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The ground state of an even-even nucleus has $J^P=0^{+}$, and the emitted alpha particle also has $0^{+}$. In an alpha decay to daughter spin $J_Y$, addition of angular momentum permits total zero only when the relative orbital angular momentum is $L=J_Y$. Parity conservation gives $+1={\pi }_Y(-1{)}^L$, hence the daughter has natural parity, $0^{+},1^{-},2^{+},3^{-},\dots$. This rule does not itself restrict $J_Y$ to even integers. An even-$J$ sequence such as $0^{+},2^{+},4^{+}$ requires additional nuclear structure, for example a positive-parity ground-state nuclear rotational band.

A complete set of commuting observables consists of mutually commuting quantum observables whose joint eigenvalues distinguish all vectors of a simultaneous eigenbasis, up to phase. For a finite-dimensional Hilbert space, commuting self-adjoint operators admit a simultaneous orthonormal basis of eigenvectors; completeness means that every joint eigenspace is one-dimensional. An additional quantum observable that is already a function of the others supplies no independent quantum number.

For the addition of angular momentum, the uncoupled labels $(j_1,j_2,m_1,m_2)$ correspond to $\{J_1^2,J_2^2,J_{1z},J_{2z}\}$, whereas the coupled labels $(j_1,j_2,j,m)$ correspond to $\{J_1^2,J_2^2,J^2,J_z\}$. Within a fixed pair of irreducible angular momentum spaces each choice gives a natural eigenbasis. The Clebsch-Gordan coefficients implement the change between them.

An electric quadrupole transition is a radiative transition of an atomic nucleus with electric multipole order two. The emitted photon carries two units of angular momentum and positive parity, so the atomic nucleus has unchanged parity.^[ref]

Addition of angular momentum imposes $|I_i-I_f|\le 2\le I_i+I_f$. Thus $2^{+}\to 0^{+}$ and $2^{+}\to 2^{+}$ can be E2, whereas $0^{+}\to 0^{+}$ cannot emit one E2 photon. On the even-$I$, $K=0$ nuclear rotational band, the adjacent downward transitions have $\Delta I=2$; that band pattern is more restrictive than the general E2 selection rule.

For dimensionless angular momentum operators and a Hamiltonian $H=a\mathbf{I}\cdot \mathbf{J}$ at fixed $I,J$, addition of angular momentum gives $E_F=a[F(F+1)-I(I+1)-J(J+1)]/2$. Subtracting neighbouring energy levels proves the displayed rule. For fine structure substitute $\mathbf{L},\mathbf{S}$ and total $J$; for magnetic-dipole hyperfine structure use electronic $J$, nuclear spin $I$ and total $F$. Electric-quadrupole interactions and mixing with nearby energy levels can violate the rule.

An atomic addition of angular momentum scheme in which the individual orbital angular momentum vectors first combine into $\mathbf{L}$ and the individual spins into $\mathbf{S}$, then these combine into total angular momentum $\mathbf{J}$. When spin-orbit coupling is weak compared with the interactions between electrons described by electrostatics, $L$ and $S$ remain useful approximate quantum numbers. The Landé g-factor describes the resulting weak-field magnetic splitting.

In a simple constituent quark model, a spin-dependent contribution $A{\mathbf{S}}_1\cdot {\mathbf{S}}_2/(m_1{m}_2)$ splits a spin-zero and spin-one meson having the same spatial configuration. For two spin-half constituents, addition of angular momentum gives ${\mathbf{S}}_1\cdot {\mathbf{S}}_2=[S(S+1)-3/2]/2$, equal to $-3/4$ for $S=0$ and $+1/4$ for $S=1$. ThusThe spin-averaged mass is $(M_0+3M_1)/4=M_{\mathrm{base}}$. Positive $A$ places the vector state above the pseudoscalar. Determining $A$ from just $M_0$ needs an additional assumption about $M_{\mathrm{base}}$; setting it equal to $m_1+m_2$ neglects all other spin-independent binding terms. A splitting below the pion mass closes a one-pion strong decay to the lower meson, often making the allowed electromagnetic transition with photon emission dominant.

Why should adding one neutron be easy for one atomic nucleus and much less favorable for its neighbor? In the nuclear shell model, nucleons occupy discrete energy levels in a nuclear mean field. Filling all available states below a sufficiently large gap produces a nuclear shell closure. This connects nuclear separation energies with the counting of quantum states, and connects the spin of an odd atomic nucleus with its last unpaired nucleon.

