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The ground state of an even-even nucleus has $J^P=0^{+}$, and the emitted alpha particle also has $0^{+}$. In an alpha decay to daughter spin $J_Y$, addition of angular momentum permits total zero only when the relative orbital angular momentum is $L=J_Y$. Parity conservation gives $+1={\pi }_Y(-1{)}^L$, hence the daughter has natural parity, $0^{+},1^{-},2^{+},3^{-},\dots$. This rule does not itself restrict $J_Y$ to even integers. An even-$J$ sequence such as $0^{+},2^{+},4^{+}$ requires additional nuclear structure, for example a positive-parity ground-state nuclear rotational band.

For a homogeneous harmonic polynomial $H_l$ of degree $l$ in three dimensions, the Euler theorem for homogeneous functions gives $D{H}_l=l{H}_l$, where $D=\mathbf{x}\cdot \nabla$. The orbital angular momentum identity ${\hat{L}}^2=-{\hslash }^2(r^2\Delta -D^2-D)$ therefore gives ${\hat{L}}^2{H}_l=l(l+1){\hslash }^2{H}_l$. This constructs angular momentum eigenstates directly from polynomials.

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 $j$.

Each orbital angular momentum operator annihilates a radial factor and $r^2$. Direct differentiation gives $L^2{x}_3^2={\hslash }^2(6x_3^2-2r^2)$ and $L_3{x}_3^2=0$. Hence the displayed trace-free quadratic polynomial, times any admissible radial factor, has orbital labels $l=2,m=0$. Its angular dependence is proportional to the spherical harmonic $Y_{20}$; subtracting $r^2/3$ removes the rotational scalar part.

The two-state calculation is not a general formula for a ferromagnet. Exchange interactions, magnetic domains, magnetic domain-wall pinning, magnetic anisotropy, and hysteresis can determine which response is accessible. Both spin angular momentum and orbital angular momentum can contribute; a single constant gyromagnetic ratio may be only a restricted effective description. Rotation sets an internal bias, while the material and its preparation determine magnetization. The Barkhausen effect illustrates why a smooth bias can produce jumps instead of the toy smooth curve.

For a limited, reversible, isotropic linear model, suppose the same intrinsic volume magnetic susceptibility ${\chi }_v$ responds to the applied bias and to $B_{\Omega }/{\mu }_0$. For a uniformly magnetized ellipsoid along a principal axis, include its demagnetizing factor $N_d$:Solving this linear equation yieldsHere $H_{\mathrm{int}}$ and $H_{\mathrm{app}}$ are actual magnetic field intensities; the added rotational bias is an equivalent drive in the constitutive response. In SI, dimensionless ${\chi }_v$ multiplies an intensity in $A/m$, so $M_z={\chi }_v\Omega /\gamma$ would omit ${\mu }_0$ and have the wrong units. A susceptibility defined instead as $\partial M/\partial B$ has different units. State the convention before comparing formulas.

At zero applied magnetic field intensity, a stipulated ${\chi }_v=3$ gives prefactor $3/2$ in a sphere, with $N_d=1/3$, and a prefactor approaching $3$ in a long axial ellipsoid. This is a geometry check of the assumed linear model. A real cylinder does not have an exactly uniform demagnetizing field, and an irreversible ferromagnet cannot in general be represented by this single linear equation. Inferring an intrinsic gyromagnetic ratio requires an independently calibrated material response and a geometry-dependent conversion of the measured stray magnetic field.

For an isospin singlet quark $h$ of charge $q_h$, the antisymmetric light-flavour pair gives an $I=0,J=1/2$ state with charge $q_h+1/3$. The symmetric light-flavour pairs $uu,\{ud\},dd$ give $I=1$ triplets of spins $1/2$ and $3/2$, with charges $q_h+4/3,q_h+1/3,q_h-2/3$. These follow from light-diquark spin-isospin correlation and addition of constituent electric charges, with zero orbital angular momentum.

