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Orbital angular momentum operators annihilate a radial multiplier $H(r)$. Their squared sum sends every degree-one homogeneous polynomial to $2{\hslash }^2$ times that polynomial, so a nonzero normalizable state $H(r)(\mathbf{a}\cdot \mathbf{x})$ has angular momentum quantum number $l=1$. With real $H$ and real coefficients, its third-component expected value is zero by angular symmetry, though it need not be an eigenstate of that component.

Labels an angular momentum eigenstate through the eigenvalue ${\hslash }^{2}j(j+1)$ of the squared angular momentum operator. The allowed projections are $m\hslash$, with $m=-j,\dots ,j$. Orbital angular momentum has integer $j$; spin can also have half-integer $j$.

A distribution of quantum state populations over atomic magnetic sublevels with a nonzero expectation value of the angular momentum operator vector.

Let the specimen rotate uniformly with angular velocity $\mathbf{\Omega }=\Omega \hat{\mathbf{z}}$. In its rotating reference frame, the internal Hamiltonian containsHere $J_z$ is the internal angular momentum operator about the rotation axis, and ${\mathcal{H}}_0$ contains the internal interactions and any imposed magnetic field expressed in that frame. It does not include the separately treated rigid body's rotational kinetic energy. For a sector with magnetic moment $m_z=\gamma J_z$, the rotational term has exactly the form of a magnetic dipole potential energy:This Barnett equivalent magnetic field has tesla units, with the angular convention for the signed gyromagnetic ratio: $\Omega$ is in radians per second, not revolutions per second. Its sign matters. For an electronic contribution with $\gamma <0$, positive $\Omega$ favors positive $J_z$ and negative $m_z$. A sector with $\gamma >0$ has the opposite magnetic sign. Equal $\Omega$ need not correspond to equal Barnett fields for sectors having different gyromagnetic ratios.

A short derivation fixes the sign independently of any material model. Put $U(t)=\exp (-\mathrm{i}\Omega t{J}_z/\hslash )$ and transform a laboratory quantum state by $|{\psi }_{\mathrm{rot}}⟩=U^{†}|{\psi }_{\mathrm{lab}}⟩$. The Schrödinger equation givesThe transformed laboratory term is denoted by ${\mathcal{H}}_0$; a laboratory perturbation not stationary in the rotating frame can make it time dependent. Establishing a steady thermal response also requires exchange of internal angular momentum with the rotating body, for example through spin-lattice relaxation. A mere relabeling of coordinates cannot polarize isolated, unrelaxed spins.

The Barnett field is a useful description of spin-rotation coupling, not a claim that empty laboratory space acquires a uniform magnetic flux density $\Omega /\gamma$. In particular, inserting it into the Lorentz force on an unrelated test electric charge is unjustified. A nearby magnetometer detects the specimen's actual stray magnetic field and backgrounds, rather than directly detecting this internal Hamiltonian parameter.

For dimensionless angular momentum operators and a Hamiltonian $H=a\mathbf{I}\cdot \mathbf{J}$ at fixed $I,J$, addition of angular momentum gives $E_F=a[F(F+1)-I(I+1)-J(J+1)]/2$. Subtracting neighbouring energy levels proves the displayed rule. For fine structure substitute $\mathbf{L},\mathbf{S}$ and total $J$; for magnetic-dipole hyperfine structure use electronic $J$, nuclear spin $I$ and total $F$. Electric-quadrupole interactions and mixing with nearby energy levels can violate the rule.

The magnetic dipole moment associated with nuclear spin, using dimensionless angular momentum operators and the nuclear magneton ${\mu }_N$. The quoted scalar moment is its expectation in the stretched quantum state, ${\mu }_I=g_I{\mu }_{N}I$. In the same electronic energy level, the leading magnetic-dipole hyperfine structure constant is proportional to ${\mu }_I/I$, apart from isotope-dependent electronic and nuclear-size corrections.

