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In strict axial gauge, the gauge-boson propagator obeys $n^{\mu }{D}_{\mu \nu }(k)=0$ and contains poles at $n\cdot k=0$. Its contraction with conserved external currents removes the gauge-choice terms. Consistent calculations require an axial-gauge pole prescription.

For a constant unitary matrix $h$, ${\varphi }^{\prime }=h\varphi$ gives ${\varphi }^{\prime †}{\varphi }^{\prime }={\varphi }^{†}{h}^{†}h\varphi ={\varphi }^{†}\varphi$. Also ${\partial }_{\mu }{\varphi }^{\prime }=h{\partial }_{\mu }\varphi$, so the kinetic contraction is unchanged. The mass term and the squared norm in the potential are unchanged for the same reason. The original scalar action has the stated global symmetry. For a spacetime-dependent $h$, however, ${\partial }_{\mu }(h\varphi )=h{\partial }_{\mu }\varphi +({\partial }_{\mu }h)\varphi$, and the extra term prevents local invariance of the ordinary-derivative kinetic term.

Choose the gauge covariant derivative $D_{\mu }={\partial }_{\mu }+ig{A}_{\mu }$, with $A_{\mu }=A_{\mu }^{\alpha }{T}_{\alpha }$ Hermitian. Demand $D_{\mu }^{\prime }(h\varphi )=h{D}_{\mu }\varphi$. Expanding both sides yieldsso the necessary Yang-Mills gauge transformation isUnitarity makes the transformed connection Hermitian. This formula includes the inhomogeneous derivative term; mere conjugation would be insufficient for a local transformation.

Define the Yang-Mills field strength by $[D_{\mu },D_{\nu }]=ig{F}_{\mu \nu }$. Direct expansion givesChoose an orthonormal Hermitian generator basis with $\mathrm{tr}(T_{\alpha }{T}_{\beta })=\kappa {\delta }_{\alpha \beta }$. The compact simple Lie algebra has an invariant positive trace metric, so this basis can be chosen; its structure constants are totally antisymmetric. With these conventions a locally invariant dynamical action isEach matter contraction is invariant because $D\varphi$ transforms like $\varphi$, and the gauge kinetic term is invariant by cyclicity of the trace. It is $-\frac{1}{4}{F}_{\mu \nu }^{\alpha }{F}_{\alpha }^{\mu \nu }$ in components. Choosing $\kappa =1/2$ gives the familiar $-\frac{1}{2}\mathrm{tr}{F}^2$ convention.

Formally introduce a gauge-field source $J_{\alpha }^{\mu }$ and a normalized generating functionalA gauge-field Green function is obtained by source differentiation:Here $\mathcal{N}$ is the unnormalized integral at zero source, so the normalized functional obeys $Z[0]=1$. This expression needs gauge fixing: the integral repeats every physical configuration over its whole gauge orbit, producing an irrelevant but divergent gauge-group volume, and the free gauge kinetic operator has longitudinal zero modes, so it has no inverse propagator on that unrestricted space. Ordinary gauge-field Green functions are gauge-dependent; physical gauge-invariant quantities must be obtained consistently from the gauge-fixed theory.

Use the Lorenz condition ${\chi }^{\alpha }[A]={\partial }^{\mu }{A}_{\mu }^{\alpha }$. To keep signs explicit, parametrize a small transformation by $h=1-ig{\omega }^{\alpha }{T}_{\alpha }+O({\omega }^2)$, equivalently the original parameter is ${\theta }^{\alpha }=g{\omega }^{\alpha }$. The finite transformation above gives the adjoint gauge variation for a positive-sign covariant derivative:Therefore $\delta {\chi }^{\alpha }={\partial }^{\mu }{\mathcal{D}}_{\mu }^{\alpha \beta }{\omega }^{\beta }$. Define the Faddeev-Popov operator as $\mathcal{M}=-{\partial }^{\mu }{\mathcal{D}}_{\mu }$, absorbing the field-independent minus sign of its determinant into normalization.

