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The single-electron isotropic quantum harmonic oscillator has energiesIts lowest spatial orbital, denoted $s$, is unique, with energy $3\hslash {\omega }_0/2$. Including spin one-half gives two spin orbitals. Two identical electrons can occupy both only with opposite spin in the unique singlet state. Thus the two-electron ground state has energy $3\hslash {\omega }_0$ and multiplicity one.

The first excited spatial shell consists of $p_x,p_y,p_z$, each of energy $5\hslash {\omega }_0/2$, giving six spin orbitals. A first-excited two-electron Slater determinant occupies one of the two ground-shell spin orbitals and one of these six excited-shell spin orbitals. HenceNo further factor of two is inserted for interchanging the electrons, since the fermionic exchange symmetry is already built into a Slater determinant.

The spatial Hamiltonian operator is invariant under simultaneous rotations of both positions, so $[H_0,L_k]=0$ for each component of the total orbital angular momentum. It is also spin independent, so $[H_0,S_k]=0$ for each total spin component. Thereforeexpressing conservation of angular momentum and conservation of the total spin magnitude for this model. The Pauli exclusion principle restricts which states occur but does not break either symmetry.

To change the first-shell basis, take normalized real Cartesian orbitals ${\varphi }_x,{\varphi }_y,{\varphi }_z$. Using spherical coordinates, $z=r\cos \theta$ and $x\pm iy=r\sin \theta e^{\pm i\phi }$. Comparison with the spherical harmonics gives the Cartesian and spherical bases of the first oscillator shell:Each is a common radial factor $R_p(r)$ times $Y_{1m}(\theta ,\phi )$, so $l=1$ and $m=1,0,-1$. Their normalizations and mutual orthogonality follow because this is a unitary change of basis. The ground spatial orbital has $l=0$.

For two ground-shell electrons, addition of angular momentum gives $L=0$, and their symmetric spatial state requires $S=0$. For one electron in each shell, $0\otimes 1$ gives only $L=1$. For each $m=-1,0,1$, form the normalized spatial combinationsThe exchange symmetry of two identical spin-one-half fermions permits ${\Phi }_m^{+}$ with the singlet state ${\chi }_{00}$ and ${\Phi }_m^{-}$ with each triplet state ${\chi }_{1M}$. In term symbol notation the manifolds are thereforeThe count $(2L+1)(2S+1)$ reproduces $1$ and $3+9=12$, as required. Both exchange symmetries are possible for the distinct occupied orbitals; dropping either spin sector would undercount the two-electron states in the first two shells of an isotropic oscillator.

For the specified spin-orbit interaction, $\mathbf{S}$ and $\mathbf{L}$ are the total two-electron operators. Set $\mathbf{J}=\mathbf{L}+\mathbf{S}$. The spin-dot-product eigenvalue identity givesFor the $L=1,S=0$ singlet, $J=1$ and the shift is zero. For the $L=S=1$ triplet, $J=0,1,2$. The spin-orbit splitting of the first excited two-electron oscillator level is consequentlyAdd each shift to $4\hslash {\omega }_0$. The multiplicities sum to twelve and their weighted shift sums to zero. For positive $\lambda$ the table is ordered in energy; for negative $\lambda$ the ordering reverses. Weak coupling means $|\lambda |{\hslash }^2\ll \hslash {\omega }_0$. Since this total-spin interaction commutes with $H_0$, $L^2$ and $S^2$, the shifts within this invariant manifold are in fact exact for the specified model, although only first order was requested.

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 $\epsilon \ll \mathrm{Ry}$.

In the idealized isolated-carbon chemical atom interpretation, both beams carry approximately $2.68\mathrm{eV}$ per incident particle, close to the ${}^1{S}_0$ threshold. However equal incident energies do not imply equal excitation channels.

