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For the left graded Leibniz rule and $s\bar{c}=-b$, $sb=0$, the displayed gauge-fixing fermion gives $b\partial \cdot A+\xi b^2/2+\bar{c}\partial \cdot Dc$. The ghost signs follow from odd parity. BRST nilpotence implies invariance of this density without using field equations. Eliminating the Nakanishi-Lautrup field gives $-(\partial \cdot A{)}^2/(2\xi )$ for nonzero $\xi$.

For adjoint fields whose coefficients have Grassmann parity, $X\times Y=-(-1{)}^{|X||Y|}Y\times X$. Thus two odd ghost fields have a symmetric adjoint bracket, and $c\times c$ need not vanish. The graded Jacobi identity and this symmetry are the algebraic mechanisms behind BRST nilpotence.

Using the graded Leibniz rule, $s^{2}A=D(sc)+g(Dc)\times c=0$ because $D(c\times c)=2(Dc)\times c$. The graded Jacobi identity gives $s^{2}c=g^2(c\times c)\times c/2=0$. BRST nilpotence on the antighost and Nakanishi-Lautrup field is immediate. No field equations are required.

The Polyakov action introduces an independent worldsheet metric $h_{ab}$:Its redundancies are worldsheet diffeomorphisms and Weyl transformations. A Polyakov path integral over all metrics without dividing by this gauge volume would count physically equivalent configurations repeatedly. Fixing conformal gauge therefore needs the Faddeev-Popov determinant, not merely the substitution of a flat metric into the action.

After using the Weyl transformation to remove the trace of a metric variation, the infinitesimal worldsheet diffeomorphism ghost operator isThe Faddeev-Popov determinant of this operator is represented by anticommuting fields: a vector Faddeev-Popov ghost $c^a$ and a symmetric traceless Faddeev-Popov antighost field $b^{ab}$. In Euclidean coordinates their worldsheet ghost action is, up to a consistent overall normalization,The two chiral copies of this bc system have conformal weights $(2,-1)$ and anticommutators $\{b_m,c_n\}={\delta }_{m+n,0}$. They are auxiliary fields for the gauge determinant, not additional spacetime particles. Their use is different from the unwanted timelike negative-norm string states of covariant matter quantization.

The chiral holomorphic stress-energy tensor of the bc system isDouble contractions in its operator product expansion give the central charge of reparameterization ghosts $c_{\mathrm{gh}}=-26$. Equivalently, a fermionic bc system with antighost weight $\lambda$ has $c=1-3(2\lambda -1{)}^2$, and $\lambda =2$ gives $-26$. The free coordinate matter contributes $c_{\mathrm{m}}=D$. Thus the total worldsheet Weyl anomaly vanishes for the flat critical bosonic string theory precisely whenThis statement assumes no additional matter or compensating Liouville field theory.

BRST symmetry expresses the gauge symmetry after gauge fixing by replacing its infinitesimal parameter with the Grassmann variable $c$. In one chiral conformal patch, its graded action has the formthere is an analogous barred copy for a closed string. The Grassmann parity makes $s^{2}X=0$: the term from $sc$ cancels the term from applying $s$ to $\partial X$, while the term with $c^2$ vanishes. The gauge-fixed chiral transformations are on shell when their conservation equations are used. At the quantum level, BRST nilpotence also tests the anomaly.

For example, an open string BRST charge can be written using the matter Virasoro generators asUse the ghost oscillator vacuum $b_{n\ge 0}|0{⟩}_{mathrmgh}=c_{n>0}|0{⟩}_{mathrmgh}=0$ and the corresponding oscillator normal ordering. In this convention the last term is $-c_0$ in the critical theory. With a differently shifted ghost $L_0$, the displayed intercept must be shifted consistently; one must not count the same shift twice. Squaring $Q_B$ shows why both critical conditions are needed. Its anomalous terms, in this convention, have coefficientsThe independent cubic and linear terms vanish at $D=26$, $a=1$. The equivalent chiral BRST current is $j_B=c(T_{mathrmm}+T_{mathrmgh}/2)+(3/2){\partial }^{2}c$, with its contour integral defining the charge.

