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Assume the station's mass remains fixed through an instantaneous kick, with subsequent motion governed only by the Earth's Newtonian gravitational field. Put $\kappa =G{M}_E$, and work per unit station mass. The initial circular orbit has specific angular momentum $j$ and specific orbital energy ${\epsilon }_0$ given byLet $\delta$ denote the fractional increase of kinetic energy. The position and potential energy are unchanged during the kick, soUnchanged angular momentum at unchanged position fixes the tangential velocity; the added kinetic energy must be radial. Thus this is a radial kick, not a purely tangential speed increase. The new orbital eccentricity follows fromAt either radial turning point, the Kepler radial energy equation gives $\epsilon =j^2/(2r^2)-\kappa /r$. Its two roots, equivalently the periapsis and apoapsis of the elliptic orbit, areFor $\delta =0.01$, $e=0.1$ andAn inward or outward kick gives the same pair of radial turning points, although it places the station at a different phase of the orbit. These two finite positive radii require $0\le \delta <1$.

Retain the unchanged-angular momentum assumption of the preceding part. The initial orbit has radius $r_o=6400+500=6900\mathrm{km}$, and the desired pericentre distance is $r_p=6400+100=6500\mathrm{km}$. Using $r_p=r_o/(1+e)$ givesThe required fractional increase in initial kinetic energy is thereforeHere $\Delta E=\Delta K>0$ because the instantaneous kick does not change potential energy. Since the initial total orbital energy is $E_0=-K_0$, an alternative convention for the requested fractional energy change givesThe total orbital energy becomes less negative. Explicitly specifying its denominator avoids interpreting a positive energy input as a decrease of energy.

The free rod has the same mass $m$ and length $l$ as in part (a). There is no pivot. Put the rod's initial center of mass at $y=0$ and the impact end at $y=l/2$. At impact the combined center of mass isConservation of momentum gives the combined center of mass velocity $V_C=v_0/2$. Evaluate angular momentum about this combined center of mass; the incident ball's perpendicular lever arm is $l/4$, so its initial magnitude is $m{v}_{0}l/4$.

After the perfectly inelastic collision, the parallel axis theorem gives the combined moment of inertiaWith negligible external angular impulse, conservation of angular momentum therefore givesIts sense is that of the incident ball's angular momentum, clockwise in the source diagram. As a consistency check, the final kinetic energy isThus $m{v}_0^2/10$, or one fifth of the initial kinetic energy, is dissipated in sticking. Treating the rod's original centre as a fixed pivot would give an incorrect angular velocity.

Define $f(r)>0$ as the magnitude of the inward central force, so $\mathbf{F}=-f(r)\hat{\mathbf{r}}$ and $V^{\prime }(r)=f(r)$. The conserved angular momentum is $J=m{r}^2\dot{\theta }$. Eliminating $\dot{\theta }$ from the mechanical energy yieldsThe added term in the effective potential is the centrifugal barrier. If instead $f$ denotes the signed radial force component, the convention becomes $V^{\prime }=-f$.

The angular-momentum commutators giveFor example $[L_x+i{L}_y,L_z]=-i\hslash L_y+i(i\hslash L_x)=-\hslash L_{+}$. In the displayed basis $|1,1⟩,|1,0⟩,|1,-1⟩$, the only nonzero upper entries of $L_{+}$ connect adjacent magnetic quantum numbers. Subtracting their column and row $L_z$ eigenvalues multiplies each by $-\hslash$, proving the first matrix commutator; the lower matrix similarly gives the second. Direct multiplication giveswhose difference is $2\hslash L_z$. Also $L^2=L_z^2+(L_{+}{L}_{-}+L_{-}{L}_{+})/2=2{\hslash }^{2}I$, confirming $\ell =1$.

A rotation changes the quantization axis without changing the spectrum, so $L_{\mathbf{n}}$ has eigenvalues $+\hslash ,0,-\hslash$. Its dimensionless matrix isSolving $(L_{\mathbf{n}}-\hslash I)v=0$ and normalizing gives the spin-one maximum-projection eigenvectorFor instance the first equation is $(\cos \theta -1){\cos }^2(\theta /2)+{\sin }^2\theta /2=0$, and the squared components sum to one. An overall phase is immaterial.