In a perturbative gauge patch, with residual zero modes removed by boundary conditions, the Faddeev-Popov gauge-orbit identity isIt follows by changing coordinates from the gauge parameter to the gauge condition: the functional Jacobian is precisely the determinant of its linearization. Insert the identity into the integral and change variables along the orbit. The action and measure are invariant, so the gauge-group volume factors out and is divided away. Averaging the gauge-slice value $f$ with Gaussian weight $\exp [-i\int f^2/(2\xi )]$ changes the delta-functional restriction to the covariant gauge-fixing termThe remaining determinant depends on $A$ through ${\mathcal{D}}_{\mu }$. Omitting it while adding only ${\mathcal{L}}_{\mathrm{gf}}$ would weight different orbits incorrectly.

For a finite matrix $M$, expansion of a Grassmann integral selects the product containing every $c$ and $\bar{c}$. Anticommuting the variables into a fixed order attaches the sign of each permutation, giving $\int d\bar{c}dc{e}^{i\bar{c}Mc}$ proportional to $\det M$, with only a matrix-independent phase. The regulated functional version therefore represents the Faddeev-Popov determinant asThe independent Faddeev-Popov ghost fields and antighost fields obey $c_{\alpha }{c}_{\beta }=-c_{\beta }{c}_{\alpha }$, with the corresponding mixed and antighost anticommutation. A commuting complex Gaussian integral instead gives an inverse determinant, so it cannot supply this Jacobian. The required local ghost term isAfter integration by parts,This exhibits the free ghost kinetic term and the ghost-gluon vertex. Ghosts are Grassmann-valued scalar auxiliary fields in the Adjoint representation, not physical fermionic particles or external asymptotic states. Their loop signs arise from anticommutation. In an Abelian theory the adjoint derivative loses the field-dependent part, so the determinant is field-independent and the ghosts decouple; in the non-Abelian theory they must be retained.

The correctly gauge-fixed source functional thus uses $S_{\mathrm{inv}}+S_{\mathrm{gf}}+S_{\mathrm{gh}}$ and integrates over $A,\varphi ,{\varphi }^{†},c,\bar{c}$. Expanding its interaction terms gives the gauge, matter and ghost vertices, and differentiating its source functional computes the gauge-field Green functions. For example, at non-null momentum the quadratic gauge kernel iswhere $P_T$ and $P_L$ are the transverse projector of a vector field and longitudinal projector of a vector field. With the Feynman prescription this gives the gauge-boson propagatorFor the chosen ghost normalization the quadratic kernel is $-{\partial }^2$, so the ghost propagator is $i{\delta }^{\alpha \beta }/(k^2+i0)$. Redefining the antighost sign changes ghost propagator and vertex signs together, without changing the determinant or physical conclusions. Gauge fixing requires the accompanying determinant, and its local representation requires anticommuting ghost and antighost fields. The gauge-orbit argument is local/perturbative: global gauge copies, the Gribov ambiguity, prevent interpreting it as a globally unique gauge slice without further treatment.

The gauge-invariant Yang-Mills action integrates over physically equivalent configurations along each gauge orbit. At the quadratic level its gauge-field kernel has longitudinal zero directions: in momentum space $K_{\mu \nu }(k)=-k^2{\eta }_{\mu \nu }+k_{\mu }{k}_{\nu }$ satisfies $K_{\mu \nu }{k}^{\nu }=0$. It therefore has no inverse from which to construct a gauge-boson propagator. Gauge fixing both removes the redundant orbit integration and supplies an invertible quadratic operator away from the ordinary physical propagator poles.

For the covariant condition ${\partial }^{\mu }{A}_{\mu }=0$, the Faddeev-Popov determinant is that of ${\partial }^{\mu }{D}_{\mu }$. Its representation by odd Faddeev-Popov ghost fields and antighost fields produces the displayed ghost term. In a nonabelian theory $D_{\mu }$ depends on the gauge field, so the determinant is not an irrelevant constant: ghost interactions must be retained. The even Nakanishi-Lautrup field $b$ writes gauge fixing in a form with off-shell BRS symmetry. For $\xi \ne 0$, its algebraic equation is $b=-\partial \cdot A/\xi$, giving $-(\partial \cdot A{)}^2/(2\xi )$ after elimination. At $\xi =0$ it imposes the gauge condition as a Lagrange multiplier.

For the nilpotence proof, write the adjoint bracket as $[X,Y{]}^a=f^{abc}{X}^b{Y}^c$, with coefficients in a Grassmann algebra. If $|X|,|Y|$ are their parities, this is a graded Lie bracket:Thus two odd fields have a symmetric bracket, in particular $[c,c]$ need not vanish. The transformations become $s{A}_{\mu }=D_{\mu }c$, $sc=-[c,c]/2$, $s\bar{c}=-b$, $sb=0$, and $s$ obeys the left graded Leibniz rule.