A single optical photon cannot strongly excite any of the listed even-parity $2p^2$ energy levels from the even-parity ${}^3{P}_0$ ground state through an electric dipole transition. In particular ${}^3{P}_0\to {}^1{S}_0$ also violates $\Delta S=0$ and the $J=0\leftrightarrow 0$ restriction. The lowest strong opposite-parity excitations lie well above $2.68\mathrm{eV}$. Thus one expects no strong resonant absorption or corresponding strong atomic fluorescence from these ground-state chemical atoms at this optical wavelength; weak higher-order processes and scattering are separate effects.

Incident electrons can undergo inelastic scattering and collisional excitation. Their orbital angular momentum and exchange with bound electrons allow transfers excluded for a single electric dipole photon. The ${}^1{D}_2$ energy level at $1.264\mathrm{eV}$ and the low fine structure excitations are energetically accessible. One therefore expects scattered electrons with discrete energy losses, possible weak forbidden spectral lines, and collisional or surface relaxation. For example ${}^1{D}_2\to {}^3{P}_2$ has wavelength approximately $10^7/(10192-43.4)=985\mathrm{nm}$; spin mixing supplies the weak spin-forbidden radiative channel.

There is a numerical threshold caveat. Literally,Neither quoted beam reaches ${}^1{S}_0$ from the exact ground state if these are taken as exact monoenergetic values. The electron value rounded to two decimal places could represent an actual energy above threshold; if so, ${}^1{S}_0$ is collisionally accessible and its decay to ${}^1{D}_2$ produces the weak electric quadrupole transition at $10^7/11456\simeq 873\mathrm{nm}$, followed by lower-level relaxation. These emissions require populations and sufficiently low collisional quenching of a metastable line.

An actual solid carbon foil has bands, collective excitations and surface losses, so it need not reproduce the isolated-chemical atom spectrum or selection rules. Its optical and electron-induced emission cannot be predicted quantitatively from the five atomic energy levels alone. This is the limitation of the foil wording in the source.

In LS coupling, first add the individual orbital angular momenta and spins separately, then add their totals:The Coulomb interaction between electrons must separate different atomic terms more strongly than the spin-orbit interaction mixes them. Then $L,S,J,M_J$ are useful quantum numbers, with $J=|L-S|,\dots ,L+S$. This approximation is usually better for lighter chemical atoms and well-separated atomic terms; strong configuration interaction or strong spin-orbit coupling can spoil it. The term symbol is ${}^{2S+1}{L}_J$.

The general electric-dipole selection rules, for isolated chemical atoms with definite parity and total electronic angular momentum, are a change of parity, $\Delta J=0,\pm 1$ excluding $J=0\leftrightarrow 0$, and $\Delta M_J=0,\pm 1$, with the accessible magnetic sublevels determined by the light's polarization. In pure LS coupling, the additional selection rules are $\Delta S=0$ and $\Delta L=0,\pm 1$, excluding $L=0\leftrightarrow 0$. In a one-electron description the electric dipole operator also requires $\Delta l=\pm 1$ for the active electron. These latter statements can fail as exact labels when atomic terms mix; the general parity and total-angular momentum restrictions still apply. Unresolved hyperfine structure is neglected here.

For strontium, the closed core contributes $L=S=0$. The Pauli exclusion principle makes the two equivalent $5s$ electrons a singlet state. Different radial orbitals permit both a singlet state and a triplet state. Thus the complete lists of atomic fine-structure levels areThe $5s5p$ electron configuration has odd parity; the other two electron configurations have even parity.

Write $D_x=y{\partial }_z-z{\partial }_y$, $D_y=z{\partial }_x-x{\partial }_z$ and $D_z=x{\partial }_y-y{\partial }_x$. The product rule applied to a smooth test function $f$ givesThe mixed second partial derivatives cancel because they commute. Since the orbital angular momentum components are ${\hat{j}}_i=-i\hslash D_i$, their commutator isThis is one of the orbital angular momentum commutation relations; the Planck constant enters through the definition of each linear operator.

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.

It represents the squared orbital angular momentum. In atomic units, $\hslash =1$, so its eigenvalues are $l(l+1)$. In ordinary units the corresponding observable is ${\hslash }^{2}l(l+1)$, with orbital angular momentum quantum number $l=0,1,2,\dots$.

