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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.

Distinct eigenvalues of the self-adjoint angular momentum operator imply $⟨{\psi }_1,{\psi }_2⟩=0$. The given combination is not normalized: its squared norm is $({\cos }^2\omega t+3{\sin }^2\omega t)/4$, which is always positive. Therefore its properly normalized expectation value isIt vanishes when ${\tan }^2\omega t=1/3$. The first positive solution has $\omega t=\pi /6$, soNormalization changes the value of the expectation value away from zero, but not its zero times.

The orbital angular momentum operators areIn particular ${\hat{L}}_3=-i\hslash (x_1{\partial }_2-x_2{\partial }_1)$. For smooth compactly supported functions $\varphi ,\psi$, integration by parts produces no boundary term, and ${\partial }_2{x}_1={\partial }_1{x}_2=0$. HenceThis proves Hermiticity on the common test-function domain, and the same calculation applies on suitable decaying domains. As an unbounded operator, it is not defined on every $L^2$ function: the usual self-adjoint operator realization is the generator of unitary rotations about the third axis. Thus the integration statement includes its necessary domain/boundary assumptions.

Use the radial exponential $e^{-\alpha r}$ shown in the original PDF; the TeX's $e^{-\alpha x}$ is a transcription defect. Every orbital angular momentum operator annihilates a radial function, so it acts only on $P=x_1+x_2+2x_3$. The angular momentum of a radial function times a linear polynomial is determined by its degree-one angular factor: direct application of the rotation derivatives givesEquivalently its restriction to the sphere is a degree-one spherical harmonic. ThusFurthermore ${\hat{L}}_3\psi =-i\hslash K(x_1-x_2)e^{-\alpha r}$. Its expected value is proportional to the integral of $(x_1+x_2+2x_3)(x_1-x_2)e^{-2\alpha r}$. The cross terms are odd in a coordinate, and the integrals of $x_1^2$ and $x_2^2$ agree by rotational symmetry. HenceThe state need not be an eigenstate of ${\hat{L}}_3$. If normalization is desired, $\int P^2{e}^{-2\alpha r}{d}^{3}x=6\pi /{\alpha }^5$, so $K={\alpha }^{5/2}/\sqrt{6\pi }$; the zero expectation holds after normalization for any nonzero initial $K$.

The total angular momentum operator isUsing the cyclic orbital angular momentum commutation relations,while $[L_z,L_z^2]=0$; hence $\boxed{[L_z,{\mathbf{L}}^2]=0}$. In Cartesian coordinates,Because $L_z(x+iy)=\hslash (x+iy)$ and it annihilates $z$ and every radial function,Replacing $i$ by $-i$ gives eigenvalue $-m\hslash$.

The six-dimensional space splits into the five trace-free quadratic spherical harmonics with $\ell =2$ and one radial state with $\ell =0$. A simultaneous eigenbasis for $L_z$ and ${\mathbf{L}}^2$ iswhere the eigenvalue $\ell (\ell +1){\hslash }^2$ of ${\mathbf{L}}^2$ was used. Every displayed state is a complex linear combination of the original six energy eigenstates, so it is still an energy eigenstate at the same level.

For $f(w)=w^M=r^M{e}^{iM\phi }$, the angular momentum operator $L_z=-i\hslash {\partial }_{\phi }$ givesThe probability density is proportional towhose maximum occurs atIn a large disk of area $A=\pi R^2$, the states that fit inside satisfy $M\lesssim qB{R}^2/(2\hslash )$. Thus the degeneracy of each Landau level is asymptoticallywhere ${\Phi }_B=BA$ is the magnetic flux. There is one state per magnetic flux quantum $h/q$, up to boundary corrections.

Since $O$ commutes with every angular momentum operator, it commutes with $J^2$. Taking a matrix element of $[O,J^2]=0$ givesThe allowed nonnegative values of $j$ make $j(j+1)$ injective, so the matrix element vanishes unless $j=j^{\prime }$.

The rotational kinetic energy is $K=L^2/(2I)$. For a three-dimensional linear rigid rotor, angular momentum operator replaces the squared magnitude by ${\hslash }^{2}l(l+1)$, where $l=0,1,2,\dots$. HenceThe three lowest distinct energy levels are thereforeTheir magnetic quantum degeneracy values are $1,3,5$. A single moment of inertia specifies this rigid rotor model; an arbitrary asymmetric body would require its other principal moments.

