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For one Grassmann variable, define the Berezin integral by $\int d\theta 1=0$ and $\int d\theta \theta =1$, extending linearly. Since ${\theta }^2=0$, these rules determine it on every polynomial. With several Grassmann variables, integrate from the rightmost differential first. Equivalently, $\int d{\theta }_n\cdots d{\theta }_1$ extracts the coefficient of ${\theta }_1\cdots {\theta }_n$ in that order. Repeated indices give zero, while permuting distinct generators gives the fermionic sign of that permutation. Hencewhere $\epsilon$ is the Levi-Civita symbol.

For the doubled variables, fix the paired measure orientation byThe exponential in the Grassmann Gaussian integral has a finite Taylor expansion. Only its term of degree $2n$ survives integration. Each barred and unbarred index must occur exactly once; the $n!$ orders of the even bilinears cancel the Taylor factorial. Reordering the unbarred indices gives the determinant permutation sign. ThereforeThis identity holds even for a singular matrix. The normalized expectations below require $B$ invertible.

Put $C=B^{-1}$. Differentiating with respect to $B_{ji}$ inserts $-{\bar{\theta }}_j{\theta }_i={\theta }_i{\bar{\theta }}_j$. The derivative of a determinant then gives the ordered Grassmann Gaussian contractionsDifferentiate once more, using $\partial C_{ij}/\partial B_{lk}=-C_{il}{C}_{kj}$. The result isThe minus sign is the interchange sign in the fermionic Wick theorem. In particular $⟨\bar{\theta }M\theta ⟩=-\mathrm{tr}(MC)$. Reordering each barred-unbarred pair in the product and contracting the indices givesThe covariance of Grassmann bilinears is the connected second term. Its Feynman diagram is one oriented fermion loop through both insertions; the first term has two separate fermion loops, one at each insertion. Each closed fermion loop contributes a minus sign, so two loops give a plus and one gives a minus.

For the Dirac field, write $D=\gamma \cdot \partial +m$ and use signature $(-,+,\dots ,+)$, consistent with the denominator in the question. A finite regulator turns the free fermionic path integral into the same Gaussian problem with $B=iD$, because $e^{iS}=e^{-i\bar{\psi }D\psi }$. The time-ordered product covariance is consequently $(i{D}_F{)}^{-1}=-i{D}_F^{-1}$, where the subscript specifies the Feynman i-epsilon prescription.

With the Fourier transform convention $\psi (x)=\int d^{d}p(2\pi {)}^{-d}{e}^{ip\cdot x}\tilde{\psi }(p)$, the derivative kernel is $i\gamma \cdot p+m$. The gamma matrices obeysince their anticommutator makes $(\gamma \cdot p{)}^2=p^{2}I$. Thus the mostly-plus Dirac propagator with a real derivative kernel isFor $p^2=-p_0^2+{\mathbf{p}}^2$, the denominator is $p_0^2-{\mathbf{p}}^2-m^2+i0$. Its positive-energy pole is below and negative-energy pole above the real axis, giving time ordering. Here $S_F$ is defined without the leading factor $i$, as in the question; conventions that include that factor in the name Dirac propagator give the same covariance.

A symmetry of a classical field theory maps solutions to solutions; for the Noether theorem we require a continuous variational symmetry of the action, allowing its density to change by a total divergence. This is stronger than simply observing an accidental symmetry of one solution. Consider first-derivative fields ${\varphi }_a$ and defineThe Euler-Lagrange field equations are $E_a=0$. For a fixed-coordinate variation ${\delta }_0{\varphi }_a=\epsilon Q_a$, the product rule gives the off-shell identityThis identity is the proof's essential integration-by-parts step.

For a general spacetime symmetry write $x^{\prime \mu }=x^{\mu }+\epsilon {\xi }^{\mu }$ and ${\varphi }_a^{\prime }(x^{\prime })={\varphi }_a(x)+\epsilon {\Delta }_a$. At a fixed point $Q_a={\Delta }_a-{\xi }^{\nu }{\partial }_{\nu }{\varphi }_a$, while the volume-element variation contributes ${\partial }_{\mu }(\mathcal{L}{\xi }^{\mu })$. Suppose action invariance takes the formCombining with the preceding identity provesThis is the Noether current for a spacetime symmetry. It also covers internal symmetries, for which $\xi =0$. A constant parameter corresponds to one Noether current; several independent parameters give several currents. Integrating the continuity equation givesThus the Noether charge is conserved when the fields have suitable falloff or no flux through the spatial boundary. These are classical identities; quantum symmetries may additionally require absence of an quantum anomaly.

