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With signature $(+---)$ and ${ϵ}_{0123}=+1$, a momentum $(k,0,0,k)$ and helicity $j_3$ give lower components $W_0=k{j}_3$ and $W_3=-k{j}_3$. They obey $W_{\mu }{P}^{\mu }=k(W_0+W_3)=0$. These longitudinal identities require only the momentum and rotation-projection eigenvalues, not a statement about transverse little-group translations.

A massless supermultiplet is a positive-energy unitary irreducible representation of the Super-Poincaré algebra with null four-momentum. For finite helicity states in four dimensions, half the supercharges act trivially and the others form a fermionic Fock space.

Choose canonical spin states by applying the rotationless Lorentz boost $L(p)$ to rest-frame spin states. Spatial inversion $\Pi =\mathrm{diag}(1,-1,-1,-1)$ commutes with rest-frame rotations, so Schur lemma makes its action on one irreducible rest spin multiplet a phase ${\eta }_P$. Since $\Pi L(p)\Pi =L(\Pi p)$, the formula follows. Canonical spin labels are unchanged; helicity reverses. With $P^2=I$, the intrinsic parity is ${\eta }_P=\pm 1$.

// Target: lie-theory.bigb

Use natural units and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. Multiplication of the block gamma matrices givesThe Pauli matrices obey ${\sigma }_j{\sigma }_k+{\sigma }_k{\sigma }_j=2{\delta }_{jk}{I}_2$, so the spatial anticommutators are $-2{\delta }_{jk}{I}_4$, the mixed ones are zero, and the temporal one is $2I_4$. ThusBecause partial derivatives commute, only the symmetric part of ${\gamma }^{\mu }{\gamma }^{\nu }$ survives in their contraction. Multiplying the Dirac equation on the left by its opposite-mass factor therefore gives the flat-space Dirac factorizationHence every component satisfies $\boxed{(\square +m^2)\psi =0}$, the Klein-Gordon equation. This implication does not remove the first-order spinor constraints imposed by the Dirac equation.

For Lorentz covariance, transform coordinates by $x^{\prime }=Lx$ and the Dirac spinor by ${\psi }^{\prime }(x^{\prime })=S(L)\psi (x)$. The chain rule gives ${\partial }_{\mu }^{\prime }=(L^{-1}{)}^{\nu }{}_{\mu }{\partial }_{\nu }$. ConsequentlyThe cancellation uses $L^{\mu }{}_{\rho }(L^{-1}{)}^{\nu }{}_{\mu }={\delta }^{\nu }{}_{\rho }$. Thus the Lorentz transformation maps every solution to another, establishing the Lorentz covariance of the Dirac operator.

The infinitesimal identity follows directly from the Clifford algebra. Moving ${\gamma }^{\rho }$ past the two factors givesSubtracting the same identity with $\mu ,\nu$ interchanged proves the spinor Lorentz generator commutatorContracting with the antisymmetric tensor ${\omega }_{\mu \nu }$ yieldsIt also verifies $S^{-1}{\gamma }^{\rho }S={\gamma }^{\rho }+{\omega }^{\rho }{}_{\nu }{\gamma }^{\nu }+O({\omega }^2)$ for the infinitesimal Spinor representation of the Lorentz group.

For a positive-energy four-momentum, put $\psi =u(p)e^{-ip\cdot x}$, $E=\sqrt{{\mathbf{p}}^2+m^2}>0$. The Dirac equation becomes $(/p-m)u=0$, where the Feynman slash notation uses $/p={\gamma }^{\mu }{p}_{\mu }=E{\gamma }^0-\mathbf{\gamma }\cdot \mathbf{p}$. Splitting $u=(\xi ,\eta {)}^T$ givesThe second equation sets $\eta =(\mathbf{\sigma }\cdot \mathbf{p})\xi /(E+m)$, and the first is then the mass shell, since $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2={\mathbf{p}}^2{I}_2$. Choosing two orthonormal two-spinors ${\chi }_s$ gives the Dirac plane waves in the standard representation:The negative-frequency solutions are independentlywith orthonormal ${\eta }_s$. They are the antiparticle modes after field quantization. The normalization gives $u_r^{†}{u}_s=v_r^{†}{v}_s=2E{\delta }_{rs}$ and ${\overline{u}}_r{u}_s=2m{\delta }_{rs}$, ${\overline{v}}_r{v}_s=-2m{\delta }_{rs}$, using the Dirac adjoint $\overline{u}=u^{†}{\gamma }^0$.

