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For real scalar multiplets with antisymmetric generators $t_a$, transforming a source $J$ in the same representation leaves $J\cdot \varphi$ invariant. With an invariant functional measure, local changes of integration variables imply the Ward identity for the quantum effective action. In the convention $Z=e^{iW}$, $\phi =W_J$, $\Gamma =-W+\int J\phi$, it isThe Legendre transform gives ${\Gamma }_{\phi }=J$ and ${\Gamma }_A{|}_{\phi }=-W_A{|}_J$. Its classical limit is $\Gamma =-S_0$, not $S_0$, for the source term $+J\varphi$. Differentiating twice in $\phi$ produces two contact terms and the momentum-space identity $p_{1\mu }{\tau }_{a,ij}^{\mu }=i(t_a{)}_{ik}{\tau }_{kj}-{\tau }_{ik}i(t_a{)}_{kj}$ with the appropriate shifted two-point momenta. Changing the overall convention for $\Gamma$ changes both two-point and mixed vertices together; changing just one violates the identity. A gauge anomaly would invalidate the assumed invariant-measure argument.

A configuration-space path integral integrates histories of the coordinates with weight $e^{iS}$. Fixed endpoint conditions specify a transition amplitude. For a quadratic kinetic energy it follows by Gaussian momentum integration from the phase-space path integral. Time-slicing fixes the normalization, and a regulator gives meaning to the otherwise formal functional measure.

Write the Hamiltonian operator as $\hat{H}={\hat{p}}^2/(2m)+V(\hat{q})$, with $[\hat{q},\hat{p}]=i$, and define the quantum-mechanical propagator by $U(\beta ,\alpha ;T)=⟨\beta |e^{-iT\hat{H}}|\alpha ⟩$. Use $N$ intervals of length $\epsilon =T/N$, endpoint coordinates $q_0=\alpha ,q_N=\beta$, and the Trotter product formula. Inserting the completeness relations of the position eigenstates between the short-time factors givesNormalize momentum eigenstates by $⟨q|p⟩=e^{ipq}/\sqrt{2\pi }$. The Gaussian integral over the intermediate momentum isThe square-root branch is fixed by continuation from $\mathrm{Im}\epsilon <0$. Consequently the precise time-sliced configuration-space path integral isThis formula defines the notation $\int \mathcal{D}q{e}^{iS[q]}$; it is not an unnormalized infinite product of ordinary integrals. For real time take the boundary value from $T-i\delta$, $\delta >0$, or the corresponding oscillatory-distribution limit. For a self-adjoint, bounded-below Hamiltonian operator under the usual product-formula hypotheses, the operator limit supplies the definition even where the coordinate kernel is a distribution. The discrete exponent approaches the action with kinetic term $m{\dot{q}}^2/2$, and the prefactor is part of the functional measure. A different consistent slicing gives the same kernel for this separated kinetic-plus-potential Hamiltonian.

For the quantum harmonic oscillator, expand the action about a classical path $x$, writing $q=x+\xi$ with homogeneous Dirichlet boundary conditions on $\xi$. The only mixed term isThe boundary term vanishes and the Euler-Lagrange equation sets the remaining integrand to zero. Thus $S[x+\xi ]=S[x]+S[\xi ]$ exactly, rather than only to quadratic approximation. Translation of each integrated coordinate has unit Jacobian determinant; in the continuum measure convention of the question this is $\mathcal{D}q=\mathcal{D}\xi$. The classical-path factorization of a quadratic path integral follows:The fluctuation action, measure and boundary values depend on $T,m,\omega$ but not on $\alpha ,\beta$. For real $T$ with $\sin \omega T\ne 0$ the boundary-value classical path exists uniquely. At the conjugate times $\omega T=k\pi$, generic endpoints admit no such path and the kernel is a delta distribution; the formula is interpreted by the same regulated boundary limit, not as an ordinary finite prefactor times a nonexistent classical solution.

