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A two-component Weyl spinor is a complex spinor transforming in one of the two inequivalent group representations $(1/2,0)$ and $(0,1/2)$ of the Lorentz spinor double cover $\mathrm{SL}(2,\mathbb{C})$. An undotted left-handed field ${\chi }_{\alpha }$ transforms by a matrix $M\in \mathrm{SL}(2,\mathbb{C})$; its complex conjugate ${\bar{\chi }}_{\dot{\alpha }}$ transforms by $M^{\ast }$. Dotted and undotted indices distinguish the two chirality representations. Raising a dotted index uses the invariant epsilon tensor and converts the complex conjugate transformation into the equivalent inverse-adjoint form. Rotations act by the same SU(2) generators on the two chiralities, while boosts have opposite generators, so the two complex representations cannot be identified by a complex-linear Lorentz invariant map.

The antisymmetric invariant tensors ${ϵ}_{\alpha \beta }$ and ${ϵ}_{\dot{\alpha }\dot{\beta }}$ raise and lower spinor indices; $M^TϵM=ϵ$ proves invariance of the contracted bilinear $\chi \psi ={\chi }^{\alpha }{\psi }_{\alpha }$. The components of a fermionic Weyl spinor are odd Grassmann variables, so this bilinear is symmetric under exchange of the two spinors: the antisymmetric epsilon sign cancels the Grassmann sign. In particular, choosing ${ϵ}_{12}=1$ and ${ϵ}^{12}=-1$ gives $\chi \chi =2{\chi }_1{\chi }_2$, generally nonzero. This Weyl spinor bilinear exchange identity permits a Majorana mass term for a single suitable Weyl spinor. With mostly-minus metric signature and ${\bar{\sigma }}^{\mu }=(I,-\mathbf{\sigma })$, the free massless kinetic term is $i{\chi }^{†}{\bar{\sigma }}^{\mu }{\partial }_{\mu }\chi$, and its field equation of motion is $\boxed{{\bar{\sigma }}^{\mu }{\partial }_{\mu }\chi =0}$.

Fix the orientation of the Berezin integral by defining its measure as the linear functional that extracts the coefficient of ${\prod }_{i=1}^N({\theta }_i{\bar{\theta }}_i)$; all monomials of lower degree integrate to zero. In particular $\int \mathcal{D}(\bar{\theta },\theta ){\prod }_i({\theta }_i{\bar{\theta }}_i)=1$. This specifies the sign convention in the notation $d^N\bar{\theta }{d}^N\theta$.

The Grassmann change-of-variables formula givesFor a unitary matrix, $\det U\det U^{\ast }=1$, so the combined measure is invariant. Diagonalize the Hermitian matrix $B$ by a unitary change of variables. If its eigenvalues are $b_i$, the integrand factors as ${\prod }_i(1-b_i{\bar{\theta }}_i{\theta }_i)={\prod }_i(1+b_i{\theta }_i{\bar{\theta }}_i)$. Extracting its top coefficient proves the Grassmann Gaussian integralThe determinant identity also holds without Hermiticity by direct expansion; Hermiticity makes the diagonalization proof immediate.

Let $C(x)={\sum }_k{x}_k{C}^{(k)}$. Integrating out the Grassmann variables first gives the exact polynomial expressionThe requested factorization requires $B$ to be invertible, an assumption omitted from the question. Under it, use the formal perturbative matrix logarithm near $g=0$:ConsequentlyExpanding $C(x)$ gives exactly the displayed coefficients $V_{k_1\dots k_m}^{(m)}=(m-1)!\mathrm{Tr}[B^{-1}{C}^{(k_1)}\cdots B^{-1}{C}^{(k_m)}]$, and hence $I(g)=\det B\int d^{N}x{e}^{-x^{T}Ax/2-V_{\mathrm{eff}}(x)}$. The logarithmic representation is a formal series, not a globally convergent logarithm for every $x$: the determinant can have zeros. The original determinant integral remains well-defined for singular $B$, but $I(0)=0$ and the requested normalized expansion then cannot be formed.

