vector coupling, it would do so differently in a carbon nucleus and an aluminium one, and the fourteen numbers would not agree.

Derives Proposition 101.20. Substitute Equation (101.11) into Equation (101.21); the factors \(F\), \(p_{e}\) and \(E_{e}\) cancel identically, leaving Equation (101.22). Writing \(\epsilon=E_{0}-E_{e}\) for the energy left to the neutrino, the bracket is \(\epsilon\sqrt{\epsilon^{2}-m_{\nu}^{2}c^{4}}\), which vanishes at \(\epsilon=m_{\nu}c^{2}\) rather than at \(\epsilon=0\), and whose derivative there is infinite. Hence the endpoint moves down by \(m_{\nu}c^{2}\) — which is Proposition 101.3 recovered from the spectrum instead of from the kinematics — and the approach to it becomes vertical. Both features are confined to the last \(m_{\nu}c^{2}\) of the spectrum, and the fraction of decays falling there is, from Equation (101.12),

so a neutrino mass of \(1\,\mathrm{eV}/c^{2}\) affects only \(2\times 10^{-13}\) of the decays of a nucleus with \(E_{0}=18.6\,\mathrm{keV}\). That cube is the whole difficulty of the measurement and the whole reason for choosing the smallest available \(E_{0}\).

Derivation. Derives Phenomenon 101.21. The bound is Proposition 101.20 applied to a measured spectrum. KATRIN scans the integral spectrum with a magnetic-adiabatic collimation spectrometer of \(1\,\mathrm{eV}\) resolution, fits the counting rate above a variable retarding potential to the predicted shape convolved with the response function and the calculable molecular final-state distribution, and treats \(m_{\nu}^{2}\) as the fit parameter — \(m_{\nu}^{2}\), and not \(m_{\nu}\), because Equation (101.22) depends on the square and because a fit constrained to positive values is biased. The result is consistent with zero, and the interval is converted to the bound Equation (101.23) [Aker:2022]. What the experiment measures is strictly an incoherent average, \(m_{\beta}^{2}=\sum_{i}\abs{U_{ei}}^{2}m_{i}^{2}\), over the mass eigenstates that make up the electron antineutrino, because the individual masses are far too close together to be resolved as separate kinks; the mixing matrix \(U\) is that of Flavour Physics and Neutrinos.

Derives Proposition 101.23. Use \(\gamma^{0}\gamma^{\mu}\gamma^{0}=\gamma_{\mu}\) — no sum, and the index is lowered, which is the statement that the time component is unchanged and the space components reverse — together with \((\gamma^{0})^{2}=\identity\) and \(\gamma^{0}\gamma^{5}\gamma^{0}=-\gamma^{5}\), the last because \(\gamma^{5}\) anticommutes with \(\gamma^{0}\). Then \(\bar{\psi}\chi=\psi^{\dagger}\gamma^{0}\chi\mapsto \psi^{\dagger}\gamma^{0}\gamma^{0}\gamma^{0}\chi=\bar{\psi}\chi\), and \(\bar{\psi}\gamma^{5}\chi\mapsto \psi^{\dagger}\gamma^{0}\gamma^{0}\gamma^{5}\gamma^{0}\chi =-\bar{\psi}\gamma^{5}\chi\), and similarly for the two vectors. For the last statement, contract two currents: the product of two quantities that both pick up \(+1\), or both \(-1\), is invariant, while a product of one of each changes sign. A theory built only from \(V\times V\) and \(A\times A\) conserves parity; one containing \(V\times A\) does not.

Derives Proposition 101.24. The pion is a pseudoscalar, \(J^{P}=0^{-}\), established from the capture of a negative pion at rest on deuterium: the capture proceeds from an atomic \(s\) state to \(nn\), the two neutrons must be in a state antisymmetric overall, and the only such state with \(J=1\) is \(^{3}P_{1}\), of orbital parity \(-1\); equating initial and final parities gives \(P_{\pi}=-1\). The parity of a system of \(n\) such particles with no relative orbital angular momentum is the product of the intrinsic parities, \((-1)^{n}\). For \(n=2\) this is \(+1\); for \(n=3\), \(-1\). If the decaying particle has a definite parity \(P_{X}\) and parity is conserved, then \(P_{X}=+1\) from the first mode and \(P_{X}=-1\) from the second, a contradiction. The escape that the three-pion state carries orbital angular momentum, and so is not in relative \(s\) waves, is closed by the observed distribution of the three pions over their kinematically allowed region: the density is uniform up to the edges, which is the signature of \(\ell=0\), since any nonzero orbital angular momentum would suppress the configurations in which two pions are collinear.

Derivation. Derives Phenomenon 101.26. Under the parity operation a polar vector such as the momentum \(\vect{p}\) reverses while an axial vector such as the angular momentum \(\vect{J}\) does not, so the scalar product \(\langle\vect{J}\rangle\cdot\hat{\vect{p}}\) appearing in Equation (101.26) is a pseudoscalar and changes sign. The mirror-image experiment therefore predicts \(W(\pi-\theta)\) wherever the original predicts \(W(\theta)\), and the two disagree unless \(A=0\). A measured \(A\neq0\) is thus in itself a proof that the interaction responsible is not invariant under \(P\): no dynamical input whatever is required, which is why one asymmetry measurement settled a question that decades of spectroscopy had not thought to ask [Lee:1956] [Wu:1957]. What the further observation \(\abs{A}\approx1\) adds is that the violation is as large as the kinematics permits — the interaction does not merely prefer one handedness, it uses one and ignores the other — and that statement is delivered by Proposition 101.27 below, once the \(V-A\) structure of Section 101.4 is in place.

Derives Proposition 101.27. The transition lowers the nuclear angular momentum by one unit along the polarization axis \(\hat{\vect{z}}\), so the lepton pair must carry away \(J_{z}=+1\): both leptons have their spin projections \(+\tfrac {1}{2}\) along \(\hat{\vect{z}}\). Now apply the chirality rule. The antineutrino, being massless to the accuracy of Phenomenon 101.21, has positive helicity exactly by Proposition 101.32, so its spin is along its momentum: with \(S_{z}=+\tfrac{1}{2}\) it must travel along \(+\hat{\vect{z}}\). The electron is emitted in the left-chiral combination, whose helicity expectation value is \(-v/c\) by Proposition 101.34; with its spin also along \(+\hat{\vect{z}}\), a negative helicity means momentum along \(-\hat{\vect{z}}\). The electron therefore prefers the hemisphere opposite the nuclear spin, and the preference is complete only in the limit \(v\to c\), being diluted by the factor \(v/c\) at lower speeds. Writing the distribution as Equation (101.26) identifies \(A=-1\) for this transition.

Comparing that with Wu's number requires one factor Equation (101.26) does not contain, and it is worth being careful, because the arithmetic does not otherwise close. The distribution is written for a fully polarized ensemble; a real source has a degree of nuclear polarization \(f=\abs{\avg{\vect{J}}}/J <1\), and only that fraction of the nuclei contributes to the front–back difference, so the measured counting-rate asymmetry between the two senses of the polarizing field is

Wu's electrons had \(v/c\approx0.6\) and the source reached \(f\approx0.6\), giving \(\abs{\varepsilon}\approx0.36\), which is the observed asymmetry of about \(0.4\). The product \(\abs{A}\,v/c\) alone would predict \(0.6\), half again too large: the polarization factor is not optional, and it is why the experiment had to measure \(f\) independently, from the anisotropy of the daughter's gamma cascade. The same factor supplies the control — the asymmetry disappears as the crystal warms and \(f\) decays to zero, on the same time constant as the gamma anisotropy, which is what makes the result an asymmetry and not an instrumental drift [Wu:1957].

Derivation. Derives Phenomenon 101.28. What has to be established is that the photon the apparatus selects carries the neutrino's helicity, since the photon's circular polarization is the only thing measured. Both steps are angular-momentum bookkeeping along a single axis.

