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

Labels an angular momentum eigenstate through the eigenvalue ${\hslash }^{2}j(j+1)$ of the squared angular momentum operator. The allowed projections are $m\hslash$, with $m=-j,\dots ,j$. Orbital angular momentum has integer $j$; spin can also have half-integer $j$.

In the isotropic three-dimensional quantum harmonic oscillator, the first excited Cartesian energy eigenstates are proportional to $x,y,z$ times the common Gaussian $e^{-r^2/(2a_0^2)}$. If ${\varphi }_x,{\varphi }_y,{\varphi }_z$ are normalized with the same real phase convention, their combinations in the title are normalized angular momentum eigenstates with $l=1$ and $m=1,0,-1$.

Indeed, $z=r\cos \theta$ and $x\pm iy=r\sin \theta e^{\pm i\phi }$ identify a common radial factor times the spherical harmonics $Y_{10},Y_{1,\pm 1}$. The transformation is unitary, so it changes the basis rather than the threefold orbital degeneracy.

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.

Use the uncoupled angular momentum eigenstates $|m_L,\uparrow ⟩$ and $|m_L,\downarrow ⟩$, with $m_L=-1,0,1$. For the Hamiltonian operator actually written in the paper, these are the energy eigenstates, withThe initial $J=3/2$ states are not all energy eigenstates of this operator. To find their energy populations, first perform the Clebsch-Gordan decomposition. The extremal states are $|3/2,3/2⟩=|1,\uparrow ⟩$ and $|3/2,-3/2⟩=|-1,\downarrow ⟩$. Applying the total angular momentum lowering operator to the first givesSimilarly,The initial preparation is an equal incoherent mixture of the four coupled states. Each extremal state contributes population $1/4$ to its one energy eigenvalue, while each nonextremal state contributes $2/3$ and $1/3$ of its own $1/4$ population to the two displayed components. The resulting Zeeman splitting is thereforefor nonzero $B_z$. The relative populations are $\boxed{3:2:2:2:3}$ in this order; the two zero-energy states each have population $1/12$. These Zeeman populations from an unpolarized angular-momentum multiplet remain constant under subsequent evolution by the stated static Hamiltonian operator, although off-diagonal quantum coherence can evolve.

There is a distinct weak-field model in which a retained spin-orbit interaction keeps $J=3/2$ isolated and one projects the magnetic perturbation into that quartet. The Landé g-factor is then $4/3$, giving four shifts ${\mu }_B{B}_z(\pm 2,\pm 2/3)$ with equal populations. That calculation requires the additional projection or dominant spin-orbit interaction; it is not diagonalization of the full magnetic Hamiltonian operator specified here, which also mixes the $J=1/2$ sector.

By addition of angular momentum, combining the first two spin one-half particles gives $0\oplus 1$. Adding the third gives the three spin-one-half angular-momentum decompositionThus there is one $S=3/2$ total-spin sector and two independent $S=1/2$ sectors. Since the particles are distinguishable, no fermionic exchange symmetry restriction removes either copy. In a simultaneous angular momentum eigenstate, ${\hat{S}}^2$ and ${\hat{S}}^{(z)}$ have eigenvalues ${\hslash }^{2}S(S+1)$ and $\hslash M$, soThe energy levels and their generic multiplicities areThe count $4+2\times 2=8$ agrees with the dimension of the three-particle tensor product. These multiplicities must be added whenever numerical energy eigenvalues coincide. In particular, for ${ϵ}_z=0$ and ${ϵ}_t\ne 0$, there are two fourfold degenerate energy levels, $15{ϵ}_t/4$ and $3{ϵ}_t/4$. For ${ϵ}_t=0$ and ${ϵ}_z\ne 0$, the multiplicities at $M=-3/2,-1/2,1/2,3/2$ are $1,3,3,1$. If both coefficients vanish, the single energy level has multiplicity eight. Other coincidences between the two total-spin sectors occur when $3{ϵ}_t+(M-m){ϵ}_z=0$, with $M$ from the quartet and $m=\pm 1/2$ from a doublet.

The squared angular momentum operator and one selected component are commuting operators, so a complete set of common eigenfunctions can be chosen: for the $z$ component these are the usual $|j,m⟩$ angular momentum eigenstates. Distinct components do not have a common complete eigenbasis, because their commutator is another nonzero component. A $j=0$ state is an exception in which all components vanish; noncommutation does not forbid every individual common eigenfunction.

In LS coupling, add all orbital and spin angular momenta separately:The quantum numbers specifyThus $M_L=-L,\dots ,L$ and $M_S=-S,\dots ,S$. The addition of angular momentum rule givesIn a product basis with individual projection labels,The magnitudes $L$ and $S$ are not generally the arithmetic sums of the individual magnitudes; they result from angular-momentum coupling. The sum rules for projections remain valid for every product component of a coupled angular momentum eigenstate.