The starting point is an approximate Hamiltonian operatorwhere $U_i$ is a confining potential energy, $a_i$ specifies nuclear spin-orbit splitting, and $V_{\mathrm{res}}$ contains interactions left out of the nuclear mean field. This is an effective organization of the nuclear many-body problem, rather than a derivation of the nuclear force. In a spherical approximation, addition of angular momentum combines the orbital angular momentum $l$ with spin $s=1/2$ to give $j=l\pm 1/2$; for $l=0$, only $j=1/2$ exists.

A level labelled $nlj$ has $2j+1$ magnetic substates $m_j=-j,-j+1,\dots ,j$. The Pauli exclusion principle permits one neutron in each such state and, separately, one proton. Thus a $j=7/2$ level holds eight neutrons, or eight protons; it does not hold sixteen identical neutrons. Here $n=1,2,\dots$ counts radial levels of the same $l$, starting with the lowest; it is not the hydrogenic principal quantum number. The letters $s,p,d,f,g,h,i$ mean $l=0,1,2,3,4,5,6$.

An atomic nucleus with odd proton number and odd neutron number. Both unpaired species can contribute to the total nuclear spin; addition of angular momentum gives possible couplings, while the nuclear residual interaction determines their ordering.

The closed shells have $L=S=0$, so only the two equivalent p electrons matter. Their addition of angular momentum allows $L=0,1,2$ and $S=0,1$, subject to the Pauli exclusion principle. Interchanging two $l=1$ orbital states has sign $(-1{)}^{2-L}=(-1{)}^L$: even L is spatially symmetric and needs the antisymmetric spin singlet; odd L is spatially antisymmetric and needs the symmetric spin triplet. Thus the allowed LS terms of a p-squared configuration are ${}^{1}S$, ${}^{1}D$ and ${}^{3}P$.

In LS coupling, $J=|L-S|,\dots ,L+S$. The complete list of term symbols is thereforeThey contain $1+5+(1+3+5)=15=\left(\genfrac{}{}{0ex}{}{6}{2}\right)$ magnetic substates, exhausting the antisymmetric two-electron space. All have even spatial parity, since each p orbital is odd.

The operators associated with the two angular momenta commute with one another, and each squared angular momentum operator commutes with every component of that same angular momentum. Two useful identities areConsequently the two maximal commuting subsets of the displayed collection are

The first set gives the uncoupled eigenbasis $|j_1{m}_1⟩\otimes |j_2{m}_2⟩$. Its $J_z$ eigenvalue is already fixed by $m=m_1+m_2$. The second gives the coupled eigenbasis $|j_1{j}_2;jm⟩$ for the addition of angular momentum. Its scalar-product eigenvalue is already fixed by $j,j_1,j_2$. Thus an independent complete set of commuting observables uses four operators in either case; literal maximality within the question's list includes the dependent fifth operator.

For generic nonzero angular momenta, no further listed operator can be added. In particular,is generally nonzero, and $[J_{1z},J^2]=2[J_{1z},{\mathbf{J}}_1\cdot {\mathbf{J}}_2]$. The same obstruction applies to $J_{2z}$. Their sum $J_z$ commutes because the two commutators cancel.

The closed $1s^2$ and $2s^2$ subshells contribute zero total orbital angular momentum and spin, so only the two $2p$ electrons matter. The first of Hund's rules maximizes their total spin, giving $S=1$. The Pauli exclusion principle then requires an antisymmetric spatial wavefunction. Two parallel-spin electrons cannot occupy the same $m_l$ orbital: the greatest possible orbital projection is $1+0=1$, so the maximal allowed $L$ within this triplet is $L=1$.

The ground term is therefore ${}^{3}P$, in term symbol notation ${}^{2S+1}{L}_J$. Since the $p^2$ subshell contains fewer than half of its six possible electrons, the third of Hund's rules places the smallest $J$ lowest. Addition of angular momentum gives $J=0,1,2$, in that energy order within the ground term. ThusThese are the fine structure partners of the same electronic term, with magnetic multiplicities $1,3,5$, rather than different principal-shell excitations. The ordering agrees with the tabulated lowest neutral-carbon levels.^[ref]

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.

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 ground electron configuration of caesium has a single $6s$ electron, with $J=1/2$. Addition of angular momentum giving $F=3,4$ fixes the nuclear spin to $I=7/2$. With dimensionless angular momentum operators, write the Fermi contact interaction asThe two hyperfine levels therefore differ by $4A$. In the convention ${\mathbf{\mu }}_I=g_I{\mu }_B\mathbf{I}$ and $g_s\simeq 2$,This exposes a missing ingredient in a prediction from the supplied data: $Z$ and $g_I$ alone do not specify the valence electron contact density. Screening effect and penetration of the closed core strongly affect it. If $g_I$ instead denotes the total nuclear moment in units of ${\mu }_B$, it must first be divided by $I$ in the displayed convention.

