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Use Minkowski metric signature $+---$ and natural units. With no lepton mixing, choose the weak charged current conventionIts conjugate convention exchanges $J$ and $J^{†}$ without changing the interaction. The relevant four-fermion interaction gives the muon decay amplitude, up to an overall phase,Here $G_F$ is the Fermi constant; the factors $1-{\gamma }_5$ are twice the left chiral projector, so there is no additional factor of two to insert.

For the final-spin sum let ${\rho }_{\mu }=(/p+m)(1+{\gamma }_5/s)/2$, where $m=m_{\mu }$. It already describes the specified initial polarization; there is no further initial-spin average. The two gamma matrix traces areThe electron-mass term vanishes in its chiral trace. In the muon trace, anticommutation with ${\gamma }_5$ givesFor the chiral fermion trace contraction, defineThe supplied trace conventions imply $L^{\alpha \beta }=8K^{\alpha \beta }(k,q)$ and $H_{\alpha \beta }=4K_{\alpha \beta }(q^{\prime },r)$, where $r=p-ms$. The factor four in the second trace follows from the polarization projector. The symmetric parts contract to $2[(a\cdot c)(b\cdot d)+(a\cdot d)(b\cdot c)]$. The Levi-Civita symbol contraction, including $i^2=-1$, contributes $2[(a\cdot c)(b\cdot d)-(a\cdot d)(b\cdot c)]$. Mixed symmetric-antisymmetric contractions vanish. Thus $K^{\alpha \beta }(a,b)K_{\alpha \beta }(c,d)=4(a\cdot c)(b\cdot d)$, and

Now neglect $m_e$ as well. Put $Q=p-k$ and $c=\hat{\mathbf{k}}\cdot \mathbf{s}$. The given moment integral over two-neutrino relativistic two-body phase space yieldsIn the muon rest frame, $k\cdot Q=mE$, $k\cdot r=mE(1+c)$, $r\cdot Q=m^2-mE(1+c)$ and $Q^2=m^2-2mE$. Substitution givesCombining this with the Lorentz-invariant phase-space measure, whose one-particle denominator is $(2\pi {)}^{3}2E$, and $d^{3}k=E^{2}dEd\Omega$, gives the polarized muon decay distributionConsequently $C=-2$, $D=-2$, $E=1$. Integration over the electron direction removes the polarization term, and ${\int }_0^1{x}^2(3-2x)dx=1/2$, recovering $\Gamma =G_F^2{m}^5/(192{\pi }^3)$.

The spin vector is axial and the momentum direction is polar: under parity, $\mathbf{s}$ stays unchanged while $\hat{\mathbf{k}}$ reverses. Their scalar product therefore changes sign. Its nonzero coefficient demonstrates parity violation in polarized muon decay, arising from the left-chiral weak charged current. At the upper endpoint the negative muon's electron is preferentially emitted opposite the spin, providing a sign check on the result.

The angles $\psi$ and $\varphi$ are cyclic. Their conjugate momentum components are $p_{\psi }=I_3(\dot{\psi }+\dot{\varphi }\cos \theta )$ and $p_{\varphi }=I_1\dot{\varphi }{\sin }^2\theta +p_{\psi }\cos \theta$, so both are conserved; consequently ${\omega }_3=p_{\psi }/I_3$ is constant. With $q=I_3{\omega }_3=p_{\varphi }>0$,Eliminating the cyclic velocities gives the nutation effective potentialapart from the constant axial spin energy. ThusThe endpoint $\theta =\pi$ is excluded for these nonzero conserved momenta: the effective potential diverges there. The apparent Euler-angle singularity at $\theta =0$ is interpreted by its smooth small-tilt limit.

Putting the unit vector along a coordinate axis gives ${\sigma }_i^2=I$. Putting it along $(e_i+e_j)/\sqrt{2}$ gives ${\sigma }_i{\sigma }_j+{\sigma }_j{\sigma }_i=0$ for $i\ne j$. Combining the resulting anticommutators with the given commutators yields ${\sigma }_i{\sigma }_j={\delta }_{ij}I+i{ϵ}_{ijk}{\sigma }_k$, and henceThe operators $S_i=(\hslash /2){\sigma }_i$ obey angular-momentum commutators and ${\mathbf{S}}^2=3{\hslash }^{2}I/4$, the spin-one-half value. In a spin-half irreducible sector they are the intrinsic spin operators; the algebra also permits direct sums of such sectors.