At $\theta =\pi /3$, $v_{+}=(3/4,\sqrt{6}/4,1/4{)}^{\mathsf{T}}$. The Born rule therefore predicts, in the requested order,This treats the Stern-Gerlach experiment as the stipulated ideal measurement of orbital projection. In an actual atom, other magnetic moments and spin-orbit coupling can change which angular momentum the apparatus resolves.

For a time-independent Hamiltonian operator, a symmetry operator $U$ satisfies $UH{U}^{†}=H$, or $[U,H]=0$. Thus applying $U$ to an energy eigenstate preserves its energy. Commuting Hermitian observables can be diagonalized within each degenerate energy eigenspace, allowing states to be labelled by symmetry quantum numbers.

A continuous unitary symmetry $U(\epsilon )=e^{-i\epsilon Q/\hslash }$ has Hermitian generator $Q$. Differentiating the invariance condition gives $[H,Q]=0$. The Heisenberg equation of motion then yieldswhen $Q$ has no explicit time dependence. spatial translations connect to momentum conservation, rotations to angular-momentum conservation, and time-translation invariance to energy conservation. A central potential energy conserves all components of angular momentum, but they cannot all be assigned sharp simultaneous values because they do not commute; $H,L^2,L_z$ are a compatible choice. Discrete reflection symmetry gives parity labels without a continuously variable generator. A quantum degeneracy can be protected by a symmetry, but a quantum degeneracy alone does not identify its physical origin.

For independent angular momenta ${\mathbf{J}}_1,{\mathbf{J}}_2$, set $\mathbf{J}={\mathbf{J}}_1+{\mathbf{J}}_2$. Since different subsystems commute, $\mathbf{J}$ satisfies the same angular-momentum algebra. The product basis $|j_1{m}_1⟩|j_2{m}_2⟩$ diagonalizes $J_1^2,J_2^2,J_{1z},J_{2z}$; the coupled basis diagonalizes $J_1^2,J_2^2,J^2,J_z$. Its labels obeyThe Clebsch-Gordan coefficients are the unitary change of basis between them. Start from the maximum-weight state $j=j_1+j_2,m=j$, apply $J_{-}=J_{1-}+J_{2-}$ with wavefunction normalization, and construct the remaining multiplets by orthogonality.

For two spin-one-half systems, the symmetric triplet state is $|\uparrow \uparrow ⟩$, $(|\uparrow \downarrow ⟩+|\downarrow \uparrow ⟩)/\sqrt{2}$, $|\downarrow \downarrow ⟩$, with $j=1$; the antisymmetric singlet state $(|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩)/\sqrt{2}$ has $j=0$. The dimension identity ${\sum }_j(2j+1)=(2j_1+1)(2j_2+1)$ checks completeness. Coupling orbital and intrinsic spin similarly gives total atomic angular momentum and organizes spin-orbit energy splittings. Exchange symmetry is an additional restriction when the constituents are identical.

For each mode in Cartesian coordinates, $[a_i,a_j^{†}]={\delta }_{ij}$ and $a_i^{†}{a}_i=H_i/(\hslash \omega )-1/2$. Consequently the two-dimensional isotropic harmonic oscillator hasThe normalized actions of the creation and annihilation operators areInvert their definitions as $x=\sqrt{\hslash /(2m\omega )}(a_x+a_x^{†})$ and $p_x=i\sqrt{m\hslash \omega /2}(a_x^{†}-a_x)$, and similarly for $y$. linear operators belonging to different coordinates commute, so substitution into $L_z=x{p}_y-y{p}_x$ givesEach term transfers one excitation between the two modes: $[N_x+N_y,a_x^{†}{a}_y]=(1-1)a_x^{†}{a}_y=0$, and likewise for its adjoint. Thus $[H,L_z]=0$ and, with no explicit time dependence, $d{L}_z/dt=(i/\hslash )[H,L_z]=0$. This is conservation of angular momentum under planar rotations.