Keep one normalized spin function fixed and write $A=1/\sqrt{24}$ and $f_m(\theta ,\varphi )=\sin \theta e^{im\varphi }$, with $m=\pm 1$. Overall constants and the spin factor pass unchanged through the spatial operators.

First, in atomic units, ${\hat{l}}_z=-i{\partial }_{\varphi }$, soFor the squared orbital angular momentum, ${\partial }_{\varphi }^2{f}_m=-f_m$ andTherefore the two terms of the angular operator giveusing $1-{\cos }^2\theta ={\sin }^2\theta$. Thus ${\hat{l}}^2\Psi =2\Psi$.

Finally, the spatial hydrogen Hamiltonian isFor $R=Ar{e}^{-r/2}$,Substitute the already established angular eigenvalue $l(l+1)=2$:Hence $E=-1/8$ hartree, consistently with $n=2$. The angular calculation must precede the radial substitution if one wants to avoid differentiating the entire product repeatedly.

Linearity and the shared $l=1$ angular eigenvalue giveUsing the relative phases in part (c),Thus each real orbital remains an energy eigenstate with $E=-1/8$ hartree and has definite orbital angular momentum magnitude $\sqrt{2}\hslash$, but neither is an eigenfunction of ${\hat{l}}_z$. A measurement of $l_z$ gives $+\hslash$ or $-\hslash$, each with probability $1/2$. Its mean is zero and its variance is ${\hslash }^2$; a zero mean does not mean that a measurement gives zero.

Using the spin values above and accounting for two metal ions per formula unit,The deviations from 2 indicate spin-orbit coupling and incomplete quenching of orbital angular momentum. The partly filled $e^{\prime \prime }$ manifold in Fe and $e^{\prime }$ manifold in Ni permit orbital degeneracy in the ideal geometry. Co's high-spin orbital occupation is more orbitally balanced, but mixing with excited ligand-field states still produces a substantial effective $g$ shift. These fitted scalar values summarize a thermally averaged response; they do not determine the individual components of an anisotropic g tensor.

For a fixed open-shell electron configuration in the LS coupling regime, Hund's rules select the free-ion ground term in the following order:

The first two rules come from the Coulomb interaction between electrons within the partially filled subshell, including its exchange interaction. Parallel spins require an antisymmetric spatial wavefunction, reducing close approach and often lowering the repulsive Coulomb interaction energy. Among the allowed high-spin terms, the angular dependence of the Coulomb interaction generally favours the largest $L$; a simple magnetic alignment argument alone does not derive this second ordering rule.

The weaker spin-orbit coupling then splits a selected $L,S$ term. In LS coupling, its term-level energy is proportional toThe effective sign appropriate to less-than-half versus more-than-half filling gives the third rule. These rules assume the electrostatic term separation dominates spin-orbit coupling; strong crystal field splitting or other coupling regimes require further analysis.

Hund's rules maximizes total spin within the partially filled shell: parallel spins require an antisymmetric spatial wavefunction, reducing Coulomb overlap through exchange. Among terms with that spin, the second rule selects the largest total orbital angular momentum. The third rule, from spin-orbit coupling within the selected term, gives $J=|L-S|$ below half filling and $J=L+S$ above half filling, with the half-filled case treated directly.

For seven $4f$ electrons, put one parallel-spin electron in each $m_l=-3,-2,-1,0,1,2,3$ orbital. Their orbital projections sum to zero and form the half-filled $L=0$ shell; all seven spins add. ThusThere is no ambiguity from applying the below- or above-half-filled formula because $L=0$. The absence of a first-order orbital moment is important for the weak magnetic anisotropy discussed in c. These are isolated-ion term rules; a crystal field or strong covalency can alter other ions' orbital behavior, but the half-filled Eu${}^{2+}$ shell is unusually robust.

For a free transition-metal ion, degeneracy among different orbital states can support an orbital contribution to its magnetic moment. In a crystal, the electric fields and bonding of the surrounding ligands split those states. If the orbital ground state becomes nondegenerate and is separated from excited orbital states, its wavefunction can usually be chosen real in zero field. Since the orbital angular momentum operator is imaginary in a real-orbital basis, its expectation value then vanishes.