Since $Q=\mathbf{n}\cdot \mathbf{\sigma }$ is Hermitian with $Q^2=I$, its operators $P_{\pm }=(I\pm Q)/2$ are Hermitian and satisfy $P_{\pm }^2=P_{\pm }$, $P_{+}{P}_{-}=0$ and $P_{+}+P_{-}=I$. Moreover $Q{P}_{\pm }=\pm P_{\pm }$. Thus $|\chi ⟩=P_{+}|\chi ⟩+P_{-}|\chi ⟩$ is its orthogonal eigenstate decomposition, omitting any zero component. The Born probabilities for a normalized state areFor spin up along $\mathbf{m}$, the anticommutator $\{\mathbf{m}\cdot \mathbf{\sigma },{\sigma }_j\}=2m_{j}I$ gives $⟨{\sigma }_j⟩=m_j$. Therefore the probability of spin up along $\mathbf{n}$ is $(1+\mathbf{n}\cdot \mathbf{m})/2$.

A Lie algebra representation is a linear map $\rho :\mathfrak{g}\to \mathrm{End}(V)$ preserving brackets: $\rho ([X,Y])=[\rho (X),\rho (Y)]$. Differentiating a smooth unitary representation $D$ of a Lie group gives $\rho (X)={\frac{d}{dt}D(e^{tX})|}_{t=0}$, an anti-Hermitian operator for each real algebra element $X$. In the physics convention one uses Hermitian generators $t_X=i\rho (X)$ and writes $D(e^{tX})=e^{-it{t}_X}$. Conversely, integration of a Lie-algebra representation gives a representation of the connected simply connected group. A representation of another group with the same algebra additionally has to be trivial on the kernel of its covering map.

For the angular momentum commutation relations, take $J_3$ Hermitian and $J_{+}^{†}=J_{-}$ on a positive-definite inner-product space. A highest-weight vector has $J_3|j,j⟩=j|j,j⟩$ and $J_{+}|j,j⟩=0$. Set $v_{\ell }=(J_{-}{)}^{\ell }|j,j⟩$. Commuting $J_3$ through each lowering operator gives $J_3{v}_{\ell }=(j-\ell )v_{\ell }$. Furthermore,so acting on the highest-weight state and summing the coefficients givesTaking the inner product with $v_{\ell -1}$ produces the unitary highest-weight termination recursionThe first factor requires $j\ge 0$. If $2j$ is not an integer, the first integer $\ell >2j+1$ would make the norm negative, while every preceding norm is positive. ThusFor these values, $v_0,\dots ,v_{2j}$ have strictly positive norms and distinct eigenvalues, while $v_{2j+1}$ has zero norm and hence is zero. They therefore span a space of dimension $2j+1$, invariant under all three generators. Its weights are $j,j-1,\dots ,-j$. This proves integer or half-integer spin and finite dimension $2j+1$. The positive-definite unitary assumption matters: an unrestricted algebraic highest-weight module need not terminate. Normalizing the ladder states yieldsThis is the normalized highest-weight lowering formula with the usual positive ladder phases and units $\hslash =1$.

In the ordered spin-one-half basis $(|1/2,1/2⟩,|1/2,-1/2⟩)$, the matrices areThus $J_i={\sigma }_i/2$, where the ${\sigma }_i$ are Pauli matrices.

Choose the active-rotation convention. The SU(2) lift of an axis-angle rotation isThe Pauli matrix multiplication law shows that conjugating $\mathbf{x}\cdot \mathbf{\sigma }$ by this matrix produces $(R\mathbf{x})\cdot \mathbf{\sigma }$, withThis is the ordinary three-dimensional rotation. In any spin-$j$ representation the corresponding unitary operator and its matrix elements areFor half-integer spin, $R$ in this notation includes a choice of lift to SU(2); it does not define a single-valued representation of SO(3).