A sequence of nuclear energy levels interpreted as different total angular momenta built on approximately the same intrinsic structure. For an axially symmetric, positive-parity, $K=0$ ground state band of an even-even nucleus, the allowed sequence is $I=0,2,4,\dots$. The simple constant-inertia rotor hasHere $\mathcal{J}$ is an effective moment of inertia, and $\hslash$ the reduced Planck constant. The even-$I$ restriction belongs to this particular intrinsic symmetry and band, rather than to all nuclear excitations. The angular momentum operator eigenvalue supplies $I(I+1)$, rather than the classical replacement $I^2$.

Set $ϵ=E_2-E_0$. Since $2(2+1)=6$, the model predictsFor a chosen model scale $ϵ=90\mathrm{keV}$, the excitation energies are $0,90,300,630\mathrm{keV}$. The corresponding successive gamma ray energies are approximately $90,210,330\mathrm{keV}$, neglecting nuclear recoil. Level energies and emitted line energies are different lists. These numbers illustrate the formula and are not a measured isotope spectrum.

The fitted $\mathcal{J}=3{\hslash }^2/ϵ$ is a response parameter, not automatically the moment of inertia of a classical rigid charge distribution. Nuclear pairing changes this response; Ben Roy Mottelson connected observed inertias with correlations beyond independent-particle motion.^[ref]

At larger angular momentum, changing intrinsic structure or nuclear pairing can invalidate a single constant $\mathcal{J}$. A useful diagnostic is to compare $(E_I-E_0)/[I(I+1)]$ across several levels: constancy tests the simple formula, while a drift identifies the need for additional physics. A successful ratio alone neither specifies every microscopic motion nor measures a literal solid-body rotation.

The operators associated with the two angular momenta commute with one another, and each squared angular momentum operator commutes with every component of that same angular momentum. Two useful identities areConsequently the two maximal commuting subsets of the displayed collection are

The first set gives the uncoupled eigenbasis $|j_1{m}_1⟩\otimes |j_2{m}_2⟩$. Its $J_z$ eigenvalue is already fixed by $m=m_1+m_2$. The second gives the coupled eigenbasis $|j_1{j}_2;jm⟩$ for the addition of angular momentum. Its scalar-product eigenvalue is already fixed by $j,j_1,j_2$. Thus an independent complete set of commuting observables uses four operators in either case; literal maximality within the question's list includes the dependent fifth operator.

For generic nonzero angular momenta, no further listed operator can be added. In particular,is generally nonzero, and $[J_{1z},J^2]=2[J_{1z},{\mathbf{J}}_1\cdot {\mathbf{J}}_2]$. The same obstruction applies to $J_{2z}$. Their sum $J_z$ commutes because the two commutators cancel.

The printed magnetic dipole moment uses dimensionless angular momentum operators: their physical counterparts are $\hslash \mathbf{L}$ and $\hslash \mathbf{S}$. Choose the magnetic field along $z$. Then the Zeeman energy is $H_Z={\mu }_{B}B(L_z+g_s{S}_z)$, whose characteristic size is ${\mu }_{B}B$. In LS coupling, treating this as a time-independent perturbation theory requires it to be much smaller than the relevant fine structure separation, rather than merely smaller than the Coulomb interaction energy.

For a concrete hydrogen atom estimate take the $2p_{1/2},2p_{3/2}$ pair. The hydrogen fine-structure correction iswhere ${\alpha }_{\mathrm{fs}}$ is the fine-structure constant, distinguished from the later potential coefficient. ThusConsequently a suitable weak-magnetic field condition for this pair isThere is no single threshold for every hydrogen energy level: the relevant fine structure separations decrease with $n$. The $1s$ energy level has no same-$n$ spin-orbit interaction partner, so it cannot itself define this comparison. This calculation neglects hyperfine structure; resolving that smaller scale requires an additional weak-magnetic field condition.

