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Take a smooth gauge transformation $g:{\mathbb{R}}^3\to SU(2)$. In the anti-Hermitian convention of the adjoint covariant derivative, setExpanding ${\partial }_j(g\Phi g^{-1})$ shows that its two differentiated-$g$ terms cancel with the extra terms in $[A_j^g,{\Phi }^g]$. Thus $D_j^g{\Phi }^g=g(D_j\Phi )g^{-1}$. The gauge field strength is $F_{jk}={\partial }_j{A}_k-{\partial }_k{A}_j+[A_j,A_k]$, and the same cancellation, or the commutator of covariant derivatives, gives $F_{jk}^g=g{F}_{jk}{g}^{-1}$.

The matrix trace is unchanged by conjugation, soEach field-strength norm, covariant-gradient norm and Higgs norm is therefore invariant pointwise. It follows that $V_{\lambda }(A^g,{\Phi }^g)=V_{\lambda }(A,\Phi )$. The displayed tensor made from these fields is also gauge invariant wherever $\Phi \ne 0$.

There is a normalization problem in the requested final identity. With the ordinary defining $2\times 2$ trace and the printed inner product, an explicit orthonormal basis isThe last identity follows from the Pauli matrix multiplication law. In particular it is not permissible to combine this orthonormality with unit structure constants of a Lie algebra.

Write $A_j=a_j^a{e}_a$ and $\Phi =\phi e_3$ with $\phi \ne 0$, and put $s=\mathrm{sgn}\phi$. Define $C_{jk}=a_j^1{a}_k^2-a_j^2{a}_k^1$. ThenOnly the bracket of the transverse terms contributes in the $e_3$ direction, givingHence the literal tensor printed in the question isThe transverse term does not cancel with this trace normalization of su(2).

A concrete counterexample to the printed closure assertion is $\Phi =e_3$, $A_1=e_1$, $A_2=x_3{e}_2$, $A_3=0$ on a ball. Directly, $F_{12}=\sqrt{2}{x}_3{e}_3$, $F_{23}=-e_2$ and $F_{31}=0$, while $D_1\Phi =-\sqrt{2}{e}_2$, $D_2\Phi =\sqrt{2}{x}_3{e}_1$ and $D_3\Phi =0$. It follows thatThe Higgs field never vanishes. If global finite energy is desired, multiply both nonzero gauge-potential components by a smooth compactly supported cutoff equal to one on this ball and keep $\Phi =e_3$ everywhere. The energy is finite and the same local counterexample remains. Thus the claimed closure is false with the ordinary trace normalization actually printed; it is not a consequence of gauge invariance alone.

The consistent 't Hooft electromagnetic tensor for the printed inner product is insteadIn the directional gauge, its commutator coefficient is $1/{\kappa }^2$, so the terms cancel exactly andOn a connected region where $\phi \ne 0$, $s$ is constant. Commutation of ordinary partial derivatives provesAt any nonzero-Higgs point, use the permitted local gauge transformation to align the field with a fixed internal direction. Since $\mathcal{F}$ is gauge invariant, the identity derived in that gauge holds in the original gauge on the same ball. These balls cover every nonzero-Higgs region, proving that the corrected tensor is a closed differential form there.

Alternatively, keeping the tensor's printed coefficient one but replacing the inner product by $-2\mathrm{tr}(XY)$ gives the orthonormal basis $e_a=-i{\sigma }_a/2$ with $\kappa =1$. The same calculation then gives $f_{jk}=s({\partial }_j{a}_k^3-{\partial }_k{a}_j^3)$ and the intended identity. Either repair makes the bracket and inner-product normalizations agree; they must not be silently mixed.

Count one neutrino species as the usual two thermally populated particle-antiparticle helicity states, with negligible chemical potential. Its entropy weight is $g_{\ast s,X}=(7/8)2=7/4$. In the ideal high-temperature Standard Model, the bosonic count is $16+8+4=28$ from gluons, electroweak gauge bosons and the Higgs field. The fermionic count is $72+12+6=90$ from quarks, charged leptons and the three ordinary neutrino species. Hence the relativistic entropy content of the Standard Model is $g_{\ast s,D}=28+(7/8)90=106.75$ for the remaining bath. Before the new species leaves, the total count is $108.5$; its separately cooling states are excluded from the subsequent bath count.

