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The remaining non-derivative component term in the Lagrangian is the Yukawa interactionEquivalently the interaction Hamiltonian has the opposite displayed sign. The first term is the two-component fermion mass, and the second its scalar-fermion interaction. A phase redefinition of $\psi$ makes the mass coefficient nonnegative, so the physical Weyl spinor mass isIt is again independent of $\alpha$, the choice of root and any sign of a real $M_v$. This distinguishes the complex bilinear coefficient from the nonnegative physical mass. In the pure-linear supersymmetry-breaking branch $W^{\prime \prime }=0$, the fermion is the massless Goldstino.

For a nonzero common mass,This agrees with supersymmetric mass degeneracy in an unbroken chiral superfield multiplet. The two real scalar modes have the same mass as the two physical polarizations of the massive fermion, as the rest-frame multiplet construction predicts.

At $\Delta =0$ both masses vanish; the numerical ratio is $0/0$ and is undefined, but the equality of masses still expresses the unbroken supersymmetry. Equal zero masses in the exceptional flat, pure-linear breaking model do not prove unbroken supersymmetry: there the nonzero auxiliary expectation and positive energy in part (c) are the decisive tests.

Take the muon mass to be $m_{\mu }$, and use the quantum electrodynamics convention ${\mathcal{L}}_{\mathrm{int}}=-e\bar{\psi }{\gamma }^{\mu }\psi A_{\mu }$ with $e>0$. Let $p_1,p_2$ be the incoming muon and antimuon momenta and $k_1,k_2$ the outgoing photon momenta, so $p_1+p_2=k_1+k_2$. External particles are on shell: $p_1^2=p_2^2=m_{\mu }^2$ and $k_1^2=k_2^2=0$. The two tree-level Feynman diagrams are the two orders in which the photon legs attach to the muon line. There is no three-photon QED vertex, and therefore no single-photon annihilation diagram into two photons at this order.

The QED Feynman rules needed here are the vertex $-ie{\gamma }^{\mu }$, the Dirac propagatoran incoming muon spinor $u(p_1)$, an incoming antimuon adjoint $\bar{v}(p_2)$, and an outgoing photon polarization ${ϵ}_{\mu }^{\ast }(k)$. At a vertex enforce four-momentum conservation. Spinor products are ordered along the fermion line; exchanging the external photons introduces no relative minus sign. Define $R(q)=(/q+m_{\mu })/(q^2-m_{\mu }^2+i0)$ and $/ϵ={\gamma }^{\mu }{ϵ}_{\mu }$.

With the convention that a diagram contributes $i\mathcal{M}$, the two muon-antimuon annihilation amplitudes areTheir denominators are respectively $t-m_{\mu }^2+i0$ and $u-m_{\mu }^2+i0$, where $t=(p_1-k_1{)}^2$, $u=(p_1-k_2{)}^2$. A different consistent overall amplitude-phase convention has no physical effect.

A physical photon polarization vector obeys $k\cdot ϵ=0$, and represents an equivalence class $ϵ\sim ϵ+\alpha k$. For a real null momentum, adding $\alpha k$ preserves transversality. One can additionally choose transverse spatial vectors with ${ϵ}^0=0$ and ${ϵ}^{\ast }\cdot ϵ=-1$; there are two independent physical polarizations. The equivalence class removes the unphysical longitudinal direction, rather than imposing four independent physical polarization states.

The invariance of the total amplitude is a two-photon fermion Ward identity. Replace ${ϵ}_1^{\ast }$ by $k_1$ and put $q_t=p_1-k_1$, $q_u=p_1-k_2=k_1-p_2$. The external Dirac equations giveAway from propagator poles, $R(q)(/q-m_{\mu })=(/q-m_{\mu })R(q)=I$; the identity extends with the common Feynman prescription. ThereforeThe two terms in the bracket then give $-\bar{v}/{ϵ}_2^{\ast }u$ and $+\bar{v}/{ϵ}_2^{\ast }u$, which cancel. Exchanging photon labels proves the second Ward identity. Thus the sum is unchanged under ${ϵ}_i\mapsto {ϵ}_i+\alpha k_i$ for either photon. Neither diagram is generally gauge-invariant separately. Physically, longitudinal pure-gauge polarization does not couple to the observable scattering amplitude; only the two transverse photon degrees of freedom contribute.

