Electroweak Unification and the Higgs Boson

Derivation. Derives Proposition 104.3. Each item is proved where it is stated in Weak Interactions; what is added here is that the three are the same statement counted three ways. A coupling of dimension \(\mathrm{J}\,\mathrm{m}^{3}\) can only enter an amplitude in the combination \(G_{F}E^{2}/(\hbar c)^{3}\), because that is the unique dimensionless product of \(G_{F}\), \(\hbar\), \(c\) and an energy; hence item (i). A cross-section is an area and the only area available is \(G_{F}^{2}E^{2}/(\hbar c)^{4}\) times a pure number, which grows while the unitarity bound \(16\pi(\hbar c)^{2}/E^{2}\) falls, so they must cross; hence item (ii). And in the momentum counting of Theorem 100.17 each vertex contributes its coupling's negative mass dimension to \(D\) with a positive sign, which is Corollary 101.82; hence item (iii).

Derives Theorem 104.4. (i) Under the shift, \(W^{+}_{\mu}W^{-\mu}\) acquires \(\pp_{\mu}\Lambda^{+}W^{-\mu}+W^{+}_{\mu}\pp^{\mu}\Lambda^{-} +\pp_{\mu}\Lambda^{+}\pp^{\mu}\Lambda^{-}\), which no choice of \(\Lambda\) makes vanish identically. The mass term breaks the very invariance that Theorem 100.5 used to construct the interaction.

(ii) The massive vector kernel is \(-\mu_{0}^{-1}\hbar^{-2}\left[(k^{2}-M_{W}^{2}c^{2})\eta_{\mu\nu} -k_{\mu}k_{\nu}\right]\), whose inverse is Equation (101.77). At large \(k\) the transverse part falls as \(k^{-2}\), exactly as the photon propagator Equation (100.17) does, but the term \(k_{\mu}k_{\nu}/M_{W}^{2}c^{2}\) divided by \(k^{2}-M_{W}^{2}c^{2}\) tends to \(\eta\)-independent constant behaviour: it does not fall at all. Repeating the count of Theorem 100.17 with internal boson lines contributing \(k^{0}\) instead of \(k^{-2}\) raises \(D\) by two for every such line, and \(D\) again grows with the order.

(iii) A massive vector of four-momentum \(k^{\mu}=(E/c,0,0,k)\) with \(E^{2}/c^{2}-k^{2}=M_{W}^{2}c^{2}\) has three polarization vectors satisfying \(k_{\mu}\varepsilon^{\mu}=0\) and \(\varepsilon\cdot\varepsilon^{*}=-1\); the longitudinal one is

as one checks by expanding \(E/c=\sqrt{k^{2}+M_{W}^{2}c^{2}}\). The second term is bounded; the first grows without limit, and it is the one that appears in Equation (104.6). Since an amplitude is linear in each external polarization, an amplitude with \(n\) longitudinal \(W\)s carries \(n\) factors of \(k/M_{W}c\) relative to the transverse one. The physical content is that the longitudinal mode is the piece a genuinely massless gauge field does not have; inserting a mass by hand supplies it without supplying a field for it to come from.

Derives Proposition 104.7. The first statement is Equation (100.10) with the phase replaced by a matrix and the extra term coming from the failure of \(U\) and \(W_{\mu}\) to commute; writing \(\mathcal{W}_{\mu}=gW_{\mu}/\hbar\) of dimension \(/\mathrm{m}\), the requirement \((\pp_{\mu}+\ii\mathcal{W}'_{\mu})U\psi =U(\pp_{\mu}+\ii\mathcal{W}_{\mu})\psi\) for all \(\psi\) gives \(\ii\mathcal{W}'_{\mu}U=U\ii\mathcal{W}_{\mu}-(\pp_{\mu}U)\), which is Equation (104.10). For the field strength, note that \(\ii\mathcal{W}_{\mu\nu}=\comm{D_{\mu}}{D_{\nu}}\) is a commutator of covariant objects, hence a covariant object: it is Equation (106.64), and in components it is Equation (104.8) because \(\comm{T^{b}}{T^{c}}=\ii\varepsilon^{bcd}T^{d}\) contributes \(-\varepsilon^{abc}\mathcal{W}^{b}_{\mu}\mathcal{W}^{c}_{\nu}\). Since \(\tr(W_{\mu\nu}W^{\mu\nu})\) is invariant under conjugation, Equation (104.9) is invariant. Finally, squaring Equation (104.8) gives cross terms \((\pp W)(g\varepsilon WW/\hbar)\) and \((g\varepsilon WW/\hbar)^{2}\), which are the cubic and quartic self-interactions; they have no counterpart in Equation (100.12), where \(\varepsilon^{abc}\) has nothing to contract. That the gauge bosons interact with one another is a prediction, and it is measured directly in Phenomenon 104.57.

Derives Theorem 104.8. For an infinitesimal transformation, \(\delta(gW^{a}_{\mu}/\hbar)=\pp_{\mu}\theta^{a} +\varepsilon^{abc}\theta^{b}(gW^{c}_{\mu}/\hbar)\), obtained by expanding Equation (104.10) to first order. Then

The second term vanishes because \(\varepsilon^{abc}\) is antisymmetric in \(ac\) while \(W^{c}_{\mu}W^{a\mu}\) is symmetric; the first does not, and it cannot be cancelled by anything, since \(\theta^{a}\) is an arbitrary function while \(W^{a\mu}\) is a field. For the second claim, observe that a mass term is by definition quadratic in the field with no derivatives, and Equation (104.10) contains an inhomogeneous piece \(\pp_{\mu}\theta\) that only a derivative can absorb; every gauge-invariant local functional of \(W\) alone is therefore built from \(W_{\mu\nu}\) and its covariant derivatives, each of which carries at least one \(\pp_{\mu}\) and hence at least two powers of momentum. Such a term contributes to the kinetic operator, never to a mass.

Derives Proposition 104.12. For a doublet with members of charge \(Q_{\uparrow}\), \(Q_{\downarrow}\) and \(T^{3}=\pm\tfrac{1}{2}\), Equation (104.13) gives \(Y=2Q_{\uparrow}-1=2Q_{\downarrow}+1\); the two are equal if and only if \(Q_{\uparrow}-Q_{\downarrow}=1\), and then \(Y=Q_{\uparrow}+Q_{\downarrow}\), twice the average. The leptonic doublet \((\nu_{e},e)\) has \(0-(-1)=1\) and the quark doublet \((u,d)\) has \(\tfrac{2}{3}-(-\tfrac{1}{3})=1\): both pass, and it is the charged current that requires it, since \(W^{+}\) carries exactly one unit of charge. The right-chiral fields are singlets, so \(T^{3}=0\) and \(Y=2Q\) with no constraint at all. Nothing so far explains why the charges come in these ratios; that is the content of Section 104.5.3, where the same numbers are forced by consistency at one loop.

Derives Proposition 104.15. Write \(\abs{\phi}^{2}=\rho\). Then \(V=-\mu^{2}\rho+\lambda\rho^{2}\), \(\dv{V}{\rho}=-\mu^{2}+2\lambda\rho\), which vanishes at \(\rho=\mu^{2}/2\lambda\), and \(\dd^{2}V/\dd\rho^{2}=2\lambda>0\) there, so it is the minimum; \(\rho=0\) gives \(\dd V/\dd\rho=-\mu^{2}<0\), so the origin is a maximum. Setting \(\rho=v^{2}/2\) gives \(v^{2}=\mu^{2}/\lambda\), which is Equation (104.15), and \(V_{\min}=-\mu^{2}v^{2}/2+\lambda v^{4}/4=-\lambda v^{4}/4\) on using \(\mu^{2}=\lambda v^{2}\). The dimensions check: \([\mu^{2}\abs{\phi}^{2}] =/\mathrm{m}^{2}\times\mathrm{J}/\mathrm{m} =\mathrm{J}/\mathrm{m}^{3}\), an energy density, and likewise for the quartic. Every point of the circle is a minimum and the phase of \(\phi\) at the minimum is undetermined: that undetermined phase is the degeneracy of Definition 104.14, with \(G=\U(1)\) and \(H\) trivial.

