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For two independent left-handed Weyl spinors $\chi$ and $\eta$, a Dirac mass term isThe four-component Dirac field ${\Psi }_D=({\chi }_{\alpha },{\bar{\eta }}^{\dot{\alpha }}{)}^T$ has independent particle and antiparticle degrees of freedom; after a phase redefinition making $m$ real, the same term is $-m{\bar{\Psi }}_D{\Psi }_D$. The two left-handed spinors can carry opposite charges under a conserved U(1) symmetry, $\chi \mapsto e^{iq\omega }\chi$, $\eta \mapsto e^{-iq\omega }\eta$. A gauge-invariant Dirac mass term contracts conjugate gauge group representations.

For one left-handed Weyl spinor, a Majorana mass term instead has the formIts self-contraction need not vanish, by the Weyl spinor bilinear exchange identity. The factor $1/2$ avoids double counting identical fields in variation. The associated four-component Majorana fermion ${\Psi }_M=({\chi }_{\alpha },{\bar{\chi }}^{\dot{\alpha }}{)}^T$ satisfies ${\Psi }_M^c={\Psi }_M$, so its particle is its own antiparticle. A Majorana mass term breaks continuous fermion-number phase symmetry for $\chi$, and is allowed only if its gauge indices admit an invariant symmetric contraction, as for a neutral field or a real gauge group representation. An isolated field of nonzero U(1) charge cannot have it. More generally, the mass matrix of several left-handed Weyl spinors is symmetric, since ${\chi }_i{\chi }_j={\chi }_j{\chi }_i$.

The barred notation denotes the antichiral supersymmetric covariant derivative. Write barred spinor indices as dotted indices, and use the stated plus-sign convention with left Grassmann derivatives:Their supercovariant derivative algebra with left derivatives gives $\{{\bar{D}}_{\dot{\alpha }},{\bar{D}}_{\dot{\beta }}\}=0$ and $\{D_{\alpha },{\bar{D}}_{\dot{\beta }}\}=2i{\sigma }_{\alpha \dot{\beta }}^{\mu }{\partial }_{\mu }$. The constraint $\boxed{{\bar{D}}_{\dot{\alpha }}\Phi =0}$ defines a chiral superfield, also called an L-superfield.

To solve the constraint, put $y^{\mu }=x^{\mu }+i\theta {\sigma }^{\mu }\bar{\theta }$. For a left Grassmann derivative, ${\partial }_{{\bar{\theta }}^{\dot{\alpha }}}(\theta {\sigma }^{\mu }\bar{\theta })=-{\theta }^{\beta }{\sigma }_{\beta \dot{\alpha }}^{\mu }$. Therefore ${\bar{D}}_{\dot{\alpha }}{y}^{\mu }=0$ and ${\bar{D}}_{\dot{\alpha }}{\theta }^{\beta }=0$, so $\bar{D}$ becomes ordinary differentiation with respect to $\bar{\theta }$ at fixed $y$. Its general solution is the chiral-superfield component expansionHere $A$ is a complex scalar, $\chi$ a Weyl spinor, and $F$ a complex auxiliary field. Before imposing field equations of motion this is four real bosonic and four real fermionic components. Chirality means independence of $\bar{\theta }$ at fixed $y$, rather than at fixed $x$. The differential supercharges $Q_{\alpha }={\partial }_{{\theta }^{\alpha }}-i{\sigma }_{\alpha \dot{\beta }}^{\mu }{\bar{\theta }}^{\dot{\beta }}{\partial }_{\mu }$ and ${\bar{Q}}_{\dot{\alpha }}={\partial }_{{\bar{\theta }}^{\dot{\alpha }}}-i{\theta }^{\beta }{\sigma }_{\beta \dot{\alpha }}^{\mu }{\partial }_{\mu }$ anticommute with both covariant derivatives, so a supersymmetry transformation preserves the chiral constraint.