For a spatial rotation, ${\sigma }^{ij}={ϵ}_{ijk}\mathrm{diag}({\sigma }_k,{\sigma }_k)$. Hence the intrinsic rotation generators areFor a massive particle in its rest frame, the two independent spinors therefore carry the two-dimensional spin one-half representation, with $J_3$ eigenvalues $\pm 1/2$ and Casimir $s(s+1)=3/4$. A $2\pi$ rotation acts as $-I$, the spinor sign under a full spatial rotation. Thus the particle has spin $\boxed{s=1/2}$; four spinor components encode particle and antiparticle sectors, not four spin states of one particle. In the massless limit the corresponding physical labels are helicities $\pm 1/2$ rather than a rest-frame spin basis.

The super-Higgs mechanism is the absorption of the goldstino by the gravitino when local supersymmetry is spontaneously broken. In rigid supersymmetry breaking, the goldstino is a physical massless fermion. In supergravity, its transformation permits a gauge choice removing it; its two degrees of freedom become the longitudinal helicity $\pm \frac{1}{2}$ states of the massive gravitino. The massless gravitino has helicities $\pm \frac{3}{2}$, whereas the massive gravitino has four spin states. This is distinct from the supersymmetric Higgs mechanism, which gives an internal gauge vector a mass and need not break supersymmetry.

For a massless supermultiplet, put the momentum along the third axis. The matrix $2\sigma \cdot p$ in the supersymmetry algebra has rank one, so only one fermionic oscillator acts nontrivially. It relates states whose helicities differ by $\frac{1}{2}$. Including the antiparticles required by CPT completion of a supermultiplet gives the following complete spectra.

A chiral multiplet contains a complex scalar $A$ and a Weyl spinor $\psi$. Its scalar particle and antiparticle have helicity $0$; its fermion particle and antiparticle have helicities $-\frac{1}{2}$ and $+\frac{1}{2}$. Therefore $n_B=2=n_F$, with maximum absolute helicity $\frac{1}{2}$.

A supersymmetric vector multiplet contains a massless gauge vector $A_{\mu }$, with helicities $+1,-1$, and a Majorana spinor gaugino $\lambda$, with helicities $+\frac{1}{2},-\frac{1}{2}$. Again $n_B=2=n_F$, now with maximum absolute helicity $1$. The chiral $F$ and vector $D$ auxiliary fields contribute no on-shell states.

The Higgs mechanism is compatible with unbroken supersymmetry. A supersymmetric vacuum may have nonzero scalar vacuum expectation values that break the internal gauge symmetry while obeying both F-flatness and D-flatness. For example, oppositely charged chiral superfields with $W=0$, no Fayet–Iliopoulos term and equal nonzero scalar magnitudes have $F=D=0$ but break the Abelian gauge group. In the supersymmetric Higgs mechanism, a massless vector multiplet combines with the chiral multiplet along the broken gauge direction. One real scalar becomes the longitudinal vector polarization; the other remains a physical real scalar, and the gaugino combines with the Higgsino into a Dirac spinor. The resulting massive vector multiplet has $3+1=4$ bosonic and $4$ fermionic states with a common mass, as in part (i). Any other uneaten chiral multiplets remain separately in the spectrum.

Use the Minkowski metric $g^{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and Feynman slash notation $/p={\gamma }^{\mu }{p}_{\mu }$. The Pauli matrices satisfy ${\sigma }_j{\sigma }_k+{\sigma }_k{\sigma }_j=2{\delta }_{jk}{I}_2$. In the Dirac representation of the gamma matrices, direct block multiplication givesandThus the spatial anticommutators are $-2{\delta }_{jk}{I}_4$, proving the Clifford relationsSince partial derivatives commute, their symmetric product selects the symmetric part of the gamma matrices. Multiplying the Dirac equation by $i{\gamma }^{\mu }{\partial }_{\mu }+m$ givesHence every spinor component satisfies the Klein-Gordon equation with the same mass.