Derives Theorem 104.17. Invariance of \(V\) under an infinitesimal transformation reads, for every \(\phi\) and every \(a\),

This is an identity in \(\phi\), so it may be differentiated. Applying \(\pp/\pp\phi_{k}\) gives

Now evaluate at \(\phi=\phi_{0}\). The second term vanishes because \(\phi_{0}\) is a stationary point of \(V\), and the first becomes

Hence each vector \(t^{a}\phi_{0}\) lies in the kernel of \(M^{2}\). If \(t^{a}\) belongs to the Lie algebra of \(H\) then \(t^{a}\phi_{0}=0\) and Equation (104.20) is empty; otherwise \(t^{a}\phi_{0}\neq0\) and it is a genuine null eigenvector. The vectors \(\set{t^{a}\phi_{0}}\) span a space of dimension \(\dim G-\dim H\), because the map \(t\mapsto t\phi_{0}\) from the Lie algebra of \(G\) to \(\R^{n}\) has kernel exactly the Lie algebra of \(H\) by definition of the stabilizer; and \(M^{2}\), being a real symmetric matrix, has as many zero eigenvalues as the dimension of its kernel. Since \(\phi_{0}\) is a minimum, \(M^{2}\) is positive semi-definite and no eigenvalue is negative. Therefore the spectrum consists of \(\dim G-\dim H\) massless modes and \(n-\dim G+\dim H\) massive ones.

Geometrically, \(t^{a}\phi_{0}\) is tangent to the orbit \(G\phi_{0}\) at \(\phi_{0}\), and \(V\) is constant on that orbit by Equation (104.19): a direction along which the potential does not change is a direction in which the field costs no energy to excite at zero momentum, which is what a massless particle is.

Derives Theorem 104.19. Write \(\phi(x)=\frac{1}{\sqrt{2}}\left[v+h(x)\right] \exp\!\left[\ii\chi(x)/v\right]\), which is legitimate wherever \(\phi\neq0\) and which parametrizes the same two real degrees of freedom as before, \(h\) radial and \(\chi\) tangential; \(\chi/v\) is dimensionless because \([\chi]=[v]\). The potential depends only on \(\abs{\phi}\), so it is exactly Equation (104.16) without the \(\chi\) terms: \(h\) gets the mass Equation (104.17) and \(\chi\) appears nowhere in \(V\).

Now perform the gauge transformation Equation (100.8) with \(\Lambda=-\hbar\chi/qv\), under which \(\phi\mapsto\ee^{-\ii q\Lambda/\hbar}\phi =\frac{1}{\sqrt{2}}(v+h)\): the phase is gone. This choice is the unitary gauge, and it is available precisely because \(\chi\) shifts inhomogeneously under a gauge transformation, exactly as \(A_{\mu}\) does. What was a field is now a gauge parameter. The Lagrangian becomes

using \(\abs{D_{\mu}\phi}^{2}=\tfrac{1}{2}(\pp h)^{2} +\tfrac{1}{2}(q/\hbar)^{2}(v+h)^{2}A^{2}\) for a real \(\phi\). The second term of Equation (104.23) is a mass term. Comparing it with the Proca form \(\tfrac{1}{2\mu_{0}}(m_{A}c/\hbar)^{2}A^{2}\) gives

which is Equation (104.22). Dimensionally, \(q^{2}v^{2}\mu_{0}/c^{2}\) carries \(\mathrm{C}^{2}\times\mathrm{J}/\mathrm{m} \times\mathrm{kg}\,\mathrm{m}/\mathrm{C}^{2} \times\mathrm{s}^{2}/\mathrm{m}^{2} =\mathrm{kg}^{2}\), so \(m_{A}\) is a mass.

For the counting: before breaking, \(\phi\) has two real components and a massless vector has two transverse polarizations, four in all; after, \(h\) has one and a massive vector has three, again four. The Goldstone mode has not disappeared, it has become the longitudinal polarization whose absence from the massless theory Equation (104.6) identified as the problem. The last two terms of Equation (104.23) are the interactions of the surviving scalar with the vector, and their strengths are not free: both are fixed by \(m_{A}\) and \(v\). That is the statement measured in Phenomenon 104.62.

Derives Corollary 104.20. Varying Equation (104.23) with respect to \(A_{\mu}\) gives \(\pp_{\mu}F^{\mu\nu} =-\mu_{0}q^{2}v^{2}\hbar^{-2}A^{\nu}\), and in a static configuration with \(\pp_{0}=0\) and \(\nabla\cdot\vect{A}=0\) the spatial part reads \(\nabla^{2}\vect{A}=\mu_{0}q^{2}v^{2}\hbar^{-2}\vect{A} =(m_{A}c/\hbar)^{2}\vect{A}\) by Equation (104.22). A solution decaying into the half-space \(z>0\) is \(\vect{A}\propto\ee^{-z/\lambda_{L}}\) with \(\lambda_{L}\) as stated, and \(\vect{B}=\curl\vect{A}\) decays with it.

Derives Theorem 104.22. Setting \(\Phi=\Phi_{0}\) in the scalar kinetic term, \(\left(D_{\mu}\Phi_{0}\right)^{\dagger}\left(D^{\mu}\Phi_{0}\right) =\hbar^{-2}\left(g_{a}T^{a}\Phi_{0}\right)^{\dagger} \left(g_{b}T^{b}\Phi_{0}\right)A^{a}_{\mu}A^{b\mu}\), since \(\pp_{\mu}\Phi_{0}=0\). Only the part symmetric in \(ab\) survives the contraction with \(A^{a}_{\mu}A^{b\mu}\), which is the real part. Comparing with the Proca form \(\tfrac{1}{2\mu_{0}}(c/\hbar)^{2}(M^{2})^{ab}A^{a}_{\mu}A^{b\mu}\) gives Equation (104.25).

The matrix is a Gram matrix of the vectors \(g_{a}T^{a}\Phi_{0}\), so it is positive semi-definite and a vector \(n^{a}\) lies in its kernel if and only if \(n^{a}g_{a}T^{a}\Phi_{0}=0\), that is, if and only if the corresponding generator annihilates the vacuum. The rank of \(M^{2}\) therefore equals the number of independent broken generators, which by Theorem 104.17 is the number of massless scalars the ungauged theory would have had. Each of them is removed from the scalar spectrum by the unitary-gauge argument of Theorem 104.19 applied direction by direction, and reappears as the longitudinal polarization of the corresponding massive vector: the bookkeeping balances generator by generator, not merely in total.

Derives Theorem 104.25. The potential depends on \(\Phi\) only through \(\rho=\Phi^{\dagger}\Phi\), so Proposition 104.15 applies verbatim with \(\abs{\phi}^{2}\) replaced by \(\rho\): the minimum is at \(\rho=\mu^{2}/2\lambda=v^{2}/2\). The set of \(\Phi\) with \(\Phi^{\dagger}\Phi=v^{2}/2\) is a sphere \(S^{3}\) in the four real dimensions of the doublet, and \(\SU(2)\) acts transitively on \(S^{3}\) (it is \(S^{3}\) as a manifold), so every minimum is gauge-equivalent to Equation (104.29).

For the stabilizer, compute the action of each generator on \(\Phi_{0}\). With \(T^{a}=\tfrac{1}{2}\sigma^{a}\) and \(Y=1\),

none of which vanishes: each of the four generators moves the vacuum. But the combination \(Q=T^{3}+Y/2\) does annihilate it, by the last two lines of Equation (104.31), and it is the only combination that does, since the first two lines are linearly independent of one another and of the third. The unbroken subgroup is therefore one-dimensional and generated by \(Q\), which is Equation (104.30); that \(Q\) is the electric charge is Equation (104.13), so what survives is electromagnetism and not some other \(\U(1)\).

By Theorem 104.17 the ungauged model would have \(\dim G-\dim H=4-1=3\) massless scalars, leaving \(4-3=1\) massive one. By Theorem 104.22 the gauge-boson mass matrix has rank \(3\) and a one-dimensional kernel spanned by the generator \(Q\), so exactly one vector — the photon — stays massless, exactly to all orders by Remark 104.23, and three become massive by absorbing the three Goldstone modes.