Take \(\hat{\vect{z}}\) along the photon's momentum. The capture \(^{152m}\mathrm{Eu}\,(0^{-})+e^{-}\to{}^{152}\mathrm{Sm}^{*}(1^{-}) +\nu_{e}\) proceeds from a \(K\)-shell \(s\) orbital, so nothing brings orbital angular momentum in, and the two products leave back to back, the neutrino along \(-\hat{\vect{z}}\) and the recoiling \(^{152}\mathrm{Sm}^{*}\) along \(+\hat{\vect{z}}\). A two-body state collinear with the quantization axis has \(L_{z}=0\), so along that axis

Take \(S_{z}^{e}=-\tfrac{1}{2}\). A negative-helicity neutrino moving along \(-\hat{\vect{z}}\) has its spin along \(+\hat{\vect{z}}\), that is \(S_{z}^{\nu}=+\tfrac{1}{2}\), and Equation (101.28) then forces \(J_{z}^{\mathrm{Sm}^{*}}=-1\). The excited state decays to a spinless ground state, \(^{152}\mathrm{Sm}^{*}(1^{-})\to {}^{152}\mathrm{Sm}\,(0^{+})+\gamma\), so the photon carries away the whole of \(J_{z}=-1\); and a photon travelling along \(+\hat{\vect{z}}\) with \(J_{z}=-1\) has helicity \(-1\). Reversing the neutrino's helicity reverses every sign in the chain, so the photon's helicity is not merely correlated with the neutrino's but equal to it.

That fixes the photons emitted opposite to the neutrino, and the resonance condition selects exactly those and no others. A photon emitted by a nucleus at rest falls short of the absorption resonance by the recoil energies of emitter and absorber; only a nucleus recoiling towards the emission direction — that is, one whose neutrino went the other way — is moving fast enough for the Doppler shift to make the deficit up, so only those photons scatter resonantly on a \(^{152}\mathrm{Sm}\) target. Their circular polarization, measured by transmission through magnetized iron, is therefore the neutrino's helicity, and it came out negative [Goldhaber:1958]. The antineutrino statement follows from \(CP\): the \(CP\) image of a negative-helicity neutrino is a positive-helicity antineutrino, and \(CP\) survives these decays even though \(C\) and \(P\) separately do not.

Derives Proposition 101.31. (i) \(\sigma^{\mu\nu}\) is a product of two distinct gamma matrices, and \(\gamma^{5}\) anticommutes with each, hence commutes with the pair; therefore \(\comm{\gamma^{5}}{\sigma^{\mu\nu}}=0\) and \(\gamma^{5}\) commutes with \(\exp(-\tfrac{\ii}{4}\omega_{\mu\nu}\sigma^{\mu\nu})\). (ii) \(\bar{\psi}\psi=\bar{\psi}(P_{\mathrm{L}}+P_{\mathrm{R}})\psi\), and \(\bar{\psi}P_{\mathrm{L}}=\overline{P_{\mathrm{R}}\psi}\) because \(\gamma^{0}\gamma^{5}=-\gamma^{5}\gamma^{0}\); the diagonal terms therefore vanish and only the cross terms survive. (iii) The free Hamiltonian is \(H=c\vect{\alpha}\cdot\vect{p}+\beta mc^{2}\) with \(\vect{\alpha}=\gamma^{0}\vect{\gamma}\); a direct computation using \(\comm{\Sigma^{i}}{\alpha^{j}}=2\ii\varepsilon^{ijk}\alpha^{k}\) and \(\comm{\vect{\Sigma}}{\beta}=0\) gives \(\comm{\vect{\Sigma}\cdot\vect{p}}{H}=0\). (iv) A boost of speed \(u>v\) along \(\hat{\vect{p}}\) sends \(\vect{p}\mapsto-\abs{\vect{p}'} \hat{\vect{p}}\) while the spin projection on \(\hat{\vect{p}}\) is unchanged, since a boost along an axis does not rotate spin about that axis; hence \(h\mapsto-h\). For \(m=0\) no such boost exists and the obstruction disappears.

Derives Proposition 101.32. Work in the Dirac representation, in which \(\gamma^{5}=\begin{pmatrix}0&\identity\\ \identity&0\end{pmatrix}\) in \(2\times2\) blocks. A positive-energy solution of definite helicity is

as follows from the Dirac equation with the lower two components given by \(c\,\vect{\sigma}\cdot\vect{p}/(E+mc^{2})\) times the upper ones and \(c\abs{\vect{p}}=\sqrt{(E+mc^{2})(E-mc^{2})}\). Its norm is \(\abs{u_{\lambda}}^{2}=N^{2}\left[(E+mc^{2})+(E-mc^{2})\right] =2N^{2}E\). Applying \(P_{\mathrm{L}}=\tfrac{1}{2}\begin{pmatrix} \identity&-\identity\\-\identity&\identity\end{pmatrix}\),

whose squared norm is

using \(\lambda^{2}=1\) and \(\sqrt{E^{2}-m^{2}c^{4}}=\abs{\vect{p}}c=\beta E\). Dividing gives Equation (101.31). For the limit, expand \(\beta=\sqrt{1-m^{2}c^{4}/E^{2}}\simeq1-m^{2}c^{4}/2E^{2}\).

Derives Proposition 101.34. By Equation (101.31) the left-chiral projection contains helicity \(-1\) with weight \(\tfrac{1}{2}(1+\beta)\) and helicity \(+1\) with weight \(\tfrac{1}{2}(1-\beta)\), and these are the relative probabilities with which the interaction produces the two. Hence \(\avg{h}=\tfrac{1}{2}(1-\beta)-\tfrac{1}{2}(1+\beta)=-\beta\). For the antiparticle the roles of the projectors are exchanged, because the negative-energy spinor \(v_{\lambda}\) carries the opposite relation between chirality and helicity, and the sign reverses. Equation (101.34) is a prediction with no free parameter: it says that a beta electron of \(v/c=0.5\) is \(50\) per cent longitudinally polarized and one of \(v/c=0.9\) is \(90\) per cent polarized, and it is measured directly, by passing the emitted electrons through a polarization-sensitive scatterer. Frauenfelder and collaborators did it for \(^{60}\mathrm{Co}\) within months of Wu's result and found the electrons longitudinally polarized, with the predicted sign and a magnitude consistent with \(-v/c\) [Frauenfelder:1957]. It is also what supplies the factor \(v/c\) in Equation (101.26), which is therefore not an empirical fudge but the electron mass making itself felt.

Derives Proposition 101.35. From Equation (101.30), \(P_{\mathrm{L}}^{2}=P_{\mathrm{L}}\) and \(\gamma^{\mu}P_{\mathrm{L}}=P_{\mathrm{R}}\gamma^{\mu}\), the latter because \(\gamma^{5}\) anticommutes with \(\gamma^{\mu}\). Also \(\overline{P_{\mathrm{L}}\psi}=\psi^{\dagger}P_{\mathrm{L}}\gamma^{0} =\bar{\psi}P_{\mathrm{R}}\), using \(P_{\mathrm{L}}^{\dagger}=P_{\mathrm{L}}\) and \(\gamma^{0}\gamma^{5}=-\gamma^{5}\gamma^{0}\). Hence

The same computation shows that the scalar bilinear does the opposite: \(\bar{\psi}_{\mathrm{L}}\chi_{\mathrm{L}} =\bar{\psi}P_{\mathrm{R}}P_{\mathrm{L}}\chi=0\), so a chirally projected theory admits vector and axial currents and forbids scalar, pseudoscalar and tensor couplings between fields of the same chirality. Equation (101.35) is therefore not one arbitrary choice among five: it is the only current a theory of purely left-chiral fields can have.

Derives Proposition 101.36. By Proposition 101.23 the \(VV\) and \(AA\) terms of the squared amplitude are \(P\)-even and the \(VA\) cross terms \(P\)-odd, and the coefficients are \(g_{V}^{2}\), \(g_{A}^{2}\) and \(2g_{V}g_{A}\) respectively. The asymmetry is therefore \(2g_{V}g_{A}/(g_{V}^{2}+g_{A}^{2})\), which by the arithmetic–geometric mean inequality is at most \(1\) in modulus with equality if and only if \(g_{V}=\pm g_{A}\). The measured asymmetries of Section 101.3.2 are of order unity, which forces the equality, and the observed sign — electrons opposite to the nuclear spin, neutrinos left-handed — selects \(g_{A}=+g_{V}\) in the convention of Equation (101.35), that is \(V-A\) and not \(V+A\).

Derives Proposition 101.38. Work in the limit \(v_{e}\to c\), where by Proposition 101.32 the electron emitted by a left-chiral coupling has helicity \(-1\) exactly and the antineutrino has helicity \(+1\) exactly; the factor \(v_{e}/c\) in Equation (101.37) is what restores the general case, being the electron's helicity by Equation (101.34). Then the spin of the antineutrino points along its momentum and the spin of the electron points against its own.