The total angular momentum operator is $\mathbf{J}=\mathbf{L}+\mathbf{S}$. Since orbital and spin operators act on different factors of the tensor product, they commute, and expanding $J^2$ givesA coupled angular momentum eigenstate $|l,s;j,m_j⟩$ is a simultaneous eigenstate of these three squared operators, with eigenvalues ${\hslash }^{2}j(j+1)$, ${\hslash }^{2}l(l+1)$ and ${\hslash }^{2}s(s+1)$. Thus the angular factor in the spin-orbit interaction is the numberThe other factor, $r^{-1}dV/dr$, is radial. It commutes with the angular operators, and a central-potential energy eigenstate separates into a radial wavefunction and an angular spinor. Taking its radial expectation value therefore givesThis spin-orbit energy shift in a central potential is independent of $m_j$, so rotational symmetry preserves the $2j+1$ magnetic degeneracy within a fixed $j$ multiplet. It is a first-order fine structure correction; the given coefficient already includes the relativistic factor needed for the electron's spin-orbit coupling.

For a particle in a ring, the arc-length coordinate is $s=R\varphi$. The tangential kinetic energy isThe constraint removes radial motion; the constant potential energy along the ring can be chosen as zero. The tangential momentum operator is ${\hat{p}}_s=-i\hslash {\partial }_s$, or equivalently the angular momentum operator is ${\hat{L}}_z=-i\hslash {\partial }_{\varphi }$. ThusIts physical domain has periodic boundary conditions, $\psi (\varphi +2\pi )=\psi (\varphi )$ and ${\psi }^{\prime }(\varphi +2\pi )={\psi }^{\prime }(\varphi )$. These make the Hamiltonian self-adjoint and reflect that $\varphi$ and $\varphi +2\pi$ describe the same point. No magnetic flux or additional angle-dependent potential energy is present.

First make the units explicit. Use angular momentum operators in units of $\hslash$, so their squared eigenvalues are $J(J+1)$ rather than ${\hslash }^{2}J(J+1)$; the coefficient $A$ then has units of energy. The operator identitygives, at fixed $J_1,J_2$,Subtracting the expression for the adjacent allowed value $J-1$ leavesThis is the Landé interval rule. Its content is that consecutive gaps grow in proportion to the higher total-angular-momentum quantum number, with the same coefficient within an isolated coupled multiplet. A negative $A$ reverses the ordering of the levels, while positive observed transition frequencies alone measure gap magnitudes.

If the source symbols instead denote physical angular momenta with units of action, the Hamiltonian written literally as $A{\mathbf{J}}_1\cdot {\mathbf{J}}_2$ gives the gap $A{\hslash }^{2}J$. Recovering the printed form then requires absorbing ${\hslash }^2$ into the coefficient or writing $H=(A/{\hslash }^2){\mathbf{J}}_1\cdot {\mathbf{J}}_2$. The convention is essential to dimensional consistency rather than an extra dynamical assumption.

The given angular momentum commutation relations imply $[L_3,L_1]=i\hslash L_2$ and $[L_3,L_2]=-i\hslash L_1$. By linearity,Similarly,For the remaining commutators, the commutator product rule gives, for each $j$,The last sum vanishes because the Levi-Civita symbol is antisymmetric under $i\leftrightarrow k$, while $L_i{L}_k+L_k{L}_i$ is symmetric. Taking $j=3$ and then linear combinations with $j=1,2$ proves

If $L^2\psi =\lambda {\hslash }^2\psi$ and $L_3\psi =m\hslash \psi$, these identities yieldOne must also prove that this candidate eigenstate is nonzero. Since the angular momentum operators are self-adjoint operators, $L_{+}^{\ast }=L_{-}$, and expanding the product givesThereforeThis is nonnegative and, by the assumed inequality, nonzero. Hence it is positive. The angular momentum raising operator really does produce a nonzero simultaneous eigenstate with the claimed eigenvalues.

To prove that raising terminates, let $v$ be any nonzero state reached after $r$ steps, with projection $q=m+r$. Positivity givesso $q^2\le \lambda$. Thus the sequence cannot remain nonzero for every integer $r\ge 0$. At its last nonzero state $v=(L_{+}{)}^k\psi$, one has $L_{+}v=0$, and the squared norm identity givesUnder the paper's initial inequality the first step is nonzero, so in fact $k\ge 1$. An angular momentum ladder endpoint must therefore be reached: bounded projection follows from positivity, and the final raising norm fixes the total angular momentum eigenvalue. Writing $j=m+k$, the complementary product $L_{+}{L}_{-}=L^2-L_3^2+\hslash L_3$ gives $\Vert L_{-}v{\Vert }^2=2j{\hslash }^2\Vert v{\Vert }^2$, also showing $j\ge 0$.