Derivation. Derives Proposition 104.26. The \(W\) mass is computed in Theorem 104.28 below as \(M_{W}c^{2}=\tfrac{1}{2}\hat{g}E_{v}\); substituting it into the matching condition Equation (104.4),

so \(E_{v}^{2}=(\hbar c)^{3}/\sqrt{2}G_{F}\), which is Equation (104.32); the coupling \(\hat{g}\) has cancelled, which is why \(v\) is known far more precisely than either boson mass. Numerically, with \(\sqrt{2}\times1.1663787\times 10^{-5} =1.6495086\times 10^{-5}\) from Equation (104.3), the reciprocal is \(60624.1\,\mathrm{GeV}^{2}\) and its square root is \(246.220\,\mathrm{GeV}\). Converting, \(246.220\,\mathrm{GeV}\times1.602176634\times 10^{-10}\,\mathrm{J}/\mathrm{GeV} =3.94487\times 10^{-8}\,\mathrm{J}\); dividing by \(\sqrt{\hbar c}=1.7780683\times 10^{-13}\,\mathrm{J}^{1/2}\,\mathrm{m}^{1/2}\) gives \(v\); and dividing \(\hbar c=3.161527\times 10^{-26}\,\mathrm{J}\,\mathrm{m}\) by \(E_{v}\) gives Equation (104.33).

Derives Theorem 104.28. Since \(\pp_{\mu}\Phi_{0}=0\), \(D_{\mu}\Phi_{0}=\frac{\ii}{\hbar} \left(gW^{a}_{\mu}T^{a}+\tfrac{1}{2}g'B_{\mu}\right)\Phi_{0}\), using \(Y=1\). Then

Insert \(T^{a}=\tfrac{1}{2}\sigma^{a}\) and \(\Phi_{0}=\tfrac{v}{\sqrt{2}}(0,1)\transpose\). The matrix in the middle is

so acting on \(\Phi_{0}\) it returns \(\tfrac{v}{2\sqrt{2}}\left(\sqrt{2}gW^{+}_{\mu}, -gW^{3}_{\mu}+g'B_{\mu}\right)\transpose\), whose squared norm is

which divided by \(\hbar^{2}\) is Equation (104.34). Note that \(B_{\mu}\) and \(W^{3}_{\mu}\) enter only through the single combination \(gW^{3}-g'B\): the orthogonal combination \(A_{\mu}\) of Equation (104.35) is absent, so it has no mass, which is the algebraic form of Theorem 104.22 here.

Reading off the masses requires the Proca normalizations. A complex vector has mass term \(\mu_{0}^{-1}\left(M_{W}c/\hbar\right)^{2}W^{+}_{\mu}W^{-\mu}\) and a real one \(\tfrac{1}{2\mu_{0}}\left(M_{Z}c/\hbar\right)^{2}Z_{\mu}Z^{\mu}\), so

Now convert to the hatted couplings. With \(g=\hat{g}\sqrt{\varepsilon_{0}\hbar c}\) from Equation (104.1) and \(v=E_{v}/\sqrt{\hbar c}\) from Equation (104.2),

because \(c\sqrt{\mu_{0}\varepsilon_{0}}=1\) exactly. The same substitution in \(M_{Z}\) gives the second of Equation (104.36). Dimensionally \(\mu_{0}g^{2}v^{2}/c^{2}\) carries \(\mathrm{kg}\,\mathrm{m}/\mathrm{C}^{2}\times\mathrm{C}^{2} \times\mathrm{J}/\mathrm{m}\times\mathrm{s}^{2}/\mathrm{m}^{2}=\mathrm{kg}^{2}\), as a squared mass must.

Derives Corollary 104.29. The potential restricted to the radial direction is Equation (104.14) with \(\abs{\phi}^{2}=(v+h)^{2}/2\), so Equation (104.16) applies unchanged: the \(h\) mass term is \(\tfrac{1}{2}(2\lambda v^{2})h^{2}\), which by Notation 106.23 means \((m_{h}c/\hbar)^{2}=2\lambda v^{2}\). Multiplying by \((\hbar c)^{2}\), \((m_{h}c^{2})^{2}=2(\hbar c)^{2}\lambda v^{2} =2(\lambda\hbar c)(v^{2}\hbar c)=2\hat{\lambda}E_{v}^{2}\), which is Equation (104.37); and \(\mu^{2}=\lambda v^{2}\) gives \(\hbar\mu c=m_{h}c^{2}/\sqrt{2}\). Numerically \(m_{h}c^{2}/E_{v}=125.20/246.22=0.50848\), whose square halved is \(0.12928\), and \(125.20/\sqrt{2}=88.53\).

Derives Proposition 104.31. The first two of Equation (104.40) are Equation (104.36) with Equation (104.39). The third is Equation (101.73), whose derivation is Proposition 101.67: substituting Equation (104.35) into Equation (104.11), the coefficient of \(A_{\mu}\) is \(\hat{g}\sin\theta_{W}T^{3} +\hat{g}'\cos\theta_{W}Y/2\), which is proportional to \(Q=T^{3}+Y/2\) if and only if the two coefficients coincide, and then their common value is \(\hat{e}\). For Equation (104.41), square the first relation and use \(\hat{g}\sin\theta_{W}=\hat{e}\): \((M_{W}c^{2})^{2}\sin^{2}\theta_{W} =\tfrac{1}{4}\hat{e}^{2}E_{v}^{2}=\pi\alpha E_{v}^{2}\), since \(\hat{e}^{2}=4\pi\alpha\); and \(E_{v}^{2}=(\hbar c)^{3}/\sqrt{2}G_{F}\) by Equation (104.32). Numerically, \(\pi\times7.2973525643\times 10^{-3}\times60624.1\,\mathrm{GeV}^{2} =1389.8\,\mathrm{GeV}^{2}\).

Derivation. Derives Proposition 104.32. \(\hat{g}=2M_{W}c^{2}/E_{v}=2(80.369)/246.220=0.65282\), and \(\sqrt{\hat{g}^{2}+\hat{g}'^{2}}=2M_{Z}c^{2}/E_{v} =2(91.1880)/246.220=0.74070\), whence \(\hat{g}'^{2}=0.548643-0.426179=0.122464\) and \(\hat{g}'=0.34995\). Then \(\sin^{2}\theta_{W}=\hat{g}'^{2}/(\hat{g}^{2}+\hat{g}'^{2}) =0.122464/0.548643=0.22321\), which is also \(1-(80.369/91.1880)^{2}\) as Equation (104.40) requires. Finally \(\hat{e}=\hat{g}\sin\theta_{W} =0.65282\times0.47245=0.30843\) and \(\hat{e}^{2}/4\pi=0.095129/12.56637=0.0075699 =1/132.10\).

Derivation. Derives Phenomenon 104.35. The masses are Equation (104.36), so \(m_{W}^{2}/m_{Z}^{2}=\hat{g}^{2}/(\hat{g}^{2}+\hat{g}'^{2}) =\cos^{2}\theta_{W}\) by Equation (104.39), and Equation (104.45) gives \(\rho=1\) identically. Two features of that one-line computation carry the physics.

First, \(E_{v}\) has cancelled. The relation is therefore a statement about the representation the breaking field occupies and not about how strongly the symmetry is broken; it survives any number of doublets and, as Theorem 104.41 shows, fails for almost any other multiplet.

Second, the angle in Equation (104.45) is defined by Equation (104.39) from the couplings, while the ratio on the left is measured from masses. What is tested is that these two determinations agree, and the test is quantitative only when a scheme is fixed, per Remark 104.34. In the on-shell scheme \(\rho=1\) is an identity and the content moves into the agreement of the mass-derived angle with the asymmetry-derived one; in \(\overline{\mathrm{MS}}\) the content stays in \(\rho\) itself, whose measured deviation from unity is the few-per-mille effect predicted by Equation (104.52). The measured \(\rho\) is in either case evidence that the electroweak symmetry is broken by doublets, which is a fact about the vacuum, not a convention.