Translations conserve energy and momentum. If the density has no explicit coordinate dependence, take the active fixed-coordinate variation $Q_a={\partial }_{\nu }{\varphi }_a$ for translation in direction $\nu$. Then ${\delta }_0\mathcal{L}/\epsilon ={\partial }_{\nu }\mathcal{L}$, and the internal-form proof with $K^{\mu }={\delta }^{\mu }{}_{\nu }\mathcal{L}$ gives the canonical energy-momentum tensorIts four spatial integrals are the conserved energy and momentum. For a real scalar field with density $\frac{1}{2}(\partial \varphi {)}^2-V(\varphi )$ this becomes $T^{\mu \nu }={\partial }^{\mu }\varphi {\partial }^{\nu }\varphi -{\eta }^{\mu \nu }\mathcal{L}$. Direct differentiation gives ${\partial }_{\mu }{T}^{\mu \nu }=(\square \varphi +V^{\prime }(\varphi )){\partial }^{\nu }\varphi$, which vanishes by the field equation.

Lorentz symmetry conserves angular momentum and boost charges. For that scalar example, $T^{\mu \nu }$ is symmetric, so the Lorentz transformation currents can be writtenTheir divergence is $T^{\rho \sigma }-T^{\sigma \rho }=0$. Spatial pairs give angular momentum, and time-space pairs give the conserved boost charges. For a spinor field or a vector field, the intrinsic field variation supplies an additional spin current; equivalently the Belinfante-Rosenfeld stress-energy tensor combines spin and orbital pieces into a symmetric stress tensor. The nonunitarity of the finite component Spinor representation of the Lorentz group does not obstruct conservation of these physical spacetime charges.

Global phase symmetry conserves particle charge. For a complex scalar field with density ${\partial }_{\mu }{\Phi }^{\ast }{\partial }^{\mu }\Phi -V({\Phi }^{\ast }\Phi )$, the constant transformation $\Phi \mapsto e^{-i\alpha }\Phi$ leaves the density invariant. Using $Q_{\Phi }=-i\Phi$ and $Q_{{\Phi }^{\ast }}=i{\Phi }^{\ast }$ gives the Noether currentFor a Dirac field, the corresponding variations $Q_{\psi }=-i\psi$, $Q_{\bar{\psi }}=i\bar{\psi }$ yield instead $j^{\mu }=\bar{\psi }{\gamma }^{\mu }\psi$, the Dirac current. Its divergence vanishes by the Dirac equation and adjoint Dirac equation. The Yukawa interaction of part 2 also preserves this phase symmetry, because each bilinear has one $\psi$ and one $\bar{\psi }$. Thus it preserves fermion number and forbids annihilation of two fermions into a single scalar. These currents express charge or particle-minus-antiparticle-number conservation, rather than conservation of total particle count in arbitrary interacting processes.

A global symmetry in field theory uses a spacetime-independent parameter and ordinarily relates physically distinct configurations. A gauge symmetry instead has arbitrary local parameters and represents redundancy in the description. For example, the Maxwell field is unchanged physically by $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\alpha (x)$: its electromagnetic field tensor is unchanged. For charged matter use $D_{\mu }={\partial }_{\mu }+iq{A}_{\mu }$ together with $\psi \mapsto e^{-iq\alpha (x)}\psi$. Differentiating shows $D_{\mu }^{\prime }{\psi }^{\prime }=e^{-iq\alpha }{D}_{\mu }\psi$, so the coupled Dirac action is locally gauge invariant. A local phase applied to free matter alone produces extra derivatives of $\alpha$ and is not an invariance; the compensating gauge field is necessary.