The supplied matrices are the chiral gamma-matrix representation. The Pauli matrix multiplication law implies $\{{\sigma }^i,{\sigma }^j\}=2{\delta }^{ij}{I}_2$. Block multiplication gives $({\gamma }^0{)}^2=I_4$, $\{{\gamma }^0,{\gamma }^i\}=0$, and $\{{\gamma }^i,{\gamma }^j\}=-2{\delta }^{ij}{I}_4$. Hence their Clifford algebra relation isSince partial derivatives commute, multiplying the Dirac equation by $i{\gamma }^{\nu }{\partial }_{\nu }+m$ givesThus every component of the Dirac spinor satisfies $\boxed{(\square +m^2)\psi =0}$, the Klein-Gordon equation with the same mass.

For a Lorentz transformation $x^{\prime }=\Lambda x$, set ${\psi }^{\prime }(x^{\prime })=S(\Lambda )\psi (x)$. The derivative transforms as ${\partial }_{\mu }^{\prime }=({\Lambda }^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$. With the given spinor-matrix identity,This proves the Lorentz covariance of the Dirac operator explicitly.

For parity, let $x_P=(x^0,-\mathbf{x})$ and ${\psi }_P(x)={\gamma }^0\psi (x_P)$. Spatial differentiation brings a minus sign, which is canceled by ${\gamma }^i{\gamma }^0=-{\gamma }^0{\gamma }^i$. ThereforeSince ${\gamma }^0$ is Hermitian and squares to one, the Dirac adjoint transforms asThe chirality matrix ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$ anticommutes with ${\gamma }^0$. The parity transformation of a Dirac bilinear consequently givesThe first bilinear is a Lorentz scalar even under parity, and the second is a pseudoscalar.

In this representation, ${\gamma }^5=\mathrm{diag}(-I_2,I_2)$. Its square is one, so the chiral projectors $P_{\pm }=(1\pm {\gamma }^5)/2$ are idempotent, mutually orthogonal and sum to the identity. They select ${\psi }_L=P_{-}\psi$ and ${\psi }_R=P_{+}\psi$, the two Weyl spinors. Because ${\gamma }^{\mu }{P}_{\pm }=P_{\mp }{\gamma }^{\mu }$, a solution of the massless Dirac equation has each chiral component solving that equation separately. In two-component form,The two chirality sectors are thus independent under massless propagation and proper Lorentz transformations. A nonzero Dirac mass term couples them, while parity exchanges them. This explains why the projectors are particularly useful in the massless theory.

Write $t=\log \mu$ and use ${\alpha }_i=g_i^2/(4\pi )$. The renormalization-group beta function impliesIntegrating the one-loop running coupling equation givesHere the sign of ${\beta }_i$ is the sign convention in the question: a negative coefficient makes the inverse coupling increase towards high energy.

For gauge coupling unification, let $L=\log (M_{\mathrm{GUT}}/M_Z)/(2\pi )$ and let $a_G$ be the common inverse coupling at the unification scale. The normalized hypercharge coupling is $(5/3){\alpha }_1$, soSubtract the second relation from the first to determine $L$, then subtract it from the third. Provided $3{\beta }_1/5-{\beta }_2\ne 0$, this provesThe factors $3/5$ belong to the inverse normalized coupling; dropping them changes the prediction.

Use the Standard Model gauge group $SU(3{)}_C\times SU(2{)}_L\times U(1{)}_Y$ and convention $Q=T_3+Y$. One fermion family and one complex Higgs doublet have the following Standard Model representations. The left-handed quark and lepton doublets contain two Weyl spinors each; the right-handed fields are single Weyl species.The row $u_R$ denotes the right-handed up quark. Equivalently an all-left-handed table replaces the three right-handed fields by $u_R^c:(\overline{3},1{)}_{-2/3}$, $d_R^c:(\overline{3},1{)}_{1/3}$, and $e_R^c:(1,1{)}_1$; these are alternative descriptions, not extra fermions. The minimal model has no right-handed neutrino. For completeness, the family-independent gauge fields are the spin-one gluons $(8,1{)}_0$, weak gauge bosons $(1,3{)}_0$, and hypercharge gauge boson $(1,1{)}_0$; they are not counted once per family.