A Fermi transition emits the pair in the spin singlet, whose two spins are antiparallel. Let the antineutrino spin be \(+\hat{\vect{z}}\) and the electron spin \(-\hat{\vect{z}}\). The antineutrino momentum is then along \(+\hat{\vect{z}}\) and the electron momentum, being opposite its spin, is also along \(+\hat{\vect{z}}\): the two particles go the same way, \(\cos\theta_{e\nu}=+1\), and the distribution is \(1+\cos\theta_{e\nu}\), that is \(a=+1\).

A Gamow–Teller transition emits the pair in the spin triplet, and the three substates must be averaged because an unpolarized nucleus populates them equally. For \(m=+1\) both spins are \(+\hat{\vect{z}}\); the antineutrino goes along \(+\hat{\vect{z}}\) and the electron along \(-\hat{\vect{z}}\), giving \(\cos\theta_{e\nu}=-1\); the same for \(m=-1\). The state \(m=0\) is \((\ket{\uparrow\downarrow}+\ket{\downarrow\uparrow})/\sqrt{2}\), which is the singlet's partner with antiparallel spins and gives \(\cos\theta_{e\nu}=+1\). Averaging, \(a=(-1-1+1)/3=-1/3\).

Replacing \(V\) by \(S\), or \(A\) by \(T\), replaces the left-chiral projector on the electron by the right-chiral one — this is Proposition 101.35 read backwards, a scalar bilinear connecting opposite chiralities — so the electron's momentum reverses relative to its spin and every sign above flips. The four entries of the table follow.

Derivation. Derives Phenomenon 101.40. The hypothesis that explains the first half is that of Feynman and Gell-Mann [Feynman:1958]: the weak vector current \(V^{\mu}=\bar{\psi}_{p}\gamma^{\mu}\psi_{n}\) and the isovector part of the electromagnetic current are three components of one isospin triplet,

with the weak charged currents the combinations \(V^{\mu}_{1}\pm\ii V^{\mu}_{2}\) and the isovector electromagnetic current \(V^{\mu}_{3}\). Isospin is a symmetry of the strong interaction, so all three currents are conserved, \(\pp_{\mu}V^{\mu}_{a}=0\), and their charges \(T_{a}=\frac{1}{c}\int\dd^{3}x\,V^{0}_{a}\) generate the algebra \(\comm{T_{a}}{T_{b}}=\ii\varepsilon_{abc}T_{c}\).

The argument is now exactly that of Proposition 100.6, transposed. A charge that generates a symmetry algebra takes eigenvalues fixed by the representation theory of that algebra — half-integers, in units set by the commutation relations — and a continuous dynamical effect cannot shift a discrete eigenvalue. The strong interaction commutes with \(T_{a}\), so it cannot alter the matrix element of \(V^{\mu}_{a}\) between states of definite isospin at zero momentum transfer, any more than the strong interaction alters the electric charge of the proton, which by Proposition 100.6 and the Ward–Takahashi identity of Section 100.5.2 is exactly the positron's. The vector coupling of a nucleon is therefore the same as that of a bare quark, and the constancy of \(\mathcal{F}t\) across the superallowed multiplet in Phenomenon 101.18 is the measurement that confirms it: a renormalization that depended on nuclear structure would show up as a spread in those fourteen numbers, and the spread is a part in \(10^{4}\).

The axial current is a different matter, because it is not conserved: nothing in the strong interaction is invariant under a chiral rotation once the quarks have mass, and, more importantly, the axial symmetry is spontaneously broken. There is accordingly no protection theorem and \(g_{A}\) is renormalized. The measured \(27\) per cent is small only because the quark masses are small, and its theoretical status belongs to the chiral symmetry of Quantum Chromodynamics. That the vector coupling is protected and the axial one is not is a structural prediction, and the two measured numbers — \(1\) and \(1.2754\) — are the test of it.

Derives Proposition 101.42. Take the matrix element of \(\pp_{\mu}A^{\mu}_{a}\) between two nucleon states of momenta \(p\) and \(p'\), and evaluate it in two ways.

On the left, use the general decomposition of the axial matrix element consistent with Lorentz invariance and parity,

with \(g_{A}\) and \(h_{A}\) dimensionless. Contracting with \(q_{\mu}\), and using \(\bar{u}(p')\gamma^{\mu}q_{\mu}\gamma^{5}u(p) =\bar{u}(p')\left(\gamma\cdot p'-\gamma\cdot p\right)\gamma^{5}u(p) =2m_{N}c\,\bar{u}(p')\gamma^{5}u(p)\) — the Dirac equation applied to each spinor, the sign of the second term flipped by \(\gamma^{5}\gamma\cdot p=-\gamma\cdot p\,\gamma^{5}\) so that the two add — gives

On the right, use Equation (101.40): the divergence is \(f_{\pi}(m_{\pi}c/\hbar)^{2}\hbar\phi_{a}\), and the nucleon matrix element of the pion field is the pion propagator times the pion–nucleon vertex, which with the coupling normalized as \(\Lag_{\pi N}=-\ii g_{\pi N} \bar{\psi}_{N}\gamma^{5}\tau_{a}\psi_{N}\phi_{a}\) reads

Equating the two expressions and cancelling the common spinor factor,

Now set \(q^{2}=0\), where the second term on the left vanishes provided \(h_{A}\) has no pole at the origin — which it has not, the pion pole sitting at \(q^{2}=m_{\pi}^{2}c^{2}\) — and the right-hand side becomes \(-2f_{\pi}g_{\pi N}\). Hence \(m_{N}c\,g_{A}(0)=-f_{\pi}g_{\pi N}\), which is Equation (101.41) up to the sign convention for \(g_{\pi N}\). The mechanism is worth naming: the right-hand side of Equation (101.44) is proportional to \(m_{\pi}^{2}\) and would vanish in the chiral limit, but the pion propagator supplies a compensating \(1/m_{\pi}^{2}\) at \(q^{2}=0\), so the product survives. That is why an exactly conserved axial current would still give a nonzero relation, and why the corrections are of relative order \(m_{\pi}^{2}/m_{N}^{2}\) rather than of order one. Numerically, with \(f_{\pi}c=92.3\,\mathrm{MeV}\), \(g_{\pi N}=13.05\), \(m_{N}c^{2}=938.9\,\mathrm{MeV}\) and \(g_{A}=1.2754\) [Navas:2024],

agreeing to \(0.6\) per cent, which is of the expected order \(m_{\pi}^{2}/m_{N}^{2}\approx0.02\). The relation is [Goldberger:1958]. What makes it worth the page is that it connects three quantities measured in three different ways — a weak decay rate, a strong scattering length and a leptonic pion decay — and is right to one per cent.

Derivation. Derives Phenomenon 101.43. The three determinations are as follows. Muon decay gives \(G_{F}\) directly by Equation (101.53), with no hadron anywhere in the problem and hence no strong correction beyond the electromagnetic ones. Superallowed nuclear decay gives \(G_{F}\abs{V_{ud}}\) by Equation (101.20), and the vector coupling entering it is unrenormalized by Phenomenon 101.40; dividing out \(\abs{V_{ud}}=0.97435\) returns the muon-decay value. Tau decay to \(\nu_{\tau}e\bar{\nu}_{e}\) is muon decay with a heavier lepton, so Equation (101.53) applies with \(m_{\mu}\to m_{\tau}\); the predicted partial width, using \(m_{\tau}c^{2}=1776.93\,\mathrm{MeV}\) and the same \(G_{F}\), reproduces the measured branching fraction of \(17.8\,\mathrm{\%}\) together with the measured lifetime \(290.3\times 10^{-15}\,\mathrm{s}\) [Navas:2024]. Three processes whose energy scales differ by a factor \(10^{4}\) and whose participants differ in every other respect share one constant. Puppi's observation that the couplings of \((\mu,\nu_{\mu})\), \((e,\nu_{e})\) and \((n,p)\) to one another form a triangle of equal sides, and Konopinski and Mahmoud's version of the same statement [Konopinski:1953], are the origin of the phrase universal Fermi interaction; the residual deficit in the strangeness-changing sector, which is what stops the triangle from closing exactly, is Section 101.5.

