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Put $f_l(\theta ,\varphi )={\sin }^l\theta e^{-il\varphi }$. For the regular, single-valued orbital angular momentum states, $l$ is a nonnegative integer. Differentiation immediately givesFor $0<\theta <\pi$, the derivatives needed for $L^2$ areConsequently the angular differential expression inside the parentheses of $-L^2/{\hslash }^2$ equalsusing ${\csc }^2\theta -{\cot }^2\theta =1$. ThereforeThe expressions extend regularly through the polar coordinate singularities: $f_l$ is the restriction of the polynomial $(x-iy{)}^l$ to the unit sphere. It is proportional to the lowest-projection spherical harmonic of degree $l$; for $l=0$ it is the constant state.

Inverting the definitions of the creation and annihilation operators givesSubstituting in the orbital angular momentum $L$ givesOperators belonging to distinct modes commute. Therefore the terms with two annihilation operators and those with two creation operators cancel, and $a_1^{\ast }{a}_2=a_2{a}_1^{\ast }$, $a_2^{\ast }{a}_1=a_1{a}_2^{\ast }$. The surviving terms give the orbital angular momentum in oscillator ladder operators:

The orbital angular momentum $L_3=X_1{P}_2-X_2{P}_1$ generates rotations in the plane. The canonical commutation relations giveUse the commutator product rule $[A,B^2]=[A,B]B+B[A,B]$. Since different coordinates commute,Thus, for arbitrary positive angular frequencies,At ${\omega }_1={\omega }_2$, this becomes $\boxed{[L_3,H]=0}$. The rotational symmetry of the two-dimensional isotropic harmonic oscillator allows a basis of simultaneous energy eigenstates and orbital angular momentum eigenstates.

Write $\alpha =m\omega /\hslash$. The first Hermite polynomials give ${\psi }_{0,0}\propto e^{-\alpha (x_1^2+x_2^2)/2}$ and ${\psi }_{1,0},{\psi }_{0,1}$ proportional, with the same constant, to $x_1{e}^{-\alpha (x_1^2+x_2^2)/2}$ and $x_2{e}^{-\alpha (x_1^2+x_2^2)/2}$. Thus in Cartesian coordinates the new wavefunctions areUsing polar coordinates $x_1=r\cos \phi$, $x_2=r\sin \phi$ givesThe orbital angular momentum is $L_3=-i\hslash {\partial }_{\phi }$, which immediately verifies its eigenvalues $0,\pm \hslash$. Multiplying these spatial wavefunctions by $e^{-i\omega t}$ for ${\chi }_0$ and $e^{-2i\omega t}$ for ${\chi }_{\pm }$ gives the corresponding full stationary states.

The angular momentum commutation relations arewhere ${\epsilon }_{ijk}$ is the Levi-Civita symbol. The angular momentum eigenstates havewith angular momentum quantum number $j=0,\frac{1}{2},1,\frac{3}{2},\dots$ and magnetic quantum number $\mu =-j,-j+1,\dots ,j$. There are $2j+1$ such states in one irreducible spin representation, so the $J^2$ eigenspace has dimension $2j+1$ within that multiplet. The use of $\mu$ avoids confusing the magnetic quantum number with the particle mass in the earlier questions.

For completeness, the ladder operators $J_{\pm }=J_1\pm i{J}_2$ satisfy $[J_3,J_{\pm }]=\pm \hslash J_{\pm }$ andThe upper and lower ends are $\mu =\pm j$; integer steps between them require $2j$ to be a nonnegative integer. This explains both the spectrum and the multiplet dimension. Pure orbital angular momentum permits only integer $j$, whereas general spin allows half-integers. If a larger Hilbert space contains several copies of the same irreducible spin representation, the full $J^2$ eigenspace includes all those copies: its dimension is $(2j+1)$ times their multiplicity, and can be infinite. The paper's dimension question is understood per irreducible multiplet.