There is a second convention to specify. The $x_i$ are real, so a complex Hermitian $A$ gives $x^{T}Ax=x^T{A}_{R}x$, where $A_R=\mathrm{Re}A=(A+A^T)/2$ is real symmetric positive definite. Its Gaussian covariance matrix is $A_R^{-1}$. It equals $A^{-1}$ only if $A$ is real symmetric. ThusThe effective Feynman rules are: a bosonic contraction joins indices $k,\ell$ with $(A_R^{-1}{)}_{k\ell }$; an $m$-leg vertex carries $-g^m\mathrm{Sym}{V}_{k_1\dots k_m}^{(m)}$; sum all closed vacuum diagrams with their symmetry factors, including disconnected diagrams. The symmetrization is necessary because the $x_k$ commute: the trace tensor is cyclically invariant but need not be fully symmetric. The minus sign comes from expanding $e^{-V_{\mathrm{eff}}}$. Taking the logarithm of the normalized integral instead keeps only connected vacuum diagrams.

Equivalently, retain the fermions: an oriented contraction is $(B^{-1}{)}_{ij}=⟨{\theta }_i{\bar{\theta }}_j⟩$, the mixed vertex has weight $+g{C}_{ij}^{(k)}$, and every closed fermion loop gives a minus sign. The same bosonic covariance joins its $x$ legs. Expanding these contractions recovers the trace vertices above. This is bosonic trace vertices from a finite fermionic determinant.

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.

A Majorana spinor equals its transform under charge conjugation, so its spinor components are not independent of their conjugates. In a real two-dimensional gamma matrix representation it can be taken real, with anticommuting Grassmann variables as components. Each target index labels a separate worldsheet Majorana fermion; it is not itself a target-space spinor.

Use worldsheet signature $(-,+)$ and the real gamma matricesTake $\bar{\psi }={\psi }^{T}C$ with $C={\gamma }^0$. Define the chirality matrix with the chosen worldsheet orientation byIt squares to one and anticommutes with each gamma matrix. Chirality is its eigenvalue, selected by the chiral projectors. Its overall sign is conventional.

Writing $\psi =({\psi }_{+},{\psi }_{-}{)}^T$, the fermion equation of motion ${\gamma }^{\mu }{\partial }_{\mu }\psi =0$ becomesHenceThe first profile travels toward increasing $\sigma$, and the second toward decreasing $\sigma$. Thus the choice of chirality orientation realizes the requested worldsheet chirality and propagation direction correspondence; reversing the orientation interchanges the labels.

For the rigid worldsheet supersymmetry variation, let $ϵ$ be a constant Grassmann-odd Majorana spinor. With the displayed conventions,The minus sign follows from $({\gamma }^{\mu }{)}^{T}C=-C{\gamma }^{\mu }$. A second identity from the Majorana Grassmann bilinear interchange is $\bar{\chi }{\gamma }^{\nu }{\gamma }^{\mu }ϵ=\overline{ϵ}{\gamma }^{\mu }{\gamma }^{\nu }\chi$. Keeping these Grassmann signs is essential.

The bosonic kinetic variation is $2i{\partial }_{\mu }{X}_a\overline{ϵ}{\partial }^{\mu }{\psi }^a$. The two fermionic kinetic variations, using the two identities above, give the full resultFor the second equality use ${\gamma }^{\nu }{\gamma }^{\mu }=2{\eta }^{\mu \nu }-{\gamma }^{\mu }{\gamma }^{\nu }$ and symmetry of the second embedding derivative. No field equation has been used. Thus the variation is an off-shell total derivative:This is the rigid worldsheet supersymmetry boundary term. It vanishes when the boundary flux vanishes, for example on a closed worldsheet with compatible fields and variations. On a spatial circle, a constant supersymmetry parameter must also respect the chosen spin structure. Periodic fermions allow this rigid transformation; antiperiodic fermions and a constant parameter would give an antiperiodic $\delta X$ and fail to preserve a periodic bosonic embedding. The local action identity remains valid, but global symmetry requires compatible boundary data. This is the spin-structure obstruction to constant worldsheet supersymmetry.

For the spectrum, assume the usual critical RNS string, physical-state constraints, a nonzero null target momentum, and the supersymmetric GSO projection. Analyze just one chiral sector. Its matter central charge is $d+d/2$; the bc system and superconformal ghosts contribute $-26+11$. Therefore $3d/2-15=0$ gives the critical dimension of the RNS superstring $d=10$. Light-cone gauge in string theory leaves eight transverse bosonic and eight transverse fermionic string oscillator directions.