The graded identity $\{Q_B,b_n\}=L_n^{\mathrm{tot}}$ recovers the gauge constraints. In particular, on $|\psi ⟩\otimes |0{⟩}_{mathrmgh}$, BRST-closed imposes $L_{n>0}^{\mathrm{m}}|\psi ⟩=0$ and $(L_0^{\mathrm{m}}-1)|\psi ⟩=0$. Physical states are classes in BRST cohomology,at the appropriate ghost number: one for the usual open-string vertex and two for the closed-string vertex, with the closed-string zero-mode and closed-string level matching conditions also imposed. Since $Q_B^2=0$, an exact shift preserves closure. BRST Ward identities make exact insertions decouple from physical amplitudes and establish gauge independence. Together with the no-ghost theorem for the critical bosonic string, this explains how the auxiliary ghosts enforce gauge invariance while the physical spectrum retains positive norm.

Choose a definite matrix convention: $A_{\mu }=A_{\mu }^a{T}_a$, $\omega ={\omega }^a{T}_a$, and $D_{\mu }={\partial }_{\mu }+e{A}_{\mu }$. Matter transforms as ${\psi }^{\prime }=U\psi$, so gauge covariance requires $D_{\mu }^{\prime }=U{D}_{\mu }{U}^{-1}$. Applying both sides to a test field yields the Yang-Mills gauge transformationDefine the Yang-Mills field strength by $[D_{\mu },D_{\nu }]=e{F}_{\mu \nu }$:Conjugating the gauge covariant derivative commutator proves $F_{\mu \nu }^{\prime }=U{F}_{\mu \nu }{U}^{-1}$: unlike the connection, the gauge field strength has no inhomogeneous derivative term. The trace normalization supplies an invariant color metric; its Lie algebra structure constants with all indices lowered are antisymmetric.

For $U=1+e\omega +O({\omega }^2)$ and $U^{-1}=1-e\omega +O({\omega }^2)$, expand the finite transformations:In components this isA convention using the opposite matter transformation or gauge covariant derivative sign changes these signs consistently. The formulas above all follow from the explicitly chosen $D_{\mu }$.

Gauge reduction and the determinant. A direct integral $\int \mathcal{D}A{e}^{iS[A]}$ integrates repeatedly over gauge-equivalent configurations. The gauge group volume is infinite, and the ungauge-fixed quadratic operator has longitudinal zero directions, so it has no ordinary inverse quantum field theory propagator. Gauge-invariant observables require dividing out this redundancy rather than treating it as an ordinary physical fluctuation.

The finite-dimensional change-of-variables identity underlying gauge fixing is precise. Suppose a condition $\chi (x^{\alpha })=f$ has one solution in the group-coordinate patch and its derivative matrix at the solution is nonsingular. ThenFor several isolated solutions it instead counts those solutions. On the slice, the Jacobian determinant can be factored into the reciprocal of the orbit integral of the delta function. An invariant measure and invariant integrand then allow the group volume to factor out. In the functional version, group coordinates carry Haar measure and the corresponding operator determinant replaces the matrix determinant.

For example, choose the Lorenz gauge ${\chi }^a(A)={\partial }^{\mu }{A}_{\mu }^a=f^a$. Its variation is $\delta {\chi }^a=-{\partial }^{\mu }(D_{\mu }^{\mathrm{ad}}\omega {)}^a$. Hence on the gauge slice the Faddeev-Popov operator and Faddeev-Popov determinant areIts extension along each local gauge orbit yields the Faddeev-Popov gauge-orbit identityFor an unoriented finite-dimensional real integral the absolute determinant is required. In a perturbative gauge patch, its sign or phase is fixed and the constant convention is absorbed in normalization. Residual zero modes must be removed by boundary conditions or separate treatment. Multiple intersections, the Gribov ambiguity, prevent interpreting this as a globally unique gauge slice without additional work.