Derives Proposition 101.45. Quantize in a box of volume \(V\) with periodic boundary conditions and normalize each one-particle state to one particle in the box,

which is correctly normalized because \(u^{\dagger}u=2E/c\) by Proposition 100.14, so that \(\int_{V}\dd^{3}x\,\psi^{\dagger}\psi=1\); note that the prefactor is dimensionally forced, \(\mathrm{m}^{-3/2}\) on the left and \(u/\sqrt{2E/c}\) dimensionless. With \(\Ham_{\mathrm{int}}=-\Lag_{\mu}\), which holds because the interaction contains no derivatives, the matrix element between the one-muon state and a three-particle final state is

the four factors \(V^{-1/2}\) coming from the four fields and the factor \(V\) from \(\int_{V}\dd^{3}x\), which also enforces momentum conservation as a Kronecker delta. Squaring, using \(\delta_{\mathrm{K}}^{2}=\delta_{\mathrm{K}}\), and inserting into the golden rule Equation (101.10) with the continuum sum \(\sum_{f}\to\prod_{k=2}^{4}V\dd^{3}p_{k}/(2\pi\hbar)^{3}\) and the continuum form of the Kronecker delta, \(\delta_{\mathrm{K}}=(2\pi\hbar)^{3}V^{-1} \delta^{3}(\vect{p}_{1}-\sum\vect{p}_{f})\), gives

every power of \(V\) having cancelled: \(V^{3}\) from the three measures against \(V^{-2}\) from the squared matrix element and \(V^{-1}\) from the delta. Now use \(\left(2\pi\hbar\right)^{4}\delta^{4}(p) =\left(2\pi\hbar\right)^{3}\delta^{3}(\vect{p})\cdot 2\pi\hbar c\,\delta(E)\), so that \(2\pi\delta(E)\left(2\pi\hbar\right)^{3}\delta^{3} =\left(2\pi\hbar\right)^{4}\delta^{4}/\hbar c\), to recognize \(\dd\Phi_{3}\) of Equation (100.39) entire:

Finally substitute \(\abs{\mathcal{A}}^{2}=2\hbar^{2}c^{2}\abs{\mathcal{M}}^{2}/G_{F}^{2}\) from Equation (101.47), which cancels \(G_{F}\) and \(\hbar\) together and leaves Equation (101.48). Nothing in the last two steps used the specific interaction: the coupling and the spinor structure enter only through \(\mathcal{M}\), while the factors of \(V\) and of \(2E/c\) belong to the external legs. Hence the formula holds for any \(1\to n\) decay, which is how it is used again in Section 101.6.4.

Derives Lemma 101.46. Write \(\mathcal{A}=L^{\alpha}M_{\alpha}\) with \(L^{\alpha}=\bar{u}_{\nu_{\mu}}\gamma^{\alpha}(1-\gamma^{5})u_{\mu}\) and \(M_{\alpha}=\bar{u}_{e}\gamma_{\alpha}(1-\gamma^{5}) v_{\bar{\nu}_{e}}\). Summing over final spins and averaging over the two muon spins, \(\overline{\abs{\mathcal{A}}^{2}} =\tfrac{1}{2}L^{\alpha\beta}M_{\alpha\beta}\) with

the spin sums being those of Equation (100.33) and the conjugation using \(\gamma^{0}\left[\gamma^{\beta}(1-\gamma^{5})\right]^{\dagger} \gamma^{0}=\gamma^{\beta}(1-\gamma^{5})\). The mass term drops, because it leaves an odd number of gamma matrices in the trace. Using \(\left(1-\gamma^{5}\right)^{2}=2\left(1-\gamma^{5}\right)\) and the trace theorems of Lemma 100.23 together with \(\tr\left(\gamma^{\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\sigma} \gamma^{5}\right)=-4\ii\varepsilon^{\mu\nu\rho\sigma}\),

and \(M_{\alpha\beta}\) is the same expression with \(p_{\nu_{\mu}}\to p_{e}\) and \(p_{\mu}\to p_{\bar{\nu}_{e}}\) and the indices lowered. Abbreviate \(P_{1}=p_{\nu_{\mu}}\), \(P_{2}=p_{\mu}\), \(P_{3}=p_{e}\), \(P_{4}=p_{\bar{\nu}_{e}}\) — numbered rather than lettered, since \(c\) is the speed of light throughout this book — and contract.

The symmetric parts contract to

the two trace terms cancelling exactly against the term in which the two metrics meet. The product of the two Levi-Civita terms, using \(\varepsilon^{\rho\alpha\sigma\beta} \varepsilon_{\kappa\alpha\lambda\beta} =-2\left(\delta^{\rho}_{\kappa}\delta^{\sigma}_{\lambda} -\delta^{\rho}_{\lambda}\delta^{\sigma}_{\kappa}\right)\) and \(\ii^{2}=-1\), contributes

while the cross terms between a symmetric and an antisymmetric tensor vanish identically. Adding, the \(\left(P_{1}\cdot P_{4}\right) \left(P_{2}\cdot P_{3}\right)\) pieces cancel and the others double:

which halved is Equation (101.50). This is the single most useful identity in weak-interaction physics: any two \(V-A\) currents, contracted and summed over spins, collapse to a product of two dot products with the coefficient \(256\), and the cancellation of the other pairing is what fixes which particle goes with which.

Derives Lemma 101.47. The integral is a Lorentz tensor built from \(q\) alone, so \(I^{\alpha\beta}=Aq^{2}\eta^{\alpha\beta}+Bq^{\alpha}q^{\beta}\). Contract with \(\eta_{\alpha\beta}\): since \(k^{2}=k'^{2}=0\) gives \(2k\cdot k'=q^{2}\), the left side is \(\tfrac{1}{2}q^{2}\Phi_{2}\) with \(\Phi_{2}=\int\dd\Phi_{2}\), so \(4A+B=\tfrac{1}{2}\Phi_{2}\). Contract with \(q_{\alpha}q_{\beta}\): \(q\cdot k=k'\cdot k=\tfrac{1}{2}q^{2}\) and likewise \(q\cdot k'\), so the left side is \(\tfrac{1}{4}q^{4}\Phi_{2}\) and \(A+B=\tfrac{1}{4}\Phi_{2}\). Hence \(A=\Phi_{2}/12\) and \(B=\Phi_{2}/6\). The total two-body phase space for massless products follows from Proposition 100.24, where \(\dd\Phi_{2}=\hbar^{-2}\abs{\vect{p}_{f}} \dd\Omega/16\pi^{2}\sqrt{s}\) with \(\abs{\vect{p}_{f}} =\tfrac{1}{2}\sqrt{s}\) here, giving \(\Phi_{2}=4\pi/32\pi^{2}\hbar^{2}=1/8\pi\hbar^{2}\); note that it carries \(\hbar^{-2}\), as every element of \(\dd\Phi\) must. Inserting gives Equation (101.52).

Derives Theorem 101.48. Combine Equation (101.48), Equation (101.47) and Equation (101.50):

The two neutrinos are not observed, so integrate over them first. Writing \(q=p_{\mu}-p_{e}\) for their total four-momentum and using Equation (101.52) with \(k=p_{\nu_{\mu}}\) and \(k'=p_{\bar{\nu}_{e}}\),

Put the muon at rest, \(p_{\mu}=(m_{\mu}c,\vect{0})\), and neglect the electron mass, \(p_{e}^{2}=0\). Then, writing \(y=p_{\mu}\cdot p_{e}=m_{\mu}E_{e}\),

so the numerator is \(\left(m_{\mu}^{2}c^{2}-2y\right)y+2y\left(m_{\mu}^{2}c^{2}-y\right) =m_{\mu}^{2}E_{e}\left(3m_{\mu}c^{2}-4E_{e}\right)\). The remaining electron integral is the one-particle element of Equation (100.39) with \(\dd^{3}p_{e}=4\pi E_{e}^{2} \dd E_{e}/c^{3}\) for a massless electron:

the upper limit being the greatest energy an electron can take when the two neutrinos go together against it. The integral is \(m_{\mu}c^{2}W^{3}-W^{4}\) with \(W=m_{\mu}c^{2}/2\), that is \(\left(m_{\mu}c^{2}\right)^{4}/16\), and \(\left(2\pi\hbar\right)^{3}=8\pi^{3}\hbar^{3}\), so the electron integral equals \(m_{\mu}^{2}\left(m_{\mu}c^{2}\right)^{4}/64\pi^{2}\hbar^{3}c^{2}\). Assembling with the prefactor \(c^{2}/2m_{\mu}c^{2}\) of Equation (101.48),

since \(64/(2\cdot96\cdot64)=1/192\). Writing \(m_{\mu}=\left(m_{\mu}c^{2}\right)/c^{2}\) and \(\hbar^{7}c^{6}=\hbar\left(\hbar c\right)^{6}\) gives Equation (101.53). The spectrum Equation (101.54) is the integrand before the last integration: with \(E_{e}=xm_{\mu}c^{2}/2\) the integrand is proportional to \(E_{e}^{2}\left(3m_{\mu}c^{2}-4E_{e}\right)\propto x^{2}\left(3-2x\right)\), and \(\int_{0}^{1}2x^{2}(3-2x)\dd x=2\left(1-\tfrac{1}{2}\right)=1\) fixes the normalization.

