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The operators $L_{\pm }=L_x\pm i{L}_y$ satisfyThey therefore change the magnetic quantum number $m$ by $\pm 1$ while preserving the total angular-momentum quantum number $\ell$.

The coordinate combinations $x_1\pm i{x}_2$ have $L_3$ eigenvalues $\pm \hslash$, so their powers carry magnetic quantum numbers $\pm m$.

The anomalous Zeeman effect is Zeeman splitting whose pattern is not the simple orbital triplet. Spin and spin-orbit interaction make the upper and lower Landé g-factors differ. The same weak-field formula still applies when its assumptions hold: “anomalous” does not mean nonlinear in the magnetic field.

For example, take an electronic transition with $L_u=1,S_u=1/2,J_u=1/2$ and $L_l=0,S_l=1/2,J_l=1/2$, neglecting hyperfine structure. With $g_s=2$, the formula above gives $g_u=2/3$ and $g_l=2$. The two allowed $q=0$ transitions have dimensionless shifts $g_u{m}_u-g_l{m}_l=\pm 2/3$, while the $q=\pm 1$ transitions give $\pm 4/3$. Thus there are four frequencies, even though each level contains only two magnetic quantum numbers. This is an independent worked deduction; it should not be substituted for the unresolved widths in the 1896 experiment.

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.

For a diatomic molecule, let $m_1,m_2$ be the two masses and $r$ their fixed separation. The distances from the centre of mass are $r_1=m_{2}r/(m_1+m_2)$ and $r_2=m_{1}r/(m_1+m_2)$. A perpendicular axis through the centre of mass therefore has moment of inertiaThe reduced mass $\mu$ here is not the electric dipole moment, for which we use $\mathbf{d}$. For a longer linear molecule, replace $\mu r^2$ by ${\sum }_i{m}_i{\rho }_i^2$, where ${\rho }_i$ is each distance along the axis from the centre of mass. In the point-mass idealization, the axial moment of inertia is zero; turning about that axis changes no nuclear positions, so no independent axial rotational degree of freedom is included.

For the field-free linear rigid rotor, assume a fixed internal quantum state with no electronic angular momentum or electronic spin, and omit couplings to nuclear spin. The rotational Hamiltonian is ${\hat{H}}_{\mathrm{rot}}={\hat{\mathbf{L}}}^2/(2I)$. Its orientation wavefunctions are spherical harmonics $Y_{JM}$, and its angular momentum eigenstates satisfywhere $J=0,1,2,\dots$ and $M=-J,-J+1,\dots ,J$. Each $E_J$ belongs to $2J+1$ angular momentum eigenstates distinguished by the magnetic quantum number $M$, an instance of degenerate energy levels. The Planck constant is $h=2\pi \hslash$; $B$ is the frequency-valued rotational constant. Notice that the energy eigenvalues scale as $J(J+1)$, not $J^2$.

Subtracting consecutive energy eigenvalues givesThus increasing level gaps produce a regularly spaced series of lines at $2B,4B,6B,\dots$, with spacing $2B$. Allowed positions do not guarantee that every line is visible: electric-dipole selection rules, occupation of the initial quantum state, and instrument response also matter.

One of the angular momentum eigenstates in a multiplet, distinguished by the magnetic quantum number along a chosen axis. An applied magnetic field can split their energy levels through Zeeman splitting.

The normal Zeeman effect is the simple three-frequency pattern obtained in the orbital limit. Assume negligible net spin, negligible hyperfine structure and quadratic shifts, and an allowed electric-dipole interaction between levels with orbital splitting factor one. A level with $J=0$ has only $m_J=0$ and no first-order shift. For the other levels, $g_J=1$ gives $\Delta E={\mu }_B{m}_{J}B$. Define $q=m_u-m_l$, rather than the final-minus-initial change of the emitting atom. The electric-dipole selection rule allows $q=0,\pm 1$, henceMany allowed transitions can share one frequency; a triplet need not mean exactly three transitions. The cleanest illustrative example is $J_u=1\to J_l=0$, with three upper quantum states and one unshifted lower quantum state. The $q=0$ component is called $\pi$, and $q=\pm 1$ components ${\sigma }^{\pm }$ with this stated convention. During emission the atom changes its magnetic quantum number by $-q$.