For Compton scattering write the incoming momenta as $p,k$ and outgoing momenta as $p^{\prime },k^{\prime }$, with $p+k=p^{\prime }+k^{\prime }$. The outgoing muon contributes $\bar{u}(p^{\prime })$, the incoming photon contributes $ϵ(k)$ without conjugation, and the outgoing photon contributes ${ϵ}^{\prime \ast }(k^{\prime })$. The two tree-level Compton amplitudes areThese are the lower two diagrams. Their intermediate muon momenta are $p+k$ and $p-k^{\prime }$, with denominators $s-m_{\mu }^2+i0$ and $u-m_{\mu }^2+i0$. The first attaches the incoming photon before the outgoing photon along fermion flow; the second reverses that attachment order. Both contribute at order $e^2$ in the amplitude.

Compton scattering has one incoming and one outgoing muon and one photon on each side; annihilation has an incoming particle-antiparticle pair and two outgoing photons. Accordingly the external spinors/polarizations and physical Mandelstam channels differ, although both arise from the same two-vertex QED matrix element. Crossing symmetry relates them by continuing $p_2\mapsto -p^{\prime }$ and one outgoing photon momentum $k_1\mapsto -k$, with the corresponding external wavefunction replacements. The annihilation $t$ channel becomes the Compton $s$ channel and the other channel becomes its $u$ channel. Identical final photons require a factor $1/2!$ in an annihilation phase-space integral over an otherwise double-counted full final-state region; there is no analogous final-state identical-particle factor for the muon-photon Compton final state. Gauge cancellation holds for the summed Compton amplitude as well.

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.

Use metric signature $+---$ and the parity matrix $P^{\mu }{}_{\nu }=\mathrm{diag}(1,-1,-1,-1)$. If a Dirac field transforms as ${\psi }^{\prime }(x)={\eta }_P{S}_P\psi (x_P)$, the chain rule changes the spatial derivatives' signs. Covariance of its Dirac equation therefore requiresThe Clifford algebra gives precisely these identities for $S_P={\gamma }^0$. In an irreducible Dirac representation any other solution differs by a scalar phase, because its ratio with ${\gamma }^0$ commutes with all gamma matrices. With $|{\eta }_P|=1$, the Dirac mass bilinear and kinetic term consequently obeyThe factor ${\gamma }^0$ compensates the sign reversal of spatial derivatives. A transformation containing only the argument change would not preserve the massive Dirac equation. This establishes parity symmetry of the free or vector-coupled Dirac theory; it does not make a chiral weak interaction parity symmetric.

For charge conjugation, transpose the adjoint Dirac equation to obtain $(i{\gamma }^{\mu T}{\partial }_{\mu }+m){\bar{\psi }}^T=0$. Multiplying by $C$ gives the original Dirac equation for ${\psi }^c=C{\bar{\psi }}^T$ ifOne may choose $C$ unitary; in four dimensions a conventional choice in a standard Dirac or chiral basis is $C=i{\gamma }^2{\gamma }^0$, with $C^T=-C$. Its overall phase and ${\eta }_C$ do not affect the following bilinear transformation. Reordering the Grassmann fermion fields gives, for a color matrix $X$ commuting with the gamma matrices,Thus a vector color current changes sign and transposes its color generator under charge conjugation.