Local arbitrariness gives identities among field equations, rather than an independent physical charge for every function $\alpha$. In the pure Maxwell field, action invariance under a compactly supported $\alpha$ impliesThus ${\partial }_{\nu }{E}^{\nu }=0$ holds off shell, directly because $F^{\mu \nu }$ is antisymmetric. This exemplifies the Noether second theorem. Gauge fixing and the Gupta-Bleuler null-state quotient remove redundant photon components; they do not remove a physical global charge. Gauge transformations that are nontrivial at a boundary can carry boundary charges, so the redundancy statement concerns transformations satisfying the chosen boundary conditions. Finally, discrete symmetries such as the sign reversal of a real scalar or parity symmetry in quantum field theory can be important without furnishing a continuous-parameter Noether current.

The CPT theorem is a structural consequence of relativistic quantum field theory. In flat spacetime, assume a positive physical state space, the positive-energy spectrum condition, Lorentz invariance, microcausality, the usual Spin-statistics theorem, and a Hermitian dynamics. For ordinary field theories these are realized by a local Hermitian Lorentz-scalar Lagrangian density. There is then an antiunitary operator $\Theta$ implementing the combined charge conjugation, parity symmetry in quantum field theory and time-reversal symmetry, with a CPT-invariant vacuum. The combined symmetry is required even when the three separate symmetries fail.

Under the combined spacetime transformation, $x\mapsto -x$. A complex scalar field can be taken to transform into its adjoint at $-x$; a Hermitian vector field transforms into minus itself at $-x$. A derivative contributes a minus sign through differentiation of its reflected argument. For a Dirac field the transformation supplied in the paper may be writtenField phases are conventional and do not affect the physical conclusions. Crucially, antiunitarity means $\Theta i{\Theta }^{-1}=-i$ and conjugates every numerical coefficient. The reversal of time is compatible with positive energies because this conjugation changes $e^{-iHt}$ to $e^{iHt}$ while leaving the Hamiltonian invariant. Treating time reversal as a unitary change of coordinates would instead give incorrect signs.

One can see explicitly how the CPT transformation of Dirac bilinears works. Let $B_{\Gamma }(x)=:{\psi }^{†}{\gamma }^0\Gamma \psi :$ be a Hermitian fermion bilinear, with ${\Gamma }^{†}={\gamma }^0\Gamma {\gamma }^0$. The colons indicate normal ordering so that the graded reordering below has no vacuum contact term. Using the reflected fields and conjugating ${\gamma }^0\Gamma$, then interchanging the two fermion factors, givesThe last step uses the Dirac-Hermiticity condition and $\{{\gamma }^0,{\gamma }_5\}=0$. The identities for the gamma matrices then giveThe pseudoscalar's sign is particularly instructive: its explicit $i$ must also be conjugated. The minus sign from exchanging fermions is equally essential. These signs depend on the number of free vector indices, rather than on a field's separate parity assignment.

This supplies a local Lagrangian argument for CPT invariance. Hermitian local field combinations with $r$ vector indices transform with sign $(-1{)}^r$, including the appropriate reordered derivative terms. A Lorentz scalar contracts those indices in pairs or with four-index invariant tensors, giving a positive total sign. For charged terms, the reflected conjugate fields and conjugated coefficients exchange a term with its Hermitian-conjugate partner. ThusThe integration variable $x\mapsto -x$ has unit absolute Jacobian. The general field-index argument is discussed in Srednicki's treatment of discrete symmetries, section 40.

For example, the scalar mass term $\overline{\psi }\psi$ is even. The Dirac electromagnetic current and vector potential are both odd, making their contracted interaction even. The Hermitian Dirac kinetic term $(i/2)\overline{\psi }{\gamma }^{\mu }{\overleftrightarrow{\partial }}_{\mu }\psi$ is also even: reflection and reordering interchange the two derivative placements, while antiunitarity conjugates its $i$. A complex Yukawa coupling and its Hermitian conjugate are exchanged rather than requiring the coupling itself to be real. Hence CP violation in the CKM matrix is compatible with exact CPT. Renormalizability is not necessary: local Hermitian higher-dimensional Lorentz-scalar operators satisfy the same argument.