For the Standard Model one-loop gauge coefficients, include three fermion families and one Higgs doublet. With $T(\mathbf{N})=1/2$, the colour Dynkin index sum in one family isThe first factor two counts the two members of $Q_L$; the colour trace is already included in the Dynkin index. There is no coloured scalar. HenceFor $SU(2{)}_L$, the three colours of $Q_L$ and the lepton doublet give $3(1/2)+1/2=2$ per family. The Higgs contributes $T_2(\varphi )=1/2$. ThusFor $U(1{)}_Y$, sum squared hypercharges over every colour and doublet component. One family contributesThe two complex Higgs components contribute $2(1/2{)}^2=1/2$. There is no Abelian gauge self-interaction term, soThis ${\beta }_1$ is for the unnormalized coupling $g_Y$ used in the question. For $g_1^{\mathrm{GUT}}=\sqrt{5/3}{g}_Y$, the corresponding coefficient is $(3/5){\beta }_1=41/10$. This distinction is required for a consistent unification calculation.

Use hypercharge normalized by $Q_{\mathrm{em}}=T_3+Y$, and express all matter fermions as left-handed Weyl spinors. Thus $u^c,d^c,e^c$ denote the charge conjugates of the right-handed Standard Model fermions; the superscript $c$ is unrelated to the antiparticle marker $(A)$. Each chiral superfield contains a complex scalar field and a Weyl spinor, while each vector superfield contains a gauge boson and a gaugino. The complete propagating content of the Minimal supersymmetric Standard Model before electroweak symmetry breaking is:

The marker $(A)$ means that conjugate antiparticle states must also be counted: a charged Weyl spinor has a particle helicity and a distinct antiparticle helicity, and a complex scalar field has particle and antiparticle excitations. The real adjoint gauge bosons and adjoint Majorana fermions already contain their conjugate states, so no extra marker is attached to them. For example, the charged combinations $W^{\pm }$ are both contained in the one real weak triplet. There are no right-handed-neutrino chiral superfields, no gravitino, and no propagating auxiliary fields in this field table. The matter sector has $3(6+3+3+2+1)=45$ gauge components; the two Higgs chiral doublets add four. Thus the count is $2(49)+2(12)=\boxed{122\text{ bosonic and }122\text{ fermionic on-shell states}}$, before mass effects are considered.

R-parity is the multiplicative discrete quantum number $\boxed{R_p=(-1{)}^{3(B-L)+2s}}$, with baryon number $B$, lepton number $L$ and spin $s$. Ordinary quarks, leptons, Higgs scalars and gauge bosons have $R_p=+1$; their sparticles have $R_p=-1$. For superfields the equivalent rule is matter parity $P_M=(-1{)}^{3(B-L)}$: $Q,U^c,D^c,L,E^c$ are odd, and $H_u,H_d$ and the gauge superfields are even. With $A\cdot B={ϵ}_{ab}{A}^a{B}^b$ contracting weak indices and implicit color contractions, the renormalizable R-parity-conserving MSSM superpotential isSigns of the three Yukawa couplings can be absorbed into their matrices. Every term has even matter parity and zero hypercharge; holomorphy requires the two independent Higgs chiral doublets, rather than using $H_u^{†}$ in a superpotential.