Write the Hermitian color connection as ${\mathcal{A}}_{\mu }=A_{\mu }^a{T}^a$. The interaction coming from $D_{\mu }={\partial }_{\mu }+ig{\mathcal{A}}_{\mu }$ is $-g\bar{\psi }{\gamma }^{\mu }{\mathcal{A}}_{\mu }\psi$. Its charge-conjugation symmetry requires ${\mathcal{A}}_{\mu }^C=-{\mathcal{A}}_{\mu }^T$. Parity acts on the Lorentz index as on a vector connection, so combining the two givesIn components, ${\mathcal{A}}_0\mapsto -{\mathcal{A}}_0^T(x_P)$ and ${\mathcal{A}}_i\mapsto +{\mathcal{A}}_i^T(x_P)$. The transpose is a color-space transpose, independent of the spinor matrix $C$. Intrinsic fermion phases cancel between a field and its adjoint.

Define the matrix gauge field strength unambiguously byThe derivative terms transform with the expected parity factors and a minus transpose. For the commutator, the crucial identity is $[X^T,Y^T]=-[X,Y{]}^T$, so the nonlinear term has the same transformation as the derivative terms. This is the CP transformation of non-Abelian field strength:For example, ${\mathcal{F}}_{0i}\mapsto +{\mathcal{F}}_{0i}^T(x_P)$ while ${\mathcal{F}}_{ij}\mapsto -{\mathcal{F}}_{ij}^T(x_P)$. With $\mathrm{tr}(T^a{T}^b)=\frac{1}{2}{\delta }^{ab}$, the color sum in the theta operator is twice a trace of two field strengths. The two charge-conjugation minus signs cancel, and transposition reverses the trace factors without changing their trace. The four parity matrices contracted with the Levi-Civita symbol contribute $\det P=-1$. HenceThe theta operator is CP odd; a generic fixed nonzero coefficient violates CP. Setting the coefficient to zero removes this term. Although the density is locally a total derivative, nontrivial gauge topology means that it need not be irrelevant to the quantum theory. If a properly normalized theta angle is identified periodically, invariance at special values equivalent to their negatives is a separate global question; it does not change the CP odd transformation of the local operator.

Put $s_W=\sin {\theta }_W$ and $c_W=\cos {\theta }_W$. The left-handed doublet haswhile the right-handed charged-lepton singlet has $T_R^3=0$ and $Q_R=-1$. There is no weakly coupled right-handed neutrino in the stated field content. For a fermion $f$, write $g_L^f=T_{3L}^f-Q_f{s}_W^2$ and $g_R^f=T_{3R}^f-Q_f{s}_W^2$, with $g_R^{\nu }=0$ here. Since ${\bar{f}}_L{\gamma }^{\mu }{f}_L=\bar{f}{\gamma }^{\mu }{P}_{L}f$ and similarly for $R$,Consequently the neutral-current vector and axial couplings in the specified overall normalization areThese coefficients accompany $g/c_W$. A convention with $g/(2c_W)$ uses twice these vector and axial coefficients; mixing the two conventions would make the widths wrong by a factor of four. For the neutrino, $c_V-c_A{\gamma }^5=\frac{1}{2}{P}_L$, so the chiral projection is already built into the vertex.

For either massless final pair with momenta $k_1,k_2$, $p=k_1+k_2$, the invariant amplitude is, up to an irrelevant overall phase,The spin-summed fermion tensor isThe symmetric part, using the four-gamma trace, isThe vector-axial interference is antisymmetric in $\mu ,\nu$, so it drops out of the symmetric $Z$ polarization sum. Also $p_{\mu }{T}_f^{\mu \nu }=0$ for massless external fermions, by their Dirac equations; hence the $p_{\mu }{p}_{\nu }/m_Z^2$ part contributes zero. Using $k_1\cdot k_2=m_Z^2/2$ givesThe massive-vector spin average supplies the factor $1/3$ for the three spin states of the massive $Z$. The massless two-body Lorentz-invariant phase space is $d{\Phi }_2=d\Omega /(32{\pi }^2)$, and so $\int d{\Phi }_2=1/(8\pi )$. The spin-averaged amplitude above is angle-independent in the rest frame, givingThe massless leptonic Z decay widths for one charged-lepton flavor and one active neutrino flavor respectively areThere is no color factor for leptons and no extra final-state identical-particle factor for a particle-antiparticle pair. The neutrino projectors already exclude sterile helicity states, so no additional factor of two is needed. These are tree-level massless widths; electroweak radiative corrections and charged-lepton masses have not been included. If desired, using $G_F=g^2/(4\sqrt{2}{m}_W^2)$ and $m_W=c_W{m}_Z$ converts the common prefactor to $G_F{m}_Z^3/(12\sqrt{2}\pi )$.

