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Gauge invariance and power counting permit the same local operators as in the original renormalizable action. With renormalized fields, a convenient counterterm parametrization isHere $D_{\mu }={\partial }_{\mu }-i{\mu }^{ϵ/2}e{A}_{\mu }$. Expanding the first term supplies kinetic, scalar-scalar-photon, and seagull counterterms with the relative coefficients required by the Ward identity. Equivalently introduce separate vertex constants first and then impose that identity. For ${\varphi }_0=Z_{\varphi }^{1/2}\varphi$, $A_0=Z_A^{1/2}A$, and $e_0={\mu }^{ϵ/2}{Z}_{e}e$, it gives $Z_e{Z}_A^{1/2}=1$. Thus charge renormalization can be represented through the photon field factor, rather than adding an independent gauge-noninvariant interaction.

The mass coefficient defined here is $\delta M^2=Z_{\varphi }{m}_0^2-m^2$; it must be distinguished from $m_0^2-m^2$. Gauge fixing may require the corresponding gauge-parameter redefinition, and vacuum diagrams can require an additive vacuum-energy counterterm. No photon mass counterterm is allowed by the unbroken gauge symmetry. The quartic scalar counterterm remains necessary for the full theory even at a scale where the renormalized $\lambda$ is zero, because gauge interactions generate scalar four-point divergences.

First remove the projector denominator explicitly:Let $A=(l+k{)}^2+m^2$ and $B=l^2$. The Feynman parameter for a simple and a squared propagator follows by differentiating the elementary two-denominator Feynman formula with respect to $B$ givesShift $q=l+xk$. Then $xA+(1-x)B=q^2+D$, with $D=x(1-x)k^2+x{m}^2$. The numerator is unchanged by this parallel shift:Therefore the correctly combined integral isThe printed weight $x/2$ is not compatible with this denominator and prefactor. Exchanging $x$ with $1-x$ changes the mass term in $D$ as well and cannot supply the missing normalization. A concrete counterexample is $m=0$ with nonzero spacelike external momentum and $2<d<4$: after Wick rotation, the remaining integral is nonzero and symmetric under $x\leftrightarrow 1-x$. The source's $4e^2\int (x/2)dx$ expression is then exactly one quarter of the correct expression. Thus the claimed equality cannot be imposed when computing the pole.

Lorentz-covariant tensor integration replaces $q_{\mu }{q}_{\nu }$ by $g_{\mu \nu }{q}^2/d$. The numerator consequently contributes $(d-1)k^2{q}^2/d$. The supplied master integral with powers $a=1,b=3$ givesInserting the corrected parameter integral givesThe pole uses $\Gamma (ϵ/2)=2/ϵ+O(1)$, $D^{-ϵ/2}=1+O(ϵ)$ and ${\int }_0^1(1-x)dx=1/2$. ThusThere is no momentum-independent pole at $\lambda =0$ in this transverse gauge after removing the photon tadpole. The two-point counterterm insertion is $-i(\delta Z_{\varphi }{k}^2+\delta M^2)$, so minimal subtraction requiresThese coefficients cancel the actual one-loop insertion derived from the action. Applying the misprinted weight in part e would give a wave-function pole four times too small. In a $d=4-2\epsilon$ convention the same answer is $\delta Z_{\varphi }=3e^2/(16{\pi }^2\epsilon )$. Finally, $\delta M^2=0$ does not mean that the bare mass parameter is unchanged: $Z_{\varphi }{m}_0^2=m^2$ gives $m_0^2-m^2=-m^2\delta Z_{\varphi }+O(e^4)$. Photon, charge and quartic counterterms require their respective Green functions and cannot be determined from this scalar two-point function alone.

Continue with $\eta =-g$ and dot products in that auxiliary metric. Odd gamma traces vanish. The two current tensors areContracting givesLet $u=2M^2-s-t$. The on-shell products are $p_1\cdot k_1=p_2\cdot k_2=(M^2-t)/2$, $p_1\cdot k_2=p_2\cdot k_1=(M^2-u)/2$, and $k_1\cdot k_2=s/2$. Substitution yieldsThe result is already averaged over incoming colours and spins and summed over outgoing ones. For $M\to 0$, it becomes $4g_s^4(t^2+u^2)/(9s^2)$.