For parity, let $x_P=(t,-\mathbf{x})$ and ${\psi }_P(x)={\gamma }^0\psi (x_P)$. The chain rule gives ${\partial }_0\psi (x_P)=({\partial }_0\psi )(x_P)$ and ${\partial }_j\psi (x_P)=-({\partial }_j\psi )(x_P)$. Combining these signs with ${\gamma }^j{\gamma }^0=-{\gamma }^0{\gamma }^j$ provesThis establishes parity symmetry in quantum field theory for the free Dirac equation, rather than merely asserting its covariance.

A positive-energy plane wave $\psi =u(p)e^{-ip\cdot x}$ solves $(/p-m)u=0$. Write $u=(\chi ,\eta {)}^T$. The two block equations areThe second gives $\eta =(\mathbf{\sigma }\cdot \mathbf{p})\chi /(E_p+m)$. In the first, $(\mathbf{\sigma }\cdot \mathbf{p}{)}^2={\mathbf{p}}^2{I}_2$ and $E_p^2-{\mathbf{p}}^2=m^2$ make the remaining coefficient zero. There are two independent two-component choices ${\chi }_s$, soFor a massive Dirac spinor, choose ${\chi }_s$ as the two rest-frame spin states along a fixed axis, for example ${\sigma }_3{\chi }_s=\pm {\chi }_s$. The label $s$ distinguishes these canonical spin states after boosting; it is not automatically a helicity label.

Now $p\cdot x_P=p_P\cdot x$ and ${\gamma }^0{u}_s(p)=u_s(p_P)$ for the same fixed ${\chi }_s$. ThereforeThe parity action on canonical massive spin states leaves the canonical spin label unchanged while reversing the momentum. Spin is an axial vector under parity; its orientation is unchanged. Consequently helicity, the projection of spin along momentum, reverses. If $s$ had instead been chosen to denote helicity, the transformed label would have the opposite sign.

The Dirac adjoint is $\bar{\psi }={\psi }^{†}{\gamma }^0$, with ${\gamma }^{0†}={\gamma }^0$. ThusThe chirality matrix has ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3=\left(\begin{array}{cc}0& I_2\\ I_2& 0\end{array}\right)$ in this representation, so $({\gamma }^5{)}^2=I_4$ and it anticommutes with every ${\gamma }^{\mu }$. In particular ${\gamma }^0{\gamma }^5{\gamma }^0=-{\gamma }^5$. The parity transformation of a Dirac bilinear therefore yieldsThe first bilinear is a scalar under parity, whereas the second is a pseudoscalar.

For the requested gamma matrix trace identities, use the cyclic property of the trace. Conjugation of a product of $n$ gamma matrices by ${\gamma }^5$ multiplies it by $(-1{)}^n$, but leaves its trace unchanged. All odd products therefore have zero trace. Taking the trace of the Clifford algebra relation and using $\mathrm{Tr}{I}_4=4$ gives $\mathrm{Tr}({\gamma }^{\alpha }{\gamma }^{\beta })=4g^{\alpha \beta }$. For four factors, move the first gamma matrix successively past the next three using the anticommutators. If $A=\mathrm{Tr}({\gamma }^{\alpha }{\gamma }^{\beta }{\gamma }^{\rho }{\gamma }^{\delta })$, this givesThus $A=4(g^{\alpha \beta }{g}^{\rho \delta }-g^{\alpha \rho }{g}^{\beta \delta }+g^{\alpha \delta }{g}^{\beta \rho })$. Contracting with the relevant four-vectors proves all four answers:

Take metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and write $M=M_W$. The outgoing electron and antineutrino momenta are $k$ and $k^{\prime }$. Up to an irrelevant overall phase, the amplitude isA fermion spin sum givesAnticommuting ${\gamma }_5$ through the massless momentum and gamma matrix, and using $(1-{\gamma }_5{)}^2=2(1-{\gamma }_5)$, reduces the trace to twice the four-gamma trace minus twice its ${\gamma }_5$ counterpart. In particular its symmetric part isThe epsilon-tensor part is antisymmetric and contributes neither to the unpolarized polarization sum nor to the real linear polarization used below. There is no extra factor for selecting the neutrino helicity: the chiral vertex has already done so.