A nuclear beta decay connects initial and final quantum states through vector and axial-vector parts of the weak charged current. At leading order in the lepton wavelength expansion, Fermi transitions carry lepton-pair spin $S=0$, while Gamow-Teller transitions carry $S=1$. If the lepton pair carries orbital angular momentum $l$, its total angular momentum is $J=l$ for $S=0$, or $J=|l-1|,\dots ,l+1$ for $S=1$. The nuclear states must obey $|J_i-J_f|\le J\le J_i+J_f$, and their parities satisfy ${\pi }_f/{\pi }_i=(-1{)}^l$ in this leading orbital classification. The least permitted $l$ labels the degree of forbiddenness. Real forbidden transitions involve several weak multipole operators; the elementary spin classification does not assert that only one operator or nuclear matrix element exists.

If an escaping stellar wind carries specific orbital angular momentum $\lambda J/M$ with constant $\lambda$, then $\dot{J}/J=\lambda \dot{M}/M$ and $J\propto M^{\lambda }$. Combining this with circular-binary orbital angular momentum gives the displayed period invariant. It is a different prescription from donor-wind angular-momentum loss, whose specific angular momentum generally varies with mass ratio.

The Bohr model assumes circular electron orbits with quantized orbital angular momentum $N\hslash$. It reproduces the hydrogen energy levels, but wavefunctions in quantum mechanics describe spatial probability distributions rather than definite trajectories. In particular, spherically symmetric Coulomb bound states have zero orbital angular momentum.

The puzzle is the coexistence of reproducible spectral lines and stable chemical atoms. A classical circular orbit in a Coulomb potential can have any negative energy: Coulomb's law alone selects no special radius or emitted frequency. Moreover, its electron has centripetal acceleration. The Larmor formula then predicts positive radiated power, so conservation of energy makes the orbit shrink. Simply drawing concentric circles explains neither stability nor a discrete spectrum.

Niels Bohr postulated nonradiating stationary states and emission with $h\nu =E_i-E_f$. His circular-orbit condition is $L=n\hslash$. These assumptions override classical radiation; the emitted frequency need not equal an orbital frequency.^[ref]

For a spherical square well of radius $a$, depth $V_0$ and reduced mass $\mu$, the regular zero-orbital angular momentum solution has reduced radial wavefunction $U=A\sin kr$ inside and $U=B{e}^{-qr}$ outside. Wavefunction matching at a finite potential step gives $k\cot (ka)=-q$, where $k^2=2\mu (V_0+E)/{\hslash }^2$ and $q^2=-2\mu E/{\hslash }^2$. The $n$th radial branch first reaches $E=0$ at $ka=(n+1/2)\pi$, for $n=0,1,\dots$. A normalizable negative-energy state on that branch requires $V_0>V_{s,n}$. Thus exactly one such state exists for $V_{s,0}<V_0\le V_{s,1}$; the new zero-energy $s$-wave solution at the upper equality is not normalizable.

For the first $\ell =1$ branch, the zero-energy exterior solution has $U\propto r^{-1}$. The regular interior solution is $U\propto r{j}_1(kr)$, where $j_1$ is a Spherical Bessel function. Matching logarithmic derivatives gives $1+z{j}_1^{\prime }(z)/j_1(z)=-1$, or $j_0(z)=0$, with $z=a\sqrt{2\mu V_0}/\hslash$. Its first threshold is $z=\pi$, so $V_{p,0}={\pi }^2{\hslash }^2/(2\mu a^2)$. Higher values of orbital angular momentum increase the positive centrifugal potential, so none can bind below this first $p$-wave threshold. Spin-dependent nuclear forces are absent from this central-well model.