Conserved R-parity forces an even number of odd particles in a process whose incoming particles are all ordinary particles, so colliders produce sparticles in pairs or another even multiplicity. Their decay chains end in the stable lightest supersymmetric particle. A neutral, weakly interacting lightest supersymmetric particle, commonly a neutralino after electroweak symmetry breaking, escapes the detector and gives missing transverse momentum; it is also a possible dark matter constituent. Stability alone does not establish its relic abundance, and a charged or colored stable lightest supersymmetric particle is not the usual viable dark matter candidate. Conserved R-parity forbids the renormalizable interactions responsible for the decay below. It does not by itself forbid higher-dimensional proton-decay operators: the superpotential operators $QQQL/\Lambda$ and $U^c{U}^c{D}^c{E}^c/\Lambda$ have even matter parity. Thus the renormalizable R-parity-conserving MSSM protects the proton, while ultraviolet physics can still induce its decay.

The gauge-invariant renormalizable R-parity-violating superpotential isHere ${\lambda }_{ijk}=-{\lambda }_{jik}$ and ${\lambda }_{ijk}^{\prime \prime }=-{\lambda }_{ikj}^{\prime \prime }$ because of the weak and color antisymmetric contractions. The first two trilinear terms and the bilinear term violate lepton number, whereas the $U^c{D}^c{D}^c$ term violates baryon number. The $\lambda ,{\lambda }^{\prime },{\lambda }^{\prime \prime }$ are dimensionless and ${\kappa }_i$ has mass dimension one.

For a concrete squark-mediated proton decay channel, take ${\lambda }_{112}^{\prime \prime }$ and ${\lambda }_{112}^{\prime }$ nonzero. Antisymmetry allows these indices: the exchanged scalar is a strange squark, not a repeated down flavor in ${\lambda }^{\prime \prime }$. Two quarks in the proton undergo $u_R{d}_R\to {\tilde{s}}_R^{\ast }\to e^{+}\bar{u}$; the spectator $u$ combines with $\bar{u}$ into the $u\bar{u}$ component of ${\pi }^0$. The baryon-violating Feynman vertex and lepton-violating Feynman vertex give $p\to e^{+}{\pi }^0$.

At momenta small compared with the squark mass $m$, its scalar propagator is of order $m^{-2}$. Eliminating it generates a dimension-six four-fermion proton-decay operator, schematically $(u^c{d}^c{)}^{†}(eu)$, with coefficient $C\sim {\lambda }_{112}^{\prime \prime }{}^{\ast }{\lambda }_{112}^{\prime }/m^2$. Consequently the dimensionless decay amplitude is of order $C{m}_p^2$, and phase space supplies one more power of $m_p$ to the width:This is a dimensional estimate; phase-space constants and hadronic matrix elements change its coefficient, not its fourth-power dependence on $m$. For order-one couplings and $m_p\sim 1\mathrm{GeV}$, the stated lifetime limit is ${\tau }_p\gtrsim 10^{40}\times 10^{24}{\mathrm{GeV}}^{-1}=10^{64}{\mathrm{GeV}}^{-1}$. Therefore $m^4\gtrsim 10^{64}{\mathrm{GeV}}^4$, giving $\boxed{m\gtrsim 10^{16}\mathrm{GeV}}$.

Such a mass is incompatible with the intended weak-scale MSSM explanation of the hierarchy problem: squarks that cancel the Higgs-sector sensitivity to high scales must not introduce a new enormous splitting from their light quark partners. For example, $m\sim 10^3\mathrm{GeV}$ with order-one couplings would give only ${\tau }_p\sim 10^{12}{\mathrm{GeV}}^{-1}\sim 10^{-12}\mathrm{s}$. The field content alone does not mathematically prohibit extremely heavy superpartners, but that choice sacrifices the weak-scale naturalness motivation. The usual resolution is conserved R-parity, or sufficiently suppressed simultaneous baryon number and lepton number violation, rather than making the exchanged squark this heavy.