The external massless equations imply $p_{\mu }{J}^{\mu }=0$ for $p=k+k^{\prime }$, so $p_{\mu }{T}^{\mu \nu }=0$. With $k\cdot k^{\prime }=M^2/2$, averaging over the three initial spin states givesIndeed $-g_{\mu \nu }{T}_S^{\mu \nu }=16k\cdot k^{\prime }=8M^2$. This is a massive-vector spin average, not an average over the final spins.

In the rest frame, eliminating the momentum delta function and integrating the remaining radial delta function $\delta (M-2|\mathbf{k}|)$ reduces the two-body Lorentz-invariant phase space to $d{\Phi }_2=d\Omega /(32{\pi }^2)$. ThereforeThis derives the massless leptonic W decay width without invoking a precomputed decay formula.

For the specified polarization there is no initial spin average. Put $\mathbf{k}=(M/2)\hat{\mathbf{n}}$, ${\mathbf{k}}^{\prime }=-(M/2)\hat{\mathbf{n}}$ and ${\hat{n}}_z=\cos \theta$. The contraction selectsIt follows thatHere $\int {\sin }^2\theta d\Omega =8\pi /3$. The total rate agrees with the spin-averaged result, but its normalized polar distribution is $(1/{\Gamma }_z)d{\Gamma }_z/d\cos \theta =3(1-{\cos }^2\theta )/4$ instead of the isotropic unpolarized distribution. This is polarized massless leptonic W decay.

For physical finite-spin one-particle states, a positive-energy Poincaré group representation is labelled by mass and spin $(m,j)$ when $m>0$, or mass and helicity $(0,h)$ when massless. The individual states also carry their four-momentum and a spin projection. For four-dimensional N=1 supersymmetry, the irreducible labels become $(m,j_{msuper})$, with $j_{msuper}$ the superspin, or $(0,h_{msuper})$, with a chosen superhelicity convention. The constituent representation labels arewith the last entry omitted for $j=0$, andA required CPT completion of a supermultiplet adds the conjugate labels $(0,-h),(0,-h-\frac{1}{2})$ and conjugates any internal charges.

In a massive four-dimensional N=1 supersymmetry representation, the two normalized raising supercharges transform as a spin-half doublet. Start with a Clifford vacuum of superspin $j=1/2$. At occupation levels zero, one and two, the ordinary spins areThe two-creation state has a spin-singlet oscillator factor because the creation operators anticommute. Thus the massive N=1 vector multiplet containsThere are four bosonic polarizations, $3+1$, and four fermionic polarizations, $2+2$, as required by boson-fermion degeneracy in a supermultiplet.

In the massless limit the helicity multiplicities become $+1,-1$, two copies each of $+1/2,-1/2$, and two zero-helicity states. They separate into the four irreducible helicity pairsWith the lower-helicity convention for superhelicity, a pair $(\kappa ,\kappa +1/2)$ has labels $\kappa =1/2,0,-1/2,-1$, respectively. A midpoint convention shifts each label by $1/4$; the explicit helicity pairs remove any naming ambiguity.

Pairing under the CPT theorem, the first and last pairs form a massless vector multiplet, and the middle two form a massless chiral multiplet. This is the massless limit of a massive N=1 vector multiplet. The longitudinal vector polarization supplies one of the chiral multiplet's two real scalar modes; the original spin-zero state supplies the other. Simply discarding the longitudinal polarization would lose a physical degree of freedom and give the wrong decomposition.