Finally $L_z|1,0⟩=i\hslash |0,1⟩$ and $L_z|0,1⟩=-i\hslash |1,0⟩$. HenceThe angular momentum is $+\hslash$; division by $\sqrt{2}$ performs wavefunction normalization on this state without changing its eigenvalue. It is one excitation of the positive circular mode.

angular momentum generates rotations. Its components obey $[J_i,J_j]=i\hslash {ϵ}_{ijk}{J}_k$, and $J^2$ commutes with each component. A compatible basis has $J^2={\hslash }^{2}j(j+1)$ and $J_z=m\hslash$, where $m=-j,-j+1,\dots ,j$. ladder operators change $m$ by one, terminate at $\pm j$, and give $2j+1$ magnetic substates. Different Cartesian components generally cannot all be sharp simultaneously.

orbital angular momentum is $\mathbf{L}=\mathbf{r}\times \mathbf{p}=-i\hslash \mathbf{r}\times \nabla$. Its angular eigenfunctions are spherical harmonics, with integer $\ell$ arising from their spatial rotation properties and single-valuedness. spin is intrinsic angular momentum, not literal rotation of a finite charged sphere. It may be half-integer; spin $1/2$ is represented by $\mathbf{S}=\hslash \mathbf{\sigma }/2$. For several contributions, $\mathbf{J}=\mathbf{L}+\mathbf{S}$ obeys the same algebra. Coupling angular momenta $j_1,j_2$ permits $J=|j_1-j_2|,\dots ,j_1+j_2$, with Clebsch-Gordan coefficients relating product and coupled bases. spin-orbit coupling can distinguish these total-angular-momentum sectors.

The Stern-Gerlach apparatus sends a collimated neutral atomic beam through an inhomogeneous magnetic field and records its deflection. A magnetic moment experiences $F_z\simeq {\mu }_z\partial B_z/\partial z$; field uniformity would produce precession but no analogous net separating force. A classically continuous distribution of moment orientations would broaden the beam continuously. Gerlach and Stern instead observed separated silver-beam components. Reversing or removing the field and comparing the undeflected profile controls the magnetic origin; collimation, velocity spread and finite resolution affect the spots. The original experiment demonstrated space quantization; its modern silver interpretation uses the unpaired electron spin with nearly zero orbital angular momentum, rather than a direct measurement of a free electron's trajectory.^[ref]

For two spin-$1/2$ particles, the triplet state sector has $S=1$ and symmetric spin statesThe spin singlet state has $S=0$ and antisymmetric spin stateTheir total $S^2$ values are $2{\hslash }^2$ and zero, while ${\mathbf{S}}_1\cdot {\mathbf{S}}_2$ has values ${\hslash }^2/4$ and $-3{\hslash }^2/4$. The singlet is rotationally invariant and gives anticorrelated spin measurements along a common axis. For identical fermions the complete state is antisymmetric: a symmetric spatial state pairs with a singlet spin state, whereas an antisymmetric spatial state pairs with triplet spin. Orbital, spin and coupled angular momenta share an algebra, but their physical origins and allowed representations differ.

electron spin is intrinsic angular momentum, independent of an orbital wavefunction's $\mathbf{r}\times \mathbf{p}$. It has $s=1/2$, soA spatial orbital therefore carries a two-component spinor degree of freedom. Noncommuting components do not describe an ordinary vector with simultaneously definite projections; measurements along one axis give only the two allowed values. Even a spherically symmetric $l=0$ orbital retains electron spin.

The associated magnetic moment is $\mathbf{\mu }=-g_{e}e\mathbf{S}/(2m_e)$ with $e>0$ and $g_e\simeq 2$. A spinor changes sign under a $2\pi$ rotation and returns under $4\pi$; that sign is observable only relative to another coherent amplitude, not as an isolated global phase. A literal classically rotating charged sphere does not account for this fixed half-integer representation, component algebra and spinor rotation law. Classical angular momentum provides useful analogies for torque and precession, but no ordinary classical orientation reproduces the full intrinsic quantum spin structure.

Exchange constraint. Exchange of two identical particles cannot change physical predictions. For the usual three-dimensional bosonic or fermionic state space, the exchange operator obeys $P_{12}^2=1$, so its allowed eigenvalues are $\pm 1$. bosons use the symmetric overall wavefunction, $P_{12}\Psi =+\Psi$; fermions use the antisymmetric overall wavefunction, $P_{12}\Psi =-\Psi$. Exchange includes the spatial coordinates and all internal labels. The identity $P_{12}^2=1$ alone does not prove which particle species has which statistics; that assignment is the Spin-statistics theorem.