An explicitly crude unscreened hydrogen-like ion model uses $|{\psi }_{6s}(0){|}^2=Z^3/(\pi 6^3{a}_0^3)$ and gives ${\nu }_0\simeq 5.76\times 10^{11}\mathrm{Hz}$, or ${\omega }_0\simeq 3.62\times 10^{12}{\mathrm{s}}^{-1}$. It describes neither the shielding nor the radial function of neutral caesium. Using a pure unit-charge $6s$ hydrogen orbital instead gives only $3.46\mathrm{MHz}$, illustrating why a unique useful estimate cannot follow from an unspecified effective charge.

For physical neutral ${}^{133}\mathrm{Cs}$, the independently known zero-magnetic field clock transition isThis measured frequency, used to define the second, is supplementary physical information, not a consequence of $Z=55$ and the rounded $g_I$ alone.^[ref] The direct electronic transition is in the microwave range and is a magnetic dipole transition, rather than a direct optical electric dipole transition. An optical implementation needs a two-field Raman transition.

Using the corrected weak-pulse Ramsey spectrum of part (b), the peak at $\delta =0$ has height $(\Omega {\tau }_p{)}^2$. The quantum interference fringes have spacing $2\pi /T$ in angular frequency, with zeros at $(2k+1)\pi /T$, and the sinc function envelope has its first zeros at $\delta =\pm 2\pi /{\tau }_p$. For $T\gg {\tau }_p$, the central peak has approximately full width at half maximum $\pi /T$ in angular frequency. The clock transition between the two $M_F=0$ states has no first-order Zeeman energy, although a quadratic one remains.

The closed $1s^{2}2s^2$ shells have $L=S=0$, leaving the equivalent $2p^2$ electrons. Addition of angular momentum and the Pauli exclusion principle permit the atomic terms ${}^{3}P,{}^{1}D,{}^{1}S$. Their term symbols use ${}^{2S+1}{L}_J$. Hund's rules put the triplet state below the singlet states; for a less-than-half-filled $p$ shell the smallest $J$ in ${}^{3}P$ is lowest. ThusAll have even parity. These assignments agree with independent tabulated carbon energy levels.^[ref] The rounded source values should not be forced to satisfy an exact Landé interval rule; pure first-order LS coupling need not account for every measured fine structure correction.

For $5s5d$, addition of angular momentum gives $L=2$ and $S=0,1$. The even-parity atomic terms and atomic fine-structure levels areThe electric-dipole selection rules exclude all connections to the even-parity $5s^2$ and $5s6s$ electron configurations. To the odd-parity $5s5p$ electron configuration, all seven allowed connections in pure LS coupling areHere a connection is listed independently of whether the $5s5d$ energy level is above or below its partner.

In a weak magnetic field, use the Landé g-factorFor the three ${}^3{D}_J$ energy levels, this gives $1/2,7/6,4/3$; for ${}^3{P}_1,{}^3{P}_2$ it gives $3/2$. The six-component spectral line is thereforeA component has frequency displacement $({\mu }_{B}B/h)(m_D/2-3m_P/2)$. The electric dipole transition with $m_D=m_P=0$ has zero matrix element, since the relevant $J=1\leftrightarrow 1$ rank-one Clebsch-Gordan coefficients vanish there. The remaining allowed pairs giveThus all six frequency positions $\pm 1/2,\pm 1,\pm 3/2$ are distinct. Perpendicular to the magnetic field, both the $\pi$ and $\sigma$ components are visible, with mutually perpendicular linear polarizations. Along the magnetic field the $\pi$ components disappear. This is the weak-field Zeeman effect, before the Paschen-Back effect changes the coupling scheme.

Take the field along $z$ and write $x={\mu }_{B}B$, with Bohr magneton ${\mu }_B$. The printed Hamiltonian uses dimensionless angular-momentum operators, with eigenvalues $L(L+1)$ and $m_L$; physical angular momenta carry an additional $\hslash$. Thus $\beta$ and $x$ are energies. Assume the usual positive spin-orbit interaction coefficient $\beta >0$. Reversing its sign reverses the zero-field ordering but not the state count.

An electron has $S=1/2$. addition of angular momentum givesThe coupled angular-momentum state count can be checked directly. The product basis contains $(2L+1)\times 2=4L+2$ states. The coupled basis containsCoupling therefore reorganizes the states rather than creating or losing them. There is no hyperfine multiplicity because $I=0$.