A partition function sums the Boltzmann weights of all states: $Z={\sum }_s{e}^{-{\beta }_{T}H(s)}$ for discrete states, or the corresponding integral for continuous fields. It normalizes equilibrium probabilities. Its physical Helmholtz free energy is $\mathcal{F}=-k_{B}T\log Z$, and its free-energy density is $\mathcal{F}/V$. With a momentum ultraviolet cutoff ${\Lambda }_0$, a scalar-field example isThe normalization of the functional measure and field-independent energy terms must be tracked when discussing absolute free energy.

Choose a smaller cutoff $\Lambda$ and split the field into retained modes ${\varphi }_{<}$ with $|q|<\Lambda$ and eliminated modes ${\varphi }_{>}$ with $\Lambda <|q|<{\Lambda }_0$. Define the Wilsonian coarse-grained statistical Hamiltonian bySubstitution into the partition function proves $Z=\int \mathcal{D}{\varphi }_{<}{e}^{-{\beta }_T{H}_{\Lambda }[{\varphi }_{<}]}$ for every intermediate cutoff. A source retained in this definition similarly preserves long-wavelength connected correlation functions and response. Thus the coefficients in a Hamiltonian density must run with $\Lambda$: the dependence induced by eliminated modes compensates the dependence of the remaining field integral. Holding the original bare coefficients fixed while changing the cutoff would generally change the system. Exact integration generates all interactions allowed by symmetry, including higher powers, derivative terms and an additive constant; a local derivative expansion or a truncation to a few couplings is an extra approximation.

Cutoff independence makes arbitrary intermediate coarse-graining scales unphysical. It allows calculations at the scale of the observable, avoids repeatedly resolving irrelevant short-distance details and exposes the renormalization-group fixed points responsible for universality. A microscopic cutoff, such as a Bravais lattice spacing, can still label a physical microscopic model; taking it to infinity as a continuum limit requires tuning appropriate bare couplings. This is distinct from lowering an intermediate Wilsonian cutoff in a fixed model.

To interpret the zero-cutoff limit, define $M=V^{-1}\int \varphi (x)d^{D}x$, the spatially uniform mode of the order parameter. Keep this mode fixed and integrate the others, rather than integrating over $M$ too. Write $\varphi =M+\eta$, with $\int \eta =0$, and define the constrained order-parameter free energy byIn a finite periodic box, once $\Lambda$ is below the smallest nonzero momentum, all nonconstant modes have been eliminated. The remaining effective energy is $H_{\Lambda }[M]=V{U}_{\Lambda }(M)$, so its defining integral is exactly $Z_M$. Therefore, with consistent normalization,When $\mathcal{H}(\Lambda ,M)$ denotes this uniform effective Hamiltonian density, this is the stated identification of $F(M)$ with its zero-cutoff limit. It requires a free-energy density constrained at fixed $M$, not the unconstrained extensive free energy; all integrated fluctuations and additive constants are included, the zero mode in field theory is retained, and a density for the uniform configuration is well defined. The finite-volume argument avoids an interchange of the thermodynamic and cutoff limits without justification. On a homogeneous pure-phase branch one may use a local potential description, while a thermodynamic limit allowing phase separation can convexify the constrained potential. A bare nonconvex Landau polynomial is accordingly a homogeneous-branch approximation, not a claim that the exact equilibrium Legendre potential stays nonconvex.

In a field $h$, the remaining integral is $Z(h)=\int dM{e}^{-{\beta }_{T}V[F_c(M)-hM]}$, up to the consistently chosen zero-mode measure. The thermodynamic saddle then gives $f(h)={\inf }_M\{F_c(M)-hM\}$ and selects the equilibrium magnetization. Landau theory assumes that the relevant uniform potential can be approximated by a regular analytic expansion in $M$, with symmetry-constrained coefficients, and minimizes it. At tree level the integrated fluctuations are neglected, recovering the familiar bare polynomial. With fluctuations retained, the coefficients run and can become singular near criticality; the zero-cutoff identification of constrained free energy alone does not justify mean-field critical exponents.