For the allowed $n=2k$ states, the normalized standing wave is a quantum superposition of two opposite angular momentum eigenstates:The Born rule therefore predictsThe expectation value is zero, but a single measurement never gives zero for these $k\ge 1$ states. Their uncertainty is $\Delta l_z=k\hslash$. The two components have the same rotor energy ${\hslash }^2{k}^2/(2I)$, so a definite energy does not imply definite angular momentum.

A good atomic qubit needs two reproducible, orthogonal internal states whose populations and relative quantum phase survive for many gate times. Both should be long lived, with weak spontaneous emission and little decoherence from fluctuating fields or atomic motion. Their transition must nevertheless permit fast, selective, coherent control with little leakage to other states. Reliable initialization and state-selective quantum measurement are also required; optical pumping and fluorescence readout can supply these functions.

For rubidium-87, use the two ground state hyperfine levels in the $5{}^2{S}_{1/2}$ electronic configuration. The electronic angular momentum is $J=1/2$, and addition of angular momentum to the given nuclear spin $I=3/2$ gives $F=|I-J|,\dots ,I+J=1,2$. A suitable Rubidium-87 clock qubit isThese are coupled electron-nucleus angular momentum eigenstates, rather than two bare nuclear-spin projections. Their hyperfine structure supplies a resolvable splitting ${\nu }_{01}\simeq 6.835\mathrm{GHz}$. Both belong to the electronic ground state, avoiding a short optical excited-state lifetime. Since the weak-field Zeeman shift is proportional to $g_F{m}_{F}B$, both have zero linear shift at $B=0$: their rubidium-87 clock transition suppresses first-order magnetic-field dephasing near zero field. A quadratic Zeeman shift of a clock transition and differential optical shifts remain. A small stable bias field can define the quantization axis and help resolve unwanted transitions, at the cost of some residual magnetic sensitivity.^[ref]

Drive the magnetic dipole transition with a resonant microwave field of suitable polarization, here $\Delta m_F=0$, or use two phase-coherent optical fields for a far-detuned stimulated Raman transition with their frequency difference resonant with ${\nu }_{01}$. The optical excited state is then only virtually populated, reducing spontaneous emission. In the rotating frame and rotating-wave approximation, define the drive phase $\varphi$ by the effective HamiltonianHere $X,Y,Z$ are the Pauli matrices on the chosen qubit, and $\Omega$ is the Rabi angular frequency. Pulse duration controls the rotation gate angle, while phase selects the axis in the equatorial plane of the Bloch sphere. A $\pi$ pulse exchanges the populations; a $\pi /2$ pulse makes an equal quantum superposition from either basis state. Controlled Rabi detuning, a differential energy shift, or compositions of these pulses give $z$ rotations. For example, $R_z(\alpha )R_y(\beta )R_z(\gamma )$ realizes any one-qubit unitary operator up to a global phase. Thus the atomic encoding admits arbitrary single-qubit quantum gates.

With $\hslash =1$, angular momentum eigenstates have $j=0,\frac{1}{2},1,\dots$ and $m=-j,-j+1,\dots ,j$. Their $J^2$ and $J_3$ eigenvalues are $j(j+1)$ and $m$. Adding spins $1/2$ and $j$ gives $J=j+1/2$ and, if $j\ge 1/2$, $J=j-1/2$. For the Clebsch-Gordan coefficients to be nonzero one requires $\boxed{M=r+m}$.

For $j=2$, write $|\pm ,m⟩$ for $|1/2,\pm 1/2⟩|2,m⟩$. The highest state and one application of the angular momentum lowering operator giveThe lowering coefficient of the highest $J=5/2$ state is $\sqrt{5}$, while the two product-state coefficients before division are $1$ and $2$. The last line is its normalized orthogonal complement. Its coefficient with maximal $m=2$ is positive, fixing the requested phase convention. All other coefficients for the allowed $M\ge 3/2$ vanish.

For the decay, the final total intrinsic spin is $1/2$. Combining it with integer orbital angular momentum $\ell$ to obtain $J=3/2$ permits only $\ell =1,2$. Parity conservation gives ${\eta }_X={\eta }_A{\eta }_B(-1{)}^{\ell }=(-1{)}^{\ell }$. Thus $\boxed{{\eta }_X=+1:\ell =2;{\eta }_X=-1:\ell =1}$.

For $\ell =2$ and initial $M=3/2$, the last displayed coupled state shows that the spin-up probability, after ignoring the orthogonal orbital states, is $1/5$. For $\ell =1$, $J=3/2$ is the maximal coupled angular momentum and its highest state is $|1/2,1/2⟩|1,1⟩$. Therefore $\boxed{P=1/5for{\eta }_X=+1,P=1for{\eta }_X=-1}$.