Because $D_{\mu }$ is even, $D_{\mu }[c,c]=[D_{\mu }c,c]+[c,D_{\mu }c]=2[D_{\mu }c,c]$. Varying both the field and the connection in $D_{\mu }c$ now givesFor the ghost,The last equality is the graded Jacobi identity applied to three copies of the odd field $c$. In components the remaining expression is $\frac{1}{2}{f}^{abc}{f}^{bde}{c}^d{c}^e{c}^c$; its coefficient is antisymmetrized over $c,d,e$, and the ordinary Jacobi identity for the structure constants makes that antisymmetrization zero. Finally $s^2\bar{c}=-sb=0$ and $s^{2}b=0$. HenceThis is off-shell nilpotence of the Yang-Mills BRST quartet.

To verify invariance of the complete Lagrangian, $s{F}_{\mu \nu }=[F_{\mu \nu },c]$, obtained from $[D_{\mu },D_{\nu }]c=[F_{\mu \nu },c]$. Thus $s(F_{\mu \nu }^a{F}^{\mu \nu a})=0$ because the product of the two curvature components is symmetric in their color labels while $f^{abc}$ is antisymmetric. The remaining terms vary asThe last term vanishes by the already proved $s^{2}A=0$, and the first and third terms cancel pointwise. ThereforeEquivalently, the covariant gauge-fixing density as a BRST variation is $s\Psi$ with $\Psi =-{\bar{c}}^a(\partial \cdot A^a+\xi b^a/2)$. The graded sign from moving $s$ past $\bar{c}$ produces the positive ghost term, and $s^2\Psi =0$ proves its invariance. No integration-by-parts qualification is needed for the density as written.

Let $Q=Q_{\mathrm{BRS}}$ be the BRS charge. The physical state space is its BRST cohomology, usually at ghost number zero:Nilpotence ensures $\mathrm{im}Q\subseteq \ker Q$. A state $Q|\chi ⟩$ is orthogonal to every closed state because $⟨\psi |Q|\chi ⟩=⟨Q\psi |\chi ⟩=0$, and its norm is $⟨\chi |Q^2|\chi ⟩=0$. Thus adding an exact state changes neither physical inner products nor amplitudes. Self-adjointness here is relative to the stated indefinite Hermitian form; a nonzero self-adjoint nilpotent operator could not exist on a positive-definite state space. Positivity must be proved on the quotient, not assumed on the gauge/ghost space.

Fix a color label and a nonzero momentum $k$, and take the usual nontrivial asymptotic charge, $\alpha \ne 0$. Nilpotence on a vector state gives $Q|c⟩=0$, because at least one component of $k$ is nonzero. Nilpotence on the antighost state then givesThe two coefficients are complex conjugates as required by self-adjointness: using the given ghost pairing,Thus the antighost-to-vector coefficient is ${\alpha }^{\ast }$, as the printed bar notation indicates. These conditions are the massless shell condition and the consistency of the charge with the inner products.

A general one-particle vector representative is $|\epsilon ⟩={\epsilon }^{\mu }|A_{\mu }⟩$. It is closed precisely when $k_{\mu }{\epsilon }^{\mu }=0$. The vector state $k^{\mu }|A_{\mu }⟩=Q|\bar{c}⟩/\bar{\alpha }$ is exact, so representatives obeyThe ghost state is also exact: choose $v^{\mu }$ with $\alpha v\cdot k=1$, then $Q(v^{\mu }|A_{\mu }⟩)=|c⟩$. The antighost state is not closed because its image is a nonzero longitudinal vector. More explicitly, closure of a general sum of vector, ghost and antighost states forces the antighost coefficient to vanish and the vector polarization to be orthogonal to $k$; its ghost component disappears in the quotient. Thus the full one-particle cohomology already reduces toWith the mostly-plus Minkowski metric, choose a Lorentz frame with $k^{\mu }=(\omega ,0,0,\omega )$, $\omega >0$. Closure gives ${\epsilon }^3={\epsilon }^0$, and the norm isSubtracting a multiple of $k$ sets ${\epsilon }^0={\epsilon }^3=0$. Consequently positive one-particle BRST cohomology consists of two transverse positive-norm polarizations per color, represented by $|A_1⟩,|A_2⟩$. A nonzero class has strictly positive norm; the zero-norm longitudinal direction was precisely the exact direction removed.