An additional check uses the centre-of-mass angle and velocity ${\beta }_t=\sqrt{1-4M^2/s}$:It tends to $4g_s^4/9$ at threshold and is positive throughout the physical region. Keeping both printed minus signs would instead give $4g_s^4[(M^2-t{)}^2+(M^2-u{)}^2-2M^{2}s]/(9s^2)$, displaying the incorrect mass contribution explicitly.

For the ghost coupling in a linear Cartesian Abelian Higgs gauge, the common factor in the scalar parametrization cancels from the gauge transformation. Separating real and imaginary parts givesConsequentlyFor the variables and gauge function actually printed, the ghost mass and Higgs-ghost vertex are thereforeFor the usual nonnegative gauge parameter, $m_c=\sqrt{\xi }g|v|$. The momentum-space ghost propagator is $-i/(k^2+m_c^2-i0)$ in this kinetic convention. There is no direct ghost-photon vertex or ghost-$b$ vertex. At $\xi =0$ the ghost becomes massless and decouples.

The scalar prefactor $1/4$ is incompatible with the stated canonical broken-vacuum interpretation. At $h=b=0$ it gives $|\varphi {|}^2=v^2/16$, whereas the potential is minimized at $|\varphi {|}^2=v^2/2$. In fact the printed parametrization givesIts $h,b$ kinetic terms have coefficient $-1/16$, not $-1/2$, and its vector mass parameter is $g^2{v}^2/8$. The given gauge function consequently does not cancel the vector-scalar mixing in those literal variables. The ghost operator computed above nevertheless follows algebraically from that gauge function without changing the printed fields.

With a canonical Abelian Higgs shift $\varphi =(v+h+ib)/\sqrt{2}$, the expansion is about a true minimum, the vector mass is $m_A=g|v|$, the gauge fixing cancels the mixing, and the same ghost formulas acquire the usual interpretation $m_c^2=\xi m_A^2$. For the literal $1/4$ parametrization, a canonical fluctuation is $h_{\mathrm{can}}=h/\sqrt{8}$, so its ghost vertex would instead be $-i\sqrt{8}\xi g^{2}v$; the non-vacuum background problem still remains. These two interpretations must not be silently mixed.

Since the hypercharge is zero, electric neutrality means $T^{3}v=0$. Choose the real-triplet vacuum $v_i=v{\delta }_{i3}$, with $v^2=m^2/(4\lambda )$. With canonical real-scalar kinetic density $-\frac{1}{2}(D_{\mu }\Phi {)}^T{D}^{\mu }\Phi$, insert this vacuum into the covariant derivative. The hypercharge field $B_{\mu }$ has no coupling to this scalar, while only $T^{1}v,T^{2}v$ are nonzero. ThusThe charged combinations $W^{\pm }=(W^1\mp i{W}^2)/\sqrt{2}$ haveBoth $T^3$ and the independent hypercharge generator annihilate the vacuum. The connected unbroken subgroup is thereforeThe electromagnetic generator $Q=T^3+Y/2$ belongs to it, but it is only one linear combination of two surviving neutral generators. This model has two massless neutral gauge fields, rather than the single massless photon and massive neutral vector of the usual scalar-doublet electroweak model. A neutral-field rotation does not create a massive eigenvalue from this zero block.

The two transverse triplet components are eaten, leaving one real, electrically neutral radial scalar $H={\varphi }_3-v$. The radial eigenvalue of the Hessian found in part i isIndeed the radial potential is a constant plus $4\lambda v^2{H}^2+4\lambda v{H}^3+\lambda H^4$, fixing the factor of two between its quadratic coefficient and its mass squared. The physical degree-of-freedom count is preserved: three real scalars plus four massless vectors give $3+4\times 2=11$, while one scalar, two massive vectors and two massless vectors give $1+2\times 3+2\times 2=11$.