Near ordinary neutrino decoupling, the still-coupled bath hasfrom photons, electron-positron pairs and the three ordinary neutrinos. After those neutrinos decouple, the electromagnetic bath falls from $2+(7/8)4=11/2$ to two when the pairs annihilate. Cosmological entropy conservation in that bath gives $T_{\nu }/T_{\gamma }=(4/11{)}^{1/3}$.

The new species has cooled as $T_X\propto a^{-1}$ since its earlier decoupling, whereas annihilation of Standard Model species heated the remaining bath. The temperature ratio of an early-decoupled neutrino is thusThese are ideal-gas estimates with the high-temperature count implicit in the question; precise electroweak thresholds change them slightly.

Entropy degrees of freedom weight temperature ratios with their cube, while energy degrees of freedom use their fourth power. The remaining Standard Model entropy today, in photon-temperature units, is $g_{\ast s,0}^{\mathrm{SM}}=2+(7/8)6(4/11)=43/11$. The new contribution is $(7/4)(T_X/T_{\gamma }{)}^3$. Therefore the fractional entropy of a relativistically decoupled species isAbout 1.6% of the total entropy is in the new species. Separate comoving entropy conservation also derives this fraction directly at decoupling. A model populating four helicity states instead would require the corresponding doubled species count.

Take $T_i={\tau }_i/2$ and the hypercharge generator $Y=I_2/2$ on the scalar. The two commuting factors of the electroweak interaction have independent invariant gauge kinetic terms and therefore independent coupling constants. Non-Abelian gauge invariance requires the same $g$ for all three $SU(2)$ generators, while the $U(1)$ factor has its own $g^{\prime }$; no symmetry of this product group requires $g=g^{\prime }$.

A nonzero Higgs doublet can be rotated by $SU(2)$ toThe fundamental representation acts transitively on normalized complex doublets, so this fixes an internal direction without changing the vacuum norm. A general infinitesimal generator annihilates this vacuum only ifFor real parameters this means ${\alpha }_1={\alpha }_2=0$ and $\beta ={\alpha }_3$. Thus the connected stabilizer subgroup is one-dimensional:This proves the claimed electroweak symmetry breaking and identifies the unbroken electric charge generator.

The gauge covariant derivative isWriting the connection as ${\mathcal{A}}_{\mu }=g{A}_{\mu }^i{T}_i+g^{\prime }{B}_{\mu }Y$, its local transformation is ${\mathcal{A}}_{\mu }^{\prime }=U{\mathcal{A}}_{\mu }{U}^{-1}-i({\partial }_{\mu }U)U^{-1}$. Direct substitution gives $D_{\mu }^{\prime }(U\varphi )=U{D}_{\mu }\varphi$. Unitarity then provesso the scalar kinetic term is gauge-invariant.

At the constant vacuum the gauge covariant derivative becomesConsequentlywhere $\mathcal{V}=(A^1,A^2,A^3,B{)}^T$ andThe factor $1/2$ is the canonical real-vector mass-term normalization; the displayed electroweak doublet gauge-boson mass matrix contains squared masses when the gauge kinetic terms are canonical.

The electric charge generator acts on gauge fields through the Adjoint representation, not through the scalar's two-by-two matrix. With $U=e^{i\alpha Q}$, the brackets $[T_3,T_1]=i{T}_2$ and $[T_3,T_2]=-i{T}_1$ give $\delta A^1=\alpha A^2$, $\delta A^2=-\alpha A^1$, and neutral $A^3,B$. Hence the electromagnetic charge matrix of electroweak gauge fields in the requested basis isChoose positive gauge couplings by the gauge-field sign convention. Set $s_W=g^{\prime }/\sqrt{g^2+g^{\prime 2}}$ and $c_W=g/\sqrt{g^2+g^{\prime 2}}$. The Weinberg angle rotation diagonalizes the neutral block, while $W^{\pm }=(A^1\mp i{A}^2)/\sqrt{2}$ obey $\delta W^{\pm }=\pm i\alpha W^{\pm }$. The physical electroweak gauge bosons areThe charges are in units of $e=g{s}_W=g^{\prime }{c}_W$. Thus the tree-level electroweak gauge-boson masses satisfy $m_W=m_Z{c}_W$, and the unbroken photon is massless.

