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Write $|m_1,m_2⟩$ for product spin states, and $J_{+}=J_{1+}+J_{2+}$. A highest-weight state is killed by $J_{+}$. For $s=2$ the unique $m=2$ state is $|1,1⟩$. In the $m=1$ subspace, $J_{+}$ kills $a|1,0⟩+b|0,1⟩$ exactly when $a+b=0$. In the $m=0$ subspace it kills $a|1,-1⟩+b|0,0⟩+c|-1,1⟩$ exactly when $a+b=b+c=0$. Normalizing givesThe left states are coupled states and the right states are product states. Lowering commutes with particle interchange, so the whole spin-$s$ multiplet has the exchange sign of its highest-weight state: $+1$ for $s=0,2$ and $-1$ for $s=1$, or $(-1{)}^s$. Integer-spin identical particles are bosons; their spatial partial wave changes by $(-1{)}^{\ell }$ under interchange. Symmetry of the full state therefore requires $\ell +s$ even.

For decay from total angular momentum one, coupling requires $|\ell -s|\le 1\le \ell +s$. The case $s=0$ forces $\ell =1$ and violates Bose symmetry. For $s=1$, the allowed range is $\ell =0,1,2$ and symmetry leaves $\ell =1$. For $s=2$, the range is $\ell =1,2,3$ and symmetry leaves $\ell =2$. The identical particles' intrinsic parity product is $+1$, so total final parity is $(-1{)}^{\ell }$. HenceThis uses the stated massive-spin angular-momentum model; no additional restrictions for massless transverse vector particles have been assumed.

Use the Clifford algebra $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2{\eta }^{\mu \nu }$ and unit-weight antisymmetrization. The duality identity between ${\gamma }_5{\gamma }_{\nu }$ and three antisymmetrized gamma matrices turns the equation intoThe overall nonzero factor in this identity depends on the orientation and ${\gamma }_5$ convention and does not change its solutions. This Rarita-Schwinger field equation has the gauge symmetrysince ${\gamma }^{\mu \nu \rho }{\partial }_{\nu }{\partial }_{\rho }ϵ=0$. It is the free-field limit of local supersymmetry, so it is suitable for the gravitino.

To check which spin it propagates, put $A={\gamma }^{\mu }{\Psi }_{\mu }$ and $B={\partial }^{\mu }{\Psi }_{\mu }$. Expanding the three gamma matrices using the Clifford algebra givesin four dimensions. Thus the Euler-Lagrange field equation requires $B=/\partial A$. Under the gauge transformation, $A\mapsto A+/\partial ϵ$; locally one may choose gamma-trace gauge fixing, $A=0$, by solving $/\partial ϵ=-A$. The Euler-Lagrange field equation then yieldsThese are the massless Dirac equation, transversality and gamma-trace constraints that remove the lower-spin part of a vector-spinor.

For a more explicit massless Rarita-Schwinger polarization count, take a positive-frequency plane wave with null momentum along the third axis, in signature $(-,+,+,+)$. Transversality relates its temporal and longitudinal components. Residual gauge transformations with $/pϵ=0$ remove both. Write the surviving components as $u_1,u_2$; their gamma-trace condition reads ${\gamma }^1{u}_1+{\gamma }^2{u}_2=0$, henceThe massless Dirac equation leaves two complex positive-frequency amplitudes in $u_1$. If $i{\gamma }^1{\gamma }^2{u}_1=s{u}_1$, $s=\pm 1$, then $(u_1,u_2)$ is proportional to $(1,-is)$ in the transverse vector index. With spatial rotation generators $J_3^{\mathrm{vec}}{}_{12}=-i$ and $J_3^{\mathrm{spin}}=-i{\gamma }^1{\gamma }^2/2$, the vector has helicity $-s$ and the Majorana spinor has helicity $-s/2$, giving total helicity $-3s/2$. The Majorana spinor condition fixes the negative-frequency coefficients from these amplitudes. Therefore a massless gravitino has spin $3/2$, with the two physical helicities $\pm 3/2$. A complex vector-spinor would also have independent antiparticle modes.

The requested count is instead the off-shell component count of minimal supergravity, before imposing any Euler-Lagrange field equations. A four-dimensional Majorana spinor has four real components. Its vector index supplies four copies, and local supersymmetry supplies four real gauge functions:One must not additionally subtract transversality and the massless Dirac equation in this count: those are on shell conditions. The gamma-trace gauge fixing already uses the same four gauge functions subtracted above. For the graviton, a symmetric metric tensor has ten real components, with four diffeomorphism gauge functions. Equivalently, a vierbein has sixteen real components, minus six local Lorentz transformation and four diffeomorphism functions:The gravitino and graviton each have two propagating states on shell, but their off-shell counts disagree by six.