Factored fermion states. If $\Psi ={\psi }_{\mathrm{space}}{\chi }_{\mathrm{spin}}$ and both factors have definite exchange symmetry, their exchange signs must multiply to $-1$. Thus a symmetric spatial factor requires an antisymmetric spin factor, and an antisymmetric spatial factor requires a symmetric spin factor. For two spin-one-half particles, the former spin factor is the singlet $(|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩)/\sqrt{2}$; the latter factors are the three triplet states. A more general entangled spin-space state need not factor, but must still be antisymmetric overall.

hydrogen spin isomers. The two protons are identical spin-one-half fermions. parahydrogen has total nuclear spin $I_{\mathrm{n}}=0$, the antisymmetric singlet; orthohydrogen has $I_{\mathrm{n}}=1$, the symmetric triplet. Here $I_{\mathrm{n}}$ is a spin quantum number, distinct from the moment of inertia used below.

Rotational restriction. In the electronic and vibrational ground state, these two factors are symmetric under nuclear exchange. Exchanging the protons reverses the internuclear axis, $(\theta ,\varphi )\mapsto (\pi -\theta ,\varphi +\pi )$. The spherical harmonic rotational state contributes the exchange sign $(-1{)}^{\ell }$. Overall antisymmetry therefore requiresHence parahydrogen has even $\ell$; orthohydrogen has odd $\ell$. The $\ell$ here labels molecular rotation, rather than the angular momentum obtained by also coupling the nuclear spins.

Conversion heat. The ortho-para conversion heat follows from the linear rigid rotor approximation, $E_{\ell }={\hslash }^2\ell (\ell +1)/(2I)$ with $I=\mu r^2\simeq m_p{r}^2/2$. Thus conversion of one molecule from $\ell =1$ to $\ell =0$ releasesNeglecting the small electron contribution to the molecular mass, a sample of mass $M$ contains $M/(2m_p)$ molecules. Only the initial orthohydrogen fraction releases this energy, givingThe estimate uses $r=74$ pm, $m_p\simeq 1.67262\times 10^{-27}$ kg and $\hslash \simeq 1.05457\times 10^{-34}$ J s. Handbook-level rounding and the rigid rotor approximation justify about three significant figures, not a high-precision thermochemical prediction.^[ref]

Storage consequence. The internal conversion is a delayed heat source even in a well insulated container, causing warming or additional evaporation of the liquid. Insulation reduces external heat transfer but does not remove this source. Converting to predominantly parahydrogen during liquefaction, while removing the released heat, reduces the subsequent storage heat load. No evaporation fraction can be calculated without a latent heat and a model for how the heat is removed.

orbital angular momentum generates spatial rotations and is $\mathbf{L}=\mathbf{r}\times \mathbf{p}$. With $\mathbf{p}=-i\hslash \nabla$, its components satisfyThus $L^2$ and one component, conventionally $L_z$, have simultaneous eigenstates, while all three components cannot generally have sharp values. On ordinary single-valued spatial wavefunctions,The spherical harmonics form an orthonormal angular basis. The integer restriction is specific to orbital wavefunctions; half-integer spin is an intrinsic degree of freedom, not the angular dependence of a scalar spatial wavefunction.

The angular momentum ladder operators obeyThe ladder ends at $m=\pm \ell$, producing $2\ell +1$ states. A state with specified $m$ has a sharp projection $\hslash m$, but the angular momentum magnitude is $\hslash \sqrt{\ell (\ell +1)}$, not $\hslash \ell$. No classical picture of a vector with simultaneously fixed Cartesian components reproduces all these measurement properties.

In a central potential, rotational symmetry gives $[H,L_i]=0$. separation of variables yields $\psi (\mathbf{r})=R_{\ell }(r)Y_{\ell m}(\theta ,\varphi )$. The radial Schrödinger equation includes a centrifugal potential ${\hslash }^2\ell (\ell +1)/(2m{r}^2)$, which suppresses small-radius amplitude for nonzero $\ell$. The energy is independent of $m$ when rotational symmetry is intact, but need not be independent of $\ell$. A magnetic field can break that degeneracy. Spatial inversion gives $Y_{\ell m}(-\hat{\mathbf{r}})=(-1{)}^{\ell }{Y}_{\ell m}(\hat{\mathbf{r}})$, connecting orbital angular momentum to parity and to the even/odd rotational restriction for the hydrogen spin isomers.