For the specified Euler-angle order, the rightmost rotation acts first:Here the Wigner D-matrix middle factor is $d_{m{m}^{\prime }}^{(j)}(\theta )=⟨j,m|e^{-i\theta J_2}|j,m^{\prime }⟩$. It can be calculated explicitly by realizing spin $j$ as the symmetric product of $2j$ spin-one-half factors. Write $c=\cos (\theta /2)$ and $s=\sin (\theta /2)$. The fundamental rotation sends $|+⟩$ to $c|+⟩+s|-⟩$ and $|-⟩$ to $-s|+⟩+c|-⟩$. Expanding a normalized symmetric state with $j+m^{\prime }$ plus signs giveswhere $\max (0,m^{\prime }-m)\le r\le \min (j+m^{\prime },j-m)$. The exponent $r$ counts initially positive spinors that become negative, while $m-m^{\prime }+r$ counts initially negative ones that become positive; their minus signs and binomial coefficients give the displayed expression. Every factorial argument is an integer even for half-integer spin.

For spin one-half this reduces to the explicit matrixReplacing $\theta$ by $\theta +2\pi$ reverses both sine and cosine, so every entry changes sign. In particular, a pure $2\pi$ rotation gives $D^{(1/2)}=-I$, while a $4\pi$ rotation gives $I$.

Finally, the Adjoint double cover from SU(2) to SO(3) is surjective by the axis-angle construction. Its kernel consists of matrices commuting with all Pauli matrices, hence scalar matrices; within SU(2) these are $\pm I$. ThusTheir Lie algebras are isomorphic, but their global topology and representations differ. In spin $j$, the central element $-I$ acts as $(-1{)}^{2j}I$, so exactly the integer-spin representations descend to ordinary SO(3) representations. This is the descent of an SU(2) representation to SO(3) criterion.

Let $V={\mathbb{C}}^3$ be the defining SU(3) representation. A mixed SU(3) tensor representation belongs to $V^{\otimes k}\otimes (V^{\ast }{)}^{\otimes l}$, with transformation lawThe unrestricted tensor has dimension $3^{k+l}$. The upper and lower counts are independent; setting them equal would lose part of the printed generality.

For an irreducible tensor, impose the appropriate Young symmetrizer on each set of indices and subtract all upper-lower contractions. Antisymmetric pairs can be converted using the invariant alternating tensor, since ${\bigwedge }^{2}3\cong \overline{3}$ and ${\bigwedge }^2\overline{3}\cong 3$. A full alternating triple is a singlet because $\det U=1$. Traces use the invariant Kronecker delta and yield lower-rank tensors. Repeating these operations gives invariant irreducible subspaces, with multiplicities retained.

An explicit algorithm for the complete decomposition starts at $(p,q)=(0,0)$, tensors with $3$ exactly $k$ times, and then with $\overline{3}$ exactly $l$ times. At each step applyomitting terms with negative labels and adding multiplicities of coincident terms. The first is the sl3 highest-weight tensor rule: a Young diagram of row lengths $(p+q,q,0)$ gains a box in one of its three rows. The third-row term loses a determinant column of height three, leaving labels $(p,q-1)$. The second rule follows by conjugation, which exchanges $p$ and $q$. Thus the procedure determines all constituents of the general tensor, not just one selected symmetry type.

The separately symmetric, traceless component has labels $(k,l)$. Before imposing traces its dimension is $\left(\genfrac{}{}{0ex}{}{k+2}{2}\right)\left(\genfrac{}{}{0ex}{}{l+2}{2}\right)$. The contraction map has target the separately symmetric tensors of degrees $(k-1,l-1)$ and is onto: in polynomial coordinates it is ${\sum }_i{\partial }_{x_i}{\partial }_{y_i}$, the adjoint of multiplication by ${\sum }_i{x}_i{y}_i$ in the monomial-factorial inner product. Multiplication is injective, so the adjoint is surjective. Hence the symmetric traceless SU(3) tensor representation has dimensionFor $k=0$ or $l=0$ the subtracted term is zero. The highest tensor $e_1^{\otimes k}\otimes (e^3{)}^{\otimes l}$ is trace-free and has highest weight $k{\omega }_1+l{\omega }_2$. The three positive-root factors in the Weyl dimension formula are $k+1$, $l+1$ and $(k+l+2)/2$, giving exactly the kernel dimension above. Thus the whole symmetric traceless kernel is that irreducible constituent.