Derivation. Derives Phenomenon 101.50. Evaluate Equation (101.53) with \(G_{F}/(\hbar c)^{3}=1.1663787\times 10^{-5}\,/\mathrm{GeV}^{2}\), \(m_{\mu}c^{2}=105.6583755\,\mathrm{MeV}\) and \(\hbar=6.582120\times 10^{-25}\,\mathrm{GeV}\,\mathrm{s}\):

that is \(\tau=2.1873\times 10^{-6}\,\mathrm{s}\), which is \(0.44\) per cent below the measurement. Two known corrections account for the difference and for nothing else.

The first is kinematic. Retaining the electron mass multiplies Equation (101.53) by

whose value here is \(f=0.99981\), a reduction of \(1.87\times 10^{-4}\): the electron mass eats into the available phase space, and the effect is second order in \(m_{e}/m_{\mu}\) because the spectrum vanishes quadratically at low \(x\) by Equation (101.54).

The second is electromagnetic. The one-loop QED correction to the contact interaction, computed once and for all by the methods of Section 100.4, multiplies the rate by

using \(\alpha(m_{\mu}c^{2})^{-1}=135.9\) for the fine-structure constant of Equation (100.3) run to the muon mass by Proposition 100.63. The combination \(f(z)\left(1+\Delta q\right)=0.99557\) gives

against the measured \(2.1969811\times 10^{-6}\,\mathrm{s}\): agreement to \(4\times10^{-5}\), which is the accuracy of the two-loop terms not included here and of the fourth digit of \(\alpha(m_{\mu}c^{2})\) used above. It is worth being explicit about the logic of the comparison. \(G_{F}\) is defined by inverting this calculation, so the agreement of the central values is a tautology; what is not a tautology, and is the actual content, is that the correction factors \(f(z)\) and \(1+\Delta q\) — computed from the electron mass and from QED, with no adjustable parameter — move the tree-level answer by exactly the \(0.44\) per cent by which it was wrong.

Derivation. Derives Phenomenon 101.52. Setting \(\rho=\tfrac{3}{4}\) in Equation (101.58) gives \(x^{2}\left[3-3x+\tfrac{1}{2}(4x-3)\right] =\tfrac{1}{2}x^{2}\left(3-2x\right)\), which is Equation (101.54) up to normalization; so the derived spectrum is the \(\rho=\tfrac{3}{4}\) member of the family and the measurement is a direct test of the Lorentz structure, independent of the rate and therefore independent of \(G_{F}\). The value \(\tfrac{3}{4}\) is not generic: a scalar or tensor contact interaction gives \(\rho=0\), and a \(V+A\) admixture moves \(\xi\) away from \(1\). What the agreement excludes is stated most sharply in the language of left–right symmetric models, where a right-handed \(W\) of mass \(M_{2}\) mixing with the ordinary one shifts \(P_{\mu}\xi\delta/\rho\) below unity; the measured value of that combination is consistent with \(1\) and bounds the mixing [Bayes:2011].

Derivation. Derives Phenomenon 101.53. What the bound establishes is that the quantum numbers \(L_{e}\) and \(L_{\mu}\) are separately conserved in charged-current processes, and not merely their sum. Were they not, the same \(W\) exchange that produces \(\mu\to e\nu\bar{\nu}\) would, with the two neutrinos joined into a loop and a photon attached, produce \(\mu\to e\gamma\) at a rate suppressed only by \(\alpha\) and by loop factors — of order \(10^{-4}\) of the total, not \(10^{-13}\). The observed suppression is therefore a statement about the structure of the current and not about the size of a coupling. In the Standard Model with massive neutrinos the amplitude is not exactly zero but is suppressed by \(\left(\Delta m_{\nu}^{2}c^{4}/M_{W}^{2}c^{4}\right)^{2}\sim 10^{-50}\), an unobservably small number, so any signal at all at the level of Equation (101.59) would be new physics; this is why the search continues. The experiment stops positive muons in a thin target and looks for a positron and a photon, back to back, each of energy \(m_{\mu}c^{2}/2=52.8\,\mathrm{MeV}\), in time coincidence [Afanaciev:2024].

Derives Proposition 101.54. The amplitude is Equation (101.47) with the external legs crossed, and Equation (101.51) applies unchanged: summing over final spins and averaging over the two electron spins — the neutrino has only one helicity, so nothing is averaged there —

using \(2p_{\nu_{\mu}}\cdot p_{e}=s\) for massless incoming particles and the same for the outgoing pair, with \(s=E_{\mathrm{cm}}^{2}/c^{2}\) a squared momentum in the conventions of Notation 100.1. Then \(\overline{\abs{\mathcal{M}}^{2}} =16G_{F}^{2}s^{2}/\hbar^{2}c^{2}\), and Equation (100.40) with \(\abs{\vect{p}_{f}}=\abs{\vect{p}_{i}}\) gives

independent of angle. Multiplying by \(4\pi\) and writing \(s=E_{\mathrm{cm}}^{2}/c^{2}\) gives Equation (101.60). The isotropy is not an accident: the contact interaction has no length scale, so it can produce only \(s\)-wave scattering, and that fact is what Section 101.7.1 turns against the theory. Applied to a nucleon rather than to an electron, with the vector and axial nucleon couplings inserted, the same calculation gives Equation (101.5): the factor \(1+3\lambda^{2}\) is the one Fermi channel plus the three Gamow–Teller channels of Proposition 101.15, each weighted by its coupling, and the factor \(\abs{V_{ud}}^{2}\) is the quark mixing of Section 101.5, which has no counterpart in the purely leptonic process computed here.

Derivation. Derives Phenomenon 101.55. Cabibbo's hypothesis is that the state entering the hadronic charged current is not the down quark but a rotation of the down-type mass eigenstates,

so that the hadronic current is

and that the same \(G_{F}\) multiplies this current and the leptonic one. The two consequences are then not independent. A \(\Delta S=0\) amplitude carries \(\cos\theta_{C}\) and a \(\Delta S=1\) amplitude carries \(\sin\theta_{C}\), so their rates stand in the ratio

which is the observed factor of about twenty; and the strangeness-conserving hadronic rate is suppressed relative to the purely leptonic one by \(\cos^{2}\theta_{C}=0.9496\), a deficit of about five per cent, which is precisely what Equation (101.20) sees as \(\abs{V_{ud}}^{2}\). The angle is therefore measured twice, in unrelated experiments: from the kaon rates as \(\sin\theta_{C}=\abs{V_{us}}=0.2243\), and from the superallowed nuclear rates as \(\cos\theta_{C}=\abs{V_{ud}}=0.97435\). The two determinations satisfy \(\sin^{2}+\cos^{2}=1\) to about a part in \(10^{3}\), and that residual is presently the sharpest unexplained tension in flavour physics; it is recorded, with the current numbers, in Phenomenon 103.7.

Derives Proposition 101.56. The strangeness-changing part of Equation (101.63) is \(\bar{\psi}_{u}\Gamma\psi_{s}\), which annihilates an \(s\) quark and creates a \(u\). The \(s\) quark carries \(S=-1\) and charge \(-\tfrac{1}{3}e\); the \(u\) carries \(S=0\) and \(+\tfrac{2}{3}e\). Hence \(\Delta S=0-(-1)=+1\) and \(\Delta Q=+\tfrac{2}{3}-(-\tfrac{1}{3})=+1\) in units of \(e\), and the two are equal. The Hermitian conjugate term gives \(\Delta S=\Delta Q=-1\). The rule is a prediction and not a convention: it forbids \(\Sigma^{+}\to n e^{+}\nu_{e}\), which has \(\Delta S=+1\) and \(\Delta Q=-1\), while allowing \(\Sigma^{-}\to n e^{-}\bar{\nu}_{e}\). The forbidden mode is bounded at the \(10^{-5}\) level and the allowed one occurs at \(10^{-3}\) [Navas:2024]: two decays of the same multiplet, differing by two orders of magnitude, with the sign of a charge as the only distinction.

