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A Lie algebra over a field $k$ is a vector space $\mathfrak{g}$ with a bilinear map $[,]:\mathfrak{g}\times \mathfrak{g}\to \mathfrak{g}$, called the Lie bracket, that is alternating and satisfies the Jacobi identity. Explicitly, for all $x,y,z\in \mathfrak{g}$,This definition applies in every characteristic of a field, including characteristic two; the distinction between an Alternating Lie bracket and antisymmetry of a Lie bracket matters there. See Pavel Etingof, Lie Groups and Lie Algebras, Definition 8.4 and Remark 8.5.

The motivating question is how to describe very small symmetries. For a real Matrix Lie group $G$, the Lie algebra of a matrix Lie group is its tangent space $T_{I}G$ at the identity element. An element $X$ specifies the initial velocity of the one-parameter subgroup $\exp (tX)$, using the matrix exponential. The Lie bracket records the first nonzero effect of changing the order of two small transformations. Even when first-order motions cancel, a second-order motion can remain: the small rotation commutator below leaves a rotation about a third axis.

This turns local symmetry questions into computations with linear maps. A basis and its structure constants of a Lie algebra determine every Lie bracket by bilinearity. For example, three matrices suffice for the sl2 matrix commutator example, while the familiar cross product describes the SO(3) Lie algebra. These descriptions help calculate rotations in three dimensions and build Lie algebra representations; the same bracket structure underlies angular momentum commutation relations. The striking feature is that a nonlinear family of motions has a local description by a vector space and one product. Global information still needs care: the Lie algebra alone does not distinguish the circle group from the additive real line.

Historically, the connection with differential equations is visible in Sophus Lie's Theorie der Transformationsgruppen, volume I (1888), prepared with Friedrich Engel. Chapter 6 interprets first-order differential operators as infinitesimal transformations, linking their action on functions to the transformations those functions admit. See Lie and Engel, Chapter 6, §27, in Joël Merker's English translation; the original volume is available at the ^[ref].

An irreducible spin representation is a nonzero Hilbert space with no proper nonzero closed invariant subspace under all three angular momentum operators. Equivalently, the associated unitary representation of SU(2) cannot be decomposed into two nonzero orthogonal invariant pieces. The irreducible possibilities are labelled by a spin $j\in \{0,\frac{1}{2},1,\frac{3}{2},\dots \}$ and have dimension $2j+1$.

Choose an orthonormal basis $|j,m_s⟩$, with $m_s=-j,-j+1,\dots ,j$. The angular momentum commutation relations imply $[{\mathbf{S}}^2,S_i]=0$, $[S_3,S_{\pm }]=\pm \hslash S_{\pm }$, $[S_{+},S_{-}]=2\hslash S_3$, and $S_{+}^{†}=S_{-}$. The complete action isIn particular, the spin raising operator annihilates $|j,j⟩$ and the spin lowering operator annihilates $|j,-j⟩$. The coefficient follows by taking the squared norm and using $S_{\mp }{S}_{\pm }={\mathbf{S}}^2-S_3^2\mp \hslash S_3$. Thus the ladder operators connect every member of the basis, while terminating at its two endpoints.

The angular momentum commutation relations are $[L_i,L_j]=i\hslash {ϵ}_{ijk}{L}_k$. The commutator product rule givessince the bracketed expression is symmetric in $j,k$ and ${ϵ}_{ijk}$ is antisymmetric. Alsoproving $\boxed{L^2=L_{-}{L}_{+}+L_3^2+\hslash L_3}$.