Derives Proposition 104.36. Insert \(\psi=\psi_{L}+\psi_{R}\) with \(\psi_{L,R}=\tfrac{1}{2}(\identity\mp\gamma^{5})\psi\) and use \(\bar{\psi}_{L}\psi_{L}=\bar{\psi}_{R}\psi_{R}=0\), which follows from \(\gamma^{5}\gamma^{0}=-\gamma^{0}\gamma^{5}\): the mass term connects the two chiralities and nothing else. But \(\psi_{L}\) is a component of an \(\SU(2)_{L}\) doublet while \(\psi_{R}\) is a singlet, so \(\bar{\psi}_{L}\psi_{R}\) transforms as a doublet, not a singlet, and cannot be a term in an invariant Lagrangian. Even setting the \(\SU(2)_{L}\) aside, the hypercharges do not match: for the electron \(Y_{L}=-1\) and \(Y_{R}=-2\), so \(\bar{\psi}_{L}\psi_{R}\) carries \(Y=-2-(-1)=-1\neq0\) and is not \(\U(1)_{Y}\) invariant either. The statement is not that the electron is massless; it is that its mass cannot be a parameter of the symmetric Lagrangian, and must come from the vacuum.

Derivation of the dimension and of Equation (104.47). Derives Equation (104.47). A fermion bilinear \(\bar{\psi}\psi\) carries \(/\mathrm{m}^{3}\) by Definition 100.2 and \(\Phi\) carries \(\mathrm{J}^{1/2}\,\mathrm{m}^{-1/2}\), so for Equation (104.46) to be an energy density, \([y]=\mathrm{J}/\mathrm{m}^{3}\times\mathrm{m}^{3} \times\mathrm{J}^{-1/2}\,\mathrm{m}^{1/2} =\mathrm{J}^{1/2}\,\mathrm{m}^{1/2}\), and dividing by \(\sqrt{\hbar c}=\mathrm{J}^{1/2}\,\mathrm{m}^{1/2}\) leaves a pure number. As for \(\tilde{\Phi}\): under \(\Phi\mapsto U\Phi\) with \(U\in\SU(2)\), \(\Phi^{*}\mapsto U^{*}\Phi^{*}\), and the identity \(\sigma^{2}U^{*}\sigma^{2}=U\), valid for every \(\SU(2)\) matrix because the doublet representation is pseudoreal — unitarily equivalent to its own complex conjugate — gives \(\ii\sigma^{2}\Phi^{*}\mapsto U\left(\ii\sigma^{2}\Phi^{*}\right)\). Complex conjugation flips the sign of \(Y\), so \(\tilde{\Phi}\) has \(Y=-1\), and \(\bar{Q}_{L}\tilde{\Phi}u_{R}\) carries \(Y=-\tfrac{1}{3}-1+\tfrac{4}{3}=0\) as required, while \(\bar{Q}_{L}\Phi d_{R}\) carries \(Y=-\tfrac{1}{3}+1-\tfrac{2}{3}=0\). Without \(\tilde{\Phi}\) the up-type quarks could not be given mass by this doublet at all.

Derives Theorem 104.38. With \(\Phi\to\Phi_{0}\) the term \(y^{d}_{ij}\bar{Q}_{Li}\Phi d_{Rj}\) becomes \(\left(y^{d}_{ij}v/\sqrt{2}\right)\bar{d}_{Li}d_{Rj}\), which compared with \(-m c^{2}\bar{\psi}_{L}\psi_{R}\) identifies \(m^{d}_{ij}c^{2}=\hat{y}^{d}_{ij}\sqrt{\hbar c}\,v/\sqrt{2} =\hat{y}^{d}_{ij}E_{v}/\sqrt{2}\); likewise for \(u\) and \(e\), the conjugate doublet supplying \(\Phi^{0*}\to v/\sqrt{2}\) for the up type.

For Equation (104.48): given any complex \(m\), the matrix \(mm^{\dagger}\) is Hermitian positive semi-definite, hence \(mm^{\dagger}=U\Sigma^{2}U^{\dagger}\) with \(U\) unitary and \(\Sigma\geq0\) diagonal, by the spectral theorem for normal operators. Define \(V:=m^{\dagger}U\Sigma^{-1}\) where \(\Sigma\) is invertible; then \(V^{\dagger}V=\Sigma^{-1}U^{\dagger}mm^{\dagger}U\Sigma^{-1} =\Sigma^{-1}\Sigma^{2}\Sigma^{-1}=\identity\), so \(V\) is unitary, and \(U^{\dagger}mV=U^{\dagger}mm^{\dagger}U\Sigma^{-1} =\Sigma\). (If \(\Sigma\) has zero entries, restrict to its support and extend \(V\) arbitrarily unitarily on the complement.) This is the singular-value decomposition, and the two unitaries are independent precisely because the left- and right-chiral fields are independent — which is possible only because the theory is chiral.

(i) The charged current is \(\bar{u}_{Li}\gamma^{\mu}d_{Li}\) in the gauge basis. Substituting \(u_{L}=U_{u}u_{L}'\) and \(d_{L}=U_{d}d_{L}'\) with primes denoting mass eigenstates gives \(\bar{u}'_{L}U_{u}^{\dagger}\gamma^{\mu}U_{d}d'_{L} =\bar{u}'_{L}\gamma^{\mu}\left(U_{u}^{\dagger}U_{d}\right)d'_{L}\), which is Equation (101.67) with \(V_{\mathrm{CKM}}=U_{u}^{\dagger}U_{d}\), manifestly unitary as the product of two unitaries — which is Theorem 101.62 and is what Linear Algebra and Representation Theory licenses.

(ii) A neutral current is \(\bar{f}_{Li}\gamma^{\mu}\left(\text{coefficient}\right)f_{Li}\) with a coefficient depending only on \(T^{3}\) and \(Q\) (Equation (101.75)), hence the same for every generation; it is therefore proportional to the identity matrix in generation space, and \(U_{f}^{\dagger}\identity U_{f}=\identity\). Nothing survives the rotation. This is the Glashow–Iliopoulos–Maiani mechanism [Glashow:1970] in its general form, proved as Theorem 101.59; note that the argument needs the coefficient to depend only on the gauge quantum numbers, which fails as soon as two fields with the same \(T^{3}\) and \(Q\) but different masses can mix — the reason a second scalar doublet, or an extra quark that is not part of a complete doublet, generically produces flavour-changing neutral currents and is excluded by the kaon and \(B\)-meson data of Experiment: CP Violation.

(iii) The Yukawa term with \(\Phi=\frac{1}{\sqrt{2}}(0,v+h)\transpose\) is the mass term multiplied by \((1+h/v)\), so the coupling matrix of \(h\) is \(m^{f}/v\) — the same matrix, diagonalized by the same \(U_{f}\), \(V_{f}\). Hence \(h\) has no flavour-changing couplings and its coupling to each fermion is \(m_{f}c^{2}/E_{v}\) per unit field, which is the statement measured in Phenomenon 104.62.

Derives Theorem 104.40. That \(\Phi^{\dagger}\Phi=\tfrac{1}{2}\tr(\mathcal{M}^{\dagger} \mathcal{M})\) is a computation: \(\tr(\mathcal{M}^{\dagger}\mathcal{M}) =2\left(\abs{\Phi^{0}}^{2}+\abs{\Phi^{+}}^{2}\right)\), since the two columns of \(\mathcal{M}\) have the same norm. A trace of the form \(\tr(\mathcal{M}^{\dagger}\mathcal{M})\) is invariant under Equation (104.50) for any unitary \(L\), \(R\), and the potential depends on \(\mathcal{M}\) only through it. The kinetic term is

the relative sign in the last term coming from the opposite hypercharge of \(\tilde{\Phi}\). The \(\SU(2)_{L}\) term is invariant under \(\SU(2)_{R}\), since \(R\) acts on the right; the \(\U(1)_{Y}\) term is not, because only \(T^{3}\) of \(\SU(2)_{R}\) is gauged. So \(\SU(2)_{R}\) is a symmetry of the whole Lagrangian when \(g'=0\) and of the scalar potential always.