There is a genuine distinction between this highest-weight constituent and the literal largest-dimensional constituent. For example, $3^{\otimes 5}$ contains $(3,1)$, obtained by antisymmetrizing one pair, with dimension $24$, whereas the fully symmetric $(5,0)$ has dimension $21$. Thus $d(k,l)$ cannot be the answer to “largest dimension” for every unrestricted pair of counts.

The maximal dimension in mixed SU(3) tensor powers can nevertheless be determined exactly. Every constituent highest weight differs from $k{\omega }_1+l{\omega }_2$ by $a{\alpha }_1+b{\alpha }_2$ with nonnegative integers $a,b$, where ${\alpha }_1=2{\omega }_1-{\omega }_2$ and ${\alpha }_2=-{\omega }_1+2{\omega }_2$. Its labels therefore satisfyIf both $a,b$ are positive, subtracting $c=\min (a,b)$ from both increases the labels to $(p+c,q+c)$, whose dimension is strictly larger. The resulting boundary labels occur in the tensor: use $a$ disjoint antisymmetric upper pairs or $b$ disjoint antisymmetric lower pairs, and then take the separately symmetric traceless component. Consequently the exact answer is the finite formulaThis supplies a practical closed finite computation for arbitrary $k,l$. To reduce it to at most four evaluations, put $K=k+1$, $L=l+1$ and $r=1+\sqrt{3}$. The first continuous maximum is at $a_{\ast }=(K-rL)/(2+r)$, clipped to $[0,k/2]$; test its two neighbouring allowed integers. The second is similarly at $b_{\ast }=(L-rK)/(2+r)$, clipped to $[0,l/2]$. Indeed differentiating the first cubic gives a derivative with the sign of $(K-2a{)}^2-2(K-2a)(L+a)-2(L+a{)}^2$, which changes sign once at that root. For equal upper and lower counts, both maxima are at zero and the familiar result is $(k+1{)}^3$.

For three quarks, the unconstrained flavour product isThe brackets indicate permutation symmetry: symmetric, mixed, and alternating. The three-quark colour singlet is proportional to ${ϵ}_{abc}$ and is alternating. The fermion wavefunction must be alternating overall. In the usual spatially symmetric ground-state S-wave, the spin-flavour part must therefore be symmetric.

The three-spin-one-half space has a symmetric spin-$3/2$ sector and mixed-symmetry spin-$1/2$ sectors; it has no fully alternating sector because ${\bigwedge }^3{\mathbb{C}}^2=0$. Symmetric flavour can pair with symmetric spin, giving the baryon decuplet of spin $3/2$. The flavour-octet multiplicity space and the spin-$1/2$ multiplicity space both transform as the two-dimensional standard representation of the symmetric group. Their product contains exactly one symmetric combination: its characters at the identity, a transposition and a three-cycle are $4,0,1$, so the symmetric multiplicity is $(4+3\cdot 0+2\cdot 1)/6=1$. This gives one baryon octet of spin $1/2$. Alternating singlet flavour would need alternating spin to make the spin-flavour part symmetric, and that sector does not exist. Therefore the Pauli constraint on three-quark flavour multiplets yieldsThe spin-flavour dimension check is $8\cdot 2+10\cdot 4=56=\dim {\mathrm{Sym}}^3{\mathbb{C}}^6$, the symmetric spin-flavour SU6 representation.

The spatial-symmetry qualification matters if the wording is interpreted as total $L=0$ alone. Rotationally scalar excited orbital functions can have mixed permutation symmetry. For example, the two independent differences among $r_{12}^2,r_{23}^2,r_{31}^2$, multiplied by a symmetric radial factor, span such a space while remaining rotational scalars. Combining this standard permutation representation with the spin-$1/2$ standard representation gives an alternating spin-orbital combination; multiplying by alternating singlet flavour makes it symmetric, and the colour factor restores overall fermionic antisymmetry. Thus total $L=0$ without a symmetric ground-state orbital assumption would not by itself exclude a singlet.