The qualifications $k\ne 0$ and $\alpha \ne 0$ are essential to this one-particle argument. The zero-momentum sector requires separate treatment, and the trivial-charge case cannot yield the asserted removal of negative-norm vector states from the stated relations alone. Abstract nilpotence and self-adjointness by themselves do not guarantee positive cohomology.

Adopt the phase convention $CP|K^0⟩=-|{\bar{K}}^0⟩$ and $CP|{\bar{K}}^0⟩=-|K^0⟩$. Changing both phases changes the convention for the mixing parameter, not physical predictions. The flavor contents are $K^0=d\bar{s}$ and ${\bar{K}}^0=s\bar{d}$. A pair of charged weak currents changes strangeness by two units through the following box contractions, with $u_i,u_j=u,c,t$ summed over.

The drawing uses unitary gauge: the W boson gauge-boson propagator contains its longitudinal part. In a gauge of a renormalizable quantum field theory, charged Goldstone boson boxes must also be included to obtain the same gauge-independent result. The external lines represent the valence quarks; the meson matrix element also contains strong interaction binding effects.

Put ${\lambda }_i=V_{is}^{\ast }{V}_{id}$, $x_i=m_i^2/m_W^2$ and $P_L=(1-{\gamma }^5)/2$. The charged weak box contribution to kaon mixing has the effective four-fermion interaction structureHere $G_F$ is the Fermi constant, $V$ is the CKM matrix, and $S$ is a dimensionless loop function from a Feynman integral; the estimate suppresses convention-dependent numerical factors. Four weak interaction Feynman vertices and a one-loop Feynman integral give the coupling and loop order suppression. Unitarity of the CKM matrix gives ${\sum }_i{\lambda }_i=0$, so the GIM mechanism cancels the mass-independent part. With a negligible up quark mass, the leading charm quark term has $S(x_c,x_c)\sim x_c$. Define $f_K$ by $⟨0|\bar{s}{\gamma }_{\mu }{\gamma }^{5}d|K^0(p)⟩=i{f}_K{p}_{\mu }$. Estimating the hadronic matrix element by $f_K^2{m}_K{B}_K$, with the dimensionless kaon bag parameter $B_K$ of order unity, givesThe hadronic normalization includes division by $2m_K$ when relativistically normalized states are converted to the effective Hamiltonian for neutral-meson mixing. Since $|V_{cs}^{\ast }{V}_{cd}|\simeq \sin {\theta }_C\cos {\theta }_C$, the Cabibbo suppression is explicit. Illustrative scales $G_F\sim 10^{-5}{\mathrm{GeV}}^{-2}$, $m_c\sim 1\mathrm{GeV}$, $f_K\sim 0.1\mathrm{GeV}$ and $m_K\sim 0.5\mathrm{GeV}$ give an order $10^{-16}$ to $10^{-15}\mathrm{GeV}$ estimate. The full sum includes charm quark-top quark and top quark-top quark terms, important for CP violation; long-distance strong interaction contributions prevent this dimensional estimate from being a precision prediction. A meson matrix element necessarily needs hadronic input in addition to electroweak parameters.

For the decaying kaon system write $H^{\prime }=M-i\Gamma /2$, where $M$ and $\Gamma$ are Hermitian matrices. The entries denoted $M_{ij}$ in the question are entries of this effective Hamiltonian for neutral-meson mixing; they need not themselves form a Hermitian matrix. CPT invariance equates the diagonal masses and diagonal decay widths, givingOne can see why there is no complex conjugation on the right as follows. In the flavor subspace let $J=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ and represent the antiunitary CPT operation by $J$ followed by complex conjugation. CPT changes a projected forward propagator into its adjoint. Indeed, if $P$ projects onto the flavor subspace and $H_{\mathrm{full}}$ is the Hermitian microscopic generator, its invariance and the antiunitary reversal of $i$ give $\Theta P{e}^{-i{H}_{\mathrm{full}}t}P{\Theta }^{-1}=P{e}^{+i{H}_{\mathrm{full}}t}P=(P{e}^{-i{H}_{\mathrm{full}}t}P{)}^{†}$. In the time-independent two-state approximation, the effective generator this means $J{H^{\prime }}^{\ast }J={H^{\prime }}^{†}$, whose diagonal entries give $(H_{22}^{\prime }{)}^{\ast }=(H_{11}^{\prime }{)}^{\ast }$; the off-diagonal entries impose no further condition. Equivalently, apply CPT separately to the dispersive and absorptive Hermitian matrices. Treating the decaying effective Hamiltonian for neutral-meson mixing as a Hermitian matrix would incorrectly discard the absorptive part.

