A gauge-invariant path integral counts all representatives of each gauge orbit. In a perturbative region with a regular gauge fixing slice, insert the Faddeev-Popov gauge-orbit identityThe Faddeev-Popov determinant is the functional Jacobian converting integration over the gauge transformation parameter into integration over the condition. The redundant gauge group volume then factors out of normalized correlation functions.

Choose $A_{\mu }^{\omega }=A_{\mu }+{\partial }_{\mu }\omega$. Its infinitesimal variation givesThere is no derivative of $A$ in this expression: only $A_{\mu }$ is varied in $A^2$. For $\lambda =0$, the determinant is field independent and can be absorbed into normalization. For $\lambda \ne 0$, it is a functional of the gauge field, so dropping it changes the path integral, even though the gauge group is Abelian. Constant residual gauge transformations are removed by boundary conditions or by dividing out their volume, as usual in perturbative gauge fixing.

For fundamental-doublet breaking of SU(2), assume $\lambda >0$ and $v>0$. The minima of the scalar potential satisfy ${\varphi }^{†}\varphi =v^2/2$, giving a three-sphere vacuum manifold. A chosen vacuum expectation value is ${\varphi }_0=(0,v{)}^T/\sqrt{2}$. No nonzero generator $a_a{\tau }^a$ annihilates ${\varphi }_0$: its squared norm is $v^2(a_1^2+a_2^2+a_3^2)/8$. At the group level, an $SU(2)$ matrix fixing the nonzero vector ${\varphi }_0$ must be the identity. Thus all three generators of the SU(2) gauge theory are broken in the usual perturbative Higgs mechanism description.

Make the radial and angular fields explicit:A local gauge transformation removes the three angular fields ${\pi }^a$. In unitary gauge, the Pauli matrices identity $\{{\tau }^a,{\tau }^b\}={\delta }^{ab}1/2$ givesThe mixed derivative term cancels between a term and its complex conjugate. The scalar potential becomesConsequently the complete physical Lagrangian density isReading the quadratic terms with their canonical factors $1/2$ givesAll physical scalar-vector and scalar self-interactions are displayed in the second line of $\mathcal{L}$. In particular the new cubic couplings proportional to $v$ are $h{A}^a{A}^a$ and $h^3$. Their Feynman vertices, together with the quartic ones, areThe three-vector and four-vector interactions are already contained in the non-Abelian gauge field strength term. There are no further physical fields or interactions in this specified model. Before fixing unitary gauge, the quadratic angular term is $\frac{1}{2}({\partial }_{\mu }{\pi }^a-m_A{A}_{\mu }^a{)}^2$. It exhibits three zero-curvature angular directions of the scalar potential, but also their mixing with the gauge fields. They supply the third polarization of each massive gauge boson, rather than massless physical particles.

The degrees of freedom count is $3\cdot 2+4=10$ before breaking and $3\cdot 3+1=10$ afterwards. If instead the symmetry were global, as in the decoupled $g=0$ scalar theory, the three angular fields would be physical massless Goldstone bosons. This is consistent with Goldstone's theorem. The language of a broken local symmetry refers to this gauge-fixed expansion; a gauge-dependent vacuum expectation value is not itself a gauge-invariant observable.

Goldstone's theorem states that in a relativistic quantum field theory, spontaneous breaking of an exact continuous global internal symmetry produces a massless Goldstone boson for each independent broken generator. More precisely, a conserved global current whose charge acts nontrivially on the vacuum expectation value has a massless pole in an appropriate current-field correlation function. The usual counting assumes a Lorentz invariant vacuum state and the standard locality and positivity hypotheses.

A local gauge symmetry is a gauge redundancy, so it does not by itself imply massless physical Goldstone bosons. In the Higgs mechanism, the corresponding scalar directions provide the longitudinal polarizations of massive gauge bosons.

For massless quark Compton scattering, the color average contributes one, since $(1/N_c){\sum }_{i,j}|{\delta }_{ji}{|}^2=1$. The spin average and incoming photon polarization average supply $1/(2\cdot 2)=1/4$. The Ward identities permit use of $-g_{\mu \nu }$ for each photon polarization completeness relation, and the massless fermion spin sum is ${\sum }_r{u}_r(p){\bar{u}}_r(p)=/p$.