For a quark-antiquark pair, colour singletness selects the contraction ${\delta }^a{}_b$, and the independent flavour product givesQuark and antiquark are distinguishable, so there is no three-identical-quark exclusion of either flavour multiplet or either spin. Both the octet and singlet admit spin singlets and spin triplets. Their baryon number is zero. The flavour octet has rows $(Y,I)=(1,1/2),(0,1),(0,0),(-1,1/2)$, while the singlet has $(Y,I)=(0,0)$. The octet's zero-isospin state is $(u\bar{u}+d\bar{d}-2s\bar{s})/\sqrt{6}$, and the singlet is $(u\bar{u}+d\bar{d}+s\bar{s})/\sqrt{3}$. Each row contains all $I_3=-I,\dots ,I$, and electric charge is $Q=I_3+Y/2$.

For ground-state mesons with $L=0$, opposite intrinsic quark-antiquark parities give $P=-1$. The neutral self-conjugate states have charge conjugation $C=(-1{)}^{L+S}$, so both flavour multiplets occur as $0^{-+}$ pseudoscalar mesons and $1^{--}$ vector mesons, forming a flavour nonet in each channel. Charged and open-flavour states have no individual charge-conjugation eigenvalue. The meson request does not restrict orbital excitation: in general $J=|L-S|,\dots ,L+S$, $P=(-1{)}^{L+1}$, and the same formula for $C$ applies where defined. Thus the flavour representations stay $1\oplus 8$, with these additional orbital-spin quantum numbers.

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.

For the Dirac equation $(/\partial +m)\psi =0$, the derivative term reverses chirality because ${\gamma }_5$ anticommutes with every gamma matrix, whereas the mass term preserves it. Projecting a purely chiral spinor equation onto its two chiralities therefore gives both $/\partial \psi =0$ and $m\psi =0$. Thus a nonzero four-component Weyl spinor satisfying this uncoupled Dirac equation must be massless. A Majorana spinor contains conjugate opposite-chirality components and can have a nonzero mass. Its charge-conjugation reality condition identifies particle and antiparticle operators, so a Majorana particle is its own antiparticle. A Weyl field does not by itself impose that identification.

This does not forbid writing a Majorana mass term for a neutral two-component Weyl field: such a term couples the field to its conjugate, and the resulting four-component massive field is Majorana, not a purely chiral solution of the uncoupled Dirac equation.

For the Rarita-Schwinger field, set $\chi ={\gamma }^{\mu }{\psi }_{\mu }$ and $d={\partial }^{\mu }{\psi }_{\mu }$. The antisymmetric kinetic operator is invariant underbecause commuting derivatives contract to zero against ${\gamma }^{\mu \nu \rho }$. Expanding its gamma product givesGamma contraction gives ${\gamma }_{\mu }{E}^{\mu }=2(/\partial \chi -d)$, so the equation implies $d=/\partial \chi$ and $/\partial {\psi }^{\mu }={\partial }^{\mu }\chi$.

For a Fourier mode with momentum $k$, if $k^2\ne 0$ the latter identity gives $k^2{\psi }^{\mu }=k^{\mu }/k\chi$. The mode is proportional to $k^{\mu }$ and hence pure gauge. Every physical nonzero-momentum mode therefore has $k^2=0$. For a null mode, choose an index with $k^{\mu }\ne 0$. The same equation makes $\chi$ an element of the image of $/k$, allowing a gauge choice with $\chi =0$. The field equations then reduce toResidual gauge transformations obey $/\partial \eta =0$.