The hierarchy problem begins with a large separation of scales: the weak scale is much smaller than possible unification or gravitational scales. A fundamental scalar mass is especially sensitive to a much heavier scale. In an effective description with ultraviolet cutoff $\Lambda$, one schematically findsThe coefficient $c$ contains gauge, scalar and Yukawa couplings, with different signs for bosonic and fermionic loops. The explicit quadratic cutoff dependence is regulator-dependent, but a heavy physical particle coupled to the Higgs produces a threshold correction of order its mass squared. That heavy-threshold sensitivity is the physical difficulty.

The technical hierarchy problem asks whether a small scale, once chosen, remains stable under such radiative corrections. Maintaining $m_H^2\ll {\Lambda }^2$ by cancelling unrelated bare and loop contributions, and readjusting the cancellation at successive orders, is the naturalness concern. It is different from explaining why the small scale was chosen in the first place. Supersymmetry primarily supplies protection against the technical instability; a complete explanation of the origin of supersymmetry-breaking scales needs additional dynamics.

The relevant comparison of mass types is:

In exact supersymmetry, each supermultiplet has matched bosonic and fermionic degrees of freedom, with their interaction strengths related. Opposite loop signs then give supersymmetric cancellation of quadratic divergences. This is not just equality of state counts: the supersymmetric relation between quartic and Yukawa couplings is also essential. A schematic paired-loop contribution has the high-momentum formFor equal partner masses the displayed contributions cancel. With a small splitting, the difference falls as $(m_F^2-m_B^2)/p_E^4$, so the remaining ultraviolet sensitivity is logarithmic, proportional to the splitting rather than to ${\Lambda }^2$.

Realistic partners need not have exactly equal masses. Soft supersymmetry breaking permits scalar squared masses, gaugino masses and suitable trilinear interactions without restoring the unwanted quadratic sensitivity. Typically,up to coefficients and threshold details. Squarks and sleptons can therefore be heavier than their chiral fermion partners while the scalar sector remains stable against the much larger ultraviolet scale. Nevertheless, very large soft masses, particularly in the Higgs-coupled sector, leave large finite or logarithmic corrections and require tuning. This soft scalar-mass sensitivity remains after the quadratic divergence has cancelled. Supersymmetry solves the quadratic radiative instability when breaking is suitably soft; it does not make arbitrarily heavy superpartners natural or explain the entire hierarchy by itself.

Assume $\lambda >0$ and $v>0$. The vacuum manifold is the sphere $|\varphi |=v$. Choose ${\varphi }_0=v{e}_3$. For the given generators, $t^3{\varphi }_0=0$ whereas $t^1{\varphi }_0$ and $t^2{\varphi }_0$ are nonzero and independent. The Adjoint representation of SU(2) rotates this sphere, and the full connected stabilizer subgroup of $v{e}_3$ is generated by $t^3$. Hence the classical symmetry pattern isThe central element of SU(2) acts trivially on the triplet and belongs to this stabilizer; it does not add a separate broken direction. The gauge-field-free representative has zero gauge field strength and constant scalar magnitude. Locally near this nonzero classical vacuum, unitary gauge aligns the triplet along the third axis.

First make the gauge normalization explicit. Write the gauge kinetic term as $-\mathcal{N}{F}_{\mu \nu }^a{F}^{a\mu \nu }/4$, where $\mathrm{Tr}(t^a{t}^b)=\mathcal{N}{\delta }^{ab}$ if $F=F^a{t}^a$ is the matrix used inside the printed trace. Canonical normalization of a gauge kinetic term givesFor a conventionally normalized component trace, $\mathcal{N}=1$ and $g_c=g$. If the trace is the ordinary matrix trace in the displayed three-dimensional representation, $\mathrm{Tr}(t^a{t}^b)=2{\delta }^{ab}$ and $\mathcal{N}=2$. The PDF does not specify which trace convention is intended; both are covered by this formula.