This suppression is orbital quenching. The remaining leading magnetic moment is the spin contribution, approximately $g{\mu }_B\mathbf{S}$ with $g\simeq 2$. Spin-orbit coupling admixes excited orbital states and produces residual anisotropy and shifts in the g-factor; quenching does not mean that every orbital matrix element or every higher-order orbital response vanishes. A degenerate ground orbital multiplet need not be quenched until a further distortion removes its degeneracy.

The orbital angular momentum operators areOn smooth wavefunctions, $[x_i,p_j]=i\hslash {\delta }_{ij}$, while position operators commute with each other and momentum operators commute with each other. Applying the product rule for commutators givesUse it on the four terms in $[L_1,L_2]$: only $[x_2{p}_3,x_3{p}_1]=-i\hslash x_2{p}_1$ and $[x_3{p}_2,x_1{p}_3]=i\hslash x_1{p}_2$ survive, with the latter entering positively from the two minus signs. ThereforeThe cyclic versions similarly give $[L_3,L_1]=i\hslash L_2$ and $[L_3,L_2]=-i\hslash L_1$. Thus, for $L_{\pm }=L_1\pm i{L}_2$,Now define $L^2=L_1^2+L_2^2+L_3^2$. For each $i$,because the Levi-Civita symbol is antisymmetric in $j,k$ and the operator pair is symmetric. Linearity then gives $\boxed{[L^2,L_{\pm }]=0}$. The PDF superscript here is the squared total angular momentum, not the component $L_2$.

Every orbital angular momentum operator annihilates a radial function: substituting ${\partial }_{i}f(r)=f^{\prime }(r)x_i/r$ makes the two terms cancel. It also annihilates $r^2$. Thus the radial factor can be carried through the angular operators. Since $x_3^2$ is independent of $x_1,x_2$, $L_3\psi =0$ for every value of $a$.

Direct differentiation givesConsequently $L^2{x}_3^2={\hslash }^2(6x_3^2-2r^2)$ and $L^2{r}^2=0$. For a nonzero admissible radial factor, comparison with $\lambda (x_3^2+a{r}^2)f(r)$ forces $\lambda =6{\hslash }^2$ and $6a=-2$. Therefore the simultaneous eigenfunction occurs atIt is the axially symmetric quadratic orbital eigenfunction, with quantum numbers $l=2$, $m=0$ and angular dependence proportional to the spherical harmonic $Y_{20}$. The subtraction of $r^2/3$ removes its rotationally invariant scalar component.

Use the position-space momentum operator $P_j=-i\hslash {\partial }_j$ and set $F=(x+iy)z$. Direct differentiation givesTo compute the squared orbital angular momentum, write $D_x=y{\partial }_z-z{\partial }_y$, $D_y=z{\partial }_x-x{\partial }_z$, $D_z=x{\partial }_y-y{\partial }_x$, so $L_j=-i\hslash D_j$. Expanding the products, including derivatives of their variable coefficients, givesFor example $D_x^2=y^2{\partial }_z^2+z^2{\partial }_y^2-2yz{\partial }_y{\partial }_z-y{\partial }_y-z{\partial }_z$; summing the three cyclic expressions proves the identity.

Here $\Delta F=0$ and $EF=2F$, so $E^{2}F=4F$. ThereforeThis is the angular momentum of a homogeneous harmonic polynomial, with angular quantum numbers $\ell =2,m=1$. The polynomial is a nonzero angular eigenfunction; by itself it is not a normalizable full-space wavefunction.

Use $|m,m^{\prime }⟩$ as shorthand for the tensor product of two total-isospin-one states with third components $m,m^{\prime }$. A total-isospin-zero state must have total third component zero, so write $a|1,-1⟩+b|0,0⟩+c|-1,1⟩$. Applying the total raising operator givesAn isospin singlet must be annihilated by this operator; hence $b=-a$ and $c=a$. Normalization gives the two-isovector singletIt is also annihilated by the total lowering operator. A raising or lowering action changes the third-component label by one; the unchanged output label in the PDF's preliminary ladder formula is a typographical omission, not the action used here.