In the Neveu–Schwarz sector, the normal-ordering intercept is $a_{\mathrm{NS}}=1/2$. The massless level is $N=1/2$ and has statesThe GSO projection retains these eight transverse vector polarizations while removing the tachyonic ground state. They are spacetime bosons, although the creating string oscillator is a worldsheet fermion. In covariant language transversality and the longitudinal null-state quotient leave $10-2=8$ polarizations.

In the Ramond sector, bosonic and fermionic zero-point energies cancel, so $a_{\mathrm{R}}=0$ and the ground states are massless. The eight transverse fermion zero modes obeyThis Ramond zero-mode Clifford algebra acts on a sixteen-dimensional ground-state space. Its two chirality subspaces each have dimension eight. The GSO projection keeps one, an $8_s$ or $8_c$ spinor of $\mathrm{SO}(8)$, the rotation subgroup of the massless little group. These are spacetime fermions. Equivalently a ten-dimensional Majorana-Weyl spinor has sixteen real components before the massless Dirac equation reduces the physical polarization count to eight.

Consequently the massless chiral RNS spectrum after GSO projection satisfiesThe GSO projection is essential: without it, the transverse Ramond sector would retain both eight-dimensional spinor chiralities. These are counts for one chiral sector, not the tensor-product counts of the full closed-string spectrum.

Spinning-particle mechanics. Keep the mostly-plus metric and place the odd multiplier $\chi$ to the left. Varying the original PDF's dotted-fermion kinetic term givesThe last two equations are multiplier constraints. In deriving the fermion equation, the variation of $i\zeta \cdot \dot{\zeta }/2$ is $i\delta \zeta \cdot \dot{\zeta }$ after integration by parts; moving $\delta \zeta$ past $\chi$ supplies the sign in the interaction variation. The TeX conversion omits the dot, which would erase the fermionic symplectic structure; the PDF fixes it.

Under an infinitesimal Lorentz rotation the target vectors $x,p,\zeta$ transform together. The scalar contractions in the action are invariant. Its antisymmetric Noether charge isThe spin bilinear is antisymmetric because the Grassmann variables anticommute. Direct differentiation makes conservation particularly transparent:while the fermion equation givesThey cancel, so ${\dot{J}}^{mn}=0$. The same charge generates Lorentz transformations through the canonical and fermionic Dirac brackets, up to the convention for the sign of the antisymmetric transformation parameter.

The pseudoclassical spinning particle uses odd classical variables rather than assigning an ordinary commuting spatial vector to spin. Eliminating the fermionic momentum constraints givesThese are symmetric graded Poisson brackets. The even bilinears $S^{mn}$ have the Lorentz transformation law of an internal angular-momentum tensor. This is spin from Grassmann bilinears.

Quantization replaces the bracket by an anticommutator,The Clifford algebra is represented on spinors. The constraint $p\cdot \zeta =0$ then becomesthe massless Dirac equation. Its square gives the massless wave equation, matching $p^2=0$. The quantized spin generator is $-i[{\gamma }^m,{\gamma }^n]/4$. This spinning-particle Dirac quantization explains why the wavefunction carries a spinor index even though the original $x$ variables are vector coordinates.

The NS phase-space action. A Fourier action compatible with the displayed constraints isThe ${\chi }_r$ are odd multipliers; an overall phase can instead be absorbed in their definition. The $b_r$ here are fermionic matter modes, not the $b$ antighost of Question3. This Neveu–Schwarz Fourier phase-space action gives the bosonic oscillator symplectic form and $\{b_r^m,b_s^n{\}}_{\mathrm{PB}}=-i{\eta }^{mn}{\delta }_{r+s,0}$.

For an open string, the two worldsheet fermion chiralities are related at each endpoint. Opposite relative signs at the two ends produce, after doubling the interval, $\psi (\sigma +2\pi )=-\psi (\sigma )$. The Neveu–Schwarz sector therefore has half-integer Fourier frequencies $r\in \mathbb{Z}+1/2$, with no matter-fermion zero mode. Equal endpoint signs give the periodic Ramond sector instead. The bosonic zero-mode normalization is again ${\alpha }_0=\sqrt{2{\alpha }^{\prime }}p$.