Insert the identity, change variables along the orbit using invariance of the action and measure, and divide out the common gauge group volume. Averaging $f$ with Gaussian weight $\exp [-i\int f^a{f}^a/(2\xi )]$ produces the covariant gauge-fixed functionalThe added term makes the quadratic gauge-field operator invertible after the stated residual-mode treatment.

Ghost representation. For a finite matrix $M$, the Grassmann Gaussian integral over independent anticommuting variables is proportional to its determinant:The constant is independent of $M$ and depends on ordering and factors of $i$. The functional counterpart introduces Faddeev-Popov ghost fields $c^a$ and antighost fields ${\overline{c}}^a$, withIntegration by parts writes its density as $({\partial }^{\mu }{\overline{c}}^a)(D_{\mu }c{)}^a$, displaying a kinetic term and a gauge-field/ghost interaction. These are Grassmann-valued fields with scalar Lorentz transformation law. They represent the orbit Jacobian determinant; their closed loops carry the fermionic minus sign, and they are not physical asymptotic particles. In the abelian case $D^{\mathrm{ad}}=\partial$, the determinant is gauge-field independent and the ghosts decouple. Nonabelian ghosts interact and are necessary for the consistent cancellation of unphysical gauge contributions.

Physical-state reduction. Let $Q$ be the Hermitian BRST charge. Its key algebraic property is BRST nilpotence, $Q^2=0$, and the gauge-fixed dynamics preserves it. Physical states are its BRST cohomology at ghost number zero:Here $|\chi ⟩$ has ghost number $-1$. BRST nilpotence ensures that exact vectors lie in the kernel, so the quotient is defined. Because $Q$ is Hermitian, an exact vector is orthogonal to every closed vector, $⟨\psi |Q\chi ⟩=⟨Q\psi |\chi ⟩=0$, and exact vectors are null, $⟨Q\chi |Q\chi ⟩=⟨\chi |Q^2\chi ⟩=0$. They therefore represent no independent physical state. Commutation of $Q$ with the Hamiltonian preserves these classes under time evolution.

The covariant gauge/ghost space before reduction has an indefinite inner product. This is essential: a Hermitian nilpotent BRST charge requires an indefinite auxiliary space if it is nonzero. On an ordinary positive-definite Hilbert space the same norm identity would force $Q=0$. Positivity of the resulting physical space uses the gauge-theory structure in addition to the abstract BRST nilpotence identities.

For a nonzero on-shell photon momentum, the one-particle conclusion isThe second relation preserves the first because $k^2=0$. In four dimensions the transverse kernel has dimension three, and quotienting by its null longitudinal direction leaves two physical photon polarizations. This is the positive one-particle BRST cohomology of the photon sector.

The Polyakov action is invariant under worldsheet diffeomorphisms and Weyl transformations, ${\gamma }_{ab}\mapsto e^{2\omega }{\gamma }_{ab}$. A two-dimensional metric has three local components: two coordinate transformations and one Weyl function remove these locally. In conformal gauge the factor $\sqrt{-\gamma }{\gamma }^{ab}$ is independent of the conformal scale, and on a flat coordinate patch the matter action isThus it is the action of $d$ free scalar fields, with their target-space contractions using the Minkowski metric. On a Euclidean sphere there are no unpunctured conformal moduli: a metric can be put into the conformal class of the round metric, and in stereographic patches the scalar action has this free-field form. This does not mean that the sphere has a globally flat metric. A spherical worldsheet here is understood after Wick rotation; it has no nonsingular global Lorentzian metric. The residual Möbius transformations and their zero modes still have to be divided out or fixed by vertex positions.