In a close binary star, the rotating effective gravitational potential defines two Roche lobes meeting at the inner Lagrange point. Stellar expansion or orbital shrinkage can bring the donor into Roche-lobe overflow. Gas then flows through the inner Lagrange point; its angular momentum can form an accretion disk around the companion. wind mass transfer in a binary star is also possible without Roche-lobe filling. For a circular, synchronized binary, the Eggleton Roche-lobe radius approximation isThis describes the lobe geometry, not the donor's dynamical response.^[ref]

conservative binary mass transfer has fixed total mass and orbital angular momentum, neglecting stellar spin. Using $J=M_d{M}_a\sqrt{Ga/(M_d+M_a)}$ givesSince ${\dot{M}}_d<0$, transfer from the more massive donor initially shrinks the orbit; after binary mass ratio reversal it expands. Stability depends on both stars and their changing lobes. Let ${\zeta }_{\ast }=d\ln R_d/d\ln M_d$ and ${\zeta }_L=d\ln R_L/d\ln M_d$. A small mass loss reduces the relative overfill if ${\zeta }_{\ast }>{\zeta }_L$. Use the adiabatic donor response for dynamical stability, and its thermal response for slower adjustment; an equilibrium mass-radius law alone cannot decide both.

Stable transfer can proceed on nuclear or thermal timescales, while dynamically unstable overflow may create a common envelope, orbital spiral-in and either envelope ejection or a merger. Escaping gas, gravitational radiation and magnetic braking make evolution nonconservative by changing orbital angular momentum. accretion onto a compact stellar object releases power of order $G{M}_a\dot{M}/R_a$, producing bright X-ray binaries; radiation-driven outflows can limit retained mass. Eclipses, radial velocities and disk emission constrain the geometry and masses, but inclination, irradiation and assumptions about mass or angular-momentum loss affect the inference. The essential feedback is the donor's radius relative to its evolving Roche lobe.

Assume a real, central, short-range potential and elastic quantum scattering. At large radius, normalize the stationary wavefunction aswhere $k=\sqrt{2mE}/\hslash$ and $\theta$ is the scattering angle. Equal incoming and outgoing speeds give the differential scattering cross-section $d\sigma /d\Omega =|f(\theta ){|}^2$.

The partial-wave scattering expansion separates the spherical harmonics. The incident plane wave has expansion ${\sum }_{\ell \ge 0}{i}^{\ell }(2\ell +1)j_{\ell }(kr)P_{\ell }(\cos \theta )$, with Spherical Bessel functions $j_{\ell }$ and Legendre polynomials $P_{\ell }$. The regular radial solution outside the potential has asymptotic form $u_{\ell }\propto \sin (kr-\ell \pi /2+{\delta }_{\ell })$. Its real scattering phase shift ${\delta }_{\ell }$ changes the ratio of outgoing to incoming waves to the Partial-wave S-matrix $S_{\ell }=e^{2i{\delta }_{\ell }}$. Matching outgoing coefficients yieldsHere $\ell$ labels the orbital angular momentum, with squared magnitude ${\hslash }^2\ell (\ell +1)$. The assumption of purely elastic scattering is what makes every $S_{\ell }$ have unit modulus.

Consider a parent atomic nucleus at rest. Write its signed nuclear polarization along a fixed axis $\hat{\mathbf{z}}$ as $P=⟨J_z⟩/(I\hslash )$, where $I\hslash$ is its maximal spin component. Let $\theta$ be the angle of the emitted electron momentum to that axis and let $\beta =v/c$. Use $\mathcal{P}$ for the parity operator, so it is not confused with $P$.

For an axially symmetric leading-dipole model, retain a constant and the term linear in the polarization vector $\mathbf{P}=P\hat{\mathbf{z}}$. Rotational symmetry makes the directional term proportional to $\mathbf{P}\cdot \hat{\mathbf{p}}$. Define its dimensionless coefficient as $A\beta$. At fixed electron energy the relative intensity and normalized probability density function per solid angle are thereforeThe factor $\beta$ is the conventional separation of the speed dependence; rotational symmetry alone does not determine that dependence or $A$. The coefficient $A$ contains the decay dynamics. This leading model omits higher angular terms and detailed recoil and radiative effects. A partially oriented higher-spin ensemble can also have alignment that is not specified by its vector nuclear polarization alone.

The unit-integral condition follows without any model of a detector:Thus $f$ has unit angular average, while the probability density function in the polar angle alone contains the Jacobian determinant factor:In particular, $f$ itself is neither a normalized probability density function in $\theta$ nor a detector count. Positivity on $-1\le u\le 1$ gives $|a|\le 1$.