For an ordinary short-range scalar thermodynamic critical point, the leading interaction is quartic. Absorb inverse temperature and the gradient coefficient into field normalization and writeScale $x=b{x}^{\prime }$ and $\varphi (x)=b^{-(D-2)/2}{\varphi }^{\prime }(x^{\prime })$ to leave the gradient term invariant. The quartic term changes to $g{b}^{4-D}\int {\varphi }^{\prime 4}/4!$, so its engineering scaling exponent is $4-D$. It is irrelevant at the Gaussian fixed point for $D>4$, relevant for $D<4$, and marginal for $D=4$. HenceThe loop expansion makes the breakdown concrete. A quartic-vertex correction is proportional to $g^2{I}_2(r)$, whereFor $0<D<4$, rescaling $q=\sqrt{r}k$ gives $I_2(r)\sim C_D{r}^{(D-4)/2}$, diverging as $r\to 0$. At $D=4$, radial integration giveswhich is logarithmically divergent. Above four dimensions its infrared limit is finite; the leading cutoff-dependent constant can be absorbed into a coupling renormalization. The effective expansion parameter is thus of order $g{\xi }^{4-D}$ below four, or $g\log (\Lambda \xi )$ at four, since $\xi \sim r^{-1/2}$. It eventually becomes large for any fixed bare interaction in the lower-dimensional critical limit, defeating the uncorrected mean-field expansion. The Ginzburg criterion gives the same threshold by comparing correlation-volume fluctuations with the squared mean order parameter.

Below four dimensions an interacting renormalization-group fixed point can replace the Gaussian fixed point and change critical exponents, provided a continuous transition exists. At four dimensions the interaction is marginal and produces logarithmic corrections: the leading power exponents can remain their mean-field values, so “breakdown” there means that pure mean-field power laws omit those corrections, rather than necessarily that every power exponent changes. Above four the stabilizing quartic interaction is dangerously irrelevant for the ordered phase, explaining why mean-field thermodynamic exponents need not obey naive hyperscaling with the spatial dimension.

Write $\epsilon ={\epsilon }_a{t}_a$ and $A_{\mu }=A_{\mu a}{t}_a$. The proposed gauge-field transformation law becomes $\delta A_{\mu }=-{\partial }_{\mu }\epsilon +[\epsilon ,A_{\mu }]$. ConsequentlyBoth $\varphi$ and its gauge covariant derivative therefore transform as the same multiplet, with no derivative of the local parameter left over. The original algebraic invariance of the Lagrangian density now applies pointwise to $\mathcal{L}(\varphi ,D\varphi )$, so this constructs the local symmetry.

Also transform $J$ by $\delta J=\epsilon J$. Antisymmetry gives $(t_{a}J)\cdot \varphi +J\cdot (t_a\varphi )=0$, so the source coupling is invariant. Let $S_{A_{\mu a}}$ denote the functional derivative at a point. Vary the action functional and integrate the derivative of ${\epsilon }_a$ by parts, taking the local parameter to have compact support:Since every ${\epsilon }_a(x)$ is arbitrary, its coefficient vanishes. This proves the requested local identity, including the derivative sign and the order of the structure-constant indices.

For the generating functional, the integral of the first term is zero as a change-of-variable identity:The divergence of this vector field in field space is proportional to $\mathrm{tr}{t}_a=0$. Thus an invariant regulated functional measure, with no gauge anomaly, leaves no Jacobian term. Inserting the classical identity into the integral givesThe same first-order differential operator annihilates the connected generating functional $W$, because $Z=e^{iW}$. Its source term is $(t_{a}J)\cdot \phi$.

For the specified Legendre transform, vary $\Gamma =-W+\int J\phi$. Since $W_J=\phi$, the terms involving $\delta J$ cancel, leavingAntisymmetry also gives $(t_{a}J)\cdot \phi =-J\cdot (t_a\phi )$. Substitution in the identity for $W$, followed by multiplication by $-1$, proves the background gauge invariance of a scalar effective action:

The derivatives of the quantum effective action are proper, amputated one-particle-irreducible vertices. At each loop order they receive contributions from one-particle-irreducible Feynman diagrams with the indicated external scalar legs, including tree vertices and the required counterterms. Diagrams which disconnect after cutting one internal line are not additional proper-vertex contributions. Translation invariance supplies the overall momentum-conserving delta function factored out in the definition. The signs of these vertices must follow the chosen overall convention for $\Gamma$.