The axiomatic form of the CPT theorem does not start by assuming a polynomial Lagrangian density. The positive-energy spectrum condition provides analytic continuations of Wightman functions, and complexified Lorentz transformations relate configurations to their total spacetime inversions. At suitable spacelike configurations, microcausality permits reversal of the order of field operators with the signs dictated by the Spin-statistics theorem. Uniqueness of analytic continuation extends this relation to the correlators generally, from which the Hilbert-space antiunitary operator is obtained. This explains why locality and positive energy are genuine hypotheses, not incidental features of a particular interaction. The elementary argument above displays the mechanism for the scalar, spinor and vector fields relevant to the Standard Model; it is not a substitute for all the analytic steps of the axiomatic proof.

For one-particle states, CPT reverses additive charges and spin projections. The spatial momentum is unchanged by the combination: parity symmetry in quantum field theory and time reversal each reverse it. Thus, up to phases,Since the energy-momentum spectrum is preserved, the CPT constraints on particle properties give equal particle-antiparticle masses, equal spin magnitudes, and opposite conserved additive charges. Relating the corresponding two-point functions also fixes equal pole positions, so unstable conjugate particles have equal total decay widths, hence equal lifetimes. Magnetic moments at the same chosen spin orientation have opposite signs. These are precise experimental tests of CPT, not tests that every individual discrete symmetry is respected.

For scattering, CPT interchanges incoming and outgoing boundary conditions. The antiunitary operator identity $⟨a|b⟩=⟨\Theta b|\Theta a⟩$ relates $⟨f,\mathrm{out}|i,\mathrm{in}⟩$ to the amplitude with transformed $f$ incoming and transformed $i$ outgoing. Thus the reversed process with antiparticles and reversed spins has the same transition probability. This is not, in general, equality with the forward CP-conjugate process. In particular, CP violation can make corresponding partial decay widths different while their complete sums remain equal. With CPT preserved, genuine CP violation requires corresponding time-reversal symmetry violation.

Finally, a proposed CPT violation requires revisiting at least one assumption: for example, a nonlocal interaction, a preferred spacetime direction violating Lorentz invariance, or loss of unitary Hermitian dynamics. A generic curved background need not possess the global spacetime inversion used here. Those observations delimit the theorem; they do not weaken its prediction for the ordinary local Standard Model in Minkowski spacetime.

The middle panel shows the Feynman diagram for the leptonic W boson decay. With the charged-current Lagrangian density sign in the question, its vertex Feynman rule isAn overall field-phase convention can change the overall amplitude sign but cannot change the decay width. With the standard Dirac field current ordering, the amplitude isThe unusual order of barred fields in the printed description should be interpreted as this physical weak charged current matrix element.

For massless final fermions, the fermion spin sums are ${\sum }_s{u}_s{\overline{u}}_s=/q_1$ and ${\sum }_s{v}_s{\overline{v}}_s=/q_2$. The massive-vector spin average divides by three. Thuswith the chiral traceAnticommuting ${\gamma }_5$ through the two following gamma matrices and using $(1-{\gamma }_5{)}^2=2(1-{\gamma }_5)$ gives its symmetric partIts remaining part is an antisymmetric epsilon tensor from the gamma matrix trace identities; it vanishes against the symmetric polarization tensor $K$. Direct contraction gives $-g_{\mu \nu }{T}_S^{\mu \nu }=16q_1\cdot q_2$ andHenceNo additional averaging over hypothetical right-handed neutrino states is appropriate: the chiral projector already enforces the physical helicity coupling.

On shell, $p=q_1+q_2$, $p^2=M_W^2$ and $q_i^2=0$, so $q_1\cdot q_2=M_W^2/2$ and $p\cdot q_i=M_W^2/2$. The squared matrix element reduces to $g^2{M}_W^2/3$. In the parent rest frame, integrate the spatial momentum delta first in two-body Lorentz-invariant phase space. The remaining radial integral isConsequently the massless leptonic W decay width isLepton universality gives $\Gamma (W^{-}\to {\mu }^{-}{\overline{\nu }}_{\mu })=\Gamma (W^{-}\to {\tau }^{-}{\overline{\nu }}_{\tau })={\Gamma }_{\ell }$ in the same massless approximation. The muon and tau lepton masses produce small corrections if retained.