In the chosen flavor phase convention CP is represented by $-J$. Its invariance requires $J{H}^{\prime }J=H^{\prime }$. Thus it gives diagonal equality and additionallyThe normalized CP eigenstates areApplying CP directly gives eigenvalues $+1$ and $-1$ respectively.

Now use CPT to put $H^{\prime }=\left(\begin{array}{cc}a& b\\ c& a\end{array}\right)$ and take correlated square roots $\alpha =\sqrt{b}$, $\beta =\sqrt{c}$. Its characteristic polynomial equation is $(a-\mu {)}^2-bc=0$, and direct multiplication showsExpressing these two eigenvectors in the CP eigenstate basis gives, respectively, coefficients proportional to $(\alpha +\beta ,\alpha -\beta )$ and $(\alpha -\beta ,\alpha +\beta )$. Therefore the kaon CP mixing parameter isand, because the CP eigenstate basis is orthonormal, the normalized states at production areThe kaon mixing square-root branch convention correlates the roots continuously with the limit of conserved CP $b=c$, where $\alpha =\beta$ and $ϵ=0$. Reversing one root interchanges the two eigenstates. The labels short and long are fixed by their decay widths: their complex eigenvalues are ${\mu }_{S,L}=m_{S,L}-i{\Gamma }_{S,L}/2$, and subsequent evolution supplies $e^{-i{\mu }_{S,L}t}$. The displayed formula assumes nonzero mixing and $\alpha +\beta \ne 0$, as in the physical kaon system; degenerate or defective matrices require a separate limiting treatment.

A Non-Abelian gauge theory describes fields with a local redundancy, not an independent physical degree of freedom for each component of its gauge field. Take a compact gauge group with Hermitian generators $T^a$ and Lie algebra structure constants defined by $[T^a,T^b]=i{f}^{abc}{T}^c$. In a mostly-plus metric signature, the pure Yang-Mills theory hasWith the matter gauge covariant derivative $D_{\mu }={\partial }_{\mu }-ig{A}_{\mu }$, a finite gauge transformation acts as $A_{\mu }\mapsto U{A}_{\mu }{U}^{-1}-(i/g)({\partial }_{\mu }U)U^{-1}$. Expanding $U=e^{ig{\omega }^a{T}^a}$ givesThe Yang-Mills action is constant along each gauge orbit. Consequently a naive functional integral $\int \mathcal{D}A{e}^{iS[A]}$ repeatedly integrates the same physical configuration and contains the infinite formal volume of the local gauge group. At quadratic order the same problem appears as noninvertibility: the free kinetic operator annihilates $A_{\mu }={\partial }_{\mu }\omega$, so no unique gauge-boson propagator exists. In canonical language the Gauss law constraint in gauge theory and the absence of an independent momentum for the time component show that the field components cannot all be independent oscillators. In covariant quantization, time-like and longitudinal polarizations must not be counted as additional physical states.

The Faddeev-Popov determinant implements a local quotient by the gauge orbit. Choose $G^a[A]={\partial }^{\mu }{A}_{\mu }^a$ and first impose $G^a[A]=f^a(x)$ for a specified function $f$. This is the Lorenz gauge condition when $f=0$. Its infinitesimal change isThe direct Faddeev-Popov operator is therefore $M_0^{ab}={\partial }^{\mu }(D_{\mu }{)}^{ab}$. Locally, if the condition cuts each orbit once and residual zero modes have been removed by boundary conditions, changing variables from $\omega$ to $G[A^{\omega }]-f$ proves the Faddeev-Popov gauge-orbit identityThe determinant is the functional Jacobian of this change of variables. A finite-dimensional version is $\int d\omega \delta (G(\omega )-f)=1/|\det (\partial G/\partial \omega )|$ near a unique root; the perturbative determinant has a fixed sign or phase, absorbed into normalization. This is the same change-of-variables mechanism used in Bastianelli's derivation of the Faddeev–Popov construction.

