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The ground level is listed as $3d^5({}^{6}S)4sa{}^7{S}_3$ at zero wavenumber. This supports the chromium configuration and term symbol; it is not a worked solution to the Cambridge exam.

Friedrich Hund organized complicated atomic spectral lines through relationships between the angular momenta and energy levels of many-electron configurations. In modern notation, a term symbol ${}^{2S+1}{L}_J$ records total electron spin $S$, total orbital angular momentum $L$ and coupled total angular momentum $J$. For example, $S=1$ permits three spin projections, whereas $S=0$ permits one. The classification reduces a large pattern of lines to relationships between fewer quantum states. This modern spin notation should be distinguished from the historical language of his original spectral analysis.^[ref]

Friedrich Hund's rules describe the usual ground-term ordering within an atomic electron configuration in the LS coupling regime. First maximize the total spin $S$ subject to the Pauli exclusion principle. Among the terms with that spin, maximize total orbital angular momentum $L$. Within the resulting term, the lowest level has $J=|L-S|$ for a less-than-half-filled open subshell and $J=L+S$ for a more-than-half-filled one. These are ordering rules, not universal replacements for diagonalizing an arbitrary atomic Hamiltonian operator.

For two equivalent $p$ electrons, parallel spins give $S=1$. Their spatial wavefunction must be antisymmetric; the highest allowed orbital projection with parallel spins is $1+0=1$, giving $L=1$. A $p^2$ subshell is less than half filled, so its lowest term symbol is ${}^3{P}_0$, followed within that term by ${}^3{P}_1$ and ${}^3{P}_2$. Closed subshells contribute zero total $L$ and $S$.

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 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.

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.

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.

The $4f^7$ shell carries a large spin (its ground atomic term under Hund's rules has $L=0,S=7/2$), and the $6s$ electron penetrates the inner shells. Its nonzero probability density at the atomic nucleus supplies a potentially large Fermi contact interaction with the nuclear magnetic dipole moment. The $6s$ shell here is singly occupied, so the obligatory paired-$6s^2$ cancellation is absent. Both stable europium isotopes have nonzero nuclear spin, as the spectrum below also establishes.

The term symbol ${}^8{D}_{11/2}$ specifies $S=7/2$, $L=2$, $J=11/2$. Its large $J$, combined with appreciable nuclear spin, permits a wide range of total angular momentum $F$ and hence a substantial total hyperfine structure span. For $I=5/2$, $F=3,\dots ,8$ and the span is $30|a|$. The penetrating $6s$ contribution and the available range of $F$ explain the large hyperfine structure; the term symbol alone does not determine its constant, which also depends on electronic recoupling, radial wavefunctions and configuration interaction.

An electron configuration specifies the occupied one-electron subshells and their populations, such as $1s^{2}2s^{2}2p^2$, without fixing how all orbital and spin angular momenta are coupled.

An atomic term groups states of a configuration with the same $L$ and $S$. An atomic fine-structure level additionally specifies $J$; spin-orbit coupling can split one term into several such levels. In zero external field, a level retains $2J+1$ magnetic substates.

The term symbol iswith letters $S,P,D,F,G,\dots$ denoting $L=0,1,2,3,4,\dots$. The superscript is the spin multiplicity. Without the $J$ subscript, ${}^{2S+1}L$ identifies an atomic term, rather than an individual fine-structure level.

The closed argon core contributes $L=S=0$. In the half-filled $3d^5$ shell, Hund's rules put one parallel-spin electron in each $m_{\ell }=-2,-1,0,1,2$ orbital. This gives $S_d=5/2$ and the orbital term $L_d=0$. Adding the parallel $4s$ electron, with $\ell =0$ and $s=1/2$, givesThe spin multiplicity is seven. The magnetic projection labels run $M_S,M_J=-3,-2,\dots ,3$, while $M_L=0$. The term symbol agrees with the observed neutral-chromium ground level.^[ref]

For chromium the configuration is $[\mathrm{Ar}]3d^{5}4s^1$. The half-filled $3d$ subshell has $S_d=5/2$ and $L_d=0$; aligning the $4s$ spin gives $S=3$, while $L=0$. Therefore the ground-state term symbol is $\mathrm{Cr}:{}^7{S}_3$. For cobalt, $[\mathrm{Ar}]3d^{7}4s^2$ has a closed $4s^2$ pair. The $d^7$ subshell has the same possible $L,S$ values as three $d$ holes; maximizing spin and then orbital angular momentum gives $S=3/2$, $L=3$. Because it is more than half full, $J=3+3/2=9/2$, giving $\mathrm{Co}:{}^4{F}_{9/2}$. For copper, $[\mathrm{Ar}]3d^{10}4s^1$ has a closed $d$ subshell, leaving $L=0$, $S=1/2$ and $\mathrm{Cu}:{}^2{S}_{1/2}$. These are neutral-atom terms, rather than the terms of the common metal ions. The tabulated zero-energy levels independently confirm all three assignments.^{[ref][ref][ref]}

The given electron configuration has an inert closed $[\mathrm{Xe}]$ core and two $4f$ electrons. The Aufbau principle identifies the occupied subshells; the free-ion term within the open $4f^2$ configuration is then selected by Hund's rules. A closed core contributes $L=S=0$.

Each $f$ electron has orbital angular momentum quantum number $l=3$. Place the two spins parallel, so $S=1$. The Pauli exclusion principle then requires distinct orbital projections. The largest allowed total projection is $M_L=3+2=5$, obtained from $m_l=3$ and $m_l=2$, giving the maximum allowed $L=5$. A larger $M_L=6$ would place two parallel-spin electrons in the same atomic orbital and is forbidden.

The $f$ subshell holds fourteen electrons; $f^2$ is less than half filled. The third of Hund's rules therefore selects $J=L-S=4$. HenceThe term symbol uses $2S+1=3$ and the spectroscopic letter $H$ for $L=5$. The praseodymium ion retains an orbital contribution, so a spin-only estimate would be inappropriate.

A chemical element of atomic number 59. Its trivalent ion has a $4f^2$ open-shell electron configuration. In LS coupling, Hund's rules select $L=5$, $S=1$, $J=4$, the ${}^3{H}_4$ term symbol. The corresponding free-ion Landé g-factor is $4/5$ and its effective paramagnetic moment is $(4/5)\sqrt{20}{\mu }_B\simeq 3.58{\mu }_B$.