To derive the worldsheet ghost action, consider an infinitesimal diffeomorphism generated by $v^a$ and a Weyl variation $\omega$. The metric variation separates asIntegration over the Weyl parameter removes the trace gauge condition. The trace-free part produces the Faddeev-Popov determinant of an operator mapping vectors to symmetric traceless tensors. This is trace elimination in string Faddeev-Popov gauge fixing. The stated Grassmann Gaussian integral exponentiates this determinant with an odd vector $c^a$ and an odd symmetric trace-free antighost $b_{ab}$.

The normalization of the operator and antighost can be chosen to match the printed convention. For example, in a flat patch take $P=-P_1/2$. It differs from the orbit operator only by a field-independent determinant factor. Tracelessness and symmetry give $b_{ab}(Pc{)}^{ab}=-b_{ab}{\partial }^a{c}^b$. Integration by parts, followed by exchanging two odd factors, gives $\int b_{ab}{\partial }^a{c}^b=\int c^a{\partial }^b{b}_{ab}$. Consequently the determinant exponent is $+(2\pi {)}^{-1}\int c^a{\partial }^b{b}_{ab}=i{I}_{\mathrm{gh}}$ withOverall constant normalization does not affect the determinant's dependence on the fields. The Faddeev-Popov ghost fields are vectors, while the antighost fields are symmetric trace-free tensors; they are not extra target-space string excitations.

Now use ${\sigma }^{\pm }=\tau \pm \sigma$, ${\partial }_{\pm }=({\partial }_{\tau }\pm {\partial }_{\sigma })/2$ and $c^{\pm }=c^0\pm c^1$. In these coordinates ${\eta }_{+-}={\eta }_{-+}=-1/2$ and ${\partial }^{\pm }=-2{\partial }_{\mp }$. Tracelessness implies $b_{+-}=0$, so directlyThereforeFor a Cartesian check, $b_{11}=b_{00}$ and $b_{++}=(b_{00}+b_{01})/2$, $b_{--}=(b_{00}-b_{01})/2$ reproduce the same result.

The original PDF repeats ${\delta }_{\eta }{c}^{+}$ for its final ghost transformation. The second occurrence must be ${\delta }_{\eta }{c}^{-}$; otherwise the displayed rules give two incompatible transformations for one ghost and none for the other. Write the corrected transformation as ${\delta }_{\eta }\Phi =\eta s\Phi$, with $\eta$ on the left. Then the odd left-acting BRST differential satisfies the graded Leibniz rule $s(AB)=(sA)B+(-1{)}^{|A|}A(sB)$. Its local rules areThe last two expressions are $s{b}_{\pm \pm }=i{\theta }_{\pm \pm }/{\alpha }^{\prime }$ with the matter-plus-ghost worldsheet stress tensor in the question.

Here is an off-shell check of the action variation, including the Grassmann signs. Temporarily set $c=c^{+}$, $f=c^{-}$, $b=b_{++}$, $h=b_{--}$, $u={\partial }_{+}X$ and $v={\partial }_{-}X$. Apart from the common factor $1/\pi$, the total Lagrangian is $\mathcal{L}=u\cdot v/{\alpha }^{\prime }+i(c{\partial }_{-}b+f{\partial }_{+}h)$. Its matter variation isThe matter terms in $sb$ and $sh$ give the ghost-action contribution ${\alpha }^{\prime -1}[c{\partial }_{-}(u^2)+f{\partial }_{+}(v^2)]$. Combining these terms with the last two matter terms produces ${\alpha }^{\prime -1}[{\partial }_{-}(c{u}^2)+{\partial }_{+}(f{v}^2)]$.

The remaining contribution from the $c,b$ sector iswhere subscripts denote derivatives. In the last equality the terms $c_{-}{c}_{+}b$ and $c_{+}{c}_{-}b$ cancel by anticommutation. The $f,h$ sector has the same identity with $+\leftrightarrow -$. ThuswhereHence the total action is BRST invariant up to a boundary term, which vanishes on a closed worldsheet or under the usual compatible boundary and time-end conditions. This calculation requires no equations of motion.