Expanding with the angular momentum commutators givesFor ${\hat{L}}_z|\psi ⟩=\alpha \hslash |\psi ⟩$, commuting ${\hat{L}}_z$ past a ladder operator givesThus the raised or lowered vector, if nonzero, is an eigenvector with the shifted eigenvalue; an angular momentum ladder endpoint can instead be annihilated. Also $[{\hat{L}}^2,{\hat{L}}_{\pm }]=0$, so ${\hat{L}}^2{\hat{L}}_{\pm }|\psi ⟩=\Lambda {\hslash }^2{\hat{L}}_{\pm }|\psi ⟩$ for every nonzero ladder result.

For orbital angular momentum, ${\hat{L}}_z=-i\hslash {\partial }_{\varphi }$ gives an azimuthal factor $e^{i\alpha \varphi }$. Single-valuedness under $\varphi \mapsto \varphi +2\pi$ requires integer $\alpha$. Positivity of ladder norms givesfor normalized states. A fixed finite $\Lambda$ therefore has upper and lower endpoints. Writing the upper endpoint as $m=\ell$, its vanishing raising norm gives $\Lambda =\ell (\ell +1)$; the lower endpoint is $m=-\ell$. Consequentlywith $\ell =0,1,2,\dots$. This integer conclusion is for orbital angular momentum; spin representations can also have half-integer labels.

For the specified Gaussian-radial state, use a Cartesian p-orbital decomposition: write $x+y+z=r(\sin \theta \cos \varphi +\sin \theta \sin \varphi +\cos \theta )$. Its angular factor isOnly $\ell =1$ occurs. The squared angular coefficients are $2,2,2$ before common normalization, so each allowed $m$ has probability one third. Therefore the simultaneous outcome isThe common normalized radial factor cancels from this ratio. The standard symmetric $m$ sequence used above resolves the duplicated central positive entry in the printed list.

Operators, quantization and rotations. Angular momentum is the generator of spatial rotations. In quantum mechanics, its components are Hermitian operators on their appropriate domains and obeyThese angular momentum commutation relations prevent all three components from having simultaneous sharp values. For example, the Robertson uncertainty principle gives $\Delta J_x\Delta J_y\ge \hslash |⟨J_z⟩|/2$. One conventionally diagonalizes ${\hat{J}}^2$ and a selected component ${\hat{J}}_z$:The allowed labels are $j=0,\frac{1}{2},1,\frac{3}{2},\dots$ and $m=-j,-j+1,\dots ,j$. An angular-momentum multiplet has $2j+1$ states. The magnitude $\hslash \sqrt{j(j+1)}$ differs from the largest possible component $\hslash j$; the remaining components fluctuate even in a state with maximal $m$.

The angular momentum lowering operator and its raising counterpart, ${\hat{J}}_{\pm }={\hat{J}}_x\pm i{\hat{J}}_y$, satisfyThe ladder terminates at $m=\pm j$. A rotation through angle $\theta$ about unit vector $\hat{\mathbf{n}}$ acts through the unitary operator $\exp [-i\theta \hat{\mathbf{n}}\cdot \hat{\mathbf{J}}/\hslash ]$. Thus quantized components and the continuous rotation of a state are compatible.

Orbital angular momentum. The orbital contribution is associated with motion in space:For a spinless spatial wavefunction, the simultaneous angular eigenfunctions are the spherical harmonics $Y_{\ell m}(\theta ,\varphi )$, with $\ell =0,1,2,\dots$ and $m=-\ell ,\dots ,\ell$. Single-valued ordinary scalar spatial wavefunctions give integer orbital labels. In a central potential energy, $[\hat{H},{\hat{L}}^2]=[\hat{H},{\hat{L}}_z]=0$, so stationary states may be labelled by these angular quantum numbers. The angular kinetic contribution appears in the radial equation as $\ell (\ell +1){\hslash }^2/(2m{r}^2)$, the centrifugal potential. Noncentral interactions need not conserve these orbital labels.

Intrinsic spin and addition. Spin angular momentum is an intrinsic internal degree of freedom with the same commutation algebra. It is not the angular momentum of a small classical charged sphere literally rotating about its centre. For an electron, $s=1/2$ and $m_s=\pm 1/2$, so ${\hat{S}}_i=(\hslash /2){\sigma }_i$ in terms of the Pauli matrices. Its squared spin is $3{\hslash }^2/4$. Other particles can have integer or half-integer spin, including spin zero.