The chiral superfield condition is $\boxed{{\bar{D}}_{\dot{\alpha }}\Phi =0}$ for both dotted components. Use left Grassmann derivatives with the displayed supersymmetric covariant derivatives. The odd product rule gives $\partial (\theta {\sigma }^{\mu }\bar{\theta })/\partial {\bar{\theta }}^{\dot{\alpha }}=-(\theta {\sigma }^{\mu }{)}_{\dot{\alpha }}$, soThe chirality constraint therefore says that $\Phi$ is independent of $\bar{\theta }$ at fixed $y$. The Grassmann algebra has only two independent $\theta$ components, so every cubic product of them vanishes. Its general even scalar superfield consequently has a scalar coefficient, a term linear in $\theta$ with a Weyl spinor coefficient, and one quadratic term. Choosing the conventional component normalization gives the chiral-superfield component expansionHere $\varphi$ is a complex scalar field, $\psi$ an odd Weyl spinor and $F_{\mathrm{aux}}$ a complex nonpropagating auxiliary field. Conversely this expansion is annihilated by $\bar{D}$ because $y$ and $\theta$ are. At fixed ordinary $x$, the same result is the terminating translation $\Phi =e^{i\theta {\sigma }^{\mu }\bar{\theta }{\partial }_{\mu }}[\varphi (x)+\sqrt{2}\theta \psi (x)+(\theta \theta )F_{\mathrm{aux}}(x)]$; this retains the printed contraction convention without changing the order in the barred square.

For a local function of scalar superfields, initially allow $f(\Phi ,{\Phi }^{†})$. The chain rule givesFor arbitrary chiral component fields the last factor is not constrained to vanish, so chirality requires $f_{{\Phi }^{†}}=0$. Thus $\boxed{f\text{ must be holomorphic in the chiral fields}}$ on its domain, with no generic conjugate-field dependence. Its ordinary holomorphic derivatives must exist where its nilpotent expansion is evaluated; a pole or branch point there does not define a regular local superfield. Using $(\theta \psi {)}^2=-\frac{1}{2}(\theta \theta )(\psi \psi )$ gives the explicit holomorphic closure of chiral superfields:All three component coefficients are evaluated at $y$. For several chiral superfields, replace $f^{\prime }\psi$ by $f_i{\psi }_i$ and $f^{\prime \prime }\psi \psi$ by $f_{ij}{\psi }_i{\psi }_j$. This requirement concerns a function that is to be chiral for arbitrary field configurations; a special constant field configuration does not impose the same functional constraint.

The ungauged two-derivative supersymmetric action is obtained from a real Kähler potential and a holomorphic superpotential:The first integral is a D-term; the second is an F-term. The Kähler metric $K_{i\bar{j}}$ must be positive for positive scalar kinetic energy. The auxiliary part is $K_{i\bar{j}}{F}_i{\bar{F}}_{\bar{j}}+W_i{F}_i+{\overline{W}}_{\bar{j}}{\bar{F}}_{\bar{j}}$, so eliminating it gives the scalar potential $V=K^{i\bar{j}}{W}_i{\overline{W}}_{\bar{j}}$ in the bosonic sector.

For the single-field Wess–Zumino model, take the canonical Kähler potential $K={\Phi }^{†}\Phi$ and spacetime signature $(+---)$. Component superspace integration givesSince $W^{\prime }=m\varphi +g{\varphi }^2$ and $W^{\prime \prime }=m+2g\varphi$, the auxiliary equation is $F_{\mathrm{aux}}=-(m\varphi +g{\varphi }^2{)}^{\ast }$. Completing its square leaves $-|W^{\prime }{|}^2$ in the Lagrangian. Hence the tree-level effective potential isFor nonzero $m,g$, its two zero-energy supersymmetric vacua are $\varphi =0$ and $\varphi =-m/g$. The potential is nonnegative because it is an absolute square.

The Trilinear scalar vertices in the Wess–Zumino model come fromWith all field legs incoming, expanding $e^{iS}$ and differentiating each cubic monomial givesThe factor $2!$ counts the two identical legs; there is no $1/2!$ in ${\mathcal{L}}_3$ to cancel it. A factor enforcing four-momentum conservation is implicit. These labels specify fields in an all-incoming Feynman rule; physical outgoing particles are assigned by crossing. There are no ${\varphi }^3$ or $({\varphi }^{\ast }{)}^3$ scalar vertices from this on-shell potential.