Nilpotence needs a qualification for these particular chiral rules. For either ghost, $s(c\partial c)=(c\partial c)\partial c-c\partial (c\partial c)=0$, because $c^2=(\partial c{)}^2=0$. Thus $s^2{c}^{\pm }=0$ algebraically. For the embedding, however, the graded calculation givesMixed second derivatives of $X$ cancel because $c^{+}{c}^{-}=-c^{-}{c}^{+}$. The remaining terms vanish on the ghost equations of motion, ${\partial }_{-}{c}^{+}={\partial }_{+}{c}^{-}=0$, which follow by varying the antighosts in the chiral ghost action. They need not vanish off shell. An explicit counterexample uses independent Grassmann variables $\xi ,\rho$: take $c^{+}=\xi$, $c^{-}=\rho {\sigma }^{+}$ and one coordinate $X={\sigma }^{-}$. Then $s^{2}X=-\xi \rho \ne 0$.

For the antighosts, use the permitted stress-tensor identity $s{\theta }_{ab}=0$ on the classical solution space. Then $s^2{b}_{\pm \pm }=(i/{\alpha }^{\prime })s{\theta }_{\pm \pm }=0$. Indeed direct variation of the displayed stress tensors giveswhich vanish using the ghost equations and the free matter wave equation. Therefore two successive BRST transformations annihilate every field on shell. This is chiral conformal-gauge BRST nilpotence is on shell, not an unrestricted off-shell identity for arbitrary ghosts. Quantum BRST nilpotence additionally requires cancellation of the worldsheet anomaly, giving $d=26$ in the critical bosonic theory.

For adjoint-valued quantities write $(X\times Y{)}_a={\epsilon }_{abc}{X}_b{Y}_c$. Substituting the infinitesimal transformation into the gauge field strength and using the Jacobi identity givesA useful way to organize the calculation is $[D_{\mu },D_{\nu }]X=g{F}_{\mu \nu }\times X$: the gauge field strength transforms covariantly because the gauge covariant derivative does. Thus $\delta (F_a^{\mu \nu }{F}_{\mu \nu a})=2g{F}_a^{\mu \nu }{\epsilon }_{abc}{F}_{\mu \nu b}{\lambda }_c=0$ by antisymmetry. This proves gauge invariance of the Yang-Mills action.

The Yang-Mills equations are $D_{\mu }{F}^{\mu \nu }=0$. They determine physical evolution only modulo gauge redundancy: gauge transformations with arbitrary spacetime-dependent parameters take solutions to equivalent solutions. In Hamiltonian mechanics, $A_0$ has no independent quadratic time-derivative term, and the corresponding equation is a Gauss law constraint in gauge theory. A gauge parameter can vanish with its necessary derivatives on the initial slice yet change the later potential energy, so the potential energy itself is not uniquely fixed by initial physical data.

In the uncorrected path integral, integration along gauge orbits overcounts equivalent fields. At the perturbative level the quadratic operator of the kinetic term has gauge zero modes and no inverse, so it does not supply a quantum field theory propagator. Gauge fixing together with the Faddeev-Popov determinant removes this obstruction in the local perturbative construction.

For the odd BRST transformation, products obey the graded Leibniz rule. The adjoint product with Grassmann variables satisfiesIn particular, $c\times c$ is not zero, and $c\times Dc=Dc\times c$. The supplied transformations givesince the even covariant derivative obeys $D(c\times c)=2(Dc)\times c$. For the ghost,The final equality is the graded Jacobi identity; in components it is the Jacobi identity for ${\epsilon }_{abc}$ contracted with the totally antisymmetric product of three ghosts. Also $s^2\bar{c}=-sb=0$ and $s^{2}b=0$. The square of an odd derivation is an even derivation, because its two mixed product terms cancel. Hence $s^2=0$ on every polynomial in the fields, off shell, without imposing field equations.