Consider a single electron of mass $m_e$ and a point atomic nucleus of mass $M$ and positive electric charge $Ze$, with $e>0$ the elementary charge. Use nonrelativistic classical mechanics, omit external fields and interactions with other electrons, and writeHere $\mu$ is the Coulomb two-body reduced mass, ${ϵ}_0$ the vacuum permittivity, and $h$ the Planck constant. The separation $r$ and relative speed $v$ describe the two-particle relative coordinate; neither body is silently pinned in place. The electron's own radius about the centre of mass is $Mr/(M+m_e)$.

For the assumed circular orbit, Coulomb's law provides the required centripetal acceleration, and the imposed orbital angular momentum condition isSubstitute $v=n\hslash /(\mu r)$ into the first equation and cancel $r^{-2}$ to obtainThe kinetic energy is $K=\mu v^2/2=\kappa /(2r)$ and the Coulomb potential energy is $U=-\kappa /r$, with zero energy at infinite separation. HenceThis derives the energy sequence from the stated circular model; the modern Schrödinger equation derivation is the distinct series-termination quantization of the Coulomb radial equation. The radius above scales as $n^2/Z$, whereas the binding energy scales as $\mu Z^2/n^2$. The Bohr radius article supplies the corresponding Coulomb potential length scale. The nonrelativistic assumption requires $v_n\ll c$, with $c$ the speed of light.

For an original numerical example choose $Z=1$, $M=1.67262192369\times 10^{-27}\mathrm{kg}$, $m_e=9.1093837015\times 10^{-31}\mathrm{kg}$, $e=1.602176634\times 10^{-19}\mathrm{C}$, $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, ${ϵ}_0=8.8541878128\times 10^{-12}\mathrm{F}{\mathrm{m}}^{-1}$, and $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$. Treat these as stipulated constants for the calculation. They give $\mu /m_e=0.99945568$, $r_1=0.0529465\mathrm{nm}$, $E_1=-13.598287\mathrm{eV}$ and $v_1/c\simeq 0.00730$, where $\mathrm{eV}$ denotes an electronvolt. At $n=2$, $r_2=4r_1$ and $E_2=E_1/4$; the ground-state ionization energy is $-E_1$.

Taking $M\to \infty$ gives $\mu \to m_e$. A finite nuclear mass reduces every binding energy by $\mu /m_e$ and increases each separation by $m_e/\mu$. This follows from relative motion, not a fitted correction to individual spectral lines. Replacing the atomic nucleus by a heavier isotope changes the same factor, providing another check of the model.

A Coulomb bound state is a physical normalizable eigenfunction for an attractive potential $V(r)=-\gamma /r$, $\gamma >0$. For the usual regular boundary condition at the origin its energies are the displayed values, $N=1,2,\dots$. The spherically symmetric states have orbital angular momentum zero; series-termination quantization of the Coulomb radial equation constructs them as a polynomial times a decaying exponential. Square integrability on the punctured radial interval alone does not replace the physical condition at the origin.

Choose the one-particle states of a hydrogen atom, or a central-field description of a chemical atom, with labelsThe principal quantum number $n$ identifies the electron shell. The orbital angular momentum label $\ell$ identifies the electron subshell; the magnetic quantum number $m_{\ell }$ labels $2\ell +1$ independent spatial orbitals. Each has two spin one-half along an axis choices, hence $2(2\ell +1)$ spin orbitals. Applying the Pauli exclusion principle once per spin orbital givesThese quantum numbers label wavefunctions and spin projections, not electron trajectories. They need not resolve exact energy eigenstates in the presence of spin-orbit interaction; changing to coupled angular momentum eigenstates preserves the count of independent one-particle states within this shell model.

For example, $n=3$ allows $\ell =0,1,2$, conventionally $3s,3p,3d$. Their spatial orbital counts are $1,3,5$; their spin orbital counts are $2,6,10$. The sum is eighteen. The algebra behind “one per complete state” is short: two-particle antisymmetrization gives $u\wedge u=(u\otimes u-u\otimes u)/\sqrt{2}=0$, and the zero vector cannot be a normalized quantum state. Opposite-spin electrons in the same spatial orbital have distinct complete states, so their antisymmetric product survives.