LetReversing the gamma matrices order gives its Dirac-conjugate matrix ${\bar{A}}^{\mu \nu }$. The squared scattering amplitude isFor clarity evaluate the four traces before dividing by their denominators. The Mandelstam variables obey $s+t+u=0$, with $t=(p_1-p_3{)}^2$, $p_1\cdot p_2=s/2$, $p_1\cdot p_4=u(-1/2)$ and $p_1\cdot p_3=-t/2$. For the squared $s$-channel contribution, the gamma matrix trace identities implyHere $/q_s/p_1/q_s=2(q_s\cdot p_1)/q_s-q_s^2/p_1=s/p_2$ and $p_3\cdot p_2=-u/2$. For the squared $u$-channel contribution, $/q_u/p_1/q_u=-u/p_4$ givesOne interference trace reduces tosince $q_s\cdot q_u=(s+t+u)/2=0$. The other interference trace is its complex conjugate and also vanishes. ThereforeEquivalently $C=512{\pi }^2{\alpha }_{\mathrm{em}}^2/81$ in terms of the fine-structure constant. In the physical noncollinear region $s>0$, $u<0$, so the result is positive. The singular massless endpoint $u=0$ reflects a collinear divergence; it is not a negative squared scattering amplitude.

Put $Q_u=2/3$, $s=q_s^2$, $u=q_u^2$, and suppress the common colour factor ${\delta }_{ji}$. Define $i\mathcal{M}$ by the product of Feynman rules. With massless external Dirac spinors the two graphs giveThe massless Dirac equation gives $/p_{1}u(p_1)=0$ and $\bar{u}(p_3)/p_3=0$. ThusBy four-momentum conservation, $p_2=p_3-q_u$, soThe two terms cancel:This is the Ward identity for the incoming photon; replacing its photon polarization vector by its four-momentum produces zero. The outgoing photon obeys the analogous identity. Each exchange graph separately fails this test, so the cancellation requires their sum. Consequently the physical scattering amplitude is invariant under ${\epsilon }^{\mu }(p_2)\mapsto {\epsilon }^{\mu }(p_2)+c{p}_2^{\mu }$.

The two tree-level Feynman diagrams are the $s$-channel and $u$-channel quark exchange graphs, drawn in the middle row of the figure. Along the directed quark line the incoming photon is absorbed before the outgoing photon is emitted in the first graph; those two insertions are reversed in the second. Their internal four-momenta are $q_s=p_1+p_2$ and $q_u=p_1-p_4$, respectively. There is no three-photon Feynman vertex in ordinary QED, and no two-photon-two-quark contact interaction in the Dirac field Lagrangian density.

The massive Dirac propagator isUse the gamma matrix trace identities $\mathrm{tr}{\gamma }^{\mu }=0$ and $\mathrm{tr}({\gamma }^{\mu }{\gamma }^{\rho })=4g^{\mu \rho }$. They giveThe integrand is odd under $k\mapsto -k$, and dimensional regularization preserves that symmetry. The same vanishing holds for $S=i{S}_F$. One must not assert that $S(0)$ itself is zero: its term proportional to $m1$ generally is nonzero and divergent, but its trace against a single gamma matrix is zero. This one-photon fermion loop is also the simplest instance of Furry's theorem.

First fix a convention that is implicit in the source. For the stated free generating functionalthe time-ordered Wick contractions are $S_F=⟨T\psi \bar{\psi }{⟩}_0=-iS$ and ${\Delta }_F=⟨TAA{⟩}_0=-i\Delta$. These follow directly by the source differentiation rules in part (i). Thus the kernels called propagators in the question are distinguished here from the contractions, which already include their factors of $i$.

At first order in the QED interaction, write $a$ for its integration point. The normalized three-point correlation function isThe zeroth-order term vanishes because the free photon one-point correlation function is zero. There are two nonzero Wick contractions of the four fields:The first term pairs the two external fermions directly and closes the two fields at the interaction into a fermion loop; its minus sign is the closed fermion loop sign. The second term joins the external fermions through the interaction, giving the connected contribution. This is also obtained by differentiating the quartic term $(i\bar{\eta }S\eta {)}^2/2$ in the free generating functional; the two choices of bilinear cancel its factor $1/2$.

Insert $S_F=-iS$ and ${\Delta }_F=-i\Delta$. Since $(-ie)(-i{)}^3=e$, the result is exactlyThe trace term is a disconnected product of a photon tadpole diagram with the free external fermion propagator; the other term is the connected three-point correlation function. In this QED calculation the disconnected term vanishes, as established next. With a convention using $S_F,{\Delta }_F$ themselves as the symbols for the propagators, the equivalent formula retains the prefactor $-ie$ and the connected-minus-trace ordering above; mixing these two conventions would give an erroneous overall phase.