Differentiate the effective-action identity twice with respect to ${\phi }_i(y)$ and ${\phi }_j(z)$ and set $\phi =A=0$. In an invariant zero-field vacuum this gives the Ward identity contact termsThe term proportional to $A$ vanishes. With the printed Fourier factors $e^{i{p}_r{x}_r}$, integration by parts turns ${\partial }_{\mu }$ into $-i{p}_{1\mu }$. The first delta function combines $p_1$ with $p_2$, and the second combines $p_1$ with $p_3$. Using $(t_a{)}_{ki}=-(t_a{)}_{ik}$ and symmetry of the scalar two-point kernel then yieldsThis derivation fixes the relative signs independently of any vertex rule.

There is an overall-sign inconsistency in the final printed example. At tree level, stationarity of the source integral gives $W=S_0[\phi ,A]+\int J\phi$, hence the specified transform gives ${\Gamma }_{\mathrm{tree}}=-S_0$. For the free scalar multiplet this implies, after factoring out the delta function,The mixed result follows by expanding $-S_0$ to first order in $A$: its contribution is $\int A_{\mu a}({\partial }^{\mu }{\phi }_i)(t_a{)}_{ij}{\phi }_j$. The two scalar derivatives yield $i{p}_2^{\mu }(t_a{)}_{ij}+i{p}_3^{\mu }(t_a{)}_{ji}$. Momentum conservation givesso both sides of the Ward identity equal $i(p_3^2-p_2^2)(t_a{)}_{ij}$. Because the scalar theory is Gaussian for prescribed $A$, its fluctuation determinant depends on $A$ but not on $\phi$; it cannot add further two-scalar vertices. Thus this quadratic verification is sufficient, not merely the first term of an omitted scalar-loop correction.

The printed mixed vertex instead has a minus sign. It satisfies the same Ward identity if one uses the opposite effective-action convention ${\Gamma }_{\mathrm{alt}}=W-\int J\phi =S_0$ at tree level, because its two-point vertex is then $-(p^2+m^2){\delta }_{ij}$ as well. Either overall convention works, but the printed Legendre transform and printed mixed vertex cannot both be used unchanged. For example, with $p_2^2\ne p_3^2$ and a nonzero generator, they give opposite sides of the claimed identity.

Fix the current sign convention by defining the localized variation as ${\delta }_{ϵ}S=-\int d^{d}x({\partial }_{\mu }{ϵ}^a)j_a^{\mu }$. For a first-derivative Lagrangian invariant without a boundary term under constant parameters, this means $j_a^{\mu }=-\partial \mathcal{L}/\partial ({\partial }_{\mu }\varphi )t_a\varphi$; include the usual improvement term if the constant variation is a total derivative. On solutions, arbitrary compactly supported parameters imply ${\partial }_{\mu }{j}_a^{\mu }=0$. The Noether charge is $Q_a=\int d^{d-1}x{j}_a^0$ and is conserved if the spatial flux vanishes. This sign convention matches the printed Ward identity; reversing the current also reverses the corresponding generator convention.

Assume an invariant regulated functional measure, invariant vacuum boundary conditions and no quantum anomaly. Changing variables in the normalized path integral gives $0=⟨{\delta }_{ϵ}X⟩+i⟨X{\delta }_{ϵ}S⟩$. Integration by parts then yieldsFor a product of scalar fields, the local Ward identity contact terms areAway from the insertions this is current conservation. Time-ordering or the distributional functional identity supplies the contact terms.