Exchange commutes with the total isospin operators. The displayed singlet is symmetric. The $I=2$ multiplet has symmetric highest state $|1,1⟩$; applying the symmetric total lowering operator preserves that exchange symmetry. The symmetric square of a three-dimensional space has dimension six, exactly $1+5$, while its antisymmetric square has dimension three. The remaining $I=1$ multiplet is therefore antisymmetric. Thus the Clebsch-Gordan decomposition isPions have zero intrinsic spin and form an isospin triplet. Bose-Einstein statistics requires the complete two-pion state to be symmetric under exchange. Its relative orbital state changes by $(-1{)}^{\ell }$, so odd orbital angular momentum requires antisymmetric isospin: an odd-$\ell$ two-pion state has $I=1$. This statement treats the charge components as states of the same isospin multiplet, rather than ignoring exchange when their charges differ.

For the decay, let $|\chi ⟩=H_I|K^0⟩$. The stated commutator with $I_3$ gives $I_3|\chi ⟩=0$. More decisively, the commuting raising operator givesThe kaon doublet cannot be raised twice. An $I=2,M=0$ component would survive two raisings, so it is excluded. Meanwhile a spinless kaon decaying into two spinless pions has $\ell =0$ by conservation of total angular momentum, independently of whether parity is conserved by the decay. The exchange condition excludes $I=1$. This proves the highest-weight weak-isospin selection in kaon decay: the two-pion final state has $I=0$.

In a consistent pion phase convention, the singlet's charge components have the structureThe normalized symmetric charged channel has coefficient $\sqrt{2/3}$ and the neutral channel coefficient $-1/\sqrt{3}$. Thus, in the isospin limit with equal pion masses and matching two-body phase space,Equivalently, ordered charged and neutral amplitudes have equal magnitudes, but neutral identical particles contribute the phase-space factor $1/2!$. This is the same counting expressed in another basis, and must not be included a second time after using the normalized charge-channel coefficients. Small pion-mass and electromagnetic effects can modify the ideal ratio.

Use units in which the proton electric charge is one. The up quark and down quark both have spin angular momentum $1/2$ and form a flavour isospin doublet:Their antiquarks have the opposite charges and third components of isospin, with the same total isospin and spin. This is strong-interaction flavour isospin, not the chiral weak isospin assignment.

The Clebsch-Gordan decomposition for SU2 says that spins $j_1,j_2$ combine into every $j$ from $|j_1-j_2|$ to $j_1+j_2$ in unit steps. It applies independently to spin and isospin. ThusBefore exchange symmetry is imposed, three light quarks have total spin $S=1/2$ or $3/2$, and total isospin $I=1/2$ or $3/2$; the doublet occurs twice in the uncoupled tensor cube. With zero orbital angular momentum, $J=S$.

Physical baryons must also be colour singlets. The three-quark colour singlet is completely antisymmetric, so the Pauli exclusion principle requires symmetric spatial-spin-flavour symmetry. For a symmetric spatial S-wave state, the spin-flavour factor is symmetric. The $I=3/2$ flavour quartet is symmetric, whereas the $I=1/2$ flavour doublets have mixed permutation symmetry; the spin factors have the analogous symmetries. Symmetric flavour therefore pairs with symmetric spin, while the two mixed factors pair in their unique symmetric combination. This is the Pauli constraint on three-quark flavour multiplets; it leavesEquivalently the symmetric cube of the four-dimensional two-flavour spin space has dimension $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20=2\cdot 2+4\cdot 4$. The unmatched spin-isospin pairings in the bare tensor product do not describe the usual symmetric-spatial ground-state baryons.

For three light quarks, charge addition gives $Q=I_3+1/2$. The observed nucleon doublet and Delta baryon quartet have the following assignments; the listed contents specify flavour counts, not the complete spin-flavour wavefunction:The proton and neutron have $J=1/2$; all four Delta baryons have $J=3/2$. These states have positive parity in the zero-orbital ground-state model.