Under full spatial reflection, $\mathbf{p}\mapsto -\mathbf{p}$ but $\mathbf{J}\mapsto \mathbf{J}$. The latter follows already for orbital angular momentum from $\mathbf{r}\times \mathbf{p}\mapsto (-\mathbf{r})\times (-\mathbf{p})=\mathbf{r}\times \mathbf{p}$; spin has the same axial vector transformation. Hence $\mathbf{P}\cdot \hat{\mathbf{p}}$ is a pseudoscalar, and $\cos \theta \mapsto -\cos \theta$. If parity conservation held for this model, equal probabilities for the inverted process would force $a=0$. Conversely, $a=0$ in an unpolarized sample proves nothing about $A$.

Integrating over the two hemispheres gives a useful original check:Choose $A=-0.8$, $P=0.5$, and $\beta =0.75$, so $a=-0.3$. Then the forward and backward hemisphere probabilities are $0.425$ and $0.575$, and their counting asymmetry is $-0.15$. These are chosen teaching values, not measured values from the Wu experiment. At a single energy, the angular shape determines the product $AP\beta$; it cannot separately determine $A$ without a polarization calibration.

An atomic orbital with orbital angular momentum quantum number $l=1$. Real $p_x,p_y,p_z$ combinations have opposite quantum phases in their two lobes and a nodal plane through the atomic nucleus. An orbital phase is not an electric charge.

The partial-wave scattering channel with orbital angular momentum quantum number one. Its angular dependence is built from degree-one spherical harmonics. For a finite-range potential away from a threshold pole, the finite-range partial-wave threshold law gives phase shift ${\delta }_1=O(k^3)$ and a partial elastic scattering cross-section proportional to $k^4$ as the relative wavenumber $k$ tends to zero.

An axially symmetric quadratic potential energy perturbation preserves parity and the $z$ component of orbital angular momentum. For $1s,2s,2p_0,2p_{\pm 1}$, its first-order energy shifts are respectively $(2\alpha +\beta )a_0^2$, $14(2\alpha +\beta )a_0^2$, $(12\alpha +18\beta )a_0^2$, and $(24\alpha +6\beta )a_0^2$. They follow by multiplying the hydrogen radial second moment by the angular average of $\alpha +(\beta -\alpha ){\cos }^2\theta$, whose ${\cos }^2\theta$ averages are $1/3,1/3,3/5,1/5$. Negative $\beta$ makes the globally extended potential unbounded below; the shifts then describe local perturbative resonance poles or a potential stabilized at large distance.

A quarkonium is a bound state of a heavy quark and its own antiquark. Examples are charmonium and the bottom-antibottom states called Upsilon mesons. Spectroscopic labels $n{}^{2S+1}{L}_J$ distinguish radial excitation, total quark spin, relative orbital angular momentum and total angular momentum.

An energy and orbital angular momentum eigenstate for a real two-dimensional central potential can be chosen with integer $k$ and real radial function $f$. Its radial equation isQuantum degeneracy permits other stationary states, including $f(r)\cos (k\varphi )/\sqrt{\pi }$, which are not single angular exponentials. Separation of variables supplies a basis rather than describing every superposition in a degenerate eigenspace.

The spin-orbit interaction couples orbital angular momentum to spin. In chemical atoms it helps determine fine structure and the coupling scale that separates weak-field Zeeman splitting from the Paschen-Back effect.

For the interaction $\lambda \mathbf{S}\cdot \mathbf{L}$ built from the total two-electron spin and orbital angular momentum, set $\mathbf{J}=\mathbf{L}+\mathbf{S}$. The spin-dot-product eigenvalue identity then gives the shift in the title. In the first excited two-electron oscillator manifold, the singlet has $L=1,S=0,J=1$ and zero shift with multiplicity three. The triplet has $L=S=1$, with $J=0,1,2$ shifts $-2\lambda {\hslash }^2,-\lambda {\hslash }^2,+\lambda {\hslash }^2$ and multiplicities $1,3,5$.

The multiplicities sum to twelve, and their weighted shift sums to zero. The specified interaction preserves both $L^2$ and $S^2$; it is distinct from summing separate single-electron spin-orbit interactions. Its shifts are exact within this model's invariant oscillator manifold, with weak coupling ensuring the original level ordering remains useful.