In unitary gauge, the gauge covariant derivative isExpanding this kinetic term and the scalar potential gives the complete physical-field LagrangianHere a square of a vector means its contraction with the Minkowski metric, in signature $(+---)$. The labels 1 and 2 represent massive vectors; 3 represents the surviving Abelian vector.

To display their charge and all interactions more clearly, put $a_{\mu }={\mathcal{A}}_{\mu }^3$, $W_{\mu }^{\pm }=({\mathcal{A}}_{\mu }^1\mp i{\mathcal{A}}_{\mu }^2)/\sqrt{2}$, and defineWith the sign of gauge field strength printed in the paper, ${\mathcal{F}}^3=f+i{g}_{c}C$ and $G^{\pm }=({\mathcal{F}}^1\mp i{\mathcal{F}}^2)/\sqrt{2}$. Thus the same Lagrangian becomesThe physical masses areIn particular, literal adjoint matrix trace gives $m_W=gv/\sqrt{2}$; the standard canonical component convention gives $m_W=gv$. These are descriptions with differently normalized couplings, not different physical spectra.

The complex vector pair carries opposite charges under the unbroken U(1) gauge symmetry. The gauge field strength terms contain $a{W}^{+}{W}^{-}$ cubic interactions, $aa{W}^{+}{W}^{-}$ quartic interactions, and four-vector interactions involving the charged fields. There is no pure Abelian cubic or quartic self-interaction. The Higgs mode $\eta$ has cubic and quartic scalar interactions and couples through $2g_c^{2}v\eta W^{+}{W}^{-}+g_c^2{\eta }^2{W}^{+}{W}^{-}$. It is neutral and has no tree-level $\eta aa$ interaction. This is the physical charged-vector Lagrangian for an adjoint SU2 Higgs model.

If the angular fields ${\pi }_1$ and ${\pi }_2$ were retained, their vanishing potential masses would identify the two Goldstone bosons of the ungauged triplet. In the gauge theory they mix with the broken-direction gauge fields and can be removed by gauge fixing; the Higgs mechanism uses them as the longitudinal polarizations of the two massive vectors. They are not additional physical massless scalars. The physical degrees of freedom are conserved: $3+3\times 2=9$ before rearrangement, and $1+2\times 3+2=9$ afterwards.

This is not the Standard Model electroweak interaction. It has three original gauge bosons, leaving two massive charged vectors and one massless neutral vector, with no massive neutral Z boson. The Standard Model instead has $SU(2{)}_L\times U(1{)}_Y$, a complex Higgs doublet, and three massive vectors plus the photon. Its charge is $Q=T_3+Y$, not just the surviving $T_3$. Adding fermions cannot supply the missing gauge generator or turn the surviving neutral vector into both a photon and a Z boson. In particular, the usual right-handed fermions in the Standard Model are SU(2) singlets; they would have zero charge if only $T_3$ were available, instead of the charges produced by hypercharge.

Particle gauge fixing and BRST. Put $C=(p^2+{\mu }^2)/2$. The classical constraint generates ${\delta }_{\epsilon }{x}^m=\epsilon p^m$, ${\delta }_{\epsilon }{p}_m=0$, ${\delta }_{\epsilon }e=\dot{\epsilon }$. Consequently the variation of $e-\bar{e}$ along a gauge orbit is the operator ${\partial }_t$ on $\epsilon$. A path integral restricted to that gauge must include its Faddeev-Popov determinant; anticommuting FP ghosts represent the determinant, rather than its inverse. Constant gauge zero modes and any proper-time modulus must be treated separately, so the relevant determinant is $\det {}^{\prime }{\partial }_t$.