For a light quark-antiquark meson, both spin and flavour doublets combine into singlet plus triplet. In a conventional choice of phases the flavour states areEach row can have $J=0$ or $1$, since the two constituent spins couple to zero or one. At zero orbital angular momentum their parity is negative, following the meson parity and charge conjugation rule $P=(-1{)}^{L+1}$. The spin-zero isotriplet is the pion multiplet, and the spin-one isotriplet the rho meson multiplet. The nonstrange spin-one isosinglet is the idealized omega meson flavour direction. Its spin-zero counterpart is the nonstrange direction contributing to the eta and eta prime mesons; it is not automatically a physical mass eigenstate once other flavours and mixing are included.

Now introduce an isospin-zero quark $h$ of spin $1/2$ and charge $q_h$. For baryons with one isospin-singlet quark, the two light quarks have either the antisymmetric flavour singlet $[ud]=(ud-du)/\sqrt{2}$ or the symmetric triplet $uu,\{ud\},dd$. Their colour pair is antisymmetric. Hence light-diquark spin-isospin correlation gives $S_{ud}=0$ for $I=0$ and $S_{ud}=1$ for $I=1$. Coupling the $h$ spin producesThe three triplet rows have $I_3=1,0,-1$. For a strange-like charge $q_h=-1/3$, these are the neutral spin-one-half Lambda baryon, the spin-one-half Sigma baryon triplet of charges $1,0,-1$, and its spin-three-half starred partners. For a charm-like charge $q_h=2/3$, they are the spin-one-half Lambda c baryon of charge $1$, and the Sigma c baryon triplets of charges $2,1,0$ with spins $1/2$ and $3/2$. Both possible new charge assignments have thus been covered; charge-conjugate antibaryons have opposite charges and unchanged $I,J$.

The meson doublets with one isospin-singlet quark are $h\overline{u},h\overline{d}$, withFor $q_h=-1/3$ their charges are $-1,0$, giving $K^{-},{\overline{K}}^0$-like kaons and their vector partners; the conjugate doublet has $u\overline{h},d\overline{h}$ and charges $1,0$. For $q_h=2/3$ the charges are $0,1$, giving $D^0,D^{+}$-like D mesons and their vector partners, with charge-conjugate partners as well. Finally a hidden-flavour pair $h\overline{h}$, if included, is neutral with $I=0$ and $J=0$ or $1$. These are quantum-number classifications; charge and isospin alone do not determine particle masses or mixing angles.

The angular momentum commutation relations are $[L_i,L_j]=i\hslash {ϵ}_{ijk}{L}_k$. The commutator product rule givessince the bracketed expression is symmetric in $j,k$ and ${ϵ}_{ijk}$ is antisymmetric. Alsoproving $\boxed{L^2=L_{-}{L}_{+}+L_3^2+\hslash L_3}$.

For orbital angular momentum, $L_i=-i\hslash {ϵ}_{ijk}{x}_j{\partial }_k$. A differentiable radial function has ${\partial }_{k}f(r)=f^{\prime }(r)x_k/r$, so its contraction with the antisymmetric symbol vanishes: $Lf(r)=0$. Put $s=x_1+i{x}_2$. The coordinate differential operators give $L_{3}s=\hslash s$, whileannihilates every power of $s$, because $({\partial }_1+i{\partial }_2)s=0$. Since all $L_i$ annihilate the radial factor, the product rule yieldsSubstitution into the expression for $L^2$ gives the highest-weight complex-coordinate orbital wavefunction resultAs $[L^2,L_{-}]=0$, lowering preserves this eigenvalue whenever the lowered function is nonzero. Explicitly,For $n=0$ the lowered function is zero, so it is not an eigenfunction. For $n\ge 0$ the original nonzero function has the usual regular angular dependence. The PDF's “any integer” differential identities also hold locally for negative $n$ where $s\ne 0$, but those negative powers are singular on the axis and are not square-integrable regular angular eigenstates. The converted TeX incorrectly writes $L_4$ in place of the PDF's $L_{+}$.

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.