The electron and neutrino form the left-chiral lepton doublet, while the right-chiral electron is a weak singlet. In four-component notation use chiral projectors $P_{L,R}=(1\mp {\gamma }_5)/2$ and packageHere each entry can be obtained by projecting a Dirac field; only the indicated chirality propagates. The hypercharges follow from $Q=T_3+Y$: $(0,-1)=(1/2,-1/2)+Y_{\Psi }$ gives $Y_{\Psi }=-1/2$, and $T_3=0,Q=-1$ gives $Y_{\psi }=-1$. The minimal theory has no right-handed neutrino.

The gauge-invariant lepton kinetic Lagrangian isIf unprojected Dirac spinors are used as notation instead, insert $P_L$ in the first kinetic term and $P_R$ in the second. The inactive components are not additional physical degrees of freedom. Each term is invariant because the gauge covariant derivative transforms in the same gauge group representation as its field.

No bare Dirac mass term is possible: $\bar{\Psi }\Psi$ and $\bar{\psi }\psi$ vanish by opposite adjoint projectors, while a cross term $\bar{\Psi }\psi$ has an uncontracted weak-doublet index and hypercharge $-1/2$. A neutrino Dirac mass would additionally require the absent right-handed field. A Majorana mass term pairs a chiral field with itself; ${\Psi }^T{C}^{-1}\Psi$ has hypercharge $-1$ and ${\psi }^T{C}^{-1}\psi$ has $-2$, so neither is invariant. These statements concern quadratic masses before electroweak symmetry breaking, not masses involving scalar insertions.

For the Higgs conjugate doublet, put $\epsilon =i{\tau }_2$. The given identity $\epsilon {\tau }_i^{\ast }=-{\tau }_i\epsilon$ exponentiates to $\epsilon U^{\ast }=U\epsilon$ for $U\in SU(2)$. ThereforeIn particular its weak isospin remains $T=1/2$ while its hypercharge reverses. Since $\Psi \mapsto e^{-i\beta /2}U\Psi$, both ${\varphi }^{\prime †}\Psi$ and $\bar{\Psi }{\varphi }^{\prime }$ transform in the trivial representation of the full gauge group. They remain a Lorentz spinor and its adjoint; gauge singlet does not mean Lorentz scalar.

Finally the Yukawa interactionis gauge invariant: the first term has total hypercharge $+1/2+1/2-1=0$, and its doublet indices are contracted. At the Higgs field vacuum it becomes $-(y_{e}v/\sqrt{2}){\bar{e}}_L{e}_R+h.c.$. Rephase $e_R$ to make the coefficient real and positive and assemble $e=e_L+e_R$. ThenThis explicitly generates the gauge-invariant electron Yukawa mass, without a bare mass in the unbroken theory.

Use the metric $g^{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2g^{\mu \nu }$. Transposing a product reverses its order but does not conjugate the factor $i$. Thus the charge-conjugation matrix identity giveswhere reversing the four distinct gamma matrices requires six exchanges. Similarly,

To fix the transpose sign, work in the Weyl representation of the gamma matrices:The Pauli identity checks $C_0{\gamma }^{\mu T}{C}_0^{-1}=-{\gamma }^{\mu }$ directly, and ${\epsilon }^T=-\epsilon$ gives $C_0^T=-C_0$, with $C_0^{†}{C}_0=I$. Any other intertwiner differs from $C_0$ by a scalar: its ratio with $C_0$ commutes with all gamma matrices, whose products span the full four-by-four matrix algebra. A matrix commuting with all matrix units is scalar. Under a spinor basis change $S$, the intertwiner becomes $S{C}_0{S}^T$, whose transpose is still its negative. Therefore in every such four-component representationThis proves the antisymmetry of the four-dimensional charge-conjugation matrix, rather than leaving the sign undetermined by the general transpose argument.