Use the path integral with source termThe sources $\eta ,\bar{\eta }$ are Grassmann variables; all derivatives below are left functional derivatives. Because $\eta$ stands to the right of $\bar{\psi }$, differentiating with respect to it introduces a minus sign. The insertion rules are thereforeFor example ${\delta }_L(\bar{\psi }\eta )/\delta \eta =-\bar{\psi }$. The order of the odd functional derivatives matters: the rightmost derivative acts first, and the displayed product inserts ${\bar{\psi }}_{\alpha }{\psi }_{\beta }$ in that order.

Expand $\exp [-ie\int \bar{\psi }{\gamma }^{\mu }\psi A_{\mu }]$ in its power series, replace each field by its source derivative, and pull the derivatives outside the free path integral. Re-exponentiating givesThis identity concerns the unnormalized generating functional. Normalized correlation functions divide by its zero-source value so that vacuum contributions cancel.

Write $H={\varphi }_1$ and $P={\varphi }_2$, both Hermitian matrices. Through quadratic order around the chosen vacuum expectation value,The trace of a commutator vanishes. The needed matrix traces are thereforeUsing $m^2=-(\alpha +N\beta )C^2$, all linear terms cancel. The quadratic scalar potential reduces towith no $P^2$ term. The kinetic term issince its mixed term is the trace of a commutator. A Hermitian matrix has $N^2$ independent real components, so $P$ describes exactly $N^2$ massless scalar modes.

To determine the other masses, take $T_0=1/\sqrt{N}$ and traceless Hermitian $T_A$ with $\mathrm{tr}(T_A{T}_B)={\delta }_{AB}$, and expand $H=h_0{T}_0+{\sum }_{A=1}^{N^2-1}{h}_A{T}_A$. These are canonically normalized real scalar fields. The scalar mass matrix is diagonal in this basis:Thus $H$ contains one massive singlet and a degenerate multiplet of $N^2-1$ massive fields in the Adjoint representation of the unbroken diagonal $SU(N)$. Both squared masses are strictly positive for the specified parameters. The $P$ fields are the $N^2$ Goldstone bosons associated with the broken generators: $N^2-1$ traceless relative left-right rotations and one overall phase rotation.

Goldstone's theorem guarantees those massless directions. It does not alone forbid additional accidental massless modes; the positive scalar mass matrix computed here supplies the further conclusion that every $H$ component is massive. The fields describe a global symmetry, so these Goldstone bosons are physical rather than being absorbed into massive gauge bosons.

It is useful first to establish the global minimum, rather than only a stationary point. The positive semidefinite matrix $X={\Phi }^{†}\Phi$ has nonnegative eigenvalues $x_1,\dots ,x_N$. For fixed $r={\sum }_i{x}_i$, the Cauchy-Schwarz inequality gives ${\sum }_i{x}_i^2\ge r^2/N$, with equality exactly when all $x_i$ are equal. Since $\alpha >0$,For $m^2<0$ and $\beta >0$, this lower bound has its minimum atThe bound is attained, so every minimum has ${\Phi }^{†}\Phi =C^{2}1$, or $\Phi =CW$ with $W$ a unitary matrix. A global transformation removes $W$ by decomposing it into an overall phase and an $SU(N)$ matrix. Thus ${\Phi }_0=C1$ is a valid representative of the vacuum manifold.

Preservation of this chosen vacuum expectation value requiresInfinitesimally, with traceless Hermitian generators $L,R$, this reads $\theta 1+L-R=0$. Taking the trace forces $\theta =0$, and then $L=R$. The surviving continuous transformations are consequentlyAll other independent continuous directions change the chosen vacuum expectation value. Thus the number of broken generators isGlobally, taking the determinant of the stabilizer equation permits the discrete phases $e^{iN\theta }=1$, paired with $U_L=e^{-i\theta }{U}_R$. They induce the same conjugation action on $\Phi$. The faithful unbroken conjugation group is $SU(N)/{\mathbb{Z}}_N$; for example, at $N=2$, $U=-1$ acts exactly like $U=1$. Thus the question's description by $SU(N)$ is correct as a parametrization and at the Lie algebra level, with this central redundancy understood. It leaves the number of broken generators unchanged.