This supplies a structural link to the periodic table: finite electron subshell capacities make repeated patterns of occupied electron configurations possible. It does not predict the exact ground state electron configuration or energy of a many-electron chemical atom. Different electron subshells overlap in energy, and electron interactions change their ordering. In particular, the periodic table period lengths are not simply successive $2n^2$ values: the third periodic table period has eight chemical elements, although the full $n=3$ electron shell has eighteen available states. A capacity counts available spin orbitals; an actual filling order requires a Quantum Hamiltonian.

A two-neutron system. At low orbital angular momentum its identical-fermion exchange symmetry forces the spin-singlet, isospin-one channel; there is no bound dineutron in vacuum at physical parameters. This does not forbid correlated neutron pairs inside larger nuclei.

A partial wave with orbital angular momentum $\ell =2$. Its orbital parity is positive. Coupling it to a spin-one pair can include total $J=1$, allowing a D wave component in the deuteron.

An electron has spin one-half. Its intrinsic spin angular momentum has two possible components, $\pm \hslash /2$, along a specified unit vector $\mathbf{n}$. These are the two eigenvalues of the corresponding spin measurement; the spin is distinct from the electron's orbital angular momentum.

An electron subshell groups the allowed electronic quantum states within an electron shell by their orbital angular momentum quantum number $\ell$. Labels $s$ and $p$ mean $\ell =0$ and $\ell =1$, respectively; their capacities are two and six electrons. The leading number in $2p$ identifies the electron shell, while $p$ identifies the electron subshell.

The spin-independent vector contribution to an allowed beta decay. The isospin operator ${\tau }_k^{\pm }$ changes a neutron into a proton or the reverse; the sign convention depends on the assigned nucleon isospin projection. Its rotational rank is zero, so $J_f=J_i$ and the parity is unchanged. The phrase Fermi-type forbidden contribution extends the $S=0$ classification to higher orbital angular momentum.

The Geiger-Nuttall law expresses the extreme sensitivity of an alpha decay half-life to its Q value. In a restricted range of heavy nuclei, a useful local form is ${\log }_{10}[T_{1/2}/T_{\ast }]\simeq a{Z}_d/\sqrt{Q_{\alpha }}+b$, where $Z_d$ is the daughter atomic number, and the constants depend on the chosen units and nuclear family. It is named for Hans Geiger and John Mitchell Nuttall.

The Coulomb barrier outside the daughter has $V(r)=2Z_d\alpha /r$ in natural units. Its outer turning point is $r_t=2Z_d\alpha /Q_{\alpha }$. The WKB approximation gives a penetration probabilitywhere $\mu$ is the daughter-alpha reduced mass. The leading large-barrier exponent is $2\pi (2Z_d\alpha )\sqrt{\mu /(2Q_{\alpha })}$, producing the approximate $Z_d/\sqrt{Q_{\alpha }}$ dependence. Finite nuclear size modifies the exponent and the slowly varying intercept. The nuclear decay constant also contains an assault frequency and an alpha-cluster formation probability. Extra orbital angular momentum raises the barrier further, and odd-nucleon structure can introduce substantial hindrance. Thus extrapolation from even-even decays gives an order-of-magnitude reference rather than a precision prediction for an odd-odd nucleus.

Suppose a transition operator obeys $[I_3,H]=H/2$ and $[I_{+},H]=0$, and acts on the lower component of an isospin-half kaon. Its output has third component zero, and $I_{+}^{2}H|K⟩=H{I}_{+}^2|K⟩=0$. This excludes an isospin-two output. A spinless parent decaying to two spinless pions has zero relative orbital angular momentum, so Bose-Einstein statistics excludes the antisymmetric isospin-one sector. The output is therefore the two-isovector singlet. In the equal-mass isospin limit, charged and neutral two-pion decay widths have ratio two, including the identical-particle counting.