Derives Proposition 101.57. The isospin assignment is immediate. For the nonleptonic case, the effective operator is a product of two hadronic currents, each of isospin \(\tfrac{1}{2}\) in the \(\Delta S=1\) sector and each of isospin \(1\) or \(0\) in the \(\Delta S=0\) sector; combining \(\tfrac{1}{2}\otimes1\) gives \(\tfrac{1}{2}\oplus\tfrac{3}{2}\), so both appear with comparable coefficients at the level of the operator. The measurement is the ratio of the two kaon decay amplitudes \(K^{0}\to\pi^{+}\pi^{-}\) and \(K^{+}\to\pi^{+}\pi^{0}\), the second of which is pure \(\Delta I=\tfrac{3}{2}\) because two pions in the \(I=0\) state cannot carry unit charge; the measured ratio of amplitudes is about \(1/22\) [Navas:2024]. Honesty requires the statement that this factor is a QCD effect — short-distance renormalization of the operators plus long-distance hadronic dynamics — and is not derived in this book; it is the reason the rule is quoted here as an observation with its magnitude, rather than proved. The related but different question of why the operator basis is what it is belongs to The Renormalization Group.

Derives Proposition 101.58. Substitute Equation (101.62) and expand. The cross terms are the last bracket.

Derives Theorem 101.59. The two rotated states are the components of \(\psi_{d'}=R\psi_{d}\) with \(R\) the orthogonal matrix of Equation (101.65). Then \(\sum_{i}\bar{\psi}_{i'}\Gamma\psi_{i'} =\bar{\psi}_{j}\left(R\transpose R\right)_{jk}\Gamma\psi_{k} =\bar{\psi}_{j}\delta_{jk}\Gamma\psi_{k}\), which is Equation (101.66). Explicitly, the cross terms are \(+\sin\theta_{C}\cos\theta_{C}\) from \(d'\) and \(-\sin\theta_{C}\cos\theta_{C}\) from \(s'\), and they cancel exactly. Nothing about the value of \(\theta_{C}\) was used: the cancellation is the orthogonality of the rotation, not a numerical accident, which is why it survives to all orders in \(\theta_{C}\) and generalizes without change to the unitary matrix of Section 101.5.4.

Derives Theorem 101.62. \(V\) is the matrix relating two orthonormal bases of one three-dimensional complex internal space — the basis of definite mass and the basis that couples to the \(W\) — and a change between orthonormal bases is unitary. That the space carries an inner product for which both bases are orthonormal is not automatic and is what has to be supplied: it holds because the flavour symmetry acting on the space can be realized by inner-product-preserving operators, which is Proposition 5.150, and because the space then decomposes cleanly, which is Proposition 5.148. Those two propositions are exactly where the licence to write the word “unitary” here comes from, and the argument is set out in Remark 5.152, which states it for this case and records the hypotheses under which each of the two fails. What unitarity buys is set out there as well: two families of relations, the row relations \(\sum_{k}V_{ik}V_{jk}^{*}=\delta_{ij}\) and the column relations \(\sum_{i}V_{ik}^{*}V_{il}=\delta_{kl}\), whose off-diagonal cases are sums of three complex numbers equal to zero and therefore closed triangles in the complex plane. The one that is measured is the subject of Experiment: CP Violation.

Derives Proposition 101.63. A unitary \(N\times N\) matrix has \(N^{2}\) real parameters, of which \(N(N-1)/2\) are the angles of the real orthogonal subgroup and the remaining \(N^{2}-N(N-1)/2=N(N+1)/2\) are phases. The quark fields may be rephased freely, \(\psi_{u_{i}}\mapsto\ee^{\ii\alpha_{i}} \psi_{u_{i}}\) and \(\psi_{d_{k}}\mapsto\ee^{\ii\beta_{k}} \psi_{d_{k}}\), which changes \(V_{ik}\mapsto \ee^{\ii(\beta_{k}-\alpha_{i})}V_{ik}\) and leaves every other term of the Lagrangian invariant, the mass terms because they are diagonal and the neutral currents because they are diagonal by Theorem 101.59. There are \(2N\) such phases, but the overall common phase \(\alpha_{i}=\beta_{k}\) acts trivially on \(V\), so \(2N-1\) of the phases in \(V\) can be removed. The number surviving is \(N(N+1)/2-(2N-1)=(N-1)(N-2)/2\), which with the angles gives Equation (101.68).

For the \(CP\) statement, note that the \(CP\) conjugate of Equation (101.67) is the same expression with \(V\) replaced by \(V^{*}\). If every entry of \(V\) can be made real by rephasing, the interaction is \(CP\) invariant; if a phase survives, it cannot, and \(CP\) is violated. With \(N=2\), \((N-1)(N-2)/2=0\): the Cabibbo matrix is real and Equation (101.62) conserves \(CP\) however \(\theta_{C}\) is chosen. With \(N=3\) one phase survives. Kobayashi and Maskawa's inference [Kobayashi:1973] was made in the reverse direction: \(CP\) violation had been observed in the neutral kaon system nine years earlier [Christenson:1964], no two-generation model could accommodate it without additional structure, and a third generation could. Two of its three quarks were subsequently found. It is one of the few occasions in this chapter on which theory ran ahead of experiment, and it is worth noting that it did so by counting parameters.

Derives Proposition 101.65. The static field equation is \(\left(\nabla^{2}-\left(Mc/\hbar\right)^{2}\right)\phi =-g\,\delta^{3}(\vect{x})\), whose spherically symmetric solution is obtained by writing \(\phi=u(r)/r\), giving \(u''=(Mc/\hbar)^{2}u\) and hence \(u\propto\ee^{-Mcr/\hbar}\) for the solution decaying at infinity. In momentum space the same statement is that the propagator is \(\left[q^{2}+(Mc)^{2}\right]^{-1}\) in the Euclidean region, which for \(q\ll Mc\) is the constant \((Mc)^{-2}\): the propagator loses all memory of the momentum transfer, and the amplitude reduces to a product of two currents at one point, which is Equation (101.6). That is why Fermi's theory works, and it also says exactly how well: the neglected terms are of relative order \(q^{2}/(Mc)^{2}\).

Derives Proposition 101.67. Substituting Equation (101.72) into the neutral part of Equation (101.70), the coefficient of \(A_{\mu}\) is \(g\sin\theta_{W}T^{3}+g'\cos\theta_{W}Y/2\). For this to be proportional to a single generator \(Q\) — which is what it means for the photon to couple to electric charge — the two coefficients must agree, \(g\sin\theta_{W}=g'\cos\theta_{W}=:e\), and then the coefficient is \(e\left(T^{3}+Y/2\right)\), giving Equation (101.73). The coefficient of \(Z_{\mu}\) is \(g\cos\theta_{W}T^{3}-g'\sin\theta_{W}Y/2\), which on eliminating \(g\) and \(g'\) in favour of \(e\) and substituting \(Y/2=Q-T^{3}\) becomes

Since \(T^{3}\) acts only on the left-chiral part, write \(T^{3}\to T^{3}P_{\mathrm{L}} =\tfrac{1}{2}T^{3}\left(\identity-\gamma^{5}\right)\); then \(T^{3}P_{\mathrm{L}}-\sin^{2}\theta_{W}Q =\tfrac{1}{2}\left[\left(T^{3}-2Q\sin^{2}\theta_{W}\right) -T^{3}\gamma^{5}\right]\), which with \(e/\sin\theta_{W}=g\) is Equation (101.74) and Equation (101.75). Finally \(e=g\sin\theta_{W}\) squared and divided by \(4\pi\varepsilon_{0}\hbar c\) gives \(\alpha=\alpha_{W}\sin^{2}\theta_{W}\).

Derives Proposition 101.68. The vertex factor read off Equation (101.76) by the rule of Proposition 100.14 — expand \(\exp(\ii S/\hbar)\) with \(S=c^{-1}\int\dd^{4}x\,\Lag\) — is \(-\ii g\gamma^{\mu}(\identity-\gamma^{5})/2\sqrt{2}\hbar\), of the same SI dimension \(\mathrm{C}/\mathrm{J}/\mathrm{s}\) as the QED vertex Equation (100.24), since \(g\) is a charge by Definition 101.66. Two vertices joined by Equation (101.77) contribute, in the limit \(q^{2} \to 0\),

the \(q_{\mu}q_{\nu}\) term being dropped because it is contracted with currents of light external fermions, on which it gives a factor \(m_{\ell}c/M_{W}c\) squared by the argument of Equation (100.18). The contact interaction Equation (101.46), treated the same way, contributes \(-\ii G_{F}/\sqrt{2}\hbar c\) times the same spinor structure. Setting the two equal,

which is the first form of Equation (101.78). For the second, substitute \(g^{2}=4\pi\varepsilon_{0}\hbar c\,\alpha_{W}\) from Equation (101.71) and \(\mu_{0}=1/\varepsilon_{0}c^{2}\):

and dividing by \((\hbar c)^{3}\) gives \(\pi\alpha_{W}/2\left(M_{W}c^{2}\right)^{2}\). The dimensional check is worth doing once: \(g^{2}\mu_{0}\hbar^{2}/M_{W}^{2}\) carries \(\mathrm{C}^{2}\times \mathrm{kg}\,\mathrm{m}/\mathrm{C}^{2}\times \mathrm{J}^{2}\,\mathrm{s}^{2}/\mathrm{kg}^{2} =\mathrm{J}\,\mathrm{m}^{3}\), which is Equation (101.7).