The scalar kinetic and potential terms depend only on the rotation-invariant combinations ${\partial }_{\mu }{\varphi }_1{\partial }^{\mu }{\varphi }_1+{\partial }_{\mu }{\varphi }_2{\partial }^{\mu }{\varphi }_2$ and ${\varphi }_1^2+{\varphi }_2^2$. They are unchanged by the constant SO(2) rotation. The Yukawa term is invariant by ii. Finally, anticommutation gives $U{\gamma }^{\mu }={\gamma }^{\mu }{U}^{-1}$, soNo derivative of $U$ appears because the parameter is global. Thus the entire classical Lagrangian density has the combined chiral symmetry.

An explicit Dirac mass term instead changes asIt is not invariant for arbitrary $\alpha$ unless $m=0$. For the concrete choice $\alpha =\pi$, the original mass term reverses sign. The invariant Yukawa term can nevertheless generate a mass after a vacuum is selected, because the scalar field transforms along with the fermions; this is spontaneous rather than explicit breaking.

The scalar potential isSince ${\mu }^2,\lambda >0$, all minima form the circle ${\varphi }_1^2+{\varphi }_2^2=v^2$, with $v=\mu /\sqrt{\lambda }>0$. Choose the representative vacuum $⟨{\varphi }_1⟩=v$, $⟨{\varphi }_2⟩=0$, and write ${\varphi }_1=v+h$, ${\varphi }_2=\pi$. A nontrivial scalar rotation moves this vacuum to another minimum, so the continuous symmetry undergoes spontaneous symmetry breaking. The invariant action and the selected ground state have different stabilizers; the circle of equivalent minima is not evidence that the action lost its symmetry.

The Yukawa terms becomeComparison with the free Dirac equation givesFor the usual choice $g>0$, this is simply $gv$. The absolute value matters if the real coupling is negative: the dispersion depends on $m_{\psi }^2$, and a field redefinition can change the mass coefficient's sign. There is no explicit symmetry-breaking fermion mass parameter in the original theory.

The scalar symmetry is broken completely. Because the spinor transformation uses a half angle, $\alpha =2\pi$ leaves the scalars fixed while sending $\psi \to -\psi$; this discrete fermion parity is unbroken in the full combined transformation group. It does not remove the broken continuous generator or the associated Goldstone boson.

At a stationary symmetry-breaking vacuum of a gauge-invariant scalar potential, the angular field has zero potential curvature. Its free kinetic term is $(\partial b{)}^2/2$, while the displayed gauge-fixing term adds $-\xi g^2{v}^2{b}^2/2$. HenceThis is the would-be Goldstone boson, whose gauge-dependent pole accompanies the longitudinal pole in i; it is not an additional physical scalar particle of that mass. For $M_A\ne 0$, fixed finite $p$ and $\xi \to +\infty$,The angular field becomes infinitely heavy in this gauge and is absent from the unitary gauge propagating description. The qualification about fixed momentum matters: loop integrations probe arbitrarily large $p$, so one cannot automatically interchange the $\xi \to \infty$ limit with ultraviolet integration. If $gv=0$, there is no massive Higgs phase and the asserted heavy-field limit does not follow. A separate radial Higgs boson may remain; its mass depends on the unspecified potential and is not determined by this gauge-fixing term.

The quadratic part of the covariant-derivative term contains $g^2{v}^2{A}_{\mu }{A}^{\mu }/2-gv{A}_{\mu }{\partial }^{\mu }b$. Its mixing term cancels the $+gv{A}_{\mu }{\partial }^{\mu }b$ in the stated gauge fixing term. Thus the vector mass and diagonal quadratic action areFor nonzero $p^2$ introduce the transverse and longitudinal transverse and longitudinal momentum projectors $P_T^{\mu \nu }=g^{\mu \nu }-p^{\mu }{p}^{\nu }/p^2$ and $P_L^{\mu \nu }=p^{\mu }{p}^{\nu }/p^2$. After integration by parts, the inverse quadratic kernel isTheir orthogonality makes the inversion immediate. The gauge-boson propagator isEquivalently,The projector formula explains the physical massive pole and the gauge-dependent longitudinal pole. The apparent $1/p^2$ in the projectors cancels in the equivalent expression; it is not an extra massless physical state. At $\xi =1$ the numerator reduces to $g^{\mu \nu }$. At fixed momentum the $\xi \to \infty$ limit gives the unitary gauge propagator $-i(g^{\mu \nu }-p^{\mu }{p}^{\nu }/M_A^2)/(p^2-M_A^2+i0)$. The relation between this limit and ultraviolet power counting is not uniform in momentum.^[ref]

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).