A convenient point-particle Faddeev–Popov ghost action at $\bar{e}=1$ isThe FP ghost normalization has been chosen to give the graded Poisson bracket $\{b,c\}=-i$, with the bracket symmetric on two odd variables. The particle BRST charge and Klein–Gordon constraint arewhere the odd constant parameter $\Lambda$ is placed on the left. The product $\Lambda Q$ is even. Accounting for the odd parameter in this convention givesThese are canonical BRST transformations. Direct variation checks the sign: the bosonic kinetic term changes by $\Lambda (\dot{c}{p}^2+cp\cdot \dot{p})$, while the FP ghost term changes by $-\Lambda C\dot{c}$. Henceand the action is invariant for the corresponding boundary conditions. The BRST charge is conserved because $C$ commutes with the gauge-fixed Hamiltonian and $c$ is constant on the FP ghost equation of motion.

Quantization gives $\{\hat{b},\hat{c}{\}}_{+}=1$ and ${\hat{c}}^2=0$. Since $C$ commutes with the FP ghosts,It generates the same variations by $\delta \hat{F}=-i[\hat{F},\Lambda \hat{Q}]$, with an ordinary commutator against the even generator. Represent $\hat{c}$ by multiplication and $\hat{b}$ by differentiation. On a ghost-number-zero wavefunction $\Psi (x)$, $\hat{Q}\Psi =0$ is exactly $\hat{C}\Psi =0$, namelythe Klein-Gordon equation for the mostly-plus metric. The specified FP ghost sector matters: for an unrestricted wavefunction ${\Psi }_0+c{\Psi }_1$, BRST closure only constrains ${\Psi }_0$, not every component independently. The complete physical prescription uses BRST cohomology, not an assertion that every vector in the entire ghost-extended kernel is a new Klein–Gordon particle.

String oscillator algebra and FP ghost Virasoro generators. Use the same bosonic brackets as above and the odd bracketsQuantization turns them intoThe $b,c$ here are worldsheet ghost fields, not the matter fermion oscillators used in the NS question.

Contract $b_m$ with the two $c$ factors in the cubic FP ghost term of the given BRST operator. The first contraction contributes $-\frac{1}{2}{\sum }_q(m-q)c_{-q}{b}_{m+q}$ and the second contributes $+\frac{1}{2}{\sum }_p(p-m)c_{-p}{b}_{p+m}$. Their sum is ${\sum }_q(q-m)c_{-q}{b}_{m+q}$. Reordering and normal ordering therefore give the ghost oscillator Virasoro generatorsFor $m\ne 0$ these formulas have no additive ambiguity. At $m=0$, moving annihilators past creators produces infinite zero-point sums. Their regularized finite constant shifts $L_0$ and corresponds to a term proportional to $c_0$ in $Q$. The convention fixed below has a ghost oscillator vacuum weight $-1$, or equivalently the usual intercept-one shift. Whether that shift is written outside the normally ordered sum or incorporated in its definition is a convention; it must not be omitted twice or counted twice. In the chosen oscillator convention this is explicitlyEquivalently, the normally ordered BRST charge contains the intercept term $-c_0$; the formal un-ordered operator in the question includes that choice only after its ordering prescription is fixed.

The oscillator brackets directly giveIf $Q^2=0$, then $[L_n,Q]=[\{b_n,Q\},Q]=[b_n,Q^2]=0$. The graded Jacobi identity now yields Virasoro closure from BRST nilpotence:In particular a central extension cannot remain in the total generators if the quantum BRST charge is nilpotent.

Low-level descendant calculation. Let $|\Omega ;p⟩$ be the oscillator ground state, annihilated by ${\alpha }_{n>0}$, $b_{n\ge 0}$ and $c_{n>0}$. All $L_{n>0}$ annihilate it. The nonzero-mode FP ghost generators act asFor example $L_1(-b_{-1}{c}_0|\Omega ⟩)=-2b_0{c}_0|\Omega ⟩=-2|\Omega ⟩$. Similarly $L_2$ on the two level-two terms gives $-8|\Omega ⟩$ and $-9|\Omega ⟩$, hence $-17|\Omega ⟩$. These signs are essential: FP ghosts have an indefinite pairing.