Derives Proposition 101.70. Repeat the matching of Proposition 101.68 with the \(Z\) vertex \(-\ii g\gamma^{\mu}(g_{V}-g_{A}\gamma^{5})/2\hbar \cos\theta_{W}\) of Equation (101.74) and the \(Z\) propagator, which is Equation (101.77) with \(M_{W}\to M_{Z}\). The effective coefficient is \(\left(g/2\cos\theta_{W}\right)^{2}\mu_{0}\hbar^{2}/M_{Z}^{2}\) times the two current structures. For the neutrino, \(T^{3}=+\tfrac{1}{2}\) and \(Q=0\) give \(g_{V}^{\nu}=g_{A}^{\nu}=\tfrac{1}{2}\) by Equation (101.75), so its factor \(\tfrac{1}{2}(\identity-\gamma^{5})\) contributes an extra \(\tfrac{1}{2}\). Hence the coefficient is

using \(M_{Z}^{2}\cos^{2}\theta_{W}=M_{W}^{2}\) and then Equation (101.78). Two independent factors have conspired: the \(Z\) is heavier than the \(W\), which weakens the interaction, and it couples more strongly by \(1/\cos\theta_{W}\), which strengthens it by exactly the same amount. That the two cancel is the content of the relation \(\rho=1\) of Equation (104.45), and it is a prediction of the doublet structure of the symmetry breaking rather than a fitted coincidence.

Derives Proposition 101.71. Split the electron current into its chiral parts, \(g_{V}-g_{A}\gamma^{5} =g_{L}\left(\identity-\gamma^{5}\right) +g_{R}\left(\identity+\gamma^{5}\right)\), which is the definition of \(g_{L}\) and \(g_{R}\). The \(g_{L}\) piece is Equation (101.46) with a different coefficient, so Equation (101.60) applies to it directly and gives an isotropic cross-section proportional to \(g_{L}^{2}\).

The \(g_{R}\) piece requires one new ingredient, and it is angular momentum rather than algebra. In the centre-of-mass frame let the neutrino move along \(+\hat{\vect{z}}\). It is left-handed, so \(S_{z}^{\nu}=-\tfrac{1}{2}\). A left-chiral electron moving along \(-\hat{\vect{z}}\) has its spin along \(+\hat{\vect{z}}\), \(S_{z}^{e}=+\tfrac{1}{2}\), so the pair has \(J_{z}=0\) and can be in a \(J=0\) state: the amplitude may be isotropic, and is. A right-chiral electron moving along \(-\hat{\vect{z}}\) has \(S_{z}^{e}=-\tfrac{1}{2}\), so \(J_{z}=-1\) and the state must have \(J\geq1\); the amplitude is then proportional to the rotation matrix element \(d^{1}_{-1,-1}(\theta^{*})=\tfrac{1}{2} \left(1+\cos\theta^{*}\right)\), which vanishes in the backward direction because a backward scatter would reverse \(J_{z}\) without anything to absorb the two units. Hence

and integrating over \(\cos\theta^{*}\in[-1,1]\) gives \(2g_{L}^{2}\) and \(2g_{R}^{2}/3\) respectively, which is the factor \(g_{L}^{2}+g_{R}^{2}/3\) in Equation (101.80) once the normalization is fixed against Equation (101.60). For the antineutrino every helicity reverses, so \(g_{L}\) and \(g_{R}\) exchange roles. Numerically, at \(\sin^{2}\theta_{W}=0.231\), \(g_{L}=-0.269\) and \(g_{R}=0.231\), giving \(0.0901\) for the neutrino and \(0.0775\) for the antineutrino, and

which is the size of the signal Gargamelle was looking for. Note that the \(g_{V}\) and \(g_{A}\) enter the two cross-sections in different combinations, so measuring both determines the sign of \(g_{A}\) as well as the value of \(\sin^{2}\theta_{W}\) — a single measurement would not.

Derivation. Derives Phenomenon 101.72. That the process exists at all is Proposition 101.70: a gauge theory containing the observed charged currents cannot avoid a neutral one, because the commutator of the two charged generators is a third generator. That it occurs at the observed rate is Proposition 101.71 applied to the leptonic channel Gargamelle actually saw, which was the antineutrino one: its predicted cross-section per unit beam energy is \(1.34\times 10^{-46}\,\mathrm{m}^{2}/\mathrm{GeV}\) at \(\sin^{2}\theta_{W}=0.231\), the neutrino channel being the larger of the two at \(1.55\times 10^{-46}\,\mathrm{m}^{2}/\mathrm{GeV}\). Gargamelle recorded a single such event in the whole exposure, which was the entire leptonic sample and which no background could produce, since an electron recoiling almost exactly along the beam direction with nothing else in the chamber has no other source [Hasert:1973a]. The hadronic channel supplied the statistics: some hundreds of muonless events, with a ratio to the charged-current rate of roughly one to five [Hasert:1973b].

The obstacle was background, and it is worth recording because the signal is defined by an absence. A neutron entering the chamber from the surrounding material produces hadrons and no muon, which is exactly the signature. The argument that settled it was the spatial distribution: neutrons are attenuated as they penetrate the liquid, so a neutron-induced sample is concentrated near the walls and falls off with depth, whereas a neutrino-induced sample is uniform because the chamber is transparent to neutrinos by Phenomenon 101.5. The observed distribution was flat. The same reasoning applied to the ratio of muonless to muonful events as a function of position gave a rate that could not be produced by neutrons at any plausible flux.

Derivation. Derives Phenomenon 101.73. The masses were predicted before they were measured, from two quantities determined in entirely different experiments. Inverting Equation (101.78),

and inserting \(G_{F}/(\hbar c)^{3} =1.1663787\times 10^{-5}\,/\mathrm{GeV}^{2}\) from the muon lifetime, \(\alpha^{-1}=127.95\) at this scale from Proposition 100.63, and \(\sin^{2}\theta_{W}=0.2312\) from the neutral-current measurements of Proposition 101.71 and Phenomenon 104.55,

that is \(M_{W}c^{2}=80.2\,\mathrm{GeV}\), and \(M_{Z}=M_{W}/\cos\theta_{W}\) gives \(M_{Z}c^{2}=91.5\,\mathrm{GeV}\). The measured values are Equation (101.82): the tree-level prediction is right to a few parts in a thousand, the residual being the electroweak radiative correction that Section 101.6.4 turns into a measurement in its own right.

Read the other way, the same relation shows why Fermi's contact theory worked so well for so long. Writing \(Q^{2}=-q^{2}c^{2}\) for the squared four-momentum transfer measured as a squared energy, the exchange contributes the propagator factor \(\left[Q^{2}+(M_{W}c^{2})^{2}\right]^{-1}\), and in a nuclear decay \(Q\) is a few \(\mathrm{MeV}\), so the neglected ratio \(Q^{2}/(M_{W}c^{2})^{2}\) is of order \(10^{-9}\). The corresponding range is \(\hbar/M_{W}c=2.5\times 10^{-18}\,\mathrm{m}\), a thousandth of the diameter of a proton, which is why no experiment before 1983 had resolved the exchange. Where it must fail is at \(Q\) approaching \(M_{W}c^{2}\), which is the unitarity problem of Section 101.7.1.