Take a null momentum in the third spatial direction. Residual gauge freedom and transversality remove the temporal and longitudinal vector components, leaving two transverse vector polarizations. Their helicities are $\pm 1$, while on-shell spinor helicities are $\pm 1/2$. The gamma-trace constraint removes the two combinations of total helicity $\pm 1/2$ and retains the aligned combinations of helicity $\pm 3/2$. Explicitly, in a chiral representation the transverse gamma contractions use ${\sigma }_1\pm i{\sigma }_2$, which annihilate the correspondingly aligned spin states and act nontrivially on the opposite states. Thus a Majorana Rarita-Schwinger field has two physical massless polarizations. The Majorana assumption is the one relevant to simple supergravity; a complex vector-spinor has these helicities for its particle and independent antiparticle modes.

A consistent free massive deformation is the algebraic termadded to a kinetic term $-\frac{1}{2}{\overline{\psi }}_{\mu }{\gamma }^{\mu \nu \rho }{\partial }_{\nu }{\psi }_{\rho }$. Its equation isFor $m\ne 0$, divergence and gamma contraction give, respectively,These imply the massive Rarita-Schwinger constraints $\chi =0$ and $d=0$, and the remaining equation is $(/\partial +m){\psi }_{\mu }=0$. Squaring gives $(\square -m^2){\psi }_{\mu }=0$ in the chosen signature. The gauge symmetry is lost. In the rest frame, transversality sets the temporal component to zero; three spatial vector components times two on-shell spinor states give six states, and gamma trace removes the two spin-one-half states. The result is four massive spin-$3/2$ states, with spin projections $\pm 3/2,\pm 1/2$. In supergravity the additional longitudinal states can be supplied by the super-Higgs mechanism.

The gravitational variable and the fields held fixed in a variation must be specified. For bosonic matter a second-order metric tensor formulation is sufficient. For spinorial matter a vierbein and a spin connection are needed to define the local Lorentz representation and its covariant derivative. “Second order” refers to a connection already expressed through derivatives of the vierbein; “first order” treats that connection as an independent variable.

For a metric tensor, a scalar field and an Abelian gauge field, consider the Einstein-Hilbert action with matter,Here $e=\sqrt{-g}$ and the connection in $R$ is Levi-Civita. The metric tensor variation uses $\delta e=-e{g}_{\mu \nu }\delta g^{\mu \nu }/2$ andMetric compatibility makes the last two terms a divergence. ThusStationarity for arbitrary compactly supported variations givesExplicitly,Scalar field and gauge-potential variations give ${\nabla }^2\varphi -V^{\prime }=0$ and ${\nabla }_{\mu }{F}^{\mu \nu }=0$. Their equations imply stress-energy conservation, consistently with the contracted Bianchi identity. If the boundary metric tensor is fixed instead of using compactly supported variations, add the Gibbons–Hawking–York boundary term, ${\kappa }^{-2}{\int }_{\partial M}{\epsilon }_{\partial M}\sqrt{|h|}K$, with the outward-normal orientation. Fixing the metric tensor alone does not otherwise remove the normal derivatives of its variation.

For spinorial matter, write $g_{\mu \nu }=e_{\mu }{}^a{e}_{\nu }{}^b{\eta }_{ab}$ and ${\nabla }_{\mu }\chi ={\partial }_{\mu }\chi +{\omega }_{\mu ab}{\gamma }^{ab}\chi /4$. In the second-order theory set $\omega =\omega (e)$ before varying. A symmetrized Dirac action, in the same no-$i$, mostly-plus convention as the earlier solutions, isVarying $\bar{\chi }$ and integrating by parts gives $(/\nabla +m)\chi =0$; varying $\chi$ gives the adjoint equation. The spinor components held fixed in a gravitational variation are components in the Lorentz frame. Both ${\gamma }^{\mu }=e_a{}^{\mu }{\gamma }^a$ and $\omega (e)$ vary with the vierbein. The former gives the canonical stress, and integrating the spin-connection variation supplies its spin-current improvement. On the Dirac equations the resulting symmetric stress tensor isThe off-shell expression also contains $g_{\mu \nu }{\mathcal{L}}_{\chi }/e$; this vanishes on the matter equations for the displayed symmetrized action. Treating the coordinate gamma matrices as fixed while varying the metric tensor would miss the displayed stress. Local Lorentz invariance removes the independent antisymmetric vierbein equation, leaving the symmetric Einstein field equations with this matter source.