For the one-loop renormalization-group beta function, write $t=\log \mu$. Separating variables givesThis expression resums the leading logarithms of the one-loop running coupling, while omitting higher-order terms in the renormalization-group beta function. Assume a positive initial coupling.

For $b_0<0$, increasing the renormalization scale decreases the running coupling; the zero-coupling renormalization-group fixed point is attractive in the ultraviolet. This is asymptotic freedom. Toward lower scales the one-loop expression has a formal pole at ${\mu }_i\exp [1/(b_0{\alpha }_i)]<{\mu }_i$. The strong-coupling neighborhood of that pole lies outside perturbative control. QCD provides the main example; the minimal Standard Model $SU(2)$ coupling is also asymptotically free at one loop.

For $b_0>0$, the running coupling grows with the renormalization scale and formally diverges at the ultraviolet Landau pole ${\mu }_i\exp [1/(b_0{\alpha }_i)]>{\mu }_i$. Toward the infrared it decreases to zero within this approximation. The Standard Model $U(1)$ weak hypercharge coupling has this sign, as does QED in a fixed charged-particle regime. A perturbative Landau pole marks failure of the approximation, rather than proving that the exact theory has that divergence.

For $b_0=0$, the running coupling is constant at this order; higher-loop terms decide whether it runs. Vanishing of the leading coefficient alone does not establish exact scale invariance.

For a concrete common convention, define ${\alpha }_i=g_i^2/(4\pi )$ with $g_Y$ in its usual, non-unified weak hypercharge normalization. In the minimal Standard Model with three generations and one Higgs doublet, above its particle thresholds,Thus the question's $b_0$ is $b_i/(2\pi )$, with the sign regimes just described. Crossing a mass threshold changes the active field content and requires the appropriate piecewise renormalization-group beta function.

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.

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.

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.

After a Wick rotation, the term in question has radial behaviorThe radial integrand is proportional to $k_E$ at the origin and to $k_E^{-3}$ at infinity. Thus it has neither an infrared divergence nor an ultraviolet divergence. In fact the area of the unit three-sphere givesThis also follows from the Feynman parameter identitywith $A=k^2+i0$, $B=k^2-m^2+i0$. Although the parameterized denominator becomes massless at $x=1$, its numerator $1-x$ cancels the apparent endpoint singularity. Since $\beta \to -2$, multiplication by $\beta$ preserves finiteness. No artificial photon mass is needed.

Keep $m>0$ and set $p=p^{\prime }=0$, which is an off shell choice. The numerator isTerms odd in $k$ integrate to zero. Lorentz covariance and dimensional regularization give, inside this integral,Applying the gamma matrices identity from part (ii) once more yieldsConsequentlyThis massive zero-momentum QED vertex integral uses an integrated tensor identity for the replacement of $N^{\mu }$, not an equality for each fixed $k$. The physical photon remains massless. The retained fermion mass supplies a scale and makes this zero-external-four-momentum calculation suitable for extracting the ultraviolet pole without a scaleless integral.

Collect the two Dirac fields into $\Psi =({\psi }_1,{\psi }_2{)}^T$, with a common nonzero mass $m$. ThenAmong complex-linear rotations of the Dirac fields, the internal symmetry of a classical field theory is $\Psi \mapsto U\Psi$, $U\in U(2)$, with $\bar{\Psi }\mapsto \bar{\Psi }{U}^{†}$. Unitarity preserves the kinetic term, and the mass matrix commutes with every $U$. This unitary flavour symmetry of degenerate Dirac fields has group $(SU(2)\times U(1))/{\mathbb{Z}}_2$, where the quotient avoids counting the common central element twice.