The 1922 interpretation concerned space quantization of atomic magnetic moments in the old quantum theory. The modern electron spin proposal came in 1925; the original observers did not identify intrinsic electron spin as the cause of their splitting. An orbital interpretation then seemed plausible, but the later spin explanation gives a different account of the same two-component evidence.^[ref]

Modern atomic data assign neutral silver the ground state electron configuration $[\mathrm{Kr}]4d^{10}5s$ and electronic term ${}^2{S}_{1/2}$. The filled electron shells contribute zero net electronic angular momentum; the outer unpaired electron has orbital angular momentum $L=0$, spin $S=1/2$, and electronic total angular momentum $J=1/2$. The magnetic quantum number has two values $m_J=\pm 1/2$.^[ref]

In the effective electronic description,The minus sign matters: for positive $G$, the component with $m_J=+1/2$ has negative electronic magnetic moment and deflects toward negative $z$ in our coordinates. Calling the upper spot “spin up” without specifying the magnetic field and channel calibration can therefore reverse the assignment. The total angular momentum operator here sums electronic orbital angular momentum and electronic spin; $J$ is not a claim that the atomic nucleus has no spin. Hyperfine structure and nuclear magnetic moments can require additional internal levels, so a two-channel electronic approximation must be justified for the field and resolution being used.

This is a neutral atomic beam, not a beam of free electrons. The deflected mass in the flight equations is the entire silver chemical atom; its dominant electronic magnetic moment supplies the state-dependent force. Replacing that atom by a free electron would also introduce the charged particle's Lorentz force and a different preparation problem. The ideal two-state model is useful because it isolates the connection between discrete internal components and spatially resolved outputs; it does not retroactively turn the 1922 deposit into a direct contemporary identification of intrinsic electron spin.

In the sudden approximation, changing the nuclear charge of a one-electron hydrogen atom from $Z_{i}e$ to $Z_{f}e$ leaves its electronic state unchanged during the change. Its subsequent probability to occupy a final state is the squared overlap with that state. Using normalized $1s$ wavefunctions ${\psi }_Z=({\beta }_Z^3/\pi {)}^{1/2}{e}^{-{\beta }_{Z}r}$, ${\beta }_Z=Z/a_0$, givesThis follows by integrating $4({\beta }_i{\beta }_f{)}^{3/2}{r}^2{e}^{-({\beta }_i+{\beta }_f)r}$ over positive $r$. A spherically symmetric initial state has zero overlap with every final state of nonzero orbital angular momentum, by the orthogonality of the spherical harmonics. Nuclear recoil and interactions with emitted particles can invalidate the perfectly spherical sudden-change model.

In LS coupling, an atomic term symbol specifies total electronic spin $S$, total orbital angular momentum $L$, and total angular momentum $J$. The superscript $2S+1$ is the spin multiplicity; the letters $S,P,D,F,\dots$ denote $L=0,1,2,3,\dots$. Coupling gives $J=|L-S|,\dots ,L+S$, and a fixed-$J$ level has $2J+1$ magnetic substates before an external field resolves them. For example, ${}^{3}P$ has $S=L=1$ and levels $J=0,1,2$, containing $1+3+5=9$ states altogether. A term symbol does not itself specify the radial electron configuration.

For an axially split $p$ multiplet with $p_z$ energy $E_0$ and equal $p_x,p_y$ energies $E_t$, a transverse $x$-directed magnetic field leaves $p_x$ unchanged and couples $p_y,p_z$ through the orbital angular momentum operator $L_x$. The matrix element has modulus ${\mu }_{B}B$, giving the two energy eigenvalues in the title. For ${\mu }_{B}B\ll |E_t-E_0|$, the orbital shifts are quadratic despite a linear Zeeman energy operator. A full quadratic-field calculation also includes the diamagnetic term; electron spin gives its own linear contribution.

For a neutral pion pair in a state of definite relative orbital angular momentum $L$, charge conjugation interchanges ${\pi }^{+}$ and ${\pi }^{-}$. Relabelling the particle slots reverses the relative momentum and contributes $(-1{)}^L$. The individual charged-pion phase conventions cancel in the pair. Thus a neutral vector rho meson decaying strongly to a pion pair has $L=1$ and $C=-1$. A charged rho meson is taken to the state of opposite electric charge, so it is not itself a charge-conjugation eigenstate.