In a first-order vierbein formulation, $e^a$ and an antisymmetric Lorentz connection ${\omega }^{ab}$ are independent. Defineand use $R(e,\omega )=e_a{}^{\mu }{e}_b{}^{\nu }{R}_{\mu \nu }{}^{ab}$. The curvature form of a connection variation is $\delta R^{ab}=D_{\omega }\delta {\omega }^{ab}$. For the Einstein-Hilbert term, integration by parts in form notation gives a connection equation proportional towhen matter has no connection dependence. For an invertible vierbein this forces $T^a=0$, so $\omega =\omega (e)$. The vierbein equation then becomes the second-order Einstein field equations. This applies to the scalar and Maxwell examples: use $F=dA$, which is independent of $\omega$ and remains gauge invariant even if torsion is present.

A Dirac field changes the connection equation. Directly varying its two covariant derivatives givesThus its spin density sources totally antisymmetric torsion. This is Einstein-Cartan theory, and the connection equation is algebraic rather than a propagation equation for an extra field. To see both its solution and its effect, writeFor the totally antisymmetric component sourced by this field, the Ricci scalar is $R(e,\omega )=R(e)-K_{abc}{K}^{abc}$ up to a divergence. The other irreducible contorsion components have no spinor source and their connection equations set them to zero. The auxiliary part of the Lagrangian is thereforeIts algebraic variation gives the algebraic torsion of a Dirac field,Here the first index of $K_{abc}$ labels the derivative, and the torsion convention is $T_{abc}=K_{bac}-K_{cab}$; specifying this convention fixes the sign. Eliminating $K$ leaves the second-order effective action with contact interactionIt is a local axial-current four-fermion interaction. Hence first-order gravity with Dirac spin density is equivalent to torsion-free second-order gravity with this additional interaction. It is not equivalent to simply declaring the spin connection torsion-free and dropping the contact term. In the effective formulation the Einstein field equations have the stress tensor of the full effective matter action. Before eliminating the connection, the antisymmetric stress and its spin current are related by the local Lorentz Noether identity.

Finally, the 1.5-order formalism is a convenient way to vary this effective action. Let $\hat{\omega }(e,\chi )$ solve the algebraic connection equation and let $S_2(e,\chi )=S_1(e,\hat{\omega }(e,\chi ),\chi )$. The chain rule givesThus vary as in first order, then insert the solved connection. One does not need to calculate the complicated induced connection variation. This procedure is justified by its connection equation, not by discarding that variation arbitrarily. In supergravity the gravitino sources a contorsion quadratic in fermions; substituting it generates the four-fermion terms, while the 1.5-order method simplifies the local supersymmetry cancellation through the lower orders considered above.

Write $q_1={\bar{Q}}^{\dot{1}}$ and $q_2={\bar{Q}}^{\dot{2}}$. In the Jacobi-consistent convention just fixed,Applying the commutator product rule twice givesHere ${\sum }_i{\sigma }_i^2=3I$ produces the constant term; moving the $J_i$ to the right accounts for it. ConsequentlyThus the physically consistent numerical coefficients areIf one mechanically retains the literal positive-sign column transformation instead, the coefficient tuple would be $(\frac{1}{2},\frac{3}{4},1,-i,1,\frac{3}{4},1,i,-1)$. It cannot describe a unitary multiplet with the supplied rotation algebra, by the explicit Jacobi counterexample in part (a).

Take the spin-zero Clifford vacuum $|\Omega ⟩$, normalized to one and annihilated by the two annihilation charges. This is a lowest Fock state of a positive-mass representation, not a zero-energy vacuum annihilated by every charge. For the first one-charge state,The ladder commutators are $[J_{+},q_1]=-q_2$, $[J_{-},q_1]=0$, $[J_{+},q_2]=0$, $[J_{-},q_2]=-q_1$. Therefore these two states are the $j=1/2$ pair. A phase choice $|\frac{1}{2},+\frac{1}{2}⟩=-N{q}_2|\Omega ⟩$ makes the usual positive ladder coefficient hold.