Derives Proposition 101.75. By Equation (101.48) the width is \(\left(c^{2}/2M_{Z}c^{2}\right)\int \overline{\abs{\mathcal{M}}^{2}}\dd\Phi_{2}\), and for massless final states \(\int\dd\Phi_{2}=1/8\pi\hbar^{2}\) by Lemma 101.47. The squared amplitude, summed over final spins and averaged over the three \(Z\) polarizations, involves the trace

contracted with the polarization sum \(\sum_{\lambda}\varepsilon^{*}_{\mu}\varepsilon_{\nu} =-\eta_{\mu\nu}+q_{\mu}q_{\nu}/M_{Z}^{2}c^{2}\). Moving the first factor \(\left(g_{V}-g_{A}\gamma^{5}\right)\) rightwards through \(\gamma\cdot p\) and then through \(\gamma^{\nu}\) — each passage flipping the sign of the \(\gamma^{5}\) term — collects the two into \(\left(g_{V}-g_{A}\gamma^{5}\right)^{2} =g_{V}^{2}+g_{A}^{2}-2g_{V}g_{A}\gamma^{5}\). The \(\gamma^{5}\) piece produces a Levi-Civita tensor, antisymmetric in \(\mu\nu\), which vanishes against the symmetric polarization sum. Hence the entire dependence on the fermion species is the factor \(g_{V}^{2}+g_{A}^{2}\), together with the colour multiplicity \(N_{c}\), and Equation (101.85) follows without evaluating anything else. Carrying the remaining common factor through, with the coupling \(g/2\cos\theta_{W}\) of Equation (101.74) re-expressed through Equation (101.78) and \(M_{W}=M_{Z}\cos\theta_{W}\), gives Equation (101.86). As a check, for a neutrino \(g_{V}=g_{A}=\tfrac{1}{2}\) and Equation (101.86) gives, as an energy width,

against the measured \(167.2\,\mathrm{MeV}\) [Schael:2006]: a tree-level formula good to one per cent, the difference being the radiative corrections. Widths are quoted throughout as energies, \(\hbar\Gamma\), which is the convention of the source; the rate itself is \(\Gamma\) and carries \(/\mathrm{s}\).

Derivation. Derives Phenomenon 101.76. Measure the total width \(\Gamma_{Z}\) from the shape of the resonance in \(e^{+}e^{-}\to\text{anything visible}\), the hadronic width from the hadronic cross-section, and the leptonic width from the lepton-pair cross-sections. The invisible width is the difference, \(\Gamma_{\mathrm{inv}}=\Gamma_{Z}-\Gamma_{\mathrm{had}} -3\Gamma_{\ell\ell}\). Then

which is arranged so that the absolute normalization of Equation (101.86) cancels and only the ratio Equation (101.85) is needed. That ratio is pure algebra: for the neutrino \(g_{V}=g_{A}=\tfrac{1}{2}\) gives \(g_{V}^{2}+g_{A}^{2}=\tfrac{1}{2}\), while for a charged lepton \(T^{3}=-\tfrac{1}{2}\) and \(Q=-1\) give, by Equation (101.75), \(g_{A}=-\tfrac{1}{2}\) and \(g_{V}=-\tfrac{1}{2}+2\sin^{2}\theta_{W}=-0.0376\) at \(\sin^{2}\theta_{W}=0.23122\), so \(g_{V}^{2}+g_{A}^{2}=0.2514\) and

The measured ratio is \(\Gamma_{\mathrm{inv}}/\Gamma_{\ell\ell}=5.943\pm0.016\) [Schael:2006], whence \(N_{\nu}=5.943/1.989=2.988\), and the full treatment with radiative corrections gives Equation (101.87). The near-cancellation of \(g_{V}\) for the charged lepton — \(-\tfrac{1}{2}+2(0.231)\) is small because \(\sin^{2}\theta_{W}\) happens to lie near \(\tfrac{1}{4}\) — is why \(\Gamma_{\ell\ell}\) is almost exactly half \(\Gamma_{\nu\bar{\nu}}\), and it is a numerical accident with no significance beyond making the arithmetic memorable.

Two caveats belong with the number. It counts only neutrinos with \(m_{\nu}<M_{Z}/2\) that couple to the \(Z\) with standard strength: a heavier or a sterile neutrino is invisible to this measurement, and Flavour Physics and Neutrinos records what other evidence says about those. And it is a counting measurement whose result is not an integer, \(2.984\pm0.008\); the deficit below three is about two standard deviations and is understood as a residual effect in the Bhabha-scattering normalization, which is exactly the kind of statement Probability and Statistics is needed to assess.

Derivation. Derives Phenomenon 101.77. This is Proposition 101.54: the cross-section is \(G_{F}^{2}E_{\mathrm{cm}}^{2}/\pi(\hbar c)^{4}\), and \(E_{\mathrm{cm}}^{2}=2m_{\mathrm{target}}c^{2}E_{\mathrm{beam}}\) for a fixed target, so \(\sigma\) is proportional to the beam energy. The growth is not a defect of the calculation but a consequence of the dimension of the coupling: with only \(G_{F}\), \(\hbar\), \(c\) and \(E_{\mathrm{cm}}\) available, and \(\sigma\) an area, the answer must be \(G_{F}^{2}E_{\mathrm{cm}}^{2}/(\hbar c)^{4}\) times a pure number. Restoring the propagator replaces this by \(\sigma\propto G_{F}^{2}E_{\mathrm{cm}}^{2}/ \left[1+E_{\mathrm{cm}}^{2}/(M_{W}c^{2})^{2}\right]\), which stops rising once \(E_{\mathrm{cm}}\) reaches \(M_{W}c^{2}\) and tends to the constant \(G_{F}^{2}(M_{W}c^{2})^{2}\) instead; the turnover is the direct observation of the propagator, and hence of the \(W\) in a scattering experiment rather than as a resonance.

Derives Theorem 101.78. The bound is the statement that the \(S\)-matrix is unitary. Writing the elastic amplitude in partial waves, \(f(\theta)=k^{-1}\sum_{J}(2J+1)a_{J}P_{J}(\cos\theta)\), unitarity of \(S_{J}=1+2\ii a_{J}\) forces \(\abs{a_{J}}\leq1\), whence \(\sigma_{J}=4\pi(2J+1)\abs{a_{J}}^{2}/k^{2} \leq4\pi(2J+1)/k^{2}\); the partial-wave apparatus is that of Scattering Theory. That the contact interaction produces only \(J=0\) is Proposition 101.54: the cross-section came out isotropic, and an isotropic amplitude is by definition pure \(s\)-wave. Setting \(G_{F}^{2}E_{\mathrm{cm}}^{2}/\pi(\hbar c)^{4} =16\pi(\hbar c)^{2}/E_{\mathrm{cm}}^{2}\) and solving gives Equation (101.90); with \(G_{F}/(\hbar c)^{3}=1.1663787\times 10^{-5}\,/\mathrm{GeV}^{2}\), \(E_{\mathrm{cm}}^{4}=16\pi^{2}/(1.1663787\times 10^{-5})^{2} \mathrm{GeV}^{4}\) and \(E_{\mathrm{cm}}=1038\,\mathrm{GeV}\).

Derives Theorem 101.81. Let \(L\) be the number of loops and \(P\) the number of internal fermion lines. Each loop contributes \(\dd^{4}\ell\) and each internal fermion propagator falls as \(\ell^{-1}\) by Equation (100.23), so \(D=4L-P\). Every vertex has four fermion ends, every internal line uses two and every external line one, giving \(4V=2P+E\); and the number of independent loops is \(L=P-V+1\). Eliminating,

which is Equation (101.91).

The contrast with Equation (100.26) is the whole point. There, \(D=4-\tfrac{3}{2}E_{e}-E_{\gamma}\) and \(V\) cancelled: the degree of divergence is a property of the amplitude, not of the order, so only finitely many amplitudes diverge and finitely many parameters absorb them, which is Definition 100.18. Here \(V\) does not cancel, and it enters with a positive sign. The four-point function has \(E=4\) and \(D=2V-2\): convergent at \(V=1\), quadratically divergent at \(V=2\), quartically at \(V=3\), and so on without end. Each new order requires a new counterterm of a new operator structure, and each new counterterm brings a new undetermined constant, so the theory has no predictive power beyond the order at which it is first fitted.

Derives Corollary 101.82. An amplitude of fixed external content has a fixed overall dimension. If a vertex carries a coupling of mass dimension \(-\delta\), then \(V\) vertices carry \(-V\delta\), and the loop integrals must supply \(+V\delta\) more powers of momentum to compensate, which is exactly \(D\to D+V\delta\). With \(\delta=2\) for \(G_{F}\) this is the \(2V\) of Equation (101.91). Conversely the QED coupling \(e/\hbar\) has \(\delta=0\) once the \(\hbar\)'s are counted — which is the content of \(\alpha\) being dimensionless, Equation (100.3) — and no such term appears.