Let $\Psi =\int d^{d}x[{\partial }^{\mu }{\bar{c}}_a{A}_{\mu a}+\xi {\bar{c}}_a{b}_a/2]$ be the gauge-fixing fermion. Since $sF=gF\times c$, $s{\mathcal{L}}_{\mathrm{YM}}=0$. The BRST-exact operator in ${\mathcal{L}}_q={\mathcal{L}}_{\mathrm{YM}}-s\psi$ has zero variation by BRST nilpotence, soTo see the resulting kinetic terms and signs, expand the exact term with $s\bar{c}=-b$:After integration by parts,The auxiliary Nakanishi-Lautrup field $b$ can be completed into a square and integrated out, giving the gauge-fixing term $-(\partial \cdot A_a{)}^2/(2\xi )$ for $\xi \ne 0$. At $\xi =0$, retain $b$ as the multiplier enforcing the gauge condition rather than divide by $\xi$. The Grassmann Gaussian integral gives the determinant of the gauge-condition operator $-\partial \cdot D$. Field-independent normalizations cancel in the normalized integral; insertions involving $b$ can be handled before integration or through its source terms.

At $g=0$ the gauge-fixed quadratic operator is invertible with the usual prescription, and the ghosts have a quadratic kinetic term. The remaining terms are cubic gauge interactions of order $g$, quartic gauge interactions of order $g^2$, and a ghost-gauge interaction of order $g$. Expanding them and using Wick contractions gives a perturbative expansion for any defined polynomial insertion $X$ of the fields. The exponent is the gauge-fixed action $S_q$, as printed in the PDF; the converted TeX incorrectly substitutes $S_A$.

At $\xi =1$, the free gauge action, up to a boundary term, is $\frac{1}{2}\int A_{\mu a}\square A_a^{\mu }$. Its Fourier kernel is $-p^2{\eta }_{\mu \nu }{\delta }_{ab}$. With the mostly-plus convention used above, the Feynman-gauge adjoint propagator isThis is consistent with the scalar-line convention in Question 2.

Finally define the odd functional $B=\frac{1}{2}\int d^{d}x{\bar{c}}_a{b}_a$. Then ${\partial }_{\xi }{S}_q=-sB$. Differentiating the normalized expectation, including its denominator, givesfor an insertion with no explicit $\xi$ dependence. A physical gauge-invariant insertion $X[A]$ is BRST-closed, since its infinitesimal gauge variation vanishes also when the parameter is replaced by $c$. The assumed BRST Ward identity gives $⟨sB⟩=0$ and $⟨s(XB)⟩=0$. For an even $X$ with $sX=0$, the latter says $⟨XsB⟩=0$. ThereforeThe same conclusion holds for any BRST-closed insertion. Mere color-singlet invariance of an arbitrary ghost-dependent expression need not imply $sX=0$; the physical-observable qualification is essential. The proof uses the invariant-measure assumptions encoded in the stated Ward 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.

For the relativistic particle phase-space action, the first-class constraint $\phi =(p^2+m^2)/2$ generatesIndeed the integrand varies by $\frac{1}{2}d[ϵ(p^2-m^2)]/dt$. The canonical gauge transformation is an invariance when the gauge parameter vanishes at fixed temporal endpoints, or when all fields and the parameter are periodic. The boundary restriction matters for the proper-time modulus.

Normalize the worldline interval to $[0,1]$. Thenis invariant because $\delta s=ϵ(1)-ϵ(0)=0$. Every allowed $e$ in its orbit can be written $e(t)=s+\dot{ϵ}(t)$: set $ϵ(t)={\int }_0^t[e(u)-s]du$. Thus $s$ remains a gauge-invariant integration variable, not another removable nonconstant mode. For a worldline circle the constant gauge parameter is a residual zero mode. Without the endpoint restriction, the assertion that $s$ is invariant would not hold.