For orbital angular momentum, $L_i=-i\hslash {ϵ}_{ijk}{x}_j{\partial }_k$. A differentiable radial function has ${\partial }_{k}f(r)=f^{\prime }(r)x_k/r$, so its contraction with the antisymmetric symbol vanishes: $Lf(r)=0$. Put $s=x_1+i{x}_2$. The coordinate differential operators give $L_{3}s=\hslash s$, whileannihilates every power of $s$, because $({\partial }_1+i{\partial }_2)s=0$. Since all $L_i$ annihilate the radial factor, the product rule yieldsSubstitution into the expression for $L^2$ gives the highest-weight complex-coordinate orbital wavefunction resultAs $[L^2,L_{-}]=0$, lowering preserves this eigenvalue whenever the lowered function is nonzero. Explicitly,For $n=0$ the lowered function is zero, so it is not an eigenfunction. For $n\ge 0$ the original nonzero function has the usual regular angular dependence. The PDF's “any integer” differential identities also hold locally for negative $n$ where $s\ne 0$, but those negative powers are singular on the axis and are not square-integrable regular angular eigenstates. The converted TeX incorrectly writes $L_4$ in place of the PDF's $L_{+}$.

In the position representation, $\hat{p}=-i\hslash \nabla$. The orbital angular momentum operator and its squared magnitude areFor example, with $D=x\cdot \nabla$, the latter is $-{\hslash }^2[r^2\Delta -D^2-D]$. Operator products here act on smooth wavefunctions in a common invariant domain.

To calculate the angular momentum commutation relations, use $[x_a,p_b]=i\hslash {\delta }_{ab}$ and the vanishing position-position and momentum-momentum commutators. Expanding the products givesMultiplication by ${ϵ}_{iab}{ϵ}_{jcd}$ then givesFinally, the product commutator rule yieldsThe final cancellation pairs the antisymmetric Levi-Civita symbol with an expression symmetric in $j,k$. Thus

The orbital angular momentum operators are $\mathbf{L}=\mathbf{x}\times \mathbf{p}$, soThe canonical commutation relations are $[x_i,x_j]=[p_i,p_j]=0$ and $[x_i,p_j]=i\hslash {\delta }_{ij}$. On a common domain of smooth wavefunctions, the product commutator identity givesContract this with ${ϵ}_{iab}{ϵ}_{jcd}$. The first term contributes $i\hslash ({\delta }_{ij}\mathbf{x}\cdot \mathbf{p}-x_j{p}_i)$ and the second contributes $i\hslash (x_i{p}_j-{\delta }_{ij}\mathbf{x}\cdot \mathbf{p})$. Their sum isThis verifies the angular momentum commutation relations explicitly. Similarly,

In the position representation $\mathbf{p}=-i\hslash \nabla$. For a radial wavefunction, $\nabla {\Psi }_0={\Psi }_0^{\prime }(r)\mathbf{x}/r$, parallel to $\mathbf{x}$, soWrite ${\Phi }_i=h(r)x_i$ with $h(r)={\Psi }_0^{\prime }(r)/r$. The operator $L_3=-i\hslash (x_1{\partial }_2-x_2{\partial }_1)$ annihilates the radial factor. It gives $L_3{x}_1=i\hslash x_2$, $L_3{x}_2=-i\hslash x_1$, and $L_3{x}_3=0$, henceThus these components, when nonzero, are the specified eigenfunctions with eigenvalues $0,+\hslash ,-\hslash$. This is an example of angular momentum of a radial function times a linear polynomial.

Represent spin by $J_i=(\hslash /2){\sigma }_i$ in the basis $|\uparrow ⟩,|\downarrow ⟩$. The Pauli matrices are Hermitian and satisfy $[{\sigma }_i,{\sigma }_j]=2i{ϵ}_{ijk}{\sigma }_k$, giving the general angular momentum commutation relations $[J_i,J_j]=i\hslash {ϵ}_{ijk}{J}_k$. The particular representation has $J^2=3{\hslash }^{2}I/4$ and $J_3$ eigenvalues $\pm \hslash /2$, the features specific to spin one-half.

In the stated order of two-particle basis vectors, the operator isIts eigenvalue-one vectors are $|\uparrow \uparrow ⟩$, $|\downarrow \downarrow ⟩$ and $(|\downarrow \uparrow ⟩+|\uparrow \downarrow ⟩)/\sqrt{2}$, all symmetric under interchange. The eigenvalue-minus-three vector is $(|\downarrow \uparrow ⟩-|\uparrow \downarrow ⟩)/\sqrt{2}$, antisymmetric under interchange. These are the triplet state and singlet state, respectively.