The derivative convention uses hypercharge Y with $Q=T_3+Y$. For each lepton family $l=e,\mu ,\tau$, the left-handed fields form an SU(2) doublet $L_l=({\nu }_l,l{)}_L^T$, with $Y=-1/2$. Their weak isospin values are $T_3=+1/2,-1/2$ and their electric charges are $0,-1$. The right-handed charged lepton $l_R$ is an SU(2) singlet, with $T_3=0$, $Y=-1$ and charge $-1$. The minimal model used here has no right-handed neutrino. A sterile right-handed neutrino, if added as a singlet with $Y=0$, would have no charged weak current interaction. Antiparticles carry the conjugate representations and opposite electric charges.

Expanding ${\bar{L}}_{l}i{\gamma }^{\mu }{D}_{\mu }{L}_l$ supplies $-g{\bar{L}}_l{\gamma }^{\mu }{A}_{\mu }^a{\sigma }_a{L}_l/2$. The off-diagonal entries of $A^a{\sigma }_a$ are $\sqrt{2}{W}^{+}$ and $\sqrt{2}{W}^{-}$. With the chiral projector $P_L=(1-{\gamma }^5)/2$, one has ${\bar{\nu }}_L{\gamma }^{\mu }{l}_L=\bar{\nu }{\gamma }^{\mu }{P}_{L}l$. Thus the two W boson interactions areFor quarks, separate rotations diagonalize the up-type and down-type mass terms. If their left-handed rotations are $U_{uL}$ and $U_{dL}$, the charged interaction contains the CKM matrix $V=U_{uL}^{†}{U}_{dL}$:There is an implicit sum over color charge. This quark mixing permits transitions between different families, rather than just the paired fields in a single doublet.

Let $J^{\mu }={\sum }_l{\bar{\nu }}_l{\gamma }^{\mu }(1-{\gamma }^5)l+J_{\mathrm{had}}^{\mu }$ and $\kappa =g/(2\sqrt{2})$. At momenta much smaller than $m_W$, neglect derivatives in the heavy W boson equation. Its mass and source terms can be completed to a square:This explicitly performs integrating out a charged vector boson. The resulting Fermi interaction isThe omitted terms contain derivatives and are relatively suppressed by momentum squared divided by $m_W^2$.

For the leptonic pion decay, Lorentz covariance restricts a vacuum-to-spin-zero current matrix element to a constant times $p^{\alpha }$. In the rest frame, rotations eliminate the spatial components. Under parity, the temporal vector current is even, whereas the vacuum is even and the pion is odd, so its temporal matrix element equals its negative and vanishes. The temporal axial current is odd and is allowed. This proves the vacuum-to-pseudoscalar current selection rule, rather than merely assuming an axial current. Define the pion decay constant byThen the vector-minus-axial charged weak current has matrix element $-i{f}_{\pi }{V}_{ud}{p}^{\alpha }$, up to an irrelevant overall phase convention. Consequently the scattering amplitude for the decay can be writtenUse $p=k+q$, the momentum-space Dirac equation ${\bar{u}}_e/k=m_e{\bar{u}}_e$, and $/q{v}_{\nu }=0$ for a massless neutrino. Since $/q(1-{\gamma }^5)=(1+{\gamma }^5)/q$, the neutrino contribution vanishes, leavingThe fermion spin sum is, using the supplied gamma matrix trace identities,Here $(k+q{)}^2=m_{\pi }^2$, $k^2=m_e^2$, and $q^2=0$. The mass term in the trace vanishes; anticommuting ${\gamma }^5$ reduces the remaining trace to twice $\mathrm{tr}(/k/q)$. It follows thatUsing two-body decay phase space, with final momentum magnitude $(m_{\pi }^2-m_e^2)/(2m_{\pi })$, gives the leptonic pseudoscalar decay widthThus the requested $m_e^2(m_{\pi }^2-m_e^2)$ factor is already present in the squared amplitude, and the phase space supplies one more threshold factor.

The physical explanation is helicity suppression. In the pion rest frame the two momenta are opposite. The massless antineutrino has positive helicity; zero total angular momentum would require the electron to have positive helicity too. The charged weak current instead creates a left-chiral electron, whose helicity is negative when its mass vanishes. A nonzero electron mass admits the required opposite helicity with an amplitude proportional to $m_e$. The decay amplitude vanishes for $m_e=0$, and the rate is suppressed by $m_e^2$.