The total angular momentum operator is $\mathbf{J}=\mathbf{L}+\mathbf{S}$. Since orbital and spin operators act on different factors of the tensor product, they commute, and expanding $J^2$ givesA coupled angular momentum eigenstate $|l,s;j,m_j⟩$ is a simultaneous eigenstate of these three squared operators, with eigenvalues ${\hslash }^{2}j(j+1)$, ${\hslash }^{2}l(l+1)$ and ${\hslash }^{2}s(s+1)$. Thus the angular factor in the spin-orbit interaction is the numberThe other factor, $r^{-1}dV/dr$, is radial. It commutes with the angular operators, and a central-potential energy eigenstate separates into a radial wavefunction and an angular spinor. Taking its radial expectation value therefore givesThis spin-orbit energy shift in a central potential is independent of $m_j$, so rotational symmetry preserves the $2j+1$ magnetic degeneracy within a fixed $j$ multiplet. It is a first-order fine structure correction; the given coefficient already includes the relativistic factor needed for the electron's spin-orbit coupling.

The angular momentum eigenstates have $J^2$ eigenvalue ${\hslash }^{2}j(j+1)$, where the angular momentum quantum number is $j=0,\frac{1}{2},1,\frac{3}{2},\dots$. For each $j$, the magnetic quantum number runs in unit steps from $-j$ to $j$. Here that magnetic quantum number is $+1/2$, so only half-integer $j$ are compatible:Integer $j$ cannot contain a $J_3$ eigenvalue $\hslash /2$. Applying the angular momentum lowering operator and using $[J_3,J_{-}]=-\hslash J_{-}$ givesThe lowered vector is nonzero, since the adjoint operator identity $J_{-}^{\ast }=J_{+}$ givesFinally, $J_3$ is a self-adjoint operator, so its eigenstates for distinct eigenvalues are orthogonal. Explicitly,and hence $\boxed{⟨\phi \mid \psi ⟩=0}$.

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.

Choose the ordered angular momentum eigenstates $(|1,1⟩,|1,0⟩,|1,-1⟩)$, so $J_3=\hslash \mathrm{diag}(1,0,-1)$. Since $[J_3,J_{\pm }]=\pm \hslash J_{\pm }$, a nonzero matrix element of $J_{+}$ must increase the magnetic quantum number by one, and one of $J_{-}$ must decrease it by one. Thus only entries $(1,2),(2,3)$ of $J_{+}$ and $(2,1),(3,2)$ of $J_{-}$ can be nonzero. The adjoint operator identity $J_{-}=J_{+}^{†}$ givesFor the $j=1$ irreducible spin representation, both adjacent transitions are present; the following calculation determines their sizes.

The Hamiltonian operator $H=\omega J_3$ is already diagonal in the angular momentum eigenstate basis of part (b). Its three distinct energy eigenvalues and stationary states areThus the three column eigenvectors are $(1,0,0{)}^{\mathsf{T}}$, $(0,1,0{)}^{\mathsf{T}}$ and $(0,0,1{)}^{\mathsf{T}}$, with energies $+\hslash \omega$, zero and $-\hslash \omega$, respectively. Each is nondegenerate since $\omega >0$. Any nonzero scalar multiple gives the same stationary state after normalization; a quantum superposition of different energies is generally not stationary.

In the ordered angular momentum eigenstate basis $(|1,1⟩,|1,0⟩,|1,-1⟩)$, with positive raising coefficients,To derive them, $[J_3,J_{+}]=\hslash J_{+}$ restricts $J_{+}$ to its two adjacent superdiagonal entries. Write these real entries as $A,B\ge 0$. Since $J_{-}=J_{+}^{†}$, the relation $[J_{+},J_{-}]=2\hslash J_3$ reads $\mathrm{diag}(A^2,B^2-A^2,-B^2)=2{\hslash }^2\mathrm{diag}(1,0,-1)$, forcing $A=B=\sqrt{2}\hslash$. Adding and subtracting the ladder operators gives the displayed matrices. Direct multiplication verifies the angular momentum commutation relations and $J^2=2{\hslash }^{2}I$.

For a central-potential Hamiltonian operator with the spin-orbit interaction $H_{\mathrm{so}}=[r^{-1}{V}^{\prime }(r)/(2m^2{c}^2)]\mathbf{L}\cdot \mathbf{S}$, use $\mathbf{J}=\mathbf{L}+\mathbf{S}$ andA coupled angular momentum eigenstate makes this angular factor a number. The radial operator commutes with all angular factors, giving the energy shift in the title. It is independent of the magnetic projection $m_j$, so a fixed-$j$ multiplet retains its $2j+1$ degeneracy.

For spin one-half, the angular bracket is $l$ when $j=l+1/2$ and $-(l+1)$ when $j=l-1/2$. Only the former exists at $l=0$, where it is zero. For the Coulomb potential energy, $r^{-1}{V}^{\prime }(r)=e^2/(4\pi {ϵ}_0{r}^3)$; the radial normalization convention must be respected when computing its expectation value.