Insert the identity into the normalized functional integral for a gauge-invariant observable $\mathcal{O}[A]$:For each fixed $\omega$, change the integration variable from $A$ to $A^{\omega }$. The action, observable and measure are invariant, and the determinant is the orbit Jacobian at the resulting gauge slice. The integrand is now independent of the separately integrated orbit parameter. Its integral is $\mathrm{Vol}\mathcal{G}$, canceling the denominator. Thus, up to normalization,This assumes an anomaly-free measure and a local perturbative gauge slice; it is not an assertion of a globally unique representative on every gauge orbit.

Average over $f$ with weight $\exp [-i\int d^{4}x{f}^a{f}^a/(2\xi )]$. The delta functional sets $f=G[A]$, so the average produces the covariant gauge termNext represent the Faddeev-Popov determinant by a Grassmann Gaussian integral. Choose $M=-M_0=-{\partial }^{\mu }{D}_{\mu }$; changing $M_0$ to $-M_0$ changes the regulated determinant by a field-independent factor that is absorbed in normalization. For a Faddeev-Popov ghost field $c$ and independent antighost field $\bar{c}$,To see why it is a determinant rather than its reciprocal, diagonalize a finite-dimensional quadratic form formally: for each Grassmann pair only the term proportional to ${\bar{c}}_j{c}_j$ survives integration, supplying one eigenvalue. Multiplying the surviving eigenvalues gives $\det M$, with an overall factor of $i$ per pair. The functional version is defined with the same regulator as the rest of the theory.

Combining these steps gives the gauge-fixed Yang–Mills LagrangianThis is the gauge-fixed Yang-Mills Lagrangian in a covariant gauge. Integrating its ghost term by parts givesThe second term is the ghost-gluon vertex in the antighost convention chosen here. The Faddeev-Popov ghosts are anticommuting Lorentz scalars introduced to represent a determinant; they are not extra asymptotic particles. A closed ghost loop has a minus sign from their Grassmann parity. In a Non-Abelian gauge theory the determinant depends on $A$, so these loops are essential. In an Abelian gauge theory $D_{\mu }$ in the adjoint sector reduces to ${\partial }_{\mu }$, the determinant is field independent, and its ghost sector decouples.

The quadratic gauge field kernel at non-null momentum isDefine the mixed-index longitudinal projector of a vector field $P_L{}^{\mu }{}_{\nu }=k^{\mu }{k}_{\nu }/k^2$ and transverse projector of a vector field $P_T=1-P_L$. The kernel is $-k^2(P_T+{\xi }^{-1}{P}_L)$ and hence has inverse $-k^{-2}(P_T+\xi P_L)$. Multiplying by $i$ gives the gauge-boson propagator, off its poles,The vacuum Feynman i-epsilon prescription completes this off-pole formula; with the mostly-plus convention its physical transverse denominator is $k^2-i0$. The choice $\xi =1$ is Feynman gauge; $\xi \to 0$ is covariant Landau gauge. The invertible quadratic kernel supplies perturbation theory, while the ghost determinant ensures the gauge-orbit measure has been accounted for.

The consistency of the unphysical sector is organized by BRST symmetry. Introduce a bosonic Nakanishi-Lautrup field $B^a$ and a left-acting BRST differentialThe graded Leibniz rule and the Jacobi identity give $s^2=0$: for example $s^{2}A=D(sc)+g[Dc,c]=0$ since $D[c,c]=2[Dc,c]$, and $s^{2}c$ vanishes by the graded Jacobi identity. Nilpotence on $\bar{c}$ and $B$ is immediate. Using the odd antighost field, the same graded Leibniz rule givesEliminating $B^a$ by its algebraic field equation $B^a=-G^a/\xi$ recovers the displayed gauge-fixing and ghost terms. Thus they are BRST-exact together, and the total action preserves BRST symmetry. At the level of states, the BRST charge $Q$ implements physical states as the ghost-number-zero BRST cohomology $\ker Q/\mathrm{im}Q$. This prescription removes longitudinal, time-like and ghost excitations from physical amplitudes while allowing them on internal lines. For a BRST-closed observable, changing $\xi$ inserts $s({\bar{c}}^a{B}^a/2)$; its expectation vanishes by the BRST change-of-variables identity when the measure preserves the symmetry. This explains the independence of physical quantities from the arbitrary gauge parameter.