Let $X={\Phi }^{†}\Phi$. Under the global transformation of the chiral matrix scalar model,The phase and $U_L^{†}{U}_L$ cancel, and $U_R^{†}{U}_R=1$. The cyclic property of the trace impliesTherefore each of the three terms in the scalar potential is invariant separately. Since $\theta ,U_L,U_R$ are constant, ${\partial }_{\mu }{\Phi }^{\prime }=e^{i\theta }{U}_L({\partial }_{\mu }\Phi )U_R^{†}$, giving invariance of the kinetic matrix trace by the same argument. Henceis a global symmetry group of the Lagrangian density. Its action has a finite central kernel, which does not affect its Lie algebra or the broken-generator counting below. For special parameter choices the symmetry can be larger; the displayed group is a symmetry for the stated potential without such extra assumptions.

Expand the matrix trace explicitly:Write ${\Phi }_{ij}=(x_{ij}+i{y}_{ij})/\sqrt{2}$ and collect the $2N^2$ real fields into a vector $X$. ThenBoth quadratic forms are preserved by every constant orthogonal matrix $R$, acting as $X\mapsto RX$. Thus the full linear internal symmetry contains $O(2N^2)$, with connected special orthogonal group $SO(2N^2)$ as requested. The enlargement beyond unitary transformations is possible because the real and imaginary components have identical kinetic and mass terms and no interactions distinguishing them.

For a Hermitian matrix $\Phi$, there are $N$ real diagonal entries and $N(N-1)/2$ independent complex off-diagonal entries, hence $N^2$ real fields. Choose an orthonormal basis $T_A$ of Hermitian matrices for the trace inner product, $\mathrm{tr}(T_A{T}_B)={\delta }_{AB}$, and put $\Phi ={\sum }_A{X}_A{T}_A/\sqrt{2}$. The same calculation giveswhere these are the connected rotation groups; their full orthogonal extensions are $O(2N^2)$ and $O(N^2)$. A purely quadratic theory with $m^2<0$ would have an unstable scalar potential, but this algebraic symmetry statement still holds.

For Møller scattering, both the $t$-channel and $u$-channel photon exchange tree-level Feynman diagrams contribute, as shown in the two remaining lower panels. Exchanging the identical final electrons reverses the sign of the scattering amplitude, so the fermionic sign fixesThe two direct squares follow from the preceding distinct-lepton calculation:The interference cannot be discarded because the final particles are identical. Summing their spins joins all four external Dirac spinors into a single trace:The result from part (i) was used with relabelled contracted indices. Its negative sign combines with the explicit subtraction of the two exchange scattering amplitudes, so the interference contribution to the averaged square is positive:ThusThis is symmetric under $t\leftrightarrow u$, as required by the identical final electrons. There is no additional $1/2!$ in the squared scattering amplitude. Such a factor is used when integrating the cross section over a final-state region that counts both orderings of the identical pair; alternatively one integrates over one representative half of that region.

Label the four-momenta as $e^{-}(p_1){\mu }^{-}(p_2)\to e^{-}(p_3){\mu }^{-}(p_4)$. There is one tree-level Feynman diagram, the $t$-channel photon exchange shown in the lower-left panel. The two lepton species are distinct, so there is no exchanged-final-lepton diagram. With $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$ and $u=(p_1-p_4{)}^2$, the Feynman rules giveThe massless fermion spin sum is ${\sum }_r{u}_r(p){\bar{u}}_r(p)=/p$. Averaging the two initial spins of each lepton supplies $1/4$, soThe gamma matrix trace identities imply, for arbitrary null external four-momenta,To see the contraction, each trace is $4(a^{\mu }{b}^{\nu }+a^{\nu }{b}^{\mu }-g^{\mu \nu }a\cdot b)$; the terms proportional to $(a\cdot b)(c\cdot d)$ cancel in four dimensions, leaving the displayed two products. Here $p_1\cdot p_2=p_3\cdot p_4=s/2$ and $p_1\cdot p_4=p_2\cdot p_3=-u/2$. ThereforeThe requested proportionality constant is $2e^4=32{\pi }^2{\alpha }_{\mathrm{em}}^2$, where ${\alpha }_{\mathrm{em}}$ is the fine-structure constant. Both numerator terms are required by the spin sum; no final-state statistical factor enters a squared scattering amplitude for distinct leptons.