For an original numerical example choose $B=0.50\mathrm{T}$ and a zero-field vacuum wavelength ${\lambda }_0=500\mathrm{nm}$. Using ${\mu }_B\simeq 9.27401\times 10^{-24}\mathrm{J}{\mathrm{T}}^{-1}$ and $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$ gives $\delta {\nu }_B\simeq 7.00\mathrm{GHz}$. The two outer components are separated by $2\delta {\nu }_B\simeq 14.0\mathrm{GHz}$. Differentiating the vacuum relation $\lambda =c/\nu$, with $c$ the speed of light, givesEach outer component lies about $0.00584\mathrm{nm}$ from the centre. Positive $q$ means higher frequency and shorter wavelength. Equal spacing in frequency is only approximately equal spacing in wavelength. These are calculated values for a chosen model, not historical measurements.

Circular polarization selects the sign of the absorbed photon's angular momentum along the beam. Choose that direction as $z$ and define ${\sigma }^{+}$ by the absorption rule $m^{\prime }=m+1$, where $m$ is the magnetic quantum number. Spontaneous emission can return an atom to a different magnetic sublevel, so absorption followed by emission is a population-transfer cycle. A sublevel that the chosen beam cannot excite is dark to that beam. Darkness is relative to the selected transition and electromagnetic polarization, rather than a claim that the atom cannot interact with any light.

A static magnetic field along $z$ defines a stable axis and gives Zeeman splitting. A transverse oscillating magnetic field can redistribute magnetic sublevel populations when its frequency matches an allowed splitting. With the appropriate optical transition, that redistribution restores photon absorption: transmitted intensity decreases, or collected fluorescence increases. These are alternative readouts whose detailed sensitivity depends on the transition, viewing direction and electromagnetic polarization.

For $\gamma >0$, setting the rate equations to zero gives the unique stationary distribution of this continuous-time Markov chain:The nonzero mean magnetic quantum number represents atomic orientation. Increasing $R/\gamma$ empties the bright state, but leaves two dark populations. For $R=0$, relaxation alone gives $p_i(t)=1/3+[p_i(0)-1/3]e^{-\gamma t}$. When $\gamma =0$ and $R>0$, every distribution with $p_{-}=0$ is stationary; the initial difference between the two dark populations survives. The equal dark populations below follow from the specified unpolarized initial condition.

Starting from $p_i(0)=1/3$, the solution for $R>0$ or $\gamma >0$ isFor pure pumping, take $p_{-}^{\ast }=0$ in this formula. Its redistribution time is $3/(2R)$, longer than the mean scattering interval $1/R$ of a bright atom, because one third of scattering events return it to the same sublevel. For arbitrary initial populations the additional mode satisfies $p_0(t)-p_{+}(t)=[p_0(0)-p_{+}(0)]e^{-\gamma t}$.

For an original numerical example choose $R=6{\mathrm{s}}^{-1}$ and $\gamma =1{\mathrm{s}}^{-1}$, with $\Gamma$ much larger than either. ThenAt $t={\tau }_{\mathrm{pump}}$, $p_{-}\simeq 0.1648$, while $p_0=p_{+}\simeq 0.4176$. If the light is then switched off, relaxation returns all three populations toward $1/3$ with time constant $1/\gamma =1\mathrm{s}$. Faster optical preparation and slower dark recovery measure different combinations of rates.

In an optically thin uniform sample of $N$ atoms, with equal collection efficiency $\eta$ for each decay branch, the detected fluorescence count rate is approximately $S_{\mathrm{fl}}=\eta NR{p}_{-}$. The same scattering removes pump photons, giving ${\Phi }_{\mathrm{out}}\simeq {\Phi }_{\mathrm{in}}-NR{p}_{-}$ when depletion is small. For the chosen example the steady scattering rate is $R/15=0.40{\mathrm{s}}^{-1}$ per atom, compared with the initial $R/3=2{\mathrm{s}}^{-1}$. Pumping lowers this signal fivefold, while increasing transmission. If rapid mixing restores nearly uniform populations, scattering rises again. Reabsorption, strong attenuation, unequal collection efficiencies or coherent RF dynamics require a more detailed model; the reset rate $\gamma$ does not stand for an exact RF Hamiltonian.