With an identity CKM matrix, the allowed quark channels are $d\overline{u}$ and $s\overline{c}$. The corresponding third-generation channel $b\overline{t}$ is closed by the top threshold. Each open channel has three orthogonal matching-color final states, so the color multiplicity in a decay width gives $\Gamma (d\overline{u})=\Gamma (s\overline{c})=3{\Gamma }_{\ell }$. There is no initial color average because the W boson is colorless. The tree-level W branching fractions are therefore:

The rounded percentages sum to 99% because of rounding. The exact fractions sum to one, with total leptonic fraction $1/3$ and hadronic fraction $2/3$. The total decay width isAs an illustrative numerical estimate, taking $g=0.65$ and $M_W=80\mathrm{GeV}$ gives ${\Gamma }_{\ell }\simeq 0.224\mathrm{GeV}$, each open quark width $\simeq 0.672\mathrm{GeV}$ and ${\Gamma }_W\simeq 2.02\mathrm{GeV}$. These are tree-level model estimates with the stated illustrative inputs, not a fit to a measured width. Strong radiative corrections increase the inclusive quark widths.

The produced quark and antiquark radiate gluons, generating a parton shower. As their virtualities fall towards the strong-interaction scale, perturbation theory gives way to hadronization: confinement converts the colored partons into color-neutral hadrons, typically seen as two particle jets, with additional radiation sometimes producing resolved extra jets. Unstable hadrons subsequently decay; the primary quarks are not observed as free particles.

Including the CKM matrix, the charged current couples an accessible up-type flavour $u_i$ to every kinematically accessible down-type flavour $d_j$. ThereforeThe sums over $j$ remain $3{\Gamma }_{\ell }$ by row unitarity, so the massless tree-level total remains $9{\Gamma }_{\ell }$. Quark mixing redistributes the hadronic branching fractions; it does not create a new total color factor. If a family is above threshold, only its accessible channels enter, and the unitarity sum must not be applied to an incomplete set of final flavours.

For completeness, CKM parameter counting starts with $N^2$ real degrees of freedom of an $N\times N$ unitary matrix. This follows by counting $2N^2$ real entries and $N^2$ independent Hermitian constraints in $V^{†}V=I$. The real-orthogonal subclass has $N(N-1)/2$ mixing angles; the remaining $N(N+1)/2$ parameters can be viewed as phases. Rephasing each up-type and down-type mass eigenfield changes $V_{ij}$ by $e^{-i{\alpha }_i}{e}^{i{\beta }_j}$. Of the $2N$ phases, the common phase does nothing, so $2N-1$ are removable. ThusFor three generic nondegenerate families this is three angles and one CP-violating phase. Degenerate masses can allow additional changes of basis and reduce the count.

For two independent left-handed Weyl spinors $\chi$ and $\eta$, a Dirac mass term isThe four-component Dirac field ${\Psi }_D=({\chi }_{\alpha },{\bar{\eta }}^{\dot{\alpha }}{)}^T$ has independent particle and antiparticle degrees of freedom; after a phase redefinition making $m$ real, the same term is $-m{\bar{\Psi }}_D{\Psi }_D$. The two left-handed spinors can carry opposite charges under a conserved U(1) symmetry, $\chi \mapsto e^{iq\omega }\chi$, $\eta \mapsto e^{-iq\omega }\eta$. A gauge-invariant Dirac mass term contracts conjugate gauge group representations.

For one left-handed Weyl spinor, a Majorana mass term instead has the formIts self-contraction need not vanish, by the Weyl spinor bilinear exchange identity. The factor $1/2$ avoids double counting identical fields in variation. The associated four-component Majorana fermion ${\Psi }_M=({\chi }_{\alpha },{\bar{\chi }}^{\dot{\alpha }}{)}^T$ satisfies ${\Psi }_M^c={\Psi }_M$, so its particle is its own antiparticle. A Majorana mass term breaks continuous fermion-number phase symmetry for $\chi$, and is allowed only if its gauge indices admit an invariant symmetric contraction, as for a neutral field or a real gauge group representation. An isolated field of nonzero U(1) charge cannot have it. More generally, the mass matrix of several left-handed Weyl spinors is symmetric, since ${\chi }_i{\chi }_j={\chi }_j{\chi }_i$.