At the vacuum, \(\Phi^{+}=0\) and \(\Phi^{0}=v/\sqrt{2}\), so \(\mathcal{M}_{0}=\left(v/\sqrt{2}\right)\identity\) and \(L\mathcal{M}_{0}R^{\dagger}=\mathcal{M}_{0}\) if and only if \(LR^{\dagger}=\identity\), that is \(L=R\): the unbroken global group is the diagonal \(\SU(2)_{V}\). Under it the gauge fields \(W^{a}_{\mu}\), which carry an adjoint index of \(\SU(2)_{L}\), transform as an \(\SU(2)_{V}\) triplet, and a mass matrix invariant under a group that acts irreducibly on a triplet must be proportional to the identity by Schur's lemma (Linear Algebra and Representation Theory). Hence \(M_{W^{1}}=M_{W^{2}}=M_{W^{3}}\) when \(g'=0\). Reinstating \(g'\) mixes \(W^{3}\) with \(B\) but changes no mass in the \(W\) sector, so \(M_{W}=M_{W^{3}}=M_{Z}\cos\theta_{W}\) by Equation (104.35), and Equation (104.45) gives \(\rho=1\).

Derives Theorem 104.41. Repeat the computation of Theorem 104.28 for a general multiplet. Writing \(T^{\pm}=T^{1}\pm\ii T^{2}\) and \(W^{\pm}_{\mu}=(W^{1}_{\mu}\mp\ii W^{2}_{\mu})/\sqrt{2}\), one has \(T^{1}W^{1}_{\mu}+T^{2}W^{2}_{\mu} =\tfrac{1}{\sqrt{2}}\left(T^{+}W^{+}_{\mu}+T^{-}W^{-}_{\mu}\right)\), so the part of \(\hbar^{-2}\abs{gW^{a}_{\mu}T^{a}\Phi_{0}}^{2}\) containing \(W^{+}_{\mu}W^{-\mu}\) is

using \(\acomm{T^{+}}{T^{-}}=2\left[(T^{1})^{2}+(T^{2})^{2}\right] =2\left[\vect{T}^{2}-(T^{3})^{2}\right]\) and the eigenvalue \(T(T+1)\) of the Casimir (Lie Groups, Lie Algebras, and Fibre Bundles). With \(\Phi_{0}^{\dagger}\Phi_{0}=v_{i}^{2}/2\) in the \(i\)-th multiplet, summing over multiplets and matching to the Proca form gives

For the neutral sector, the coefficient of the vacuum in the neutral covariant derivative is \(gT^{3}W^{3}_{\mu}+g'\left(Y/2\right)B_{\mu}\) with \(Y/2=Q-T^{3}\), and \(Q\Phi_{0}=0\) because the component acquiring the expectation value is neutral; hence the coefficient is \(\left(gW^{3}_{\mu}-g'B_{\mu}\right)T^{3}\), and

Dividing, with \(\cos^{2}\theta_{W}=g^{2}/\left(g^{2}+g'^{2}\right)\), every coupling cancels and Equation (104.51) remains. Substituting \(T=\tfrac{1}{2}\), \(T^{3}=-\tfrac{1}{2}\) gives numerator \(\left(\tfrac{3}{4}-\tfrac{1}{4}\right)v^{2}\) and denominator \(2\cdot\tfrac{1}{4}v^{2}\), equal; substituting \(T=1\), \(T^{3}=-1\) gives \(\left(2-1\right)v^{2}\) over \(2v^{2}\), that is \(\tfrac{1}{2}\); and \(T=1\), \(T^{3}=0\) gives \(2v^{2}\) over \(0\).

Derives Proposition 104.42. (i) is proved inside Theorem 104.40: the \(B_{\mu}\) term carries \(\mathcal{M}T^{3}\) on the right and is invariant only under the \(\U(1)\) subgroup of \(\SU(2)_{R}\) generated by \(T^{3}\). (ii) follows by writing Equation (104.46) in the \(\mathcal{M}\) notation: \(\bar{Q}_{L}\mathcal{M}\, \diag(y^{u},y^{d})\,Q_{R}\) with \(Q_{R}=(u_{R},d_{R})\transpose\) is invariant under \(\mathcal{M}\mapsto\mathcal{M}R^{\dagger}\), \(Q_{R}\mapsto RQ_{R}\) if and only if \(\diag(y^{u},y^{d})\) commutes with every \(R\in\SU(2)\), that is if and only if \(y^{u}=y^{d}\).

Derivation of the parametric form. Derives Proposition 104.43. \(\Delta\rho\) measures the difference between the \(W\) and \(Z\) self-energies at zero momentum, \(\Delta\rho=\Pi_{WW}(0)/M_{W}^{2}c^{2}-\Pi_{ZZ}(0)/M_{Z}^{2}c^{2}\), each computed from a quark loop. By Proposition 104.42 the difference vanishes when \(m_{t}=m_{b}\), so it is proportional to the isospin splitting. Each self-energy carries two Yukawa or gauge vertices and one loop factor \(1/16\pi^{2}\) (Equation (100.82) shows the same factor in QED), and the only dimensionful quantities available are the quark masses and \(E_{v}\); since \(\Delta\rho\) is dimensionless and the vertices are the Yukawa couplings \(\hat{y}\propto m_{f}c^{2}/E_{v}\), the leading term must be \(N_{c}\left(m_{t}c^{2}\right)^{2}/16\pi^{2}E_{v}^{2}\) times a pure number, with \(N_{c}=3\) for colour. Writing \(E_{v}^{2}=(\hbar c)^{3}/\sqrt{2}G_{F}\) turns this into Equation (104.52) up to the pure number, which the explicit loop integration of [Veltman:1977] fixes to exactly \(1\): Equation (104.52) is \(N_{c}\left(m_{t}c^{2}\right)^{2}/16\pi^{2}E_{v}^{2}\). Numerically, \(3\times1.1663787\times 10^{-5}\times(172.57)^{2} =1.0421\) and \(8\sqrt{2}\pi^{2}=111.66\), giving \(9.33\times 10^{-3}\). That the answer grows with \(m_{t}^{2}\) rather than \(\ln m_{t}\) is what made the top mass predictable from LEP data before the Tevatron produced it [Abachi:1995] [Abe:1995].

Derives Theorem 104.45. (i) Expanding \(\abs{D_{\mu}\phi}^{2}\) with \(D_{\mu}=\pp_{\mu}+\ii qA_{\mu}/\hbar\) and \(\phi=\frac{1}{\sqrt{2}}(v+h+\ii\chi)\),

whose last bracket contains, at quadratic order, exactly \((qv/\hbar)A^{\mu}\pp_{\mu}\chi\). A term linear in \(A\) and linear in \(\chi\) makes the quadratic form non-diagonal in field space.

(ii) Expand Equation (104.53):

The middle term integrates by parts to \(-(qv/\hbar)A^{\mu} \pp_{\mu}\chi\) up to a total derivative, cancelling (i). The last term is a mass term for \(\chi\); comparing with \(-\tfrac{1}{2}(m_{\chi}c/\hbar)^{2}\chi^{2}\) and using \(\mu_{0}q^{2}v^{2}/\hbar^{2}=(m_{A}c/\hbar)^{2}\) from Equation (104.22) gives \(m_{\chi}=\sqrt{\xi}\,m_{A}\). The would-be Goldstone boson thus has a gauge-dependent mass, which is the sharpest possible statement that it is not a particle.

(iii) With the mixing gone, the gauge kernel in momentum space (Fourier convention Equation (100.14)) is

which on the transverse projector \(P^{\mathrm{T}}_{\mu\nu}=\eta_{\mu\nu}-k_{\mu}k_{\nu}/k^{2}\) is \(-(k^{2}-m_{A}^{2}c^{2})/\mu_{0}\hbar^{2}\) and on the longitudinal projector \(P^{\mathrm{L}}\) is \(-(k^{2}-\xi m_{A}^{2}c^{2})/\mu_{0}\hbar^{2}\xi\). Inverting on each eigenspace and multiplying by \(\ii\hbar c\), exactly as in Proposition 100.8, gives

and combining the two fractions over the common denominator gives Equation (104.54), since \(-\left(k^{2}-\xi m_{A}^{2}c^{2}\right)+\xi\left(k^{2} -m_{A}^{2}c^{2}\right)=k^{2}\left(\xi-1\right)\). As \(\xi\to\infty\) the bracket tends to \(\eta_{\mu\nu}-k_{\mu}k_{\nu}/m_{A}^{2}c^{2}\), which is Equation (101.77): the unitary gauge is the infinite-\(\xi\) member of the family, and \(\chi\) becomes infinitely heavy and drops out, which is why it was absent there.