The nonnegative integer $l$ for which a spherical harmonic has squared orbital angular momentum ${\hslash }^{2}l(l+1)$. The angular part of a regular scalar wavefunction has $l=0,1,2,\dots$ and magnetic quantum number $m=-l,\dots ,l$.

Use $|m_1,m_2⟩$ for a product of the two individual spin-one spin eigenstates, with magnetic quantum numbers $m_i=-1,0,1$. Total $S_z=0$ requires $m_1+m_2=0$, leaving preciselyHere these are product labels, not labels of the total angular momentum. Write their coefficients in this order as $\alpha ,\beta ,\gamma$. With physical spin ladder operators $S_{\pm }=S_{1\pm }+S_{2\pm }$, each nonzero individual step has amplitude $\sqrt{2}\hslash$, soBoth must vanish for a singlet state. The independent product states therefore require $\alpha =\gamma =-\beta$. Wavefunction normalization gives the spin-one singlet stateup to an overall quantum phase. Conversely, $S_z|\psi ⟩=S_{+}|\psi ⟩=0$ implies $S^2|\psi ⟩=0$ through $S^2=S_{-}{S}_{+}+S_z^2+\hslash S_z$, so these conditions really select total spin zero. Distinguishability imposes no additional exchange restriction. The ladder hint uses dimensionless angular momentum units; restoring its common factor $\hslash$ does not change the coefficient ratios.

Neglect hyperfine structure and use the electronic weak-field Zeeman splitting regime, where ${\mu }_{B}B$ is small compared with the spin-orbit interaction splitting. The sodium levels are $3s{}^2{S}_{1/2}$, $3p{}^2{P}_{1/2}$ and $3p{}^2{P}_{3/2}$. The Landé g-factor formula givesFor example, inserting $L=1,S=1/2,J=1/2$ gives $3/2+(3/4-2)/(2\cdot 3/4)=2/3$. Each sublevel with magnetic quantum number $m_J$ shifts by $g_J{m}_J{\mu }_{B}B$. Thus the lower level has shifts $\pm {\mu }_{B}B$, the upper $J=1/2$ level has shifts $\pm {\mu }_{B}B/3$, and the upper $J=3/2$ level has shifts $\pm 2{\mu }_{B}B$ and $\pm 2{\mu }_{B}B/3$.

The electric-dipole selection rule allows $\Delta J=0,\pm 1$ except $0\leftrightarrow 0$, and $m_u-m_l=0,\pm 1$. Both fine structure branches are allowed. The emitted photon energy isAll the allowed components of the sodium D-line Zeeman pattern are listed below; the last column is the photon shift in units of ${\mu }_{B}B$:These are four components of D1 and six of D2. Their strengths need not be equal; the table and arrows specify allowed transitions, not intensities.

The $3p{}^2{P}_{3/2}$ level lies above $3p{}^2{P}_{1/2}$, so D2 has the larger zero-field photon energy. ^[ref] In the stated weak-field regime its most positive shift is $5{\mu }_{B}B/3$. Since $\lambda =hc/E_{\gamma }$, the shortest wavelength is therefore produced byThe tempting $m_u=3/2\to m_l=-1/2$ pair has a larger difference of level shifts but is forbidden because $|m_u-m_l|=2$.

Neglect hyperfine structure, spin-orbit coupling and the diamagnetic term, and set $b={\mu }_{B}B$, $\epsilon =\alpha a_0^2$. The Zeeman energy is $b(L_{\hat{B}}+2S_{\hat{B}})$. The electron spin is free, so every spatial result below is duplicated with an added $\pm b$.

For a magnetic field parallel to $z$, the old magnetic quantum number $m$ remains good and the first-order energy levels areThe two $s$ orbitals have zero orbital angular momentum, so only their electron spin contributes at this order.

For a magnetic field parallel to $x$, the expectation of $L_x$ in each individual $p_m$ orbital vanishes. One must still check the degenerate $m=\pm 1$ subspace: its projected $L_x$ is zero as well. Thus there is no first-order orbital splitting in a sufficiently weak transverse magnetic field. Spin still contributes $\pm b$.