Here $\overline{u}$ is the Dirac field's adjoint, not a second independently defined charge-conjugate field. Charge conjugation of a Dirac field gives $u\mapsto C{\overline{u}}^T$ and $\overline{u}\mapsto -u^T{C}^{-1}$; parity gives $u(x)\mapsto {\gamma }^{0}u(x_P)$ and $\overline{u}(x)\mapsto \overline{u}(x_P){\gamma }^0$. With the product convention $\widehat{CP}=\hat{C}\hat{P}$, therefore,Reversing the convention for the operator product changes an overall intrinsic phase, which does not change the bilinear transformations when used consistently. The transpose also exchanges the fundamental colour index with its conjugate.

An experimentally observed example of CP violation is the long-lived neutral kaon decay $K_L\to {\pi }^{+}{\pi }^{-}$. The two-pion spin-zero final state is CP even, whereas $K_L$ would be the CP-odd neutral-kaon state in the CP-conserving limit. Its two-pion decay reveals CP violation. The historical experimental account is cern-courier.web.cern.ch/a/cp-violations-early-days/ .

For the CKM matrix, diagonalize the up- and down-quark mass matrices independently by biunitary transformations. Writewith $L_u^{†}{M}_u^{\prime }{R}_u$ and $L_d^{†}{M}_d^{\prime }{R}_d$ positive diagonal matrices. Only the left transformations enter the weak charged current, because its chiral projector removes the right-handed fields. Substitution givesThe matrix is unitary but need not be diagonal: the two left mass diagonalizations generally differ.

Using the field transformations in the subparts and reordering the anticommuting Dirac fields, the charged bilinear transforms aswhere the fields on the right are both evaluated at $x_P$. The W boson contributes its second minus sign and Lorentz parity factor, so the product maps into its Hermitian-conjugate product. Crucially CP symmetry acts unitarily: it does not separately complex-conjugate the numerical coefficient $V_{ij}$. Thus the transformed interaction isComparing with the original interaction requires $V_{ij}=V_{ij}^{\ast }$ in this canonical CP convention. The physical condition is quark rephasing and CP conservation:A merely complex-looking matrix in an arbitrary mass-eigenstate phase convention is not itself proof of CP violation.

For CKM parameter counting, a unitary $N\times N$ matrix has $N^2$ real parameters: $N(N-1)/2$ mixing angles and $N(N+1)/2$ phases. Rephasing the $N$ up and $N$ down mass eigenfields removes $2N-1$ phases, since their common phase cancels out of $V$. For generic nondegenerate masses this leavesThere is one physical phase for three families and none for one or two. Degenerate masses would permit additional transformations, so the generic count need not apply unchanged.

For the QCD part of the gauge covariant derivative, define the Hermitian colour matrix ${\mathcal{A}}_{\mu }=A_{\mu }^a{T}^a$ and chooseThis sign convention agrees with the field-strength definition ${\mathcal{F}}_{\mu \nu }=-i[D_{\mu },D_{\nu }]/g_s$ in the source. The kinetic term contains $-g_s\overline{u}{\gamma }^{\mu }{\mathcal{A}}_{\mu }u$. A vector bilinear under CP changes sign, gains the Lorentz parity matrix, and transposes its colour matrix upon exchanging quark and antiquark. Invariance of this interaction therefore requires the CP transformation of a non-Abelian gauge connection:Both the time/spatial distinction and the colour transpose are essential; individual colour components need not all share one charge-conjugation sign.

The same gauge covariant derivative givesApply the transformation above and the chain rule ${\partial }_{\mu }f(x_P)={\Pi }_{\mu }{}^{\alpha }({\partial }_{\alpha }f)(x_P)$. The transpose reverses commutator order, $[X^T,Y^T]=-[X,Y{]}^T$, so the nonlinear term transforms with the same overall minus sign as the derivatives. This proves the CP transformation of non-Abelian field strength:Finally $\mathrm{tr}(T^a{T}^b)={\delta }^{ab}/2$ rewrites the theta density as $2\theta {ϵ}^{\mu \nu \rho \sigma }\mathrm{tr}({\mathcal{F}}_{\mu \nu }{\mathcal{F}}_{\rho \sigma })$. The two minus signs cancel, and transposition followed by cyclicity leaves the colour trace unchanged. But the four parity matrices acting on the Levi-Civita symbol contribute $\det \Pi =-1$. HenceThe density is C even and P odd, therefore CP odd. A generic fixed nonzero theta coefficient violates CP symmetry. This is the source of the Strong CP problem, separate from the charged-current phase in the CKM matrix.