The ground electron configuration of caesium has a single $6s$ electron, with $J=1/2$. Addition of angular momentum giving $F=3,4$ fixes the nuclear spin to $I=7/2$. With dimensionless angular momentum operators, write the Fermi contact interaction asThe two hyperfine levels therefore differ by $4A$. In the convention ${\mathbf{\mu }}_I=g_I{\mu }_B\mathbf{I}$ and $g_s\simeq 2$,This exposes a missing ingredient in a prediction from the supplied data: $Z$ and $g_I$ alone do not specify the valence electron contact density. Screening effect and penetration of the closed core strongly affect it. If $g_I$ instead denotes the total nuclear moment in units of ${\mu }_B$, it must first be divided by $I$ in the displayed convention.

An explicitly crude unscreened hydrogen-like ion model uses $|{\psi }_{6s}(0){|}^2=Z^3/(\pi 6^3{a}_0^3)$ and gives ${\nu }_0\simeq 5.76\times 10^{11}\mathrm{Hz}$, or ${\omega }_0\simeq 3.62\times 10^{12}{\mathrm{s}}^{-1}$. It describes neither the shielding nor the radial function of neutral caesium. Using a pure unit-charge $6s$ hydrogen orbital instead gives only $3.46\mathrm{MHz}$, illustrating why a unique useful estimate cannot follow from an unspecified effective charge.

For physical neutral ${}^{133}\mathrm{Cs}$, the independently known zero-magnetic field clock transition isThis measured frequency, used to define the second, is supplementary physical information, not a consequence of $Z=55$ and the rounded $g_I$ alone.^[ref] The direct electronic transition is in the microwave range and is a magnetic dipole transition, rather than a direct optical electric dipole transition. An optical implementation needs a two-field Raman transition.

Using the corrected weak-pulse Ramsey spectrum of part (b), the peak at $\delta =0$ has height $(\Omega {\tau }_p{)}^2$. The quantum interference fringes have spacing $2\pi /T$ in angular frequency, with zeros at $(2k+1)\pi /T$, and the sinc function envelope has its first zeros at $\delta =\pm 2\pi /{\tau }_p$. For $T\gg {\tau }_p$, the central peak has approximately full width at half maximum $\pi /T$ in angular frequency. The clock transition between the two $M_F=0$ states has no first-order Zeeman energy, although a quadratic one remains.

Let the dimensionful angular momentum operators obey $\mathbf{F}=\mathbf{I}+\mathbf{J}$ and take the interaction Hamiltonian as $H^{\prime }=A\mathbf{I}\cdot \mathbf{J}$. Sincethe energy eigenvalues at fixed $I,J$ arewhere $F=|I-J|,\dots ,I+J$. Subtraction gives the Landé interval ruleMany atomic conventions use dimensionless angular momentum operators and call the energy-valued constant itself $A$; then the right-hand side is $AF$. Each energy level has $2F+1$ magnetic sublevels with the same energy in zero field.

For fine structure, the effective interaction is $a_{LS}\mathbf{L}\cdot \mathbf{S}$ with dimensionless angular momentum operators. It originates in the spin-orbit interaction, including the magnetic interaction of electron spin with orbital motion and its relativistic correction. Within an isolated LS coupling atomic term, the labels are the electron configuration, $L,S,J,M_J$, and the interval between adjacent $J$ energy levels is $a_{LS}J$. This is smaller than the electrostatic separation of atomic terms.

For magnetic-dipole hyperfine structure, the nuclear magnetic dipole moment interacts with the magnetic field generated by the electrons. Contributions include the Fermi contact interaction of penetrating $s$ electrons, orbital magnetic fields, and the dipolar field of electron spin. Projected onto one electronic $J$ energy level, their rotationally invariant magnetic-dipole part has the form $a_{IJ}\mathbf{I}\cdot \mathbf{J}$. The labels are $I,J,F,M_F$, together with the approximate electronic labels such as $L,S$. This splitting is generally smaller than fine structure, reflecting the much smaller nuclear magnetic dipole moment. An electric nuclear quadrupole moment can add a different interaction and violate the simple Landé interval rule.