The transverse orbital Zeeman mixing can be seen without assuming its size. Put $E_0=-\mathrm{Ry}/4+3\epsilon$, $E_t=-\mathrm{Ry}/4+21\epsilon$, and $\Delta =E_t-E_0=18\epsilon$. In the real $p_x,p_y,p_z$ orbital basis, $p_x$ is unaffected while $p_y,p_z$ have off-diagonal matrix elements of modulus $b$. Diagonalizing this two-dimensional Hamiltonian gives the spatial energy eigenvaluesWhen $b\ll \Delta$, their shifts relative to the earlier levels are $0,+b^2/\Delta ,-b^2/\Delta$, with $\pm b$ added for electron spin. The displayed square-root expression also handles $b$ comparable to $\Delta$, provided coupling to other principal shells stays negligible. The $2s$ orbital retains zero orbital correction.

These quadratic orbital estimates keep the enhanced contribution from mixing the split $2p$ levels. Minimal magnetic minimal coupling also produces $e^2{B}^2{r}_{\perp }^2/(8m_e)$; its expectation value is positive and must be added for a full order-$B^2$ calculation. For example it gives $e^2{B}^2{a}_0^2/(4m_e)$ for $1s$. It is parametrically smaller than the $2p$ mixing term when $\epsilon \ll \mathrm{Ry}$.

The spherical harmonic is also an eigenfunction of the $z$ component of orbital angular momentum:These are atomic units; restoring units gives $m\hslash$. The integer $m$ is the magnetic quantum number.

In atomic units the time-independent Schrödinger equation isThe displayed equation is the Radial Schrodinger equation for the hydrogen atom. Its radial derivative terms come from the Laplacian in spherical coordinates, the Coulomb term is the potential $-1/r$, and separation into spherical harmonics uses the orbital angular momentum eigenvalue $l(l+1)$, producing the centrifugal term $-l(l+1)R/r^2$.

For $E=-1/(2n^2)$, substitute $R=r^{\alpha }{e}^{-r/n}$. Dividing the equation by $R$ and comparing powers of $r$ givesThe solution regular at the origin has $\alpha =l$, and when $l=n-1$ both equations agree. This is the circular Coulomb bound state. ThusThe radial probability measure is $|R(r){|}^2{r}^{2}dr$. With $a=2/n$ and the gamma integral, the mean radius of a circular Coulomb bound state is

At fixed principal quantum number $n$, the allowed orbital angular momentum quantum numbers are $l=0,\dots ,n-1$, and each has magnetic quantum number $m=-l,\dots ,l$. Hence the orbital quantum degeneracy isIncluding electron spin would double this to $2n^2$, but spin is absent from the stated wavefunctions.

The canonical commutation relations are $[X_i,X_j]=[P_i,P_j]=0$ and $[X_i,P_j]=i\hslash {\delta }_{ij}$. Substitution into the definitions of the creation and annihilation operators gives

The normalized ground state is defined byRepeated use of $[A_i,A_i^{†}]=1$ gives the Cartesian number state of the three-dimensional isotropic harmonic oscillatorwith energy eigenvalue

Inverting the ladder-operator definitions yieldsThe antisymmetry of the Levi-Civita symbol then cancels the two-creation and two-annihilation terms in $\mathbf{L}=\mathbf{X}\times \mathbf{P}$, leaving the orbital angular momentum in oscillator ladder operatorsWrite $|j⟩=A_j^{†}|0⟩$. ThenSince the orbital angular momentum eigenvalue is $\ell (\ell +1){\hslash }^2$, every first-excited state hasFinally,Therefore the normalized magnetic quantum number $m=+1$ state can be chosen aswhich obeys $L_z|\psi ⟩=\hslash |\psi ⟩$ and $\Vert |\psi ⟩\Vert =1$.

For the two inline operator labels, (a) $L_z$ has eigenvalue $m_{\ell }\hslash$, and (b) $L^2$ has eigenvalue $\ell (\ell +1){\hslash }^2$, with $\ell =0,1,\dots$ and $m_{\ell }=-\ell ,\dots ,\ell$. The subscript distinguishes the magnetic quantum number from the electron mass.