An auxiliary field has an algebraic Euler-Lagrange field equation and no propagating kinetic term. In old-minimal supergravity, a complex scalar $M$ contributes two real bosonic components and a real vector $b_{\mu }$ contributes four, so the bosonic count becomes $6+2+4=12$, matching the gravitino. Their supersymmetry transformations also absorb the terms that would otherwise require the propagating Euler-Lagrange field equations to close the local algebra. Thus auxiliary fields allow off-shell supersymmetry closure and matching component counts without adding physical particles. Eliminating them recovers an on shell formulation.

A periodic lattice Hamiltonian commutes with lattice translations. The Bloch theorem therefore classifies its states by characters of the translation group:where $u$ is lattice periodic. For each crystal momentum $\mathbf{k}$, solving on a primitive cell with the corresponding quasiperiodic boundary condition gives discrete eigenvalues indexed by $n$. Their continuous dependence on $\mathbf{k}$ produces the energy bands $E_n(\mathbf{k})$ as $\mathbf{k}$ ranges through the Brillouin zone. Bragg mixing opens gaps at suitable degeneracies, although bands can overlap in three dimensions and gaps are not guaranteed between every pair.

The quantum numbers are band index, wavevector modulo reciprocal lattice vectors, and spin where spin is conserved. In a finite crystal the allowed wavevectors are discrete; in the thermodynamic limit they become dense and the band description emerges. Spin-orbit coupling can replace a separately conserved spin label by an appropriate internal band label.

Dilute randomly placed impurities break exact translation symmetry, so crystal momentum is no longer an exact quantum number. Most extended states retain approximately the host band structure but acquire impurity scattering and disorder broadening. An impurity potential can also produce impurity levels in a crystal, often in a band gap: levels associated with a donor impurity lie near the conduction-band edge and can supply electrons, while those associated with an acceptor impurity lie near the valence-band edge and can create holes. The impurity's species and potential determine whether such levels exist and how deep they are; not every impurity necessarily makes a gap state. At low density these levels are nearly isolated, while overlap at increasing density can broaden them into an impurity band. Added or removed carriers shift the chemical potential as well.

At zero temperature, fill all momentum states inside the Fermi sphere, with two spin states per momentum. Phase-space counting givesFor dispersion $E=cp$, the ratio of energy to particle-number integrals isLet $\epsilon$ be the common chemical potential in the weak field, and let $+$ denote the lower Zeeman-energy branch $cp-\mu B$. Its populations areKeeping $N_{+}+N_{-}=N$ changes $\epsilon$ only at order $B^2$, so to first order it equals ${\epsilon }_F=c{p}_F$. Subtracting the two populations givesEach excess lower-energy spin contributes moment $\mu$ along the field. Hence the Pauli moment of a massless electron gas isThis includes the spin contribution specified in the problem. The field is weak enough that both spin branches remain occupied, and orbital magnetic effects are not part of the assumed energy model.

The Adjoint double cover from SU(2) to SO(3) is obtained by writing a traceless Hermitian matrix as $X=x_i{\sigma }_i$, where the ${\sigma }_i$ are the Pauli matrices, and defining $AX{A}^{†}=(R(A)x{)}_i{\sigma }_i$. Explicitly,Conjugation preserves $\frac{1}{2}\mathrm{tr}{X}^2=|x{|}^2$, so $R(A)$ is orthogonal; connectedness of SU(2) gives $\det R(A)=1$. Composition of conjugations makes $R$ a group homomorphism. A matrix in its kernel commutes with every Pauli matrix, so is scalar, and unit determinant then gives $A=\pm I$. Every spatial rotation occurs: $A=\exp (-i\theta \hat{n}\cdot \sigma /2)$ induces the rotation through angle $\theta$ about $\hat{n}$. ThusSince SU(2) is the simply connected three-sphere, this is the universal double cover; the two Lie algebras are isomorphic, although the groups are not.