Identical spin-half particles are fermions. With no other degrees of freedom to carry antisymmetry, only the singlet is allowed. Indeed, for total angular momentum $J=(\hslash /2)({\sigma }^{(A)}+{\sigma }^{(B)})$,This gives $2{\hslash }^2$ on the triplet and zero on the singlet, corresponding to total spin one and zero, as required by addition of angular momentum.

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.

It is essential to distinguish a finite-dimensional spinor matrix from its action on a family of operator components. Take the Pauli matrices and ${\bar{\sigma }}^{\mu \nu }$ in the supplied convention, with $\eta =\mathrm{diag}(1,-1,-1,-1)$. They giveIf both displayed transformations are read literally as left multiplication on the column ${\bar{Q}}^{\dot{\alpha }}$, expanding the operator conjugation and comparing its linear term gives the formal answer $[M^{\mu \nu },\bar{Q}]=+{\bar{\sigma }}^{\mu \nu }\bar{Q}$.

There is a sign inconsistency in that literal reading. With the usual angular momentum commutation relations, coefficient matrices $R_i$ in $[J_i,q_a]=(R_i{)}_{ab}{q}_b$ must satisfyIndeed, the Jacobi identity gives $[J_i,[J_j,q]]-[J_j,[J_i,q]]=(R_j{R}_i-R_i{R}_j)q$, whereas $[[J_i,J_j],q]=i{ϵ}_{ijk}{R}_{k}q$. The formal $R_i=+{\sigma }_i/2$ violates this identity. For example it would give $[J_3,q_1]=q_1/2$, $[J_{+},q_1]=q_2$ and $[J_3,q_2]=-q_2/2$; the two sides of the Jacobi identity for $J_3,J_{+},q_1$ are then $-q_2/2$ and $3q_2/2$.

A consistent canonical column-operator convention uses the inverse adjoint action for the displayed spinor transformation, equivalently the contragredient operator-component prescription. It givesThis is the convention used for the physical states below. Reversing the adjoint action is the minimum needed sign correction in the literal column reading; a dual row notation can encode the same correction by its index contractions. The formal printed-sign coefficients are also recorded below, so the distinction is explicit rather than silently changing a convention. The representation matrices themselves remain $J_i={\sigma }_i/2$ and $K_i=i{\sigma }_i/2$ in Question 3: the minus sign here concerns commutators of operator components.

The product rule for a commutator gives $[A,B^2]=[A,B]B+B[A,B]$. Using the permitted cyclic angular momentum commutation relations,The first two expressions cancel, even though the factors need not commute. HenceThe squared angular momentum operator commutes with each of its components by the same cyclic calculation.

The angular momentum commutation relations implyThus $S_{\pm }|s,\sigma ⟩$ has the same total-spin eigenvalue and magnetic quantum number $\sigma \pm 1$. Its squared norm follows fromChoosing the conventional positive phases gives the spin ladder operator formula

For $s=1$, in the ordered basis $|1,1⟩,|1,0⟩,|1,-1⟩$,Since $S_x=(S_{+}+S_{-})/2$,

The incident state is the spin coherent state obtained by rotating $|1,1⟩$ through $\theta$ about the $y$ axis. Its amplitudes in the $S_z$ basis areup to an irrelevant phase convention. The Born rule therefore gives

The spin raising operator and spin lowering operator areThe angular momentum commutation relations implyso $S_{\pm }|s,\sigma ⟩$ is proportional to $|s,\sigma \pm 1⟩$. Sinceits squared norm isChoosing the conventional positive phase gives

Using the spin ladder operators,and thereforeIn the ordered product basis of the tensor product of quantum systemsthe matrix representation isThe two parallel-spin states are already eigenvectors. Diagonalizing the central block gives the four energy eigenvaluesThis is the spectrum of the Two-spin Heisenberg Hamiltonian in an opposing longitudinal field.