In old covariant string quantization, all target components are retained and obeyOther oscillator brackets vanish. The momentum-labelled oscillator vacuum satisfies ${\alpha }_{k>0}|0;p⟩=b_{r>0}|0;p⟩=0$, ${\hat{p}}_m|0;p⟩=p_m|0;p⟩$, and ${\alpha }_0=\sqrt{2{\alpha }^{\prime }}p$; take its matter norm positive. The zero-point convention stated in the question givesA physical oscillator ground state has ${\alpha }^{\prime }{p}^2=a$, soIt is a scalar, not a spinor, since the NS sector has no Clifford zero modes. It is a tachyon if $a>0$, massless if $a=0$, and massive if $a<0$. In the usual limiting $a=1/2$ theory it is the unprojected NS tachyon, removed by the standard GSO projection. Calling it tachyonic before specifying the sign of $a$ would be too strong.

The half-level vector and its norm. Let $|A;p⟩=A_m(p)b_{-1/2}^m|0;p⟩$. The level is $N_{\mathrm{NS}}=1/2$. Its $L_0$ condition and the only potentially nonzero positive supercurrent condition giveIndeed $\{G_{1/2},b_{-1/2}^m\}={\alpha }_0^m$ implies $G_{1/2}|A;p⟩=\sqrt{2{\alpha }^{\prime }}p\cdot A|0;p⟩$. Higher positive $G_r$ annihilate the state. Likewise $[L_n,b_{-1/2}^m]=(1/2-n/2)b_{n-1/2}^m$ is either zero or an annihilator, so all $L_{n>0}$ conditions hold. This is the half-level Neveu–Schwarz vector state.

Its matter norm is proportional to $A^{\ast }\cdot A$. For a real, nonzero on-shell momentum in ordinary $D\ge 2$ Minkowski space, three cases exhaust the possibilities:

Thus, with the usual nonzero particle-momentum hypothesis,This is a level-specific result, not a proof of the full NS no-ghost theorem in arbitrary dimension.

At the limiting intercept the vector becomes massless, and the longitudinal polarization is a null state. It is generated by $G_{-1/2}|0;p⟩=\sqrt{2{\alpha }^{\prime }}p\cdot b_{-1/2}|0;p⟩$. Factoring out that null direction identifies $A\sim A+\kappa p$ and leaves $D-2$ positive physical polarizations of a gauge vector. In the consistent critical NS string one additionally has $D=10$; the half-level positivity argument alone does not derive that dimension.

There is a genuine zero-momentum exception if the printed word “all” is read literally. At $a=1/2$ and $p=0$, the polarization $A=(1,0,\dots )$ obeys every displayed positive-mode constraint and the mass shell, but has norm $-1$. Transversality is vacuous, and $A\sim A+\kappa p$ removes nothing. This nonzero-momentum condition in the massless vector norm test is therefore necessary for the intended endpoint assertion. Including zero momentum changes the unrestricted elementary test to strict $a<1/2$; ordinary propagating massless particle states use nonzero null momentum.

Finally, an oscillator vacuum used to build the excited state is labelled by that state's momentum. It need not separately satisfy the scalar-ground-state mass shell: imposing both ${\alpha }^{\prime }{p}^2=a$ and ${\alpha }^{\prime }{p}^2=a-1/2$ at the same momentum would incorrectly exclude all such excitations.

The sources are independent odd Grassmann variables and anticommute with the Dirac field. Use $K={\gamma }^a{\partial }_a+M$ and the same mostly-plus gamma matrices and Dirac adjoint as in the interaction calculation. The inverse $K_F^{-1}$ is selected with vacuum Feynman i-epsilon prescription boundary conditions. With integral kernels and spinor contractions understood, the exponent can be completed to a square:Translations preserve the Berezin integral, so the Gaussian generating functional for a Dirac field isThe positions of the sources matter. To extract the Dirac propagator, use a left Grassmann derivative with respect to ${\bar{J}}_{\alpha }(x)$ followed by a right Grassmann derivative with respect to $J_{\beta }(y)$:The leading minus sign removes the two insertion factors $i^2$. It can also be checked by differentiating the quadratic source exponential: its ordered second derivative is $-iZ[0,0]K_F^{-1}$.