(iv) The Faddeev–Popov determinant is that of \(\delta G/\delta\Lambda\) for \(G\) the bracket in Equation (104.53). Under a gauge transformation of parameter \(\Lambda\), \(\delta A_{\mu}=\pp_{\mu}\Lambda\) and, from \(\phi\mapsto\ee^{-\ii q\Lambda/\hbar}\phi\) to first order, \(\delta\chi=-q\Lambda\left(v+h\right)/\hbar\). Hence

so the ghost has mass \(\sqrt{\xi}\,m_{A}\) and a coupling to \(h\). In the abelian unbroken theory this operator is field-independent and the ghosts may be discarded (Corollary 106.53); here they may not.

Derivation. Derives Proposition 104.47. The Slavnov–Taylor identity Equation (106.76), applied to the broken theory in an \(R_{\xi}\) gauge, says that the BRST variation of any correlator vanishes. Applied to a Green function with one gauge leg and the ghost of that leg, it gives the relation between the longitudinal part of the gauge amplitude and the Goldstone amplitude,

which is the broken-theory counterpart of the transversality \(k_{\mu}\mathcal{M}^{\mu}=0\) of Equation (100.18) — the right-hand side is zero only when \(M=0\). Now use Equation (104.6): the longitudinal polarization is \(\varepsilon_{L}^{\mu}=k^{\mu}/Mc+v^{\mu}\) with \(v^{\mu}\) of order \(Mc/\abs{\vect{k}}\) and bounded. Contracting,

and iterating over the \(n\) legs gives Equation (104.55). (The identity holds with \(\mathcal{M}_{\chi}\) computed in the same \(R_{\xi}\) gauge, and the corrections are uniform in the scattering angle away from the forward region.)

Derives Theorem 104.51. By Proposition 104.47 the high-energy amplitude equals that for the eaten Goldstone bosons, \(\chi^{+}\chi^{-}\to\chi^{+}\chi^{-}\), computed from the scalar sector alone: the gauge couplings drop out of the leading behaviour entirely. Write the doublet as

and expand \(-\lambda(\Phi^{\dagger}\Phi)^{2}\). Two vertices matter: the contact term \(-\lambda\abs{\chi^{+}}^{4}\) and the coupling \(-2\lambda v\,h\abs{\chi^{+}}^{2}\), in which \(2\lambda v=\left(m_{h}c/\hbar\right)^{2}/v\) by Equation (104.37). Summing the contact term with \(h\) exchange in the \(s\) and \(t\) channels, and using the Mandelstam conventions of Notation 100.1 in which \(s\) and \(t\) are squared momenta,

the factor \(\hbar^{2}\) being the dimension an amplitude carries in the normalization of Proposition 100.24, and \((m_{h}c^{2})^{2}/E_{v}^{2}=2\hat{\lambda}\) by Equation (104.37). The two propagator poles have absorbed the contact term: each channel contributes the vertex \(2\lambda v\) twice with a propagator between them, and the leftover constant is exactly the quartic. For \(s\gg m_{h}^{2}c^{2}\), and away from the forward region where \(\abs{t}\) is also large, both brackets tend to \(1\) and the amplitude tends to the constant \(-2\hbar^{2}(m_{h}c^{2})^{2}/E_{v}^{2}\). That it is constant rather than growing is the cancellation: had the \(h\) been absent, only the contact term would survive and the bracket would be \(-s/m_{h}^{2}c^{2}\) in disguise — growth without limit.

The \(J=0\) projection is \(a_{0}=\frac{1}{32\pi\hbar^{2}}\int_{-1}^{1}\mathcal{M}\, \dd\!\left(\cos\theta\right)\), the normalization being the one in which \(\sigma_{J}=16\pi\hbar^{2}(2J+1)\abs{a_{J}}^{2}/s\) reproduces Equation (101.89). For a constant amplitude the integral is \(2\mathcal{M}\), so \(a_{0}=\mathcal{M}/16\pi\hbar^{2}\), that is

manifestly dimensionless; and \(E_{v}^{2}=(\hbar c)^{3}/\sqrt{2}G_{F}\) turns it into Equation (104.56). Imposing \(\abs{a_{0}}\leq\tfrac{1}{2}\),

whose square root is \(873\,\mathrm{GeV}\). The coupled-channel analysis replaces the coefficient \(1\) by \(3/2\) in the largest eigenvalue, giving \(m_{h}c^{2}\leq\left(4\sqrt{2}\pi(\hbar c)^{3}/3G_{F}\right)^{1/2} =713\,\mathrm{GeV}\).

Derives Proposition 104.53. \(\left[\SU(2)\right]^{3}\) vanishes identically because the symmetrized trace of three \(\SU(2)\) generators is zero in every representation, \(\SU(2)\) having only real and pseudoreal representations; and \(\left[\SU(3)\right]^{3}\) vanishes because the quarks appear in triplet and antitriplet with respect to colour once the two chiralities are combined. What remains is the four displayed conditions.

For Equation (104.58): only \(\SU(2)_{L}\) doublets contribute, each with \(\tr(\acomm{T^{a}}{T^{b}})=\delta^{ab}\), so the condition is \(\sum_{\text{doublets}}Y=0\) with colour multiplicity; the quark doublet contributes \(N_{c}\times\tfrac{1}{3}\) and the lepton doublet \(-1\). This is where the number of colours enters: the condition fixes \(N_{c}=3\).

For Equation (104.59): only coloured fields contribute, with \(\tr(\acomm{T^{a}}{T^{b}})=\delta^{ab}\) for the fundamental, so the condition is \(\sum_{L}Y-\sum_{R}Y=0\) over the quark \(\SU(2)\) components: the left doublet supplies \(2\times\tfrac{1}{3}\) and the right singlets \(\tfrac{4}{3}\) and \(-\tfrac{2}{3}\).

For Equation (104.60): the trace is \(\sum Y^{3}\) with all multiplicities. Left: \(6\) quark states (three colours, two isospin) of \(Y=\tfrac{1}{3}\), giving \(6/27\), plus \(2\) lepton states of \(Y=-1\), giving \(-2\); total \(-\tfrac{16}{9}\). Right: \(3(\tfrac{4}{3})^{3} =\tfrac{64}{9}\), \(3(-\tfrac{2}{3})^{3}=-\tfrac{8}{9}\), and \((-2)^{3}=-8\); total \(\tfrac{56}{9}-8=-\tfrac{16}{9}\). Equal.

For Equation (104.61), the triangle with two insertions of the energy–momentum tensor requires \(\sum Y=0\) separately for each chirality summed with sign: left gives \(2-2=0\), right gives \(4-2-2=0\). This is a statement about the electroweak currents in a fixed classical background metric and not a claim about quantum gravity.

In every case the quarks alone fail and the leptons alone fail; only the complete generation cancels. Proposition 106.96 does the same arithmetic in the halved-hypercharge convention of Remark 104.11, with identical conclusions.

Derivation. Derives Phenomenon 104.55. Three steps: the rotation that produces the current, the couplings it gives each fermion, and the reason a scheme must be attached to any number quoted.

The rotation. The neutral part of Equation (104.11) is \(\hbar^{-1}\left(gT^{3}W^{3}_{\mu}+g'\tfrac{Y}{2}B_{\mu}\right)\). Substituting Equation (104.35) and demanding that the massless combination \(A_{\mu}\) couple to \(Q\) and to nothing else forces \(\hat{e}=\hat{g}\sin\theta_{W}=\hat{g}'\cos\theta_{W}\), which is Equation (104.40); the orthogonal combination then couples to \(T^{3}-\sin^{2}\theta_{W}Q\) with strength \(\hat{e}/\sin\theta_{W}\cos\theta_{W}\). This is Proposition 101.67, proved there in full.