At the larger electrostatic energy scale, an effective two-spin exchange interaction can also be written $a_s{\mathbf{s}}_1\cdot {\mathbf{s}}_2$ within fixed orbitals. Its origin is the Coulomb interaction combined with the Pauli exclusion principle, rather than a large direct magnetic interaction. It separates singlet state and triplet state atomic terms; before spin-orbit coupling, useful labels include the electron configuration, $L,M_L,S,M_S$. General electrostatic atomic term separations need not obey a single Landé interval rule as $L$ varies.

Use the commutator derivation identity $[A,BC]=[A,B]C+B[A,C]$ and the orbital angular momentum commutation relations. They giveThe first two expressions cancel on addition, soThe same argument works for each component of the angular momentum operator.

The squared angular momentum operator and one selected component are commuting operators, so a complete set of common eigenfunctions can be chosen: for the $z$ component these are the usual $|j,m⟩$ angular momentum eigenstates. Distinct components do not have a common complete eigenbasis, because their commutator is another nonzero component. A $j=0$ state is an exception in which all components vanish; noncommutation does not forbid every individual common eigenfunction.

The Weiss temperature $\Theta$ is the extrapolated zero of $1/\chi$ in a Curie–Weiss law fit. It summarizes the leading high-temperature effect of interactions between magnetic moments. For the sign conventionwhere ${\mathbf{J}}_i$ are dimensionless angular momentum operators, mean-field theory givesThus positive net exchange interaction in this convention favours parallel magnetic moments, giving $\Theta >0$ and tendencies towards ferromagnetism; negative net exchange interaction favours antiparallel magnetic moments, giving $\Theta <0$ and antiferromagnetic tendencies. Competing interactions can partially cancel in this sum. The Weiss temperature need not equal the actual ordering temperature, and its sign alone is not a complete determination of the low-temperature ordered structure.

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.

On the tensor product of spins $j_1$ and $j_2$, the combined angular momentum operator is $\mathbf{J}={\mathbf{S}}^{(1)}\otimes I+I\otimes {\mathbf{S}}^{(2)}$. The Clebsch-Gordan decomposition iswhere $J$ increases by one and each irreducible spin representation occurs once. The dimensions agree: ${\sum }_J(2J+1)=(2j_1+1)(2j_2+1)$.

The uncoupled orthonormal basis diagonalizes $S_3^{(1)},S_3^{(2)}$; the coupled orthonormal basis diagonalizes ${\mathbf{J}}^2,J_3$. The Clebsch-Gordan coefficients are their change-of-basis entries, with conventionThey vanish unless $m_1+m_2=M$. Since both bases are orthonormal, this change-of-basis matrix obeys unitarity. Overall phases of the coupled basis vectors are fixed by a convention and do not change the energy eigenvalues.

The operator commutator is $[A,B]=AB-BA$, acting on a common domain where both products are defined. Use the canonical commutation relation $[x_i,p_j]=i\hslash {\delta }_{ij}$, with positions mutually commuting and momenta mutually commuting. On smooth test functions the only nonzero terms in the expansion areThus the orbital angular momentum commutation relations giveThe product rule for a operator commutator givesbecause the last operator expression is symmetric in $i,k$, whereas ${ϵ}_{iwk}$ is antisymmetric. Hence $\boxed{[L^2,L_w]=0}$ for every component.

Write $r=\sqrt{x^2+y^2+z^2}$ and $P=x+y+z$. The differential angular momentum operator $L=-i\hslash x\times \nabla$ annihilates every radial function, so $L_i(P{e}^{-r})=e^{-r}{L}_{i}P$ and the same holds for its square. For example $L_{x}x=0$, $L_y^{2}x={\hslash }^{2}x$ and $L_z^{2}x={\hslash }^{2}x$. Cyclic symmetry gives $L^{2}y=2{\hslash }^{2}y$ and $L^{2}z=2{\hslash }^{2}z$ as well. ThereforeThe wavefunction is an $l=1$ orbital angular momentum state; its nontrivial radial factor does not change its angular eigenvalue. The calculation holds away from the origin and defines the same angular-operator identity in the square-integrable state space.