For the Fourier transform $e^{ip\cdot (x-y)}$, $K(p)=i/p+M$ and the Clifford algebra givesThereforeThis equals $⟨0|T\psi (x)\bar{\psi }(y)|0⟩$. The fermionic time ordering is explicitlywith the minus sign supplied by exchanging odd fields. The poles put positive energy forward in time and negative energy backward, which is the antiparticle contribution. Finally, $K_x{S}_F(x,y)=iI{\delta }^{(4)}(x-y)$ checks the numerator and overall sign. All formulas are distributional limits with the indicated boundary prescription.

A Grassmann variable is an odd generator of a Grassmann algebra: ${\eta }_i{\eta }_j=-{\eta }_j{\eta }_i$, so ${\eta }_i^2=0$. For one generator, any function is $f(\eta )=a+b\eta$. The Berezin integral is the linear operationThus integration extracts a coefficient, rather than assigning a length or volume. For many generators it extracts the coefficient of the highest-degree monomial with the sign fixed by the order of the measure. Odd coefficients and Grassmann derivatives must retain their order; exchanging two odd objects changes the sign.

This operation is invariant under odd translations, because a translation only changes terms of lower degree. For an invertible ordinary matrix $S$ and $\eta =S\xi$, the Grassmann change-of-variables formula isThe inverse Jacobian determinant, rather than the ordinary commuting-variable Jacobian, compensates for the factor $\det S$ multiplying the top monomial. Integration agrees with the appropriate ordered Grassmann derivatives, but the orientation must be specified when combining barred and unbarred variables.

Use left Grassmann derivatives and the usual contractions ${\partial }^2={\partial }^{\alpha }{\partial }_{\alpha }$ and ${\bar{\partial }}^2={\bar{\partial }}_{\dot{\alpha }}{\bar{\partial }}^{\dot{\alpha }}$. Set $a={\theta }^1,b={\theta }^2,c={\bar{\theta }}^{\dot{1}},d={\bar{\theta }}^{\dot{2}}$. The printed epsilon convention givesThe reversal of the barred contraction is essential. The left Grassmann derivative obeys the graded Leibniz rule, so ${\partial }_a(ab)=b$ but ${\partial }_b(ab)=-a$. With the index-raising convention,Applying these to the displayed Grassmann algebra monomials gives $A=B=-4$. Since every antisymmetric two-index product is proportional to epsilon, the $12$ and $\dot{1}\dot{2}$ components giveThus $C=-2$ and $D=2$.

In the remaining contraction, moving the first barred Grassmann variable past the second unbarred one introduces a minus sign. Inserting the two spinor identities then givesThe trace normalization is the one explicitly supplied in the paper. These two-component superspace contraction signs therefore giveAs a direct check in the mostly-minus convention, ${\sigma }^0=I$ gives $(ac+bd{)}^2=-2abcd$, whereas $(\theta \theta )(\bar{\theta }\bar{\theta })=-4abcd$. Their ratio is $+1/2$. A convention with $\mathrm{Tr}({\sigma }^{\mu }{\bar{\sigma }}^{\nu })=-2{\eta }^{\mu \nu }$ would change this last metric-relative sign; it is not the printed trace convention.

Introduce $y^{\mu }=x^{\mu }+i\theta {\sigma }^{\mu }\bar{\theta }$ and retain the same Grassmann variables. The supersymmetric derivatives in chiral coordinates follow from the left Grassmann derivative chain rule. Differentiating $y$ givesThe minus sign in the second relation comes from moving the odd Grassmann derivative past $\theta$. Hence, as operators on a superfield expressed in $(y,\theta ,\bar{\theta })$,Substitution into the two supersymmetric covariant derivatives adds the two unbarred spacetime terms and cancels the two barred ones:These operator equalities prove both requested actions on $V$. In particular, a chiral superfield becomes independent of $\bar{\theta }$ at fixed $y$. The TeX aid corrupts the second formula by replacing its ordinary barred derivative with a covariant one; the original PDF has the ordinary $-{\bar{\partial }}_{\dot{\alpha }}$.