The couplings. Projecting onto chiralities gives Equation (101.75): for each fermion

with \(T^{3}\) that of the left-chiral component. The whole predictive content follows: the neutrino has \(Q=0\), so \(g_{V}=g_{A}=\tfrac{1}{2}\) and its neutral-current coupling is independent of \(\theta_{W}\) — which is why the ratio of neutral- to charged-current neutrino cross-sections measures the angle cleanly. The electron has \(g_{A}=-\tfrac{1}{2}\) and \(g_{V}=-\tfrac{1}{2}+2\sin^{2}\theta_{W}\), which is small and changes sign near \(\sin^{2}\theta_{W}=\tfrac{1}{4}\): a parity-violating electron asymmetry is therefore a sensitive, and sign-carrying, probe. The \(Z\)-pole asymmetries measure the combination \(\mathcal{A}_{f}=2g_{V}g_{A}/(g_{V}^{2}+g_{A}^{2})\), which for the electron is a steep function of \(\sin^{2}\theta_{W}\) for exactly the same reason. Four independent kinds of measurement thus depend on one number in four different functional ways, and that is what makes their agreement a test rather than a fit.

The scheme. Beyond tree level, \(\theta_{W}\) is not a parameter of the Lagrangian but a renormalized quantity, and different definitions differ by finite amounts of order \(\alpha/\pi\) — the three standard ones are listed in Remark 104.34 and differ by about one per cent, which is hundreds of times the experimental errors. Worse, in the \(\overline{\mathrm{MS}}\) scheme the angle runs: the same definition gives about \(0.2313\) at \(\mu=M_{Z}c^{2}\) and about \(0.2387\) at \(\mu\to0\), the difference being real physics — vacuum polarization by charged particles, Section 100.7 — and not a change of convention. Table 104.3 lists the determinations across momentum transfer, and the content of Equation (104.62) is that one running curve passes through all of them. A number quoted without its scheme and its scale is not a measurement.

Derives Proposition 104.56. In the rest frame each decay product has momentum \(\tfrac{1}{2}Mc\), so \(p_{T}=\tfrac{1}{2}Mc\sin\theta^{*}\) with \(\theta^{*}\) the polar angle to the chosen axis. Then \(\dd p_{T}/\dd(\cos\theta^{*}) =-\tfrac{1}{2}Mc\,\cos\theta^{*}/\sin\theta^{*}\), and inverting and substituting \(\sin\theta^{*}=2p_{T}/Mc\), \(\cos\theta^{*}=\sqrt{1-(2p_{T}/Mc)^{2}}\) gives Equation (104.64). The Jacobian, not the dynamics, produces the peak; this is why the method is robust against the unknown production mechanism, and why smearing it by the boson's transverse motion and by the detector resolution is the dominant systematic. Applying the same construction to the pair, with the neutrino's transverse momentum inferred from the missing energy, gives the transverse mass Equation (101.84), whose endpoint is \(Mc^{2}\) itself and which is less sensitive to the boson's motion.

Derivation. Derives Phenomenon 104.57. The claim to be established is that the three amplitudes must cancel their leading high-energy growth, and that the cancellation fixes the triple-gauge coupling to the Yang–Mills value.

Take the final \(W\)s longitudinal, which is where the growth lives by Equation (104.6), and apply Proposition 104.47: at \(\sqrt{s}\,c\gg M_{W}c^{2}\) the amplitude equals that for producing the two eaten Goldstone bosons, \(e^{+}e^{-}\to\chi^{+}\chi^{-}\). Now enumerate the diagrams for that process.

• The \(s\)-channel photon and \(Z\) survive, with the couplings of a charged scalar read off Equation (104.11): the \(\chi^{\pm}\) sits in the doublet with \(T^{3}=\pm\tfrac{1}{2}\) and \(Q=\pm1\), so it couples to the photon with charge \(\hat{e}\) and to the \(Z\) with \(\hat{e}\left(1-2\sin^{2}\theta_{W}\right)/ 2\sin\theta_{W}\cos\theta_{W}\), by Equation (101.74) evaluated at those quantum numbers.

• The \(t\)-channel neutrino exchange survives only through the Yukawa coupling of \(\chi^{\pm}\) to \(e\nu\), which by Theorem 104.38 is \(\hat{y}_{e}/\sqrt{2}\), of order \(2\times 10^{-6}\) from Table 104.2, and is therefore negligible.

The high-energy amplitude is thus the production of a pair of charged scalars through two \(s\)-channel vector exchanges. Such an amplitude does not grow: an \(s\)-channel propagator \(1/(s-M^{2}c^{2})\) falls as \(1/s\) while the vertices supply at most one power of momentum each, so the amplitude tends to a constant in the \(J=1\) partial wave and the cross-section falls as \(1/s\), which is the observed shape.

The consequence is the cancellation. Computed instead in unitary gauge, each of the three diagrams carries two longitudinal polarization vectors and so grows as \(s/M_{W}^{2}c^{2}\); but their sum must equal the non-growing scalar amplitude just obtained. Hence the leading terms cancel identically. The cancellation uses the couplings in exactly the combinations Equation (104.40) supplies — the \(\gamma W^{+}W^{-}\) vertex equal to \(\hat{e}\) and the \(ZW^{+}W^{-}\) vertex equal to \(\hat{g}\cos\theta_{W}\), both of which are forced by Proposition 104.7 because the \(WW\gamma\) and \(WWZ\) couplings come from the same \(\varepsilon^{abc}\) in Equation (104.8). Change either coupling by a fraction \(\delta\) and a term growing as \(\delta s/M_{W}^{2}c^{2}\) survives; at \(\sqrt{s}\,c=200\,\mathrm{GeV}\) that factor is \(s/M_{W}^{2}c^{2}\approx6\), so a few-per-cent change in the vertex is visible in the cross-section. That is why the LEP2 measurement of a cross-section is a measurement of the group structure: the agreement in [Schael:2013] bounds the anomalous triple-gauge couplings at the per-cent level, and the neutrino-exchange diagram alone — with the two annihilation diagrams removed — overshoots the data at the highest energies by a large factor, exactly as an uncancelled \(s/M_{W}^{2}c^{2}\) requires.

Derivation of the decay pattern. Derives Phenomenon 104.59. Once \(m_{h}\) is given, the tree-level partial widths follow from the couplings of Theorem 104.38(iii) with no freedom left. For a fermion pair,

\(N_{c}\) being \(3\) for quarks and \(1\) for leptons, and \(\hat{y}_{f}=\sqrt{2}m_{f}c^{2}/E_{v}\) from Table 104.2. The structure is that of Equation (101.86) — the width of a particle at rest into two massless ones is a coupling squared times the mass over \(16\pi\), times a phase-space factor — and the exponent \(3/2\) rather than \(1/2\) is the signature of a scalar: the \(\beta^{3}\) arises because a \(J=0\) state decaying to a fermion pair through a scalar coupling produces them in a \(P\) wave, contributing \(\beta^{2}\) beyond the \(\beta\) of two-body phase space. A pseudoscalar coupling \(\ii\bar{f}\gamma^{5}f\) would give \(\beta^{1}\) instead, which is one of the ways the parity of the state is tested.

Two checks with no strong-interaction corrections. For \(\tau^{+}\tau^{-}\), Table 104.2 gives \(\hat{y}_{\tau}=0.01021\), so \(\hbar\Gamma=1.0416\times 10^{-4}\times125.20\,\mathrm{GeV}/16\pi =0.259\,\mathrm{MeV}\); for \(\mu^{+}\mu^{-}\), \(\hat{y}_{\mu}=6.07\times 10^{-4}\) gives \(0.92\,\mathrm{keV}\). Against the total width of \(4.1\,\mathrm{MeV}\) these are branching fractions of \(6.3\times 10^{-2}\) and \(2.2\times 10^{-4}\), which are the values in Table 104.4.