In the ordered orthonormal basis $(|\frac{1}{2},\frac{1}{2}⟩,|\frac{1}{2},-\frac{1}{2}⟩)$, the spin one-half angular momentum operators are $J_i=(\hslash /2){\sigma }_i$, with Pauli matricesThe ladder operators areExplicit multiplication gives $J_{+}{J}_{-}={\hslash }^2\mathrm{diag}(1,0)$ and $J_{-}{J}_{+}={\hslash }^2\mathrm{diag}(0,1)$. Their commutator is therefore

Recover the angular momentum operators from $J_1=(J_{+}+J_{-})/2$ and $J_2=(J_{+}-J_{-})/(2i)$. The spin-one angular momentum matrices in the chosen descending-$J_3$ orthonormal basis areThey are self-adjoint operators, satisfy $[J_i,J_j]=i\hslash {\epsilon }_{ijk}{J}_k$, and give $J^2=2{\hslash }^{2}I$, as required for $j=1$.

Replacing $L$ by the angular momentum operator gives the Hamiltonian operatorThe time-dependent Schrödinger equation is thereforeFor a stationary state, separate variables as $\Psi (\varphi ,t)=\psi (\varphi )T(t)$. Dividing by the product separates a function of $t$ from one of $\varphi$:Thus $T=e^{-iEt/\hslash }$ up to a constant, and the time-independent Schrödinger equation is $-{\hslash }^2{\psi }^{\prime \prime }/(2I)=E\psi$.

Diagonalize the commuting angular momentum operator at the same time: $-i\hslash {\psi }^{\prime }=\ell \psi$ has solutions $A{e}^{i\ell \varphi /\hslash }$. A single-valued wavefunction requires $e^{2\pi i\ell /\hslash }=1$, hence $\ell =n\hslash$ with $n\in \mathbb{Z}$. Normalizing with the angular probability measure $d\varphi$ givesThese form an orthonormal basis under ${\int }_0^{2\pi }{\psi }_m^{\ast }{\psi }_{n}d\varphi ={\delta }_{mn}$. The periodic boundary conditions apply to both the wavefunction and its first derivative, ensuring the physical self-adjoint Hamiltonian operator. For probability per arc length $ds=rd\varphi$, the normalization instead reads ${\psi }_n=e^{in\varphi }/\sqrt{2\pi r}$; the two conventions describe the same probabilities. General states are superpositions of these energy eigenstates.

A rigid rotor models rotational motion with fixed internal separations. Its use in the synthetic Stark effect figure omits vibrational motion and additional interactions in real molecules.

Rigidity fixes the molecular geometry, not the orientation, and does not require a spherical mass distribution. Remove motion of the centre of mass, choose principal axes of the inertia tensor, and write the classical rotational kinetic energy asHere $I_a,I_b,I_c$ are the principal moments of inertia, ${\omega }_a,{\omega }_b,{\omega }_c$ components of angular velocity, and $L_a,L_b,L_c$ components of angular momentum. Quantization replaces these last components by angular momentum operators. Different principal moments of inertia produce different spectra: the single-constant linear rigid rotor is a special case, not a formula for every molecule.^[ref]

The restriction of a component of the angular momentum operator to discrete eigenvalues; the Stern-Gerlach experiment tested its magnetic manifestation.

A rank-$k$ spherical tensor operator has components $T_q^{(k)}$, $-k\le q\le k$, that transform under rotation like an angular momentum $k$. Equivalently, their commutators with the angular momentum operator satisfyFor a vector operator, the rank is one and a conventional choice is $T_0=V_z$, $T_{\pm 1}=\mp (V_x\pm i{V}_y)/\sqrt{2}$. The Wigner-Eckart theorem separates the matrix elements of a spherical tensor operator into Clebsch-Gordan coefficients and a reduced matrix element.

In the ordered spin-one basis $(m=1,0,-1)$, the second angular momentum operator satisfies $J_2^3=J_2$. Therefore its matrix exponential reduces to $I-i\sin \theta J_2+(\cos \theta -1)J_2^2$. A half-turn exchanges the two outer weights and negates the middle weight. Applied to the pion isospin triplet, this fixes the rotation factor in G parity.

A spin rotation acts on a quantum state through the unitary representation generated by its angular momentum operators. With $\hslash =1$, rotation by angle $\theta$ about axis $\hat{\mathbf{n}}$ is $e^{-i\theta \hat{\mathbf{n}}\cdot \mathbf{S}}$. Simultaneously rotating every site is a global symmetry of an isotropic Heisenberg model.