Let $U(t,t_0)$ be the unitary Schrödinger propagator. The Heisenberg state is fixed at its initial value, while $A_H=U^{†}{A}_{S}U$. Since ${\psi }_S(t)=U{\psi }_H$, matrix elements and probabilities agree exactly in the two pictures. Differentiating, for an operator with no explicit Schrödinger-time dependence, givesFor constant Hamiltonian operator, $U=e^{-iH(t-t_0)/\hslash }$ and $H_H=H$.

A spatial rotation is represented by $U_R(\theta )=\exp [-i\theta n\cdot J/\hslash ]$, where $J$ is the total angular momentum operator. For a particle $J=L+S$. The commutators $[J_i,x_j]=i\hslash {ϵ}_{ijk}{x}_k$ and $[J_i,S_j]=i\hslash {ϵ}_{ijk}{S}_k$ giveThese are the correct infinitesimal vector rotations, verifying the generator and sign.

For the specified Hamiltonian operator, $p^2$ and $x^2$ commute with every $J_i$. Moreoverby antisymmetry. Thus $[J_i,H]=0$ and every total-angular-momentum component is constant in the Heisenberg picture. The separate torques areThey cancel but need not vanish individually, so orbital and spin angular momentum are generally not separately conserved.

Let $\overline{\mathbf{r}}=a^2\mathbf{r}/r^2$ and $\hat{\mathbf{r}}=\mathbf{r}/r$. Its Jacobian isThe chain rule therefore yieldsTaking its cross product with $\mathbf{r}$ kills the radial term and proves that spherical inversion preserves angular derivatives:This is also the invariance of the differential expression underlying the angular momentum operator.

A scalar spherical harmonic $Y$ depends only on direction, so $Y(\hat{\mathbf{r}})=Y(\widehat{\overline{\mathbf{r}}})$ and its gradient is tangential. Thus $\hat{\mathbf{r}}\cdot {\nabla }_{\overline{\mathbf{r}}}Y=0$, andEach of the three normalized vector spherical harmonic families is consequently unchanged as an angular basis function:The normalization requires $n\ge 1$ for $\mathbf{B}$ and $\mathbf{C}$; at $n=0$ the gradient vanishes and only $\mathbf{P}$ remains.

This substitution into the basis functions is distinct from the pushforward of a vector field of a physical vector field by inversion. Under that Jacobian map the radial vector picks up $-a^2/r^2$ and the two tangential vectors pick up $+a^2/r^2$. Likewise, the weighted Kelvin transform of a scalar includes the extra factor $a/r$. Neither operation changes the angular differential identity just proved.

Applying the angular momentum operator directly givesFor an eigenvalue $\lambda$, comparison with $\lambda (f(x)+iy)z$ gives $\hslash x=\lambda f(x)$ and $\hslash f^{\prime }(x)=\lambda$. The first identity excludes $\lambda =0$; differentiating it and using the second gives ${\lambda }^2={\hslash }^2$. No additive constant in $f$ is permitted. The possibilities are $f(x)=x$, $\lambda =\hslash$, and $f(x)=-x$, $\lambda =-\hslash$.

For $f=x$, the polynomial $\psi =(x+iy)z$ is homogeneous of degree two and satisfies $\Delta \psi =0$. Expanding the three squared orbital angular momentum operators gives, with $D=x{\partial }_x+y{\partial }_y+z{\partial }_z$,Here $D\psi =2\psi$ and $D^2\psi =4\psi$. Therefore ${\hat{L}}^2\psi =6{\hslash }^2\psi$. This is the degree-two case of angular momentum of a homogeneous harmonic polynomial; its angular labels are $l=2$, $m=1$.