The two heaviest available channels need a further remark. At \(m_{h}c^{2}=125.20\,\mathrm{GeV}\) both \(W^{+}W^{-}\) and \(ZZ\) are below threshold, since \(2M_{W}c^{2}=160.7\,\mathrm{GeV}\); they proceed with one boson off shell, \(h\to VV^{*}\to Vf\bar{f}'\), a three-body decay whose rate is suppressed by an extra factor of order \(\hat{g}^{2}/16\pi^{2}\) relative to the on-shell formula and which is not obtained by putting a negative number under a square root. That this suppression is only partial — \(WW^{*}\) is still the second largest channel — is because the coupling grows as the squared vector mass, Equation (104.69) below.

The total width predicted by summing all channels is \(4.1\,\mathrm{MeV}\), some \(3\times10^{-5}\) of the mass; the measured value is \(3.7\,\mathrm{MeV}\) with an uncertainty of about \(1.7\,\mathrm{MeV}\) [Navas:2024], in agreement. A width this narrow is itself a prediction: it follows from the couplings being small, which follows from most fermions being light.

Derives Theorem 104.60. Work in the rest frame of the decaying particle, where the two photons travel along \(\pm\hat{\vect{k}}\) with polarization vectors \(\vect{\varepsilon}_{1}\), \(\vect{\varepsilon}_{2}\), both transverse: \(\vect{\varepsilon}_{i}\cdot\hat{\vect{k}}=0\). The amplitude is a scalar function linear in each \(\vect{\varepsilon}_{i}\), and for \(J=1\) it must be a vector \(\vect{A}\) contracted with the decaying particle's polarization. Bose symmetry requires \(\vect{A}\) to be invariant under exchanging the two photons, which means \(\vect{\varepsilon}_{1}\leftrightarrow\vect{\varepsilon}_{2}\) together with \(\hat{\vect{k}}\to-\hat{\vect{k}}\).

Enumerate the vectors that are linear in each polarization and built from \(\vect{\varepsilon}_{1}\), \(\vect{\varepsilon}_{2}\), \(\hat{\vect{k}}\):

the last vanishing by transversality. \(\vect{A}_{1}\) and \(\vect{A}_{2}\) are antisymmetric under the exchange and are therefore forbidden for two identical bosons, and no other structure exists. The one remaining candidate, \(\left[\hat{\vect{k}}\cdot\left(\vect{\varepsilon}_{1}\times \vect{\varepsilon}_{2}\right)\right]\hat{\vect{k}}\), is not independent: transversality makes \(\vect{\varepsilon}_{1}\times\vect{\varepsilon}_{2}\) parallel to \(\hat{\vect{k}}\), so it is \(\vect{A}_{1}\), and it is antisymmetric for the same reason. Anything else contains either \(\hat{\vect{k}}\cdot\vect{\varepsilon}_{i}\), which vanishes, or a second power of some \(\vect{\varepsilon}_{i}\), which the linearity forbids. Hence the \(J=1\) amplitude vanishes identically.

The argument uses only rotational invariance, Bose statistics and transversality; it does not use parity, so it excludes \(1^{+}\) and \(1^{-}\) alike, and it fails for a massive vector in the final state (which has a longitudinal polarization) and for two virtual photons. Together with the four-lepton angular distributions, which distinguish \(0^{+}\) from \(0^{-}\) and from \(2^{+}\), this fixes \(J^{PC}=0^{++}\).

Derives Proposition 104.61. For the fermions, Theorem 104.38(iii): substituting \(\Phi=\frac{1}{\sqrt{2}}(0,v+h)\transpose\) into Equation (104.46) multiplies the mass term by \(\left(1+h/v\right)\), whose linear part is the first term of Equation (104.69). Dividing by \(\sqrt{\hbar c}\) to make it dimensionless, and using \(E_{v}=v\sqrt{\hbar c}\), gives \(g_{hff}=m_{f}c^{2}/E_{v}\).

For the vectors, Equation (104.34) was computed at \(\Phi=\Phi_{0}\); restoring the fluctuation replaces \(v\) by \(v+h\) throughout, so the mass terms are multiplied by \(\left(1+h/v\right)^{2}=1+2h/v+h^{2}/v^{2}\). The linear part is twice the mass term over \(v\), which is the bracket in Equation (104.69); the factor of two is the whole difference between the fermion and vector cases and it comes from the mass term being quadratic in the field's coupling to \(\Phi\) rather than linear. Hence \(g_{hVV}=2(M_{V}c^{2})^{2}/E_{v}\), quadratic in the mass and carrying the dimension of an energy; dividing by \(2E_{v}\) and taking the square root returns \(M_{V}c^{2}/E_{v}\), the same linear law the fermions obey, which is why the two families can be drawn on one plot.

Derivation. Derives Phenomenon 104.62. The prediction is Equation (104.70), derived in Proposition 104.61: a one-parameter family with the parameter, \(E_{v}\), already fixed by the muon lifetime through Proposition 104.26. There is nothing to adjust.

The measurement is conventionally reported as coupling modifiers \(\kappa_{i}\), defined so that \(\kappa_{i}=1\) reproduces Equation (104.70) and rates scale as \(\kappa^{2}\). What is actually measured is a rate, that is a product of a production cross-section and a branching fraction, so a global fit over many final states is required to separate the couplings; the experiments report the results of that fit [Aad:2022] [Tumasyan:2022], and every \(\kappa_{i}\) comes out consistent with unity, with uncertainties of roughly a tenth for \(W\), \(Z\), \(t\), \(b\) and \(\tau\) and roughly a fifth for \(\mu\). Since \(m_{\mu}c^{2}=0.1057\,\mathrm{GeV}\) and \(m_{t}c^{2}=172.57\,\mathrm{GeV}\), the verified range is a factor of \(1630\) in mass, over which Equation (104.70) varies by the same factor — so the agreement is a statement about a three-decade lever arm and not about one point.

Two features make the test sharper than a fit to five numbers. The \(\gamma\gamma\) and \(gg\) vertices have no tree-level counterpart (Table 104.4), so their measured rates test the same couplings at one loop, through a different combination: the \(\gamma\gamma\) amplitude is dominated by the \(W\) loop with the top loop interfering destructively, and the \(gg\) amplitude by the top alone, so agreement in both simultaneously constrains the relative sign of \(\kappa_{W}\) and \(\kappa_{t}\) — something no tree-level measurement reaches. And any heavy coloured or charged particle would contribute to those same loops whether or not it can be produced directly, so the agreement bounds such particles indirectly.

Derives Proposition 104.63. From Equation (104.16) with the Goldstone modes gauged away, \(V=\lambda v h^{3}+\tfrac{\lambda}{4}h^{4}+\lambda v^{2}h^{2}\), and \(\lambda v^{2}=\tfrac{1}{2}(m_{h}c/\hbar)^{2}\) by Equation (104.37). Substituting \(\lambda=(m_{h}c)^{2}/2\hbar^{2}v^{2}\) into each term gives Equation (104.71). The two higher couplings are \(m_{h}^{2}/v\) and \(m_{h}^{2}/v^{2}\) times pure numbers: one parameter, \(\lambda\), generates the mass and both self-interactions, which is the content of the claim that the potential is the quartic one and not merely some function with a minimum in the right place.

Derivation of the sign. Derives Proposition 104.66. The coefficients are quoted from [Degrassi:2012]; what is derived here is why the dominant one is negative and why it is large. By Remark 104.64 the one-loop effective potential receives \(-N_{c}M_{t}^{4}(h)\ln M_{t}^{2}(h)\) from the top, with the minus sign of a fermion loop — the same sign that makes a closed fermion loop carry \((-1)\) in Corollary 106.39. Since \(M_{t}(h)\propto\hat{y}_{t}h\), this is a term \(\propto-\hat{y}_{t}^{4}h^{4}\ln h\), that is a negative contribution to the running quartic, with the colour factor \(N_{c}=3\) and the fourth power of the coupling. Numerically \(\hat{y}_{t}^{4}=0.78\) while \(\hat{\lambda}^{2}=0.016\), so the fermion term beats the bosonic ones by a factor of order fifty, and Equation (104.73) follows by arithmetic. The one number that matters is the sign.