Hermitian conjugating the charged Dirac equation, using real $A_{\mu },e,m$ and ${\gamma }^0{\gamma }^{\mu †}{\gamma }^0={\gamma }^{\mu }$, gives the adjoint Dirac equationTranspose it and multiply by $C$. Since $C{\gamma }^{\mu T}=-{\gamma }^{\mu }C$, the field ${\psi }^c=C{\bar{\psi }}^T$ obtained by charge conjugation obeysThus algebraic charge conjugation reverses the electromagnetic coupling.

If ${\gamma }_5\psi =\eta \psi$, $\eta =\pm 1$, Hermiticity of ${\gamma }_5$ and its anticommutation with ${\gamma }^0$ implyUsing the first transpose identity,Hence $\boxed{\bar{\psi }{\gamma }_5=-\eta \bar{\psi },{\gamma }_5{\psi }^c=-\eta {\psi }^c}$: algebraic charge conjugation reverses chirality. This concerns conjugating the already projected spinor, not the distinct operation of keeping a chiral projector outside an operator conjugation.

For the Grassmann variation of a chiral Majorana mass term, variations of odd fields must be moved past odd fields with a minus sign. If $A^T=-A$, componentwise reordering givesBoth $C$ and $C^{-1}$ are antisymmetric. Treating $\psi ,\bar{\psi }$ independently, integrating the kinetic variation by parts, and placing varied Grassmann fields on the left givesThe factors two from variation cancel the mass-term factors $1/2$. The first coefficient gives one equation; multiplying the second by $C$ gives the other:The variations can be restricted to their chiral subspaces: the two terms in each equation have the same opposite output chirality, so no inactive component equation has been imposed.

Applying $i/\partial$ to the first equation and substituting the second, using $(i/\partial {)}^2=-\square$, givesA phase rotation removes the phase of $m$, and combining the chiral field with its conjugate then gives a Majorana fermion of this mass, not two independent chiral species.

Let $\chi ={\varphi }^{\prime †}\Psi$ and $\bar{\chi }=\bar{\Psi }{\varphi }^{\prime }$. Question 1 shows they are gauge singlets, so both terms in ${\mathcal{L}}_G=G{\chi }^T{C}^{-1}\chi -G^{\ast }\bar{\chi }C{\bar{\chi }}^T$ preserve electroweak gauge invariance. The original PDF has $C^{-1}$ in the first term; the TeX aid incorrectly replaces it by $G^{-1}$. At the Higgs field vacuum,soComparing with the just-derived chiral mass equations yields the Majorana neutrino mass from the Weinberg operator:For nonzero $G$ this generates a mass without introducing a right-handed neutrino. Under lepton number rotation $\Psi \mapsto e^{i\alpha }\Psi$, the first operator acquires $e^{2i\alpha }$ and its conjugate acquires $e^{-2i\alpha }$. Thus the interaction violates lepton number by $\boxed{\Delta L=2}$. Each composite contains one scalar of mass dimension one and one fermion of dimension $3/2$, so the interaction is dimension five and $G$ has dimension $-1$. This is the Weinberg operator, an effective extension of the minimal renormalizable Standard Model, consistent with the absence of bare quadratic masses in Question 1.

Put $h=|\Phi {|}^2$, $D_i={\partial }_i-i{A}_i$, $j_i=\mathrm{Im}(\bar{\Phi }{D}_i\Phi )$, and $W(h)=h(1-h{)}^2/4$. Because the inverse-density Abelian Higgs model contains $B^2/h$, its finite-energy admissible class matters at zeros of the Higgs field. First derive the formal equations on the open set where $h>0$; on the first-order solutions below the potentially singular quotients extend across the zeros.

For variations $\psi =\delta \Phi$ and $a_i=\delta A_i$,Covariant integration by parts in the scalar term, and ordinary integration by parts in the magnetic term, givewhere ${ϵ}_{12}=1$. HereThus the Euler-Lagrange field equations areEquivalently the gauge equation is ${\partial }_j(F_{ji}/h)+j_i=0$. The current sign follows from the explicit choice $D=\partial -iA$.

For the Bogomolny bound for the inverse-density Abelian Higgs model, computeusing $[D_1,D_2]\Phi =-iB\Phi$. Expanding the first square givessoAlso,The current boundary integral vanishes by the assumed rapid decay. Combining the identities therefore givesFor fixed positive flux the minimum is $\boxed{V_{\min }=\pi N}$, attained when the Bogomolny equations areThe quotient at a zero is understood through the finite-energy limit, not by dividing a literal nonzero numerator by zero.