The Abelian field-strength chiral projection is linear in the vector superfield, so isolate the terms containing the gaugino $\lambda$ and the $D$ auxiliary field. Replacing $x$ by $y-i\theta \sigma \bar{\theta }$ leaves these terms unchanged: the shift of the gaugino term would contain three barred Grassmann variables, and the shift of the $D$ term would contain three of each chirality, hence both vanish. Their contribution is thereforeThe $2i({\sigma }^{\mu }\bar{\theta }{)}_{\alpha }{\partial }_{\mu }$ part of the supersymmetric covariant derivative also adds a third barred factor, so it vanishes on these two terms. The ordinary left Grassmann derivative, with ${\partial }_{\alpha }(\theta \theta )=2{\theta }_{\alpha }$, givesAt fixed $y$, $\bar{D}=-\bar{\partial }$ and ${\bar{D}}^2(\bar{\theta }\bar{\theta })=-4$. Thus the chiral field-strength superfield hasThe remaining components come from the other independent terms in the vector superfield; linearity ensures they cannot alter the two coefficients just calculated. This proves the requested components without computing the omitted terms. The gaugino phase and the sign of the vector component are those in the printed Wess-Zumino gauge expansion.

$W_{\alpha }$ is a fermionic chiral spinor superfield. Its chirality follows directly from the Abelian field-strength chiral projection:There are only two independent barred supersymmetric covariant derivatives, and their equal-chirality anticommutators vanish. Every product of three barred derivatives therefore vanishes. Equivalently, the previous calculation has no independent barred Grassmann variable at fixed $y$. The free undotted index makes this a chiral spinor superfield, rather than a scalar chiral superfield; its lowest component is the odd gaugino ${\lambda }_{\alpha }$.

Use the Minkowski metric, path-integral weight $e^{iS}$ and the Abelian gauge theory transformation $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\omega$. The gauge functional $F[A]=\partial \cdot A+A^2$ varies asThus the Faddeev-Popov operator is ${\mathcal{M}}_A=\square +2A\cdot \partial$. It depends on the gauge field despite the gauge group being Abelian: the ghosts interact because this gauge condition is nonlinear.

Choose the gauge-fixing fermion $\Psi =\int d^{4}x\bar{c}(F[A]+\xi h/2)$. The gauge-fixed action $S=S_{\mathrm{Maxwell}}+s\Psi$ isHere the tensor $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$ is distinct from the scalar gauge functional $F[A]$. With $\xi =0$, integrating over $h$ imposes the exact printed constraint. For nonzero $\xi$, eliminating $h=-F[A]/\xi$ instead givesThis version displays the additional gauge-dependent cubic and quartic gauge-field vertices as well as the ghost interaction; the strict condition is its $\xi \to 0$ limit.

For the Fourier transform convention $c(x)=\int d^{4}p(2\pi {)}^{-4}{e}^{-ipx}c(p)$, the quadratic ghost kernel is $p^2$. With the ordering $⟨c(p)\bar{c}(q)⟩$,The term $-2\bar{c}{A}^{\mu }{\partial }_{\mu }c$ has Fourier coefficient $2i{p}_{\mu }$, where $p$ is the incoming ghost momentum. Multiplication by $i$ in the Feynman rule givesThese signs refer to the displayed action, Fourier convention and ghost ordering. Reversing the ghost/antighost convention changes corresponding signs consistently. A closed ghost loop has the additional minus sign from Grassmann variables.

With left Grassmann derivatives, differentiating $\theta {\sigma }^{\mu }\bar{\theta }$ with respect to $\bar{\theta }$ introduces a minus sign. ConsequentlyAt fixed $y$, the chirality condition becomes independence of $\bar{\theta }$. There are only two independent Grassmann variables ${\theta }^{\alpha }$, so the chiral-superfield component expansion terminates:Here $\varphi$ is a complex scalar field, $\psi$ a Weyl spinor, and $F$ a complex auxiliary field. Its four real off-shell bosonic components, two in $\varphi$ and two in $F$, match the four real off-shell fermionic components. Equivalently, in ordinary coordinates the complete expansion is fixed without ambiguous contraction signs byThe exponential terminates because of the Grassmann algebra.