Choose the four Hermitian matrices $T^0=1_2/2$, $T^a={\sigma }^a/2$ for $a=1,2,3$, with ${\sigma }^a$ the Pauli matrices. Promoting the four parameters to functions gives $\delta S=-\int d^{4}x({\partial }_{\mu }{\alpha }^A)\bar{\Psi }{\gamma }^{\mu }{T}^A\Psi$. The four Noether currents are thereforeTheir divergences vanish by the equal-mass Dirac equations. The singlet is half the total Dirac current; the other three currents rotate the two species.

Free-theory qualification. If “global internal symmetry” allows mixing the Dirac fields with charge-conjugate fields, the completely free theory written here has a larger enhanced internal symmetry of free Dirac fields. Write ${\psi }_i=({\lambda }_{2i-1}+i{\lambda }_{2i})/\sqrt{2}$ using four Majorana fermions. Up to a total derivative,The full orthogonal group $O(4)$ acts on the four real species; its connected part is $SO(4)$. Its six Noether currents can be chosen as $J_{AB}^{\mu }=i{\bar{\lambda }}_A{\gamma }^{\mu }{\lambda }_B$ for $A<B$, up to the common sign convention. The antisymmetry of this Majorana fermion vector bilinear makes these the six independent rotations. A concrete rotation outside the stated complex-linear $U(2)$ is ${\lambda }_2\mapsto \cos \theta {\lambda }_2+\sin \theta {\lambda }_3$, ${\lambda }_3\mapsto -\sin \theta {\lambda }_2+\cos \theta {\lambda }_3$: it preserves the displayed Lagrangian density while mixing ${\psi }_i$ with their charge-conjugate fields. The usual $U(2)$ answer specifies the subgroup preserving the chosen common fermion number.

In the Weyl representation of the gamma matrices, write each Dirac field as ${\psi }_i=({\ell }_{i\alpha },r_i^{\dot{\alpha }}{)}^T$, where ${\ell }_i$ and $r_i$ have opposite chirality. With the Weyl sigma matrices ${\sigma }^{\mu }=(1,\mathbf{\sigma })$ and ${\bar{\sigma }}^{\mu }=(1,-\mathbf{\sigma })$,The Dirac mass term couples the two Weyl fields. Under $\ell \mapsto U_L\ell$, $r\mapsto U_{R}r$, its first term acquires $U_L^{†}{U}_R$, so a common nonzero mass requires $U_L=U_R$. At zero mass there is no such constraint:This is the chiral flavour symmetry of massless Dirac fields. Its eight Noether currents are $j_{L,A}^{\mu }={\ell }^{†}{\bar{\sigma }}^{\mu }{T}^A\ell$ and $j_{R,A}^{\mu }=r^{†}{\sigma }^{\mu }{T}^{A}r$. The sum $j_R+j_L$ is the vector current, and $j_R-j_L$ is the axial current, for each flavour generator. This statement concerns the free theory in flat spacetime; an interacting theory can have a chiral anomaly.

For completeness, allowing the extra transformations described in part e gives four left-handed Weyl fields, $q=({\ell }_1,{\ell }_2,r_1^c,r_2^c{)}^T$, where $r_i^c$ is the left-handed charge-conjugate field of $r_i$. After integration by parts their kinetic term is $i{q}^{†}{\bar{\sigma }}^{\mu }{\partial }_{\mu }q$, whose maximal linear unitary group is $U(4)$. The symmetric mass matrix for equal Dirac masses iswith the antisymmetric spinor contraction implicit. A unitary change of basis makes $M=m{1}_4$; preserving both the kinetic term and $M$ requires $V^{†}V=V^{T}V=1_4$, hence real $V\in O(4)$. Thus the unrestricted free-theory enhancement is $U(4)$ at zero mass and $O(4)$ at equal nonzero mass; the familiar $U(2{)}_L\times U(2{)}_R$ and diagonal $U(2)$ are their common-fermion number-preserving subgroups.