The worldline gauge-orbit determinant is the Jacobian from gauge-orbit coordinates $ϵ$ to the nonconstant part of $e$ is the Faddeev-Popov determinant of ${\partial }_t$. Equivalently, the gauge-fixing identity has the formThe determinant is taken between the appropriate boundary-condition spaces, with the modulus removed; on a circle the prime also removes the constant parameter. Gauge fixing therefore leaves a factor $\det {}^{\prime }{\partial }_t$ and a modulus measure, after division by any residual gauge volume. Even though this determinant is field independent in the present Abelian example, it is the required change-of-variables Jacobian. A Grassmann integral over the Faddeev-Popov ghosts exponentiates it:The overall determinant phase depends on the integration convention and can be absorbed into normalization. Zero modes and the same endpoint restrictions must be treated separately rather than included in an invertible determinant.

For the free-ended open string, take $0\le \sigma \le \pi$. A canonical cosine expansion at a fixed time isIt implements the Neumann boundary conditions and has ${\alpha }_n^{†}={\alpha }_{-n}$. With ${\alpha }_0^m=p^m/\sqrt{\pi T}$, its Nambu-Goto phase-space action, up to a total time derivative, isReality requires ${\lambda }_{-n}={\lambda }_n^{\ast }$. Numerical factors can be absorbed into these Lagrange multipliers. In this covariant quantization of the bosonic string the oscillators retain all $D$ spacetime components, in contrast to the transverse oscillators in the preceding solution. The canonical commutation relations areThe oscillator vacuum $|0;p⟩$ is annihilated by ${\alpha }_n^m$ for $n>0$. Its momentum label will sometimes be suppressed.

Define the matter Virasoro algebra generators using normal ordering:No additive intercept is included in this definition of $L_0$. For $n>0$ the indices of the two factors in each term add to $n$; they cannot both be negative. After normal ordering there is a positive-mode annihilation operator on the right, possibly accompanied by the zero mode. Hence $L_n|0;p⟩=0$ for every $n>0$. In $L_0$, commuting positive modes past negative modes formally adds $\frac{D}{2}{\sum }_{k>0}k$. This divergent constant needs a prescription, and a finite shift is an ordering ambiguity. Our convention instead puts the physical string intercept into the constraint $L_0-a$.

With this convention the matter Virasoro algebra isA different additive constant in $L_0$ would change the linear-in-$m$ central term, so stating the convention is essential.

For the worldsheet ghost fields, useand choose a ghost oscillator vacuum withThen $c_0|0{⟩}_{\mathrm{gh}}\ne 0$ since $\{b_0,c_0\}=1$. This choice specifies the ghost zero-mode doublet; it is not a claim that both zero modes annihilate one state. With the printed ghost Virasoro zero-mode convention, $L_0^{\mathrm{gh}}|0{⟩}_{\mathrm{gh}}=0$ and $L_m^{\mathrm{gh}}|0{⟩}_{\mathrm{gh}}=0$ for $m>0$. The latter follows by putting positive ghost modes on the right; a possible contraction requires $m=0$ and is absent here.

Apply the supplied BRST charge to the matter state times this ghost oscillator vacuum. Terms with a rightmost $c_m$, $m>0$, vanish, as do the positive-mode ghost generators. ThusThese one-ghost states are independent, as can also be seen by applying $b_0,b_m$. Therefore the BRST physical-state constraints of an open string areFor the matter oscillator vacuum, $L_0={\alpha }^{\prime }{p}^2$, so $p^2=1/{\alpha }^{\prime }$ and $M^2=-1/{\alpha }^{\prime }=-2\pi T$: the physical ground state is a tachyon. The momentum must satisfy this equation; the zero-momentum oscillator vacuum by itself would not be BRST-closed.