For the gauge theory write $[X,Y{]}^a=f^{abc}{X}^b{Y}^c$ and use the left-acting BRST differential $s$, with ${\delta }_{ϵ}=ϵs$. It obeys the graded Leibniz rule andAssume the structure constants satisfy the Jacobi identity and the dot product is invariant; antisymmetry alone would not be enough. Since $c$ and $D_{\mu }c$ are odd, their Lie brackets are symmetric in these two arguments. ConsequentlyAlso $s{F}_{\mu \nu }=D_{\mu }{D}_{\nu }c-D_{\nu }{D}_{\mu }c=[F_{\mu \nu },c]$. The graded Jacobi identity gives $s[c,c]=0$, so $s^{2}c=0$; the other three fields have zero second variation immediately. For independent odd parameters, ${\delta }_{{ϵ}^{\prime }}{\delta }_{ϵ}=-{ϵ}^{\prime }ϵs^2=0$. These are off-shell BRST nilpotence identities because the Nakanishi-Lautrup field $b$ is retained.

The Yang-Mills theory variation is proportional to $F^{\mu \nu }\cdot [F_{\mu \nu },c]=0$. The gauge-fixing variation is $({\partial }^{\mu }b)\cdot D_{\mu }c$, while the ghost variation is its negative; the $b^2$ term does not vary. Equivalently these terms are $s\Psi$ for the gauge-fixing fermion $\Psi =({\partial }^{\mu }\bar{c})\cdot A_{\mu }+\xi \bar{c}\cdot b/2$. BRST nilpotence makes this expression invariant. Ghost-number scaling also leaves every term invariant: the ghost and antighost factors carry opposite weights.

Here are the two explicit currents in the same sign convention as the Ward identity. Localizing the even ghost parameter gives coefficient $({\partial }_{\mu }\theta )[\bar{c}\cdot D^{\mu }c-({\partial }^{\mu }\bar{c})\cdot c]$. Localizing the odd parameter, keeping it on the left, gives coefficientThe last sign comes from moving the odd parameter through the odd antighost derivative. Since the current was defined as minus this coefficient,These are the ghost-number Noether current and the BRST current in derivative-b gauge fixing. If the opposite Noether sign is used, both displayed currents acquire an overall minus sign. Integrating the gauge-fixing term by parts changes the Noether representative by the associated boundary improvement; mixing the two Lagrangian conventions without that improvement gives incorrect signs.

With no BRST anomaly, the conserved odd BRST charge has $Q_B^2=0$. The BRST cohomology identifies closed states $Q_B|\psi ⟩=0$ modulo exact states $Q_B|\chi ⟩$. Nilpotence puts every exact state in the closed space. In the usual indefinite gauge-fixed state space, a Hermitian BRST charge makes exact states orthogonal to closed states; the standard no-ghost/positivity assumptions then give a physical inner product on the quotient. The physical sector is its ghost-number-zero component,The ghost number assigns $+1$ to $c$, $-1$ to $\bar{c}$ and zero to gauge and auxiliary fields. Gauge-invariant observables and the chosen vacuum have ghost number zero; unphysical ghost excitations are removed in BRST pairs. Thus physical representatives are expected to satisfy $Q_G|\psi ⟩=0$. This zero-grading selection is part of the physical-state prescription, not a consequence of nilpotence alone.

Use natural units and the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$. A real scalar field assigns a real variable $\varphi (t,\mathbf{x})$ to each spatial point. Its Lagrangian density can be taken to beThe principle of stationary action gives the Euler-Lagrange equation $\square \varphi +V^{\prime }(\varphi )=0$. With $V(\varphi )=m^2{\varphi }^2/2$, this is the Klein-Gordon equation. An additional nonlinear part of $V$ describes interactions.

The canonical momentum is $\pi =\partial \mathcal{L}/\partial \dot{\varphi }=\dot{\varphi }$. The Legendre transform in mechanics gives the canonical Hamiltonian density of a real scalar fieldThe Hamiltonian equations $\dot{\varphi }=\pi$ and $\dot{\pi }={\nabla }^2\varphi -V^{\prime }(\varphi )$ recover the same field equation. In canonical quantization, the fields become operators satisfying the equal-time canonical commutation relationsA spatial lattice makes the analogy with many coupled quantum-mechanical coordinates precise. Each lattice field value is a coordinate, with its own conjugate momentum. The path integral is another representation of the same quantum evolution.