Finally, a local condition such as Lorenz gauge can have several solutions on the same gauge orbit, the Gribov ambiguity. The preceding Faddeev-Popov gauge-orbit identity and perturbative gauge fixing construction apply in a neighborhood where the linearized operator is invertible after residual modes are excluded. They resolve perturbative gauge overcounting and supply consistent covariant Feynman rules; they do not by themselves prove a globally unique nonperturbative gauge fixing.

Take metric signature $(+---)$ and the Fourier transform convention $e^{-ipx}$. Introduce an even source $J$ for the gauge field and odd sources $\eta ,\bar{\eta }$ for the Faddeev-Popov ghost field:At tree level the two-point functions come from the quadratic action. Integrating the gauge kinetic term by parts gives its momentum-space kernelIts inverse follows from transverse and longitudinal projectors: $K=-p^2(P_T+{\zeta }^{-1}{P}_L)$, so $K^{-1}=-p^{-2}(P_T+\zeta P_L)$. Completing the bosonic Gaussian square gives $Z_0[J]=\exp [-iJ{K}_F^{-1}J/2]$, with the Feynman i-epsilon prescription in the inverse. Two source derivatives and their insertion factors give the gauge-boson propagator $D_F=i{K}_F^{-1}$. HenceFor $\zeta =1$ this reduces to the Feynman-gauge adjoint propagator. The longitudinal double pole is understood with the compatible Feynman boundary prescription.

The two tree-level Feynman diagrams have t-channel W boson exchange. The antiquark process reverses the fermion-flow arrows on the lower line.

Each vertex contributes $g/(2\sqrt{2})$ multiplying a chiral current. At momentum transfer $|q^2|\ll M_W^2$, the gauge-boson propagator is approximated by its momentum-independent term $g_{\alpha \beta }/M_W^2$, up to the overall phase. The longitudinal term does not contribute to these conserved massless currents. Thus the coefficient is $g^2/(8M_W^2)=G_F/\sqrt{2}$, giving the Fermi interaction. For the quark and antiquark amplitudes, apart from irrelevant overall phases,andThe low-energy operator form is ${\mathcal{L}}_{\mathrm{eff}}=-(G_F/\sqrt{2})(\bar{e}{\gamma }^{\alpha }(1-{\gamma }^5){\nu }_e)(\bar{u}{\gamma }_{\alpha }(1-{\gamma }^5)d)+\text{h.c.}$; its external spinor matrix elements are the displayed products.

For an unpolarized initial parton, average over its two spin states. There is no factor $1/2$ for the incoming active neutrino, which has one allowed helicity. The initial color average cancels the final color sum, so there is no surviving color multiplicity in this parton cross section. DefineThe symmetric and antisymmetric parts have zero cross contraction. Their separate contractions are $128[(a\cdot c)(b\cdot d)+(a\cdot d)(b\cdot c)]$ and $128[(a\cdot c)(b\cdot d)-(a\cdot d)(b\cdot c)]$. Thus the given Levi-Civita symbol identity yieldsIncluding the coefficient $G_F^2/2$ and the initial-parton spin average givesMassless kinematics and $q=p-p^{\prime }=k^{\prime }-k$ imply $p\cdot k=p^{\prime }\cdot k^{\prime }=s/2$. From $y_q=k\cdot q/(k\cdot p)$, $p^{\prime }\cdot k=p\cdot k^{\prime }=s(1-y_q)/2$. ThereforeTo reduce the supplied phase-space formula, use the massless center-of-mass flux $2s$ and $d{\Phi }_2=d\Omega /(32{\pi }^2)$. This gives $d\sigma /d\Omega =\overline{|\mathcal{M}{|}^2}/(64{\pi }^{2}s)$. In that frame $y_q=(1-\cos \vartheta )/2$, so $d\Omega /d{y}_q=4\pi$. HenceThe antiquark channel's angular factor is a consequence of the chiral current and the reversed lower-line spinor order, rather than a color factor.

The quadratic gauge-field action in momentum space isIn terms of the transverse projector of a vector field and longitudinal projector of a vector field,the operator is $k^2(P_T+{\xi }^{-1}{P}_L)$. Its inverse, the gauge-boson propagator, isTherefore $X=1$ and $Y=\xi -1$.