The supplied gamma matrices use the mostly-minus metric signature. The Pauli matrix multiplication law gives ${\sigma }^1{\sigma }^2{\sigma }^3=i{I}_2$, hence block multiplication yieldsThe off-diagonal blocks of every ${\gamma }^{\mu }$ connect opposite eigenspaces of this diagonal matrix. Explicitly,Thus $\{{\gamma }^5,{\gamma }^{\mu }\}=0$, and $({\gamma }^5{)}^2=I_4$. The chiral projectors $(1\mp {\gamma }^5)/2$ select the upper and lower two-component blocks, respectively.

Write the Dirac spinor as $\psi =({\chi }_L,{\chi }_R{)}^T$. Substituting the block matrices into the massless Dirac equation gives two independent Weyl spinor equations:They are decoupled because the mass term, which would join the two chirality components, is absent.

For a left-handed plane wave, use $p\cdot x=p^{0}t-\mathbf{p}\cdot \mathbf{x}$. Its coefficient $\xi$ obeysSince $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2=|\mathbf{p}{|}^2{I}_2$, the determinant is $(p^0{)}^2-|\mathbf{p}{|}^2$. A nonzero coefficient therefore requires the massless mass shell $\boxed{p^2=0}$. For positive energy and nonzero $\mathbf{p}$, put $P=|\mathbf{p}|=p^0$ and parameterize its direction by angles $\vartheta ,\phi$. ThenMultiplication gives $(\mathbf{\sigma }\cdot \hat{\mathbf{p}}){\chi }_{-}=-{\chi }_{-}$ and ${\chi }_{-}^{†}{\chi }_{-}=1$. Hence the left-handed massless plane-wave spinor can be writtenHere $N(p)$ is an arbitrary complex normalization; $|N|=\sqrt{2P}$ gives ${\lambda }_L^{†}{\lambda }_L=2P$. Away from the negative $z$ axis one can equivalently takeAt $\mathbf{p}=-P\hat{\mathbf{z}}$ choose $(1,0{)}^T$ instead. These are phase patches for the same one-dimensional kernel. The other frequency branch $p^0=-P$ uses the positive eigenvector ${\chi }_{+}=(\cos (\vartheta /2),e^{i\phi }\sin (\vartheta /2){)}^T$ for the direction of $\mathbf{p}$. At $p^{\mu }=0$, the algebraic equation vanishes and any two-component coefficient is possible, but no momentum direction or helicity is defined there.

A spatial rotation by $\alpha$ about $\hat{\mathbf{p}}$ acts on the two-component coefficient by $\exp (-i\alpha \mathbf{\sigma }\cdot \hat{\mathbf{p}}/2)$. The negative eigenvector therefore acquires the phase $e^{i\alpha /2}$. Since a state with helicity $h$ acquires $e^{-i\alpha h}$, the positive-energy left-handed particle has $\boxed{h=-1/2}$. There is only one such spin state at a fixed nonzero momentum, rather than two independent polarizations of the same chiral particle.

The helicity content of a quantized Weyl field also includes an antiparticle with opposite helicity. In the field expansion, the negative-frequency term multiplies an antiparticle fermionic creation operator, while the positive-frequency term multiplies a particle fermionic annihilation operator. Their adjoint roles give conjugate phases for the corresponding created states, so a left-handed field creates particles with $h=-1/2$ and antiparticles with $h=+1/2$. The right-handed Weyl spinor has the opposite assignments. Quantizing the full massless Dirac field, rather than one chiral component, therefore gives both particle helicities and both antiparticle helicities; every excitation has spin one-half, while chirality determines the helicity of each massless sector.

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.