For one particle with orbital and spin degrees of freedom, $\hat{\mathbf{J}}=\hat{\mathbf{L}}+\hat{\mathbf{S}}$; the two component sets commute with each other. Addition of angular momentum permits $j=|\ell -s|,|\ell -s|+1,\dots ,\ell +s$. The coupled states $|j,m_j⟩$ are linear combinations of the uncoupled $|\ell ,m_{\ell }⟩|s,m_s⟩$ with $m_j=m_{\ell }+m_s$, determined by Clebsch-Gordan coefficients. A rotationally invariant spin-orbit interaction can conserve $J^2,J_z$ while mixing the separate component labels.

Angular momentum also carries a magnetic moment. For an electron's orbital motion and intrinsic spin,The minus signs reflect negative electric charge. With the Bohr magneton ${\mu }_B=e\hslash /(2m_e)$, a spin projection $m_s$ has ${\mu }_z=-g_s{\mu }_B{m}_s$. A magnetic field can therefore distinguish angular-momentum states through their interaction energy $-\mathbf{\mu }\cdot \mathbf{B}$.

Stern–Gerlach measurement. In the Stern-Gerlach experiment, atoms emerge from a heated source, pass through collimating apertures, traverse a strongly inhomogeneous magnetic field, and strike a position-sensitive screen. The beam is neutral, so its separation is interpreted through the atomic magnetic moment rather than the ordinary charged-particle Lorentz force. Near a chosen quantization direction, the interaction givesA uniform field can exert a torque but would not separate the beam by this translational force; the gradient is essential. Different allowed magnetic-moment components become correlated with different trajectories. A classical ensemble with continuously varying orientations would yield a continuous spread, whereas discrete quantum components yield separated channels.

In the modern electronic description of ground-state silver, filled shells cancel their electronic angular momenta and the unpaired $5s$ electron has $L=0$, $S=1/2$, $J=1/2$. The effective electronic measurement therefore has two components, approximately ${\mu }_z=\pm {\mu }_B$, rather than an orbital $\ell =1/2$, which is not allowed. For positive field gradient the electronic $m_J=+1/2$ component has negative ${\mu }_z$ and deflects downward in this coordinate convention. Nuclear spin and hyperfine structure are neglected in this simplified two-channel account. Historically the original splitting preceded the electron-spin interpretation, so the modern explanation should not be confused with the original investigators' interpretation.

More generally a prepared spin state $a|+z⟩+b|-z⟩$ gives the two outcomes with probabilities $|a{|}^2,|b{|}^2$. Selecting one channel prepares a state for further measurements. An ideal second analyser along the same axis then gives a definite result, whereas one along a perpendicular axis gives two equally likely outcomes for a spin-one-half eigenstate. This connects the apparatus directly to incompatible spin components and the Born rule. Angular momentum is quantized through its operator algebra; the Stern–Gerlach gradient turns that discrete internal information into spatially distinguishable outcomes.

The quantum rotor has orientation angle $\varphi$ of period $2\pi$, so physical wavefunctions satisfy $\psi (\varphi +2\pi )=\psi (\varphi )$. The exponential solutions of $Hu=Eu$ are thereforeThey are normalized and orthogonal on $[0,2\pi )$. The corresponding angular momentum operator is $L_z=-i\hslash {\partial }_{\varphi }$, with eigenvalue $m\hslash$. The $m=0$ ground level is nondegenerate, and every $|m|>0$ energy level has the twofold degeneracy $m\leftrightarrow -m$. This uses ordinary periodic boundary conditions, with no flux or twisted quantum phase.

Expand the initial wavefunction in the angular momentum eigenbasis using ${\cos }^2\varphi =1/2+(e^{2i\varphi }+e^{-2i\varphi })/4$. With the normalized $u_m$ from part (a), this givesThe coefficient squares sum to one, also verifying the stated normalization. The Born rule therefore gives angular momentum outcomesFor an energy measurement the two distinct $m=\pm 2$ states belong to the same energy eigenspace. Sum their probabilities, rather than adding their amplitudes, to obtainTheir relative probabilities are $4:1:1$ for the angular momenta and $2:1$ for the energies. In particular, zero mean angular momentum does not imply zero rotational energy.