Put $b=a_0(\ell +1)$ and $\kappa =e^2/(4\pi {ϵ}_0)$. For the proposed radial shape,The angular-momentum term cancels the inverse-square part. The inverse-radius coefficient vanishes if ${\hslash }^2(\ell +1)/(mb)=\kappa$. ThereforeThis nodeless Coulomb radial eigenstate has principal quantum number $n=\ell +1$, not the complete set of hydrogen radial states. Since the spherical harmonic is normalized, radial normalization requiresup to an arbitrary phase.

For the hydrogen ground-state nuclear-volume probability, use $R=2a_0^{-3/2}{e}^{-r/a_0}$. With $u=r_p/a_0\ll 1$,No numerical proton radius is supplied or required. A sharp proton boundary is an approximation to a charge distribution, and the point-Coulomb potential changes inside an extended proton. Finite proton mass, relativistic and electromagnetic corrections also limit the ideal wavefunction's precision.

Classically the kinetic energy must satisfy $E-V(r)\ge 0$. For $E=-\kappa /(2a_0)$ this demands $r\le 2a_0$. The hydrogen ground-state probability beyond the classical turning point isA substantial probability lies beyond the classical turning radius; the bound-state wavefunction decays there rather than vanishing.

In spherical coordinates, the angular Laplacian has spherical harmonics as eigenfunctions, $L^2{Y}_{lm}={\hslash }^{2}l(l+1)Y_{lm}$ and $L_z{Y}_{lm}=\hslash m{Y}_{lm}$. Separation leavesRegularity and normalizability give principal quantum number $n=1,2,\dots$, orbital quantum number $l=0,\dots ,n-1$, and magnetic quantum number $m=-l,\dots ,l$. Here the $m$ in the kinetic coefficient is the electron mass, distinct from the magnetic label.

For the proposed angular factor, $\cos \theta \propto Y_{10}$, so $l=1$ and magnetic quantum number zero. Direct differentiation gives ${\nabla }^2(r{e}^{-\alpha r}\cos \theta )=({\alpha }^2-4\alpha /r)\psi$. The $1/r$ term in the Schrödinger equation vanishes when $2{\hslash }^2\alpha /m=C$. With Bohr radius $a_0={\hslash }^2/(mC)$,wavefunction normalization uses $\int d\Omega {\cos }^2\theta =4\pi /3$ and ${\int }_0^{\infty }{r}^4{e}^{-2\alpha r}dr=24/(2\alpha {)}^5$, givingAn arbitrary constant phase is allowed. This is the $2p_z$ state. Under spatial inversion, $r$ is unchanged and $\cos \theta$ changes sign, so its parity eigenvalue is $-1$, also given by $(-1{)}^l$.

Rotational invariance means rotated states have the same energy; it does not require each member of a degenerate energy eigenspace to be invariant. rotations mix the three $l=1$ magnetic substates. An orbital oriented along $z$ can therefore have an anisotropic probability density while the Hamiltonian remains spherically symmetric.

Insert hydrogen energy eigenstates on either side of the commutator. Because $L_z$ is Hermitian and has eigenvalue $\hslash m$,ThusThis is a necessary electric-dipole selection rule, not a sufficient condition for a nonzero matrix element. For example, $⟨200|z|200⟩=0$ by parity even though $m=m^{\prime }=0$. The angular factor $z=r\cos \theta$ additionally requires $l^{\prime }=l\pm 1$: multiplication of a spherical harmonic by $\cos \theta$ gives only harmonics with the same $m$ and adjacent $l$. The commutator alone establishes only the magnetic quantum number restriction; radial integrals and the remaining electric-dipole selection rules for hydrogen must still be considered.

The spherical harmonics satisfyCommuting $L^2$ or $L_z$ past $L_{\pm }$ therefore givesFor a nonzero result these are the eigenvalue equations with unchanged $l$ and shifted magnetic quantum number. Taking the squared norm, using $L_{\mp }{L}_{\pm }=L^2-L_z^2\mp \hslash L_z$, fixes the conventional coefficient:At the appropriate endpoint the coefficient is zero. The zero vector is not itself an eigenstate.