These equations imply the second-order equations directly. Applying $D_1-i{D}_2$ to the first gives $(D_i{D}_i+B)\Phi =0$. On the second equation,which is exactly the scalar equation. The first equation also implies $j_1=-\frac{1}{2}{\partial }_{2}h$ and $j_2=\frac{1}{2}{\partial }_{1}h$. Since $B/h=(1-h)/2$, the gauge equations follow too. All these coefficients are regular across zeros for smooth first-order fields.

For the scalar reduction, write $\Phi =e^{u/2+i\chi }$ away from its zeros. Separating the real and imaginary parts of the first Bogomolny equation givesThus, with the oriented planar Hodge star $\ast d{x}^1=d{x}^2$, $\ast d{x}^2=-d{x}^1$,away from the zeros. A zero $p_j$ of multiplicity $n_j$ has $u=2n_j\log |x-p_j|+\text{smooth}$ and phase winding $2\pi n_j$. The multiplicities are positive: locally the first-order equation is a covariant Cauchy–Riemann equation, and a nonvanishing integrating factor turns its solutions into holomorphic functions. Distributionally,Consequently the logarithmic equation for inverse-density Abelian Higgs vortices isThe delta terms are essential for multi-vortex reconstruction; omitting them would lose the specified zeros and flux.

Here is an explicit regular reconstruction, which also removes all artificial singularities. Given distinct locations $p_j$ and integer multiplicities, putSince $\Delta u_0=4\pi {\sum }_j{n}_j{\delta }_{p_j}-g$, solving the smooth nonlinear elliptic problem for $v=u-u_0$ amounts toLet $z=x^1+i{x}^2$ and let $z_j$ represent $p_j$. A smooth solution with the topological decay reconstructsThese formulas are smooth at every prescribed zero and agree with the phase formula off the zeros. Taking a curl and using the equation for $v$ verifies $B=e^u(1-e^u)/2$; substitution verifies the other first-order equation. At infinity the phase winds $N$ times, producing flux $2\pi N$. Thus multi-vortex construction reduces to solving one scalar elliptic boundary-value problem for specified positions and multiplicities, followed by these explicit formulas.

For the topological branch, $u\le 0$: a positive interior maximum would have $\Delta u\le 0$, whereas the equation there gives $\Delta u=e^u(e^u-1)>0$. At infinity, the linearized scalar equation is $\Delta u-u=0$. Comparison barriers and elliptic estimates give the decaying tail for regular topological solutions, and hence the required decay of $B,D\Phi$ and $1-h$. The nonlinear elliptic problem determines $v$; the reconstruction formulas do not constitute an arbitrary superposition of single-vortex profiles.

The coupling is allowed, with an on-shell threshold condition. The Higgs field kinetic term producesand hence a tree-level Feynman vertex. A decay into two real W bosons requires $m_h>2m_W$ for nonzero phase space. Below that threshold the vertex still mediates $h\to W{W}^{\ast }$ or decays through two virtual bosons into fermions. Thus an allowed Lagrangian coupling does not by itself guarantee an on-shell two-body decay.

Both the left-handed Electron and the electron neutrino participate in weak charged currents through $W^{\pm }$, and both have weak neutral currents mediated by the Z boson. Their electromagnetic interactions differ: the Electron has electric charge $-e$ and couples to the photon, while the neutrino is electrically neutral and has no tree-level electromagnetic coupling.

In the minimal massless-neutrino Standard Model, the active neutrino has only a left-handed particle state and a right-handed antiparticle state. The Electron has both chiralities: its right-handed component has electromagnetic interactions and a weak neutral current, but no weak charged current. Its nonzero mass comes from a Yukawa coupling to the Higgs field; the minimal massless-neutrino model contains no right-handed neutrino for an analogous Dirac mass term.