Each energy eigenstate evolves by its own quantum phase $e^{-i{E}_{m}t/\hslash }$. The two $m=\pm 2$ components have the same energy, while the $m=0$ quantum phase stays one. ThusIn angle representation this isAt $t=0$, $1+\cos 2\varphi =2{\cos }^2\varphi$ recovers the initial state. The fixed coefficient magnitudes keep the angular momentum and energy probabilities constant even though the density in angle changes.

A Stern-Gerlach experiment uses an oven and slits to produce a collimated atomic beam, conventionally of neutral silver atoms, and sends it through an inhomogeneous magnetic field onto a position-sensitive detector. The unpaired outer electron supplies the effective spin one-half magnetic dipole moment; neutral atoms avoid the ordinary Lorentz deflection of a charged beam. For an analysing direction $z$, the spatial gradient produces a force approximately $F_z={\mu }_z{\partial }_z{B}_z$. The resulting separated beam positions measure the allowed projections of angular momentum. A spin one-half beam gives two branches, rather than the continuous distribution expected from classical dipoles with arbitrary orientations. A uniform magnetic field can rotate a spin but does not itself provide this spatial separation.

To measure the rotation, first use a $z$-directed Stern-Gerlach experiment and an aperture to transmit only the $+z$ branch. Pass this prepared spin eigenstate through a region of approximately uniform field in the $x$ direction, then use a second $z$ analyser to separate and count the two populations. For a field region of length $\ell$ and beam speed $v$, the precession angle is $\theta =\lambda B\ell /v$, and the predicted fractions areVarying $B$ or the transit time gives complementary oscillations, demonstrating coherent Larmor precession. Velocity selection and a sufficiently uniform rotation field keep different particles from accumulating different phases and washing out the contrast. Detection should compare the branch fractions after allowing for losses and analyser efficiency.

The orbital angular momentum of a particle is the operator $\mathbf{L}=\mathbf{r}\times \mathbf{p}$, with $\mathbf{p}=-i\hslash \nabla$ in the position representation. Its angular momentum commutation relations areThus one may specify $L^2$ and one component, conventionally $L_z$, simultaneously. One generally cannot assign sharp values to all three components, because they do not commute. The common angular eigenfunctions are the spherical harmonics:For a scalar orbital wavefunction, single-valuedness in the azimuthal angle gives integer $m$, and regularity on the sphere and termination of the ladder give integer $l\ge |m|$. Half-integer angular momenta describe intrinsic spin rather than scalar orbital motion.

The ladder operators $L_{\pm }=L_x\pm i{L}_y$ obey $[L_z,L_{\pm }]=\pm \hslash L_{\pm }$. Their normalized action isThe coefficient vanishes at the top or bottom of the multiplet. Its squared magnitude follows from $L_{\mp }{L}_{\pm }=L^2-L_z^2\mp \hslash L_z$, and positivity of the resulting norm constrains the allowed $m$ values.

For a central potential, the Hamiltonian commutes with $L^2$ and every component of $\mathbf{L}$. Stationary states can be separated as $R_{nl}(r)Y_l^m(\theta ,\varphi )$, and rotational symmetry makes the energy independent of $m$, giving a $2l+1$ quantum degeneracy at fixed radial label and $l$. Additional degeneracy between distinct $l$ values requires additional structure, as in the Coulomb potential. The parity of $Y_l^m$ is $(-1{)}^l$. These results organize atomic orbitals and their selection rules. They describe quantized rotation and symmetry, without assigning the electron a classical orbit or a simultaneously definite three-component angular momentum vector.

The specified energy requires $n_x+n_y=2$, so the relevant energy eigenspace has the orthonormal basisApplying the creation and annihilation operators givesThus its matrix in this ordered basis isWriting $q=\hslash \lambda$, the characteristic polynomial gives $\lambda ({\lambda }^2-4)=0$. The possible angular momentum eigenvalues are thereforeFor completeness, normalized simultaneous energy and angular momentum eigenstates areEach value occurs once in this energy eigenspace. Equivalently, circular ladder operators of a two-dimensional oscillator give $\hat{H}=\hslash \omega ({\hat{N}}_{+}+{\hat{N}}_{-}+1)$ and ${\hat{L}}_z=\hslash ({\hat{N}}_{+}-{\hat{N}}_{-})$. With $N_{+}+N_{-}=2$, the difference can only be $2,0,-2$.