First derive the contracted gamma matrices identities from the Clifford algebra in four dimensions. Moving the outer gamma matrix through two factors givesMoving it through three factors, and using this two-factor result, givesConsequently ${\gamma }^{\nu }/b{\gamma }_{\mu }/c{\gamma }_{\nu }=-2/c{\gamma }_{\mu }/b$ and ${\gamma }_{\mu }/b/d{\gamma }^{\mu }=4(b\cdot d)1$. Apply these in succession:The final step uses $\mathrm{tr}({\gamma }^{\alpha }{\gamma }^{\beta })=4g^{\alpha \beta }$. This is a genuinely four-dimensional instance of the gamma matrix trace identities; using it unchanged inside a $D$-dimensional loop calculation would require separate justification.

Let $ϵ$ here be a constant infinitesimal transformation parameter, unrelated to the dimensional regularization parameter in question 1. The matter and Maxwell terms undergo an ordinary infinitesimal gauge transformation with $\alpha =ϵ\varphi$, and their variations cancel as before. The gauge fixing contribution isThe extra real scalar field kinetic term varies byAdding the two gives a total derivative:The last equality uses vanishing boundary contributions, for example sufficiently fast decay of the fields. No field equations were used. This gauge-scalar global symmetry of covariantly fixed QED leaves the action invariant even though the density need not be invariant pointwise; the scalar's variation supplies exactly the missing boundary-term cancellation.

For a real local parameter $\alpha (x)$, the Dirac adjoint varies as $\delta \bar{\psi }=ie\alpha \bar{\psi }$. The matter kinetic term varies byThe QED interaction varies by $-e({\partial }_{\mu }\alpha )\bar{\psi }{\gamma }^{\mu }\psi$, since the phases in $\bar{\psi }{\gamma }^{\mu }\psi$ cancel. These two variations cancel pointwise. The mass term is invariant for the same reason, and $\delta F_{\mu \nu }={\partial }_{\mu }{\partial }_{\nu }\alpha -{\partial }_{\nu }{\partial }_{\mu }\alpha =0$.

Only the gauge fixing term varies. Put $B={\partial }_{\mu }{A}^{\mu }$; then $\delta B={\partial }^2\alpha$, givingFor arbitrary local $\alpha$, the gauge-fixed action is therefore not invariant under this transformation of its dynamical fields. The restricted parameters satisfying ${\partial }^2\alpha =0$ give a residual invariance of the displayed density.

The Slavnov-Taylor identity implies transversality of the full perturbative gluon self-energy:The quark loop is separately transverse. This follows without evaluating its Feynman integral: contract its vector Feynman vertex with $k_{\mu }$ and use $/k=(/l+/k-m)-(/l-m)$ between its two Dirac propagators. One obtains a difference of two loop integrals that cancel after a shift of $l$ in dimensional regularization. The colour factor is proportional to $\mathrm{tr}(T^a{T}^b)$ and does not obstruct this argument.

The four-gluon tadpole diagram vanishes after removal of $m_g$. Thus the nontransverse parts of the two-three-gluon-vertex loop and the ghost loop must cancel. Contracting the preceding ghost-loop pole of the gluon self-energy givesFor QCD, set $C_A=3$. No evaluation of the remaining three-gluon Feynman diagram is required. Its uncontracted transverse part is not fixed by this identity.

There is one four-gluon tadpole diagram. Temporarily replace its internal gluon propagator by $-i{g}_{\rho \sigma }{\delta }^{cd}/(l^2-m_g^2+i0)$. Contract the supplied four-gluon Feynman vertex with $g_{\rho \sigma }{\delta }^{cd}$. Its first colour factor vanishes because $f^{cce}=0$. Each of the other two terms contributes $C_A{\delta }^{ab}(D-1)g^{\mu \nu }$, soInclude the symmetry factor $1/2$ and the factor $-i$ from the internal gluon propagator:The exact regulated tadpole integral isIn particular,The entire expression is proportional to $(m_g^2{)}^{1-ϵ/2}$ and vanishes as the auxiliary mass tends to zero near four dimensions. Equivalently, its expanded terms are proportional to $m_g^2/ϵ$, $m_g^2\log m_g^2$ and $m_g^2$, all of which disappear when the regulator is removed. HenceThis is the massless scaleless integral result obtained directly in dimensional regularization. The artificial mass is only an intermediate device for this graph; retaining it in the full theory would modify the gauge symmetry and cannot be used when applying the massless transversality identity in part (iv).