Equivalently, if $m,g$ are real and $\varphi =(A+iB)/\sqrt{2}$ with canonically normalized real scalars $A,B$, then ${\mathcal{L}}_3=-mgA(A^2+B^2)/\sqrt{2}$. Differentiation gives $\boxed{V_{AAA}=-3\sqrt{2}img,V_{ABB}=-\sqrt{2}img}$, with vanishing $AAB$ and $BBB$ vertices. This is the same pair of complex-field interactions in a real basis.

For the non-renormalization argument, promote $m,g$ to nondynamical chiral spurions. Define one ordinary $U(1)$ and one R-symmetry by the Wess–Zumino spurion charge assignment:

Each superpotential term is neutral under the ordinary symmetry. Under the R-symmetry, both terms have charge two, which is canceled by the charge $-2$ of $d^2\theta$. In particular, $m$ is R-neutral as required. The scalar, spinor and auxiliary components of $\Phi$ have R-charges $1,0,-1$. These are formal symmetries when the couplings transform as spurions; replacing them by fixed nonzero numbers need not leave an actual continuous field symmetry.

A local loop-corrected Wilsonian effective action superpotential $F_{\mathrm{eff}}(\Phi ,m,g)$ must be holomorphic in all three chiral variables and obey the same charges. The ratio $z=g\Phi /m$ is neutral under both symmetries, while $m{\Phi }^2$ has the required charges and mass dimension three. Therefore, on a patch with $m\ne 0$, the most general symmetry-allowed form iswith $h$ holomorphic on its domain. This form follows directly for a monomial ${\Phi }^a{m}^b{g}^c$: ordinary charge gives $a-2b-3c=0$ and R-charge gives $a-c=2$, hence $a=c+2$, $b=1-c$. Each allowed monomial is $g^c{m}^{1-c}{\Phi }^{c+2}$ and already has mass dimension three. Charge symmetry alone leaves $h$ undetermined.

To constrain perturbative loops, retain the original elementary chiral superfield and a nonzero infrared cutoff, rather than integrating out the entire field. A regular perturbative superpotential can be expanded about $g=0$, so $h(z)={\sum }_{n\ge 0}{c}_n{z}^n$ andTerms with $n\ge 2$ have poles at $m=0$. They are excluded by regularity of the local Wilsonian coefficients in the massless limit with the infrared cutoff fixed. At $g=0$ the theory is Gaussian, so the quadratic coefficient remains $c_0=1/2$. The term linear in $g$ is the single original cubic vertex: a diagram with three external fields and only that one interaction vertex has no internal loop. Holomorphy disallows coefficients involving $g^{\ast }$ or $m^{\ast }$, so it cannot conceal the usual nonholomorphic wave-function corrections in $c_1$. Matching the one-vertex term gives $c_1=1/3$. Thus the holomorphy argument for superpotential non-renormalization givesThe infrared qualification matters: a singular one-particle-irreducible expression or the action after eliminating an entire massive multiplet is not the same local Wilsonian effective action for the retained field.

Consequently the holomorphic superpotential and its parameters m and g receive no perturbative vertex renormalization. The Kähler potential does undergo wave-function renormalization. If its quadratic term is $Z{\Phi }^{†}\Phi$, then canonical normalization gives ${\Phi }_c=Z^{1/2}\Phi$ andThese canonically normalized running couplings can therefore be renormalized despite the non-renormalization theorem for the holomorphic superpotential. Loop corrections can also change the effective scalar potential through the corrected Kähler metric. The distinction between holomorphic coefficients and canonically normalized couplings is essential to answering the renormalization question.