Use $\mathbf{J}=\mathbf{L}+\mathbf{S}$ to obtainThe two spin-orbit interaction energies are $\beta L/2$ and $-\beta (L+1)/2$. In a weak field, project the Zeeman effect onto a fixed $J$ multiplet:This is the first-order energy formula, not a statement that $H_Z$ has no matrix elements between different $J$ values. For a common $m=m_J$ and a conventional choice of relative phase, its complete coupled-basis block isThe inter-multiplet Zeeman coupling terms give second-order weak-field corrections because the two spin-orbit interaction multiplets are already separated.

In the strong-field basis the Zeeman part is diagonal, $E_Z=x(m_L+2m_S)$. WriteIts diagonal matrix elements are $\beta m_L{m}_S$. Its off-diagonal matrix elements connect $(m_L,+1/2)$ to $(m_L+1,-1/2)$ with coefficientand the reverse spin change has $m_L+1$ replaced by $m_L-1$ in the corresponding ladder coefficient. These terms preserve $m_J$ but mix the two product states. To first order in $\beta /x$, the energy correction is simply $\beta m_L{m}_S$.

For the P state, $L=1$. Measured relative to its gross-structure centroid, the six low-field levels areThus the upper quartet starts at $\beta /2$ with slopes $+2,+2/3,-2/3,-2$ in a plot against $x$; the lower doublet starts at $-\beta$ with slopes $\pm 1/3$. Their zero-field separation is $3\beta /2$.

At high field all six product-state energies, retaining the term linear in $\beta$, areThe two zero-slope levels have different $m_J$ and no mutual matrix element, so their first-order degeneracy does not require an additional degenerate perturbation. At finite field they split at order ${\beta }^2/x$.

An exact interpolation makes the requested whole-field sketch unambiguous. The stretched levels areFor $m=\pm 1/2$, use product states $|m-1/2,+1/2⟩$ and $|m+1/2,-1/2⟩$. Their Hamiltonian block issoThese energies reduce to the quartet and doublet at zero field and to the six strong-field expressions above. Within a fixed $m_J$, the nonzero off-diagonal element prevents a crossing of the two branches. Levels in different $m_J$ sectors can cross without hybridizing. For the S state, $L=0$ and the spin-orbit interaction term vanishes, leavingthroughout this Hamiltonian's range of validity.

For electric dipole transitions at low field, $\Delta J=0,\pm 1$ and $\Delta m_J=0,\pm 1$ apply, with the usual exclusion of $J=0\to 0$. Each P state has opposite parity to the S state. The $P_{1/2}\to S_{1/2}$ transition has four allowed sublevel pairs. For $P_{3/2}\to S_{1/2}$, the two stretched upper states each have one destination and the other two each have two, giving six. Hence there are 10 allowed state-to-state channels at weak nonzero field. Relative to the two zero-field transition energies, their Zeeman shifts are

In the decoupled high-field limit, an electric dipole acts on orbital coordinates, so $\Delta m_S=0$. Each of the six P product states connects to the S state with the same spin and $m_L^{\prime }=0$. Thus there are 6 allowed channels in the limiting product-state description. Their photon energies areThis Paschen-Back transition-channel count is not the count of distinct frequencies: to first order in $\beta$ the two $m_L=0$ channels coincide, giving 5 distinct frequencies, with shifts $0,\pm (x+\beta /2),\pm (x-\beta /2)$. If the residual spin-orbit interaction splitting is unresolved there are the three usual orbital Zeeman groups.

The apparent loss of four channels is a change in eigenstate mixing, not a change in Hilbert-space dimension. At finite large field, four weak spin-forbidden Paschen-Back lines arise from small opposite-spin admixtures, with intensities vanishing as $(\beta /x{)}^2$. A generic finite-field calculation therefore retains ten nonzero channels; six carry finite strength in the complete-decoupling limit. Observation direction and polarization select which of these channels a detector actually receives.

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.

Let $X$ exchange the two equal-spin factors, so $X^{†}=X=X^{-1}$. For the unperturbed Hamiltonian operator, addition of angular momentum givesFor nonzero $\alpha$, different $J$ have distinct energy eigenvalues, while each eigenspace has the $2J+1$ values of $M$. The stated symmetry of the Clebsch-Gordan coefficients gives the exchange parity of two coupled identical spins:However, $D=(S_3^{(1)}-S_3^{(2)})/\hslash$ changes sign: $XDX=-D$. For any two basis vectors in the same $J$ eigenspace, insertion of $X^2=I$ therefore givesThus the entire projected perturbation $P_J\beta D{P}_J$ is zero, including its off-diagonal matrix elements. Degenerate perturbation theory now proves that every first-order-in-$\beta$ energy correction vanishes. This is first-order cancellation for an exchange-odd perturbation; checking only the individual diagonal expectations would not by itself settle a degenerate problem.

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.