The asserted ground-state ordering requires the antiferromagnetic case $\alpha >0$. As $B\to 0$,The corresponding state at $E_{-}$ is the unique spin-one-half singlet statewhereas the three states at the first excited energy form the spin-one-half triplet state.

Indeed, for the total angular momentum operatorone has at $B=0$The Hamiltonian then has rotational symmetry, so energy depends only on the total-spin sector. The singlet has total spin $s=0$ and multiplicity one; the triplet has $s=1$ and multiplicity $2s+1=3$. This rotational symmetry explains why the triplet energy eigenspace has dimension three.

The angular momentum commutation relations are $[L_x,L_y]=i\hslash L_z$ and cyclic permutations. The product rule yields $[L_z,L_x^2]=i\hslash (L_y{L}_x+L_x{L}_y)$ and $[L_z,L_y^2]=-i\hslash (L_x{L}_y+L_y{L}_x)$, so $[L_z,L^2]=0$. Simultaneous eigenstates have eigenvalues ${\hslash }^{2}l(l+1)$ and $\hslash m$; orbital angular momentum has integer $l\ge 0$ and $m=-l,\dots ,l$.

By cyclic symmetry, $L^2$ commutes with every component and hence every squared component. Therefore $[H,L^2]=0$: an energy eigenbasis can be chosen with definite $l$. For unequal transverse coefficients, $L_z$ need not commute with $H$, so definite $m$ is not generally available.

Expanding the ladder operators gives $L_{+}^2+L_{-}^2=2(L_x^2-L_y^2)$ and $L^2-L_z^2=L_x^2+L_y^2$. Comparing coefficients givesFor $l=1,m=0$, both squared ladder operators vanish, since two raising or lowering steps would leave the multiplet. Thus $E_0=2\beta {\hslash }^2={\hslash }^2(a_x+a_y)$. In the remaining basis $\{|1,1⟩,|1,-1⟩\}$ the Hamiltonian isIts symmetric and antisymmetric combinations have energies ${\hslash }^2(a_x+a_z)$ and ${\hslash }^2(a_y+a_z)$, respectively. ConsequentlyThe energies are related by permutation of the principal axes, not generally equal. Two coincide when the corresponding coefficients coincide; all three become $2a{\hslash }^2$ in the isotropic limit.

Operators, quantization and rotations. Angular momentum is the generator of spatial rotations. In quantum mechanics, its components are Hermitian operators on their appropriate domains and obeyThese angular momentum commutation relations prevent all three components from having simultaneous sharp values. For example, the Robertson uncertainty principle gives $\Delta J_x\Delta J_y\ge \hslash |⟨J_z⟩|/2$. One conventionally diagonalizes ${\hat{J}}^2$ and a selected component ${\hat{J}}_z$:The allowed labels are $j=0,\frac{1}{2},1,\frac{3}{2},\dots$ and $m=-j,-j+1,\dots ,j$. An angular-momentum multiplet has $2j+1$ states. The magnitude $\hslash \sqrt{j(j+1)}$ differs from the largest possible component $\hslash j$; the remaining components fluctuate even in a state with maximal $m$.

The angular momentum lowering operator and its raising counterpart, ${\hat{J}}_{\pm }={\hat{J}}_x\pm i{\hat{J}}_y$, satisfyThe ladder terminates at $m=\pm j$. A rotation through angle $\theta$ about unit vector $\hat{\mathbf{n}}$ acts through the unitary operator $\exp [-i\theta \hat{\mathbf{n}}\cdot \hat{\mathbf{J}}/\hslash ]$. Thus quantized components and the continuous rotation of a state are compatible.