To see its origin, first consider one coordinate with $H=p^2/(2M)+V(q)$. Split a time interval into $N$ steps of length $\epsilon$ and insert position and momentum resolutions of the identity. The short-time kernel isMultiplying the kernels and integrating over intermediate positions gives the phase-space path integralThe endpoints of $q$ are fixed. The momentum integrals are Gaussian integrals; completing the square produces the configuration-space path integralAt finite slicing its normalization contains $(M/(2\pi i\epsilon ){)}^{N/2}{\prod }_{j=1}^{N-1}d{q}_j$. This fixes the composition law and the initial delta-function kernel. One sums over all paths, not merely solutions of the classical equation. Restoring $\hslash$ replaces the weight by $e^{iS/\hslash }$; stationary phase explains the emergence of classical trajectories.

For the field, use scalar field configuration eigenstates $|\phi ⟩$, satisfying $\hat{\varphi }(\mathbf{x})|\phi ⟩=\phi (\mathbf{x})|\phi ⟩$. Insert their completeness relations on every time slice. This givesThe endpoint field configurations are fixed. Integrating the Gaussian momentum variables leaves the scalar field path integral $\mathcal{N}\int \mathcal{D}\varphi e^{iS[\varphi ]}$. The functional measure means a regulated product over the field variables. A spacetime lattice or another ultraviolet cutoff makes this product finite before the continuum limit; interacting continuum calculations may require renormalization. The oscillatory Minkowski weight is an amplitude, not a positive probability density.

For vacuum expectation values, the boundaries must select the vacuum rather than arbitrary field configurations. Long imaginary-time evolution suppresses excited states: $e^{-TH}|n⟩=e^{-T{E}_n}|n⟩$, so after normalization only the lowest-energy component remains as $T\to \infty$. This is vacuum projection by imaginary time. The corresponding Feynman i-epsilon prescription in the real-time integral specifies the vacuum boundary conditions and the poles of the propagator. With $t=-i\tau$, the Euclidean path integral has the weight $e^{-S_E}$, whereIt is often a useful regulated starting point; analytic continuation returns the vacuum time-ordered quantities.

Introduce a classical source $J(x)$ and define the normalized vacuum generating functionalwith the same vacuum prescription in numerator and denominator. A functional derivative brings down $i\varphi (x)$. The order of the time slices makes the operator insertion time-ordered. Thus source differentiation inserts time-ordered field operators:The denominator removes vacuum diagrams and gives normalized expectation values. It is essential that these are time-ordered products; differentiating this vacuum functional does not directly give every possible operator ordering.

The free theory illustrates the method. Its quadratic kernel is $K=-\square -m^2$ with the vacuum pole prescription, and completing the square gives the Gaussian evaluation of a free scalar generating functionalTwo source derivatives give ${\Delta }_F=⟨\Omega |T\{\hat{\varphi }(x)\hat{\varphi }(y)\}|\Omega ⟩$. Higher derivatives give all pairings, the content of Wick theorem. For an interaction $S_{\mathrm{int}}$, one may use path-integral perturbation by source derivatives:Expanding this expression generates Feynman diagrams and their Wick contractions. The connected generating functional $W[J]=-i\log Z[J]$ retains connected contributions; in particular $\delta W/\delta J=⟨\varphi {⟩}_J$. These functionals turn the computation of field-operator expectations into source differentiation of an ordinary regulated integral.

With Euclidean source convention $+\int d^{d}x{J}_a{\varphi }_a$, the generating functional isHere $J_a(x)$ is a nondynamical source. Its functional derivatives insert fields:This is why $Z$ “generates” correlation functions.