The supplied gamma matrices use the mostly-minus metric signature. The Pauli matrix multiplication law gives ${\sigma }^1{\sigma }^2{\sigma }^3=i{I}_2$, hence block multiplication yieldsThe off-diagonal blocks of every ${\gamma }^{\mu }$ connect opposite eigenspaces of this diagonal matrix. Explicitly,Thus $\{{\gamma }^5,{\gamma }^{\mu }\}=0$, and $({\gamma }^5{)}^2=I_4$. The chiral projectors $(1\mp {\gamma }^5)/2$ select the upper and lower two-component blocks, respectively.

Write the Dirac spinor as $\psi =({\chi }_L,{\chi }_R{)}^T$. Substituting the block matrices into the massless Dirac equation gives two independent Weyl spinor equations:They are decoupled because the mass term, which would join the two chirality components, is absent.

For a left-handed plane wave, use $p\cdot x=p^{0}t-\mathbf{p}\cdot \mathbf{x}$. Its coefficient $\xi$ obeysSince $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2=|\mathbf{p}{|}^2{I}_2$, the determinant is $(p^0{)}^2-|\mathbf{p}{|}^2$. A nonzero coefficient therefore requires the massless mass shell $\boxed{p^2=0}$. For positive energy and nonzero $\mathbf{p}$, put $P=|\mathbf{p}|=p^0$ and parameterize its direction by angles $\vartheta ,\phi$. ThenMultiplication gives $(\mathbf{\sigma }\cdot \hat{\mathbf{p}}){\chi }_{-}=-{\chi }_{-}$ and ${\chi }_{-}^{†}{\chi }_{-}=1$. Hence the left-handed massless plane-wave spinor can be writtenHere $N(p)$ is an arbitrary complex normalization; $|N|=\sqrt{2P}$ gives ${\lambda }_L^{†}{\lambda }_L=2P$. Away from the negative $z$ axis one can equivalently takeAt $\mathbf{p}=-P\hat{\mathbf{z}}$ choose $(1,0{)}^T$ instead. These are phase patches for the same one-dimensional kernel. The other frequency branch $p^0=-P$ uses the positive eigenvector ${\chi }_{+}=(\cos (\vartheta /2),e^{i\phi }\sin (\vartheta /2){)}^T$ for the direction of $\mathbf{p}$. At $p^{\mu }=0$, the algebraic equation vanishes and any two-component coefficient is possible, but no momentum direction or helicity is defined there.

A spatial rotation by $\alpha$ about $\hat{\mathbf{p}}$ acts on the two-component coefficient by $\exp (-i\alpha \mathbf{\sigma }\cdot \hat{\mathbf{p}}/2)$. The negative eigenvector therefore acquires the phase $e^{i\alpha /2}$. Since a state with helicity $h$ acquires $e^{-i\alpha h}$, the positive-energy left-handed particle has $\boxed{h=-1/2}$. There is only one such spin state at a fixed nonzero momentum, rather than two independent polarizations of the same chiral particle.

The helicity content of a quantized Weyl field also includes an antiparticle with opposite helicity. In the field expansion, the negative-frequency term multiplies an antiparticle fermionic creation operator, while the positive-frequency term multiplies a particle fermionic annihilation operator. Their adjoint roles give conjugate phases for the corresponding created states, so a left-handed field creates particles with $h=-1/2$ and antiparticles with $h=+1/2$. The right-handed Weyl spinor has the opposite assignments. Quantizing the full massless Dirac field, rather than one chiral component, therefore gives both particle helicities and both antiparticle helicities; every excitation has spin one-half, while chirality determines the helicity of each massless sector.

Use mostly-plus Lorentzian signature and $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$, with antisymmetrized gamma products of unit weight. A Majorana spinor obeys a reality condition under charge conjugation:A Weyl spinor instead has definite chirality,A four-dimensional Majorana spinor has four real components, while a Weyl spinor has two complex components. Charge conjugation reverses chirality in this signature, so a nonzero spinor cannot satisfy both conditions simultaneously.