Observed neutrino oscillations establish nonzero neutrino mass-squared differences. Atmospheric muon-neutrino disappearance depends on travel distance and energy, and solar electron neutrinos convert into other active flavors; reactor and accelerator oscillations provide complementary evidence. In propagation, the relative phase of two mass eigenstates is $\Delta m_{ij}^{2}L/(2E)$. If all masses were zero there would be no such relative phase. Thus at least two neutrino mass eigenstates are nonzero in the three-neutrino description with two independent measured splittings. A flavor state is a superposition of mass eigenstates, so the evidence does not assign a single definite mass to a flavor label.

Use $t_{\mu }=(1,-1,-1,-1{)}_{\mu }$ without summing in individual component transformations. The Dirac current transforms asSince the coupling is real, invariance of its contraction fixes the time reversal of an electromagnetic gauge field:These equations use a compatible gauge; an additional pure gauge transformation would not change the gauge field strength.

Coordinate differentiation at $x_T$ introduces $r_{\mu }=-t_{\mu }$, because the time coordinate is reversed and the spatial coordinates are unchanged. HenceThus the electric field is even and the magnetic field is odd. The chirality matrix is even by part (a). The two tensor signs cancel in the dipole contraction, but antiunitarity conjugates its explicit $i$. Therefore the time-reversal parity of a fermion electric dipole operator isIt is a time-reversal symmetry violating interaction, and, under the usual local relativistic quantum field theory hypotheses of the CPT theorem, it violates CP symmetry. A nonzero coefficient is absent from the renormalizable tree-level Standard Model Lagrangian: the broken-phase fermion electric dipole moment operator has mass dimension five, and its electroweak-invariant completion requires a Higgs field and has dimension six. However, it can arise as a radiatively induced effective interaction in the Standard Model, whose CKM matrix contains a CP-violating phase. It is not forbidden to all orders. An explicit primary calculation of quark dipoles is Czarnecki and Krause's Standard Model calculation.

Take ${\tau }^a={\sigma }^a/2$, $Y_{\varphi }=1/2$, metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, and a Higgs potential with ${\mu }_h^2>0$, $\lambda >0$. The gauge-scalar electroweak interaction isHere $B_{\mu \nu }={\partial }_{\mu }{B}_{\nu }-{\partial }_{\nu }{B}_{\mu }$. With the printed plus-sign convention for $D_{\mu }$, define $[D_{\mu },D_{\nu }]=ig{W}_{\mu \nu }^a{\tau }^a+i{g}^{\prime }{Y}_{\varphi }{B}_{\mu \nu }$; thus $W_{\mu \nu }^a={\partial }_{\mu }{W}_{\nu }^a-{\partial }_{\nu }{W}_{\mu }^a-g{ϵ}^{abc}{W}_{\mu }^b{W}_{\nu }^c$. The minus sign in the nonlinear gauge field strength follows from this convention.

The minima have ${\varphi }^{†}\varphi =v^2/2$, with $v^2={\mu }_h^2/\lambda$. A gauge transformation rotates the vacuum to $(0,v{)}^T/\sqrt{2}$. In unitary gauge, the three angular Goldstone bosons are removed, leaving $\varphi =(0,v+H{)}^T/\sqrt{2}$ and one real Higgs boson. The generator $Q=T^3+Y$ annihilates the vacuum, because its lower component has $T^3=-1/2$ and hypercharge $1/2$. This identifies the unbroken electromagnetic gauge group.

The quadratic terms from the Higgs field kinetic term areDefine the Weinberg angle and physical fields byThe neutral combinations areThe neutral gauge-boson mass matrix is $(v^2/4)\left(\begin{array}{cc}{g}^2& -g{g}^{\prime }\\ -g{g}^{\prime }& g^{\prime 2}\end{array}\right)$. It has eigenvalues zero and $(g^2+g^{\prime 2})v^2/4$, with the zero eigenvector giving $A_{\mu }$. Expanding the Higgs potential about its minimum gives $V=\text{constant}+\lambda v^2{H}^2+\lambda v{H}^3+\lambda H^4/4$. Therefore the tree-level electroweak gauge-boson masses and scalar mass areAlso $m_W=m_Z\cos {\theta }_W$ and $e=g\sin {\theta }_W=g^{\prime }\cos {\theta }_W$. The three removed Goldstone bosons supply the longitudinal polarization vector degrees of freedom of the massive electroweak gauge bosons.