With the notation of the question, $W[J]=-\log Z[J]$ is the Euclidean connected generating functional; it is distinct from a Wilsonian effective action. Define the classical fieldThe quantum effective action is the Legendre transformwhere $J$ is eliminated in favor of $\Phi$. Its first derivative isso at zero source the quantum expectation value is a stationary point of $\Gamma$. Moreover,where $G_{ab}=⟨{\varphi }_a{\varphi }_b{⟩}_{J,\mathrm{conn}}$ is the exact connected correlation function. Thus the second derivative of $\Gamma$ is the inverse exact propagator.

Perturbatively, $Z[J]$ sums all Feynman diagrams, including disconnected ones. The exponential formula for combinatorial structures says that its logarithm selects connected Feynman diagrams, so $W$ sums connected diagrams with the sign dictated by the convention above. The Legendre transform removes diagrams that disconnect when one internal line is cut; consequently $\Gamma$ sums one-particle-irreducible Feynman diagrams. Equivalently, every connected diagram is a tree assembled from one-particle-irreducible vertices and full propagators, and the Legendre transform inverts that tree construction.

Now let $J_a^{\prime }=J_b{U}_{ba}$. ThenChanging variables to ${\varphi }_b^{\prime }=U_{ba}{\varphi }_a$ and using invariance of both the action and functional measure gives $Z[J^{\prime }]=Z[J]$, hence $W[J^{\prime }]=W[J]$. In the Legendre transform, the pairing obeysChanging the source variable and using the invariance of $W$ therefore givesThe symmetry of the classical action and measure is inherited by the full quantum effective action when it has no quantum anomaly.

A quantum anomaly occurs when a symmetry of the classical action cannot be preserved by the regulator or functional measure of the quantum theory. An anomaly in a global symmetry is physically allowed: it changes a classical current conservation law and can mediate observable processes. A gauge anomaly is fatal unless it cancels, because gauge symmetry is the redundancy that removes negative-norm and longitudinal unphysical states; its loss spoils the gauge Ward identities and unitarity.

Let $\Lambda$ be a chiral gauge parameter. One consistent Abelian supergauge transformation convention isfor which ${\bar{\Phi }}_i{e}^{q_{i}V}{\Phi }_i$ is invariant. Wess-Zumino gauge uses the nonordinary components of $\Lambda$ to remove the superfluous scalar and spinor components of $V$, leaving the photon, gauginos, complex scalar, and real auxiliary field of the two-dimensional vector multiplet. Ordinary $U(1)$ gauge transformations remain.

Define the field-strength multiplet, up to conventional normalization, byGauge invariance follows because chirality, antichirality, ${\bar{D}}_{+}^2=D_{-}^2=0$, and $\{{\bar{D}}_{+},D_{-}\}=0$ annihilate the variation of $V$. The same identities giveso $\Sigma$ is a twisted chiral superfield. The Fayet–Iliopoulos term is a twisted F-term. In the standard axial convention its measure $d{\theta }^{+}d{\bar{\theta }}^{-}$ has axial charge $-2$, so invariance requiresReversing all axial-charge conventions reverses both signs but leaves this statement unchanged: the field and measure have opposite charges.

The charged matter fermions are chiral with respect to the axial symmetry. In a background with gauge flux, their functional measure has the two-dimensional axial anomalyup to orientation and current normalization. Equivalently, a Fujikawa transformation multiplies the torus path integral by a phase proportional to $\alpha ({\sum }_i{q}_i)\int F$. Since the background flux may be nonzero, the continuous axial $U(1)$ survives quantum mechanically precisely when

Perform the infinitesimal local change of integration variablesin the Euclidean path integral for $⟨j^{\mu }(x)\psi (x_1)\bar{\psi }(x_2)⟩$. The vector transformation has no quantum anomaly, so its functional measure is invariant. The action variation supplies $-i\int \alpha {\partial }_{\mu }{j}^{\mu }$, while varying the two charged insertions supplies contact terms at $x_1$ and $x_2$ with opposite signs. Since a change of integration variables cannot change the integral, the coefficient of the arbitrary function $\alpha (x)$ vanishes:This Schwinger-Dyson equation is the position-space Ward-Takahashi identity.