Define the conjugate fundamental spinor of SU(2) by ${\bar{\eta }}_{\alpha }=({\eta }^{\alpha }{)}^{\ast }$. Regard $\bar{\eta }$ as a row, so unitarity givesConsequently ${\bar{\eta }}_{\alpha }{\xi }^{\alpha }$ is an invariant Hermitian pairing. The basic invariant tensors are the Kronecker delta ${\delta }^{\alpha }{}_{\beta }$ and the alternating tensors ${ϵ}_{\alpha \beta }$ and ${ϵ}^{\alpha \beta }$. Choose ${ϵ}_{12}=1$, ${ϵ}^{12}=-1$, so that ${ϵ}^{\alpha \gamma }{ϵ}_{\gamma \beta }={\delta }^{\alpha }{}_{\beta }$. The unit-determinant identity $A^TϵA=ϵ$ makes $[\eta ,\xi ]={ϵ}_{\alpha \beta }{\eta }^{\alpha }{\xi }^{\beta }$ invariant; its inverse gives the corresponding upper-index invariant. These contractions generate the tensor invariants of the defining representation. In particular ${\tilde{\eta }}^{\alpha }={ϵ}^{\alpha \beta }{\bar{\eta }}_{\beta }$ transforms by $A$, showing that the conjugate representation is equivalent to the defining one. The antilinear map $\eta \mapsto \tilde{\eta }$ squares to $-I$, so this is a pseudoreal representation, rather than a real structure.

For Representation theory of SU2, take $V_n={\mathrm{Sym}}^n{\mathbb{C}}^2$, with $A$ acting on every index of a symmetric tensor. Its independent components are determined by the number $r$ of indices equal to $2$, so there are $n+1$ of them. Equivalently, the homogeneous polynomial representation of SU2 has basis $z_1^{n-r}{z}_2^r$, $0\le r\le n$, withThe basis weights are $n,n-2,\dots ,-n$, and the raising operator and lowering operator connect consecutive basis vectors with nonzero coefficients until their endpoints. Any nonzero invariant subspace contains a weight vector, by applying polynomial spectral projections in $H$. Raising it reaches $z_1^n$, and lowering then obtains the entire basis. Thus $V_n$ is an irreducible representation, of spin $j=n/2$. The classification of finite-dimensional representations of SU2 gives every finite-dimensional complex irreducible representation in this way. The central element $-I$ acts on a rank-$n$ tensor as $(-1{)}^n$, so descent of an SU(2) representation to SO(3) gives

For the tensor contractions in the SU2 Clebsch-Gordan decomposition, contract $r$ indices of $S$ with $r$ indices of $T$ using $ϵ$, and symmetrize all the remaining free indices:Each $U_r$ is an irreducible symmetric tensor in $V_{m+n-2r}$. To prove these are precisely the constituents, use two sets of formal variables $z,w$. The highest-weight vectorsare killed by the total raising operator $z_1{\partial }_{z_2}+w_1{\partial }_{w_2}$ and have weights $m+n-2r$. The determinant factor is invariant; lowering the remaining factor gives an irreducible string of dimension $m+n-2r+1$. Distinct highest weights give inequivalent summands. The unitary representation on this tensor product makes invariant complements available, and the dimension checkexhausts the space. Hence the contractions above are its irreducible projections, up to normalization, and the Clebsch-Gordan decomposition for SU2 is

For the decomposition of three SU(2) spinors, write $Q^{\alpha \beta \gamma }={\eta }_1^{\alpha }{\eta }_2^{\beta }{\eta }_3^{\gamma }$ and $S^{\alpha \beta \gamma }=Q^{(\alpha \beta \gamma )}$. The contractionstransform as defining spinors. Alternation in three indices in dimension two vanishes, giving $d_1-d_2+d_3=0$; there are exactly two independent contractions. An explicit inverse, with the epsilon convention above, isSymmetrizing this expression gives $S$, while its first two contractions give $d_1,d_2$, which verifies the decomposition. Thus

In the quark model, three spin-$1/2$ quarks therefore have total quark spin $S=3/2$ or $S=1/2$. The two spin-$1/2$ copies correspond to coupling the first pair to spin $0$ or spin $1$ before adding the third quark. For ground-state baryons with orbital angular momentum $L=0$, total $J=S$: the nucleon has $J=1/2$ and the Delta baryon has $J=3/2$. These two coupling copies alone do not assert two unrestricted physical baryon families. The antisymmetric three-quark colour singlet, the spatial wavefunction and flavor symmetry must also satisfy the Pauli constraint on three-quark flavour multiplets; orbital excitations can give other total angular momenta.

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.