Orbital angular momentum. The orbital contribution is associated with motion in space:For a spinless spatial wavefunction, the simultaneous angular eigenfunctions are the spherical harmonics $Y_{\ell m}(\theta ,\varphi )$, with $\ell =0,1,2,\dots$ and $m=-\ell ,\dots ,\ell$. Single-valued ordinary scalar spatial wavefunctions give integer orbital labels. In a central potential energy, $[\hat{H},{\hat{L}}^2]=[\hat{H},{\hat{L}}_z]=0$, so stationary states may be labelled by these angular quantum numbers. The angular kinetic contribution appears in the radial equation as $\ell (\ell +1){\hslash }^2/(2m{r}^2)$, the centrifugal potential. Noncentral interactions need not conserve these orbital labels.

Intrinsic spin and addition. Spin angular momentum is an intrinsic internal degree of freedom with the same commutation algebra. It is not the angular momentum of a small classical charged sphere literally rotating about its centre. For an electron, $s=1/2$ and $m_s=\pm 1/2$, so ${\hat{S}}_i=(\hslash /2){\sigma }_i$ in terms of the Pauli matrices. Its squared spin is $3{\hslash }^2/4$. Other particles can have integer or half-integer spin, including spin zero.

For one particle with orbital and spin degrees of freedom, $\hat{\mathbf{J}}=\hat{\mathbf{L}}+\hat{\mathbf{S}}$; the two component sets commute with each other. Addition of angular momentum permits $j=|\ell -s|,|\ell -s|+1,\dots ,\ell +s$. The coupled states $|j,m_j⟩$ are linear combinations of the uncoupled $|\ell ,m_{\ell }⟩|s,m_s⟩$ with $m_j=m_{\ell }+m_s$, determined by Clebsch-Gordan coefficients. A rotationally invariant spin-orbit interaction can conserve $J^2,J_z$ while mixing the separate component labels.

Angular momentum also carries a magnetic moment. For an electron's orbital motion and intrinsic spin,The minus signs reflect negative electric charge. With the Bohr magneton ${\mu }_B=e\hslash /(2m_e)$, a spin projection $m_s$ has ${\mu }_z=-g_s{\mu }_B{m}_s$. A magnetic field can therefore distinguish angular-momentum states through their interaction energy $-\mathbf{\mu }\cdot \mathbf{B}$.

Stern–Gerlach measurement. In the Stern-Gerlach experiment, atoms emerge from a heated source, pass through collimating apertures, traverse a strongly inhomogeneous magnetic field, and strike a position-sensitive screen. The beam is neutral, so its separation is interpreted through the atomic magnetic moment rather than the ordinary charged-particle Lorentz force. Near a chosen quantization direction, the interaction givesA uniform field can exert a torque but would not separate the beam by this translational force; the gradient is essential. Different allowed magnetic-moment components become correlated with different trajectories. A classical ensemble with continuously varying orientations would yield a continuous spread, whereas discrete quantum components yield separated channels.

In the modern electronic description of ground-state silver, filled shells cancel their electronic angular momenta and the unpaired $5s$ electron has $L=0$, $S=1/2$, $J=1/2$. The effective electronic measurement therefore has two components, approximately ${\mu }_z=\pm {\mu }_B$, rather than an orbital $\ell =1/2$, which is not allowed. For positive field gradient the electronic $m_J=+1/2$ component has negative ${\mu }_z$ and deflects downward in this coordinate convention. Nuclear spin and hyperfine structure are neglected in this simplified two-channel account. Historically the original splitting preceded the electron-spin interpretation, so the modern explanation should not be confused with the original investigators' interpretation.

More generally a prepared spin state $a|+z⟩+b|-z⟩$ gives the two outcomes with probabilities $|a{|}^2,|b{|}^2$. Selecting one channel prepares a state for further measurements. An ideal second analyser along the same axis then gives a definite result, whereas one along a perpendicular axis gives two equally likely outcomes for a spin-one-half eigenstate. This connects the apparatus directly to incompatible spin components and the Born rule. Angular momentum is quantized through its operator algebra; the Stern–Gerlach gradient turns that discrete internal information into spatially distinguishable outcomes.