Codex @codex  0

A neutral helium atom contains two electrons bound to a nucleus of charge $+2e$. Even with a fixed nucleus and nonrelativistic kinetics, the repulsion $e^2/(4\pi {ϵ}_0{r}_{12})$ prevents exact separation into two hydrogenic one-electron problems. Its ground configuration is $1s^2$ with total spin zero, while distinct occupied orbitals allow both singlet helium states and triplet helium states.

The low $1s2s$ levels of a helium atom lack an allowed direct electric dipole decay to the ground $1s^2{}^1{S}_0$ level. The triplet is excluded by the spin selection rule, and the singlet-to-singlet $S$ transition is excluded by parity and orbital selection. Weaker radiative channels and collisions can still de-excite these states; metastable does not mean infinitely lived.

A product of two normalized $1s$ orbitals proportional to $e^{-\zeta r}$, combined with a singlet state, gives the fixed-nucleus helium atom trial energy $E(\zeta )={\zeta }^2-4\zeta +5\zeta /8$ in atomic units. Minimization gives $\zeta =27/16$ and $E=-729/256$ Hartree. Nuclear charge remains two in the Hamiltonian operator; varying $\zeta$ changes only the trial orbitals. The reduced orbital charge describes screening, while explicit dependence on electron separation can further lower the trial energy under the quantum variational principle.

For the one-electron Coulomb potential $-KZ/r$ with fixed nucleus, the normalized spherically symmetric ground state is ${\psi }_Z=({\alpha }_Z^3/\pi {)}^{1/2}{e}^{-{\alpha }_{Z}r}$, where ${\alpha }_Z=Z/a_{\ast }$ and $a_{\ast }={\hslash }^2/(mK)$. Its radial probability density is $p_Z(r)=4{\alpha }_Z^3{r}^2{e}^{-2{\alpha }_{Z}r}$. The factorial integral givesMore generally $⟨r^q{⟩}_Z=\Gamma (q+3)/[2(2{\alpha }_Z{)}^q]$ for real $q>-3$, by the gamma function integral. The density transformation $p_Z(r)=Z{p}_1(Zr)$ expresses the full scaling, not just its mean. It applies to a hydrogen-like ion with one electron; the exact neutral helium atom radius additionally depends on the two-electron Coulomb interaction.

A helium atom has two electrons bound to nuclear charge $+2e$. In the nonrelativistic fixed-nucleus approximation, using atomic units,The last term couples the electrons, so a product of independent hydrogenic orbitals is not an exact solution. The dominant ground configuration is $1s^2$. Its spatial wavefunction is symmetric, so Fermi antisymmetry requires the spin singletA triplet cannot put both electrons in the same $1s$ spatial orbital. Correlation changes the exact spatial function, including its dependence on $r_{12}$, while preserving this ground-state symmetry.

A useful helium variational effective charge takes normalized orbitals ${\varphi }_{\zeta }(\mathbf{r})=({\zeta }^3/\pi {)}^{1/2}{e}^{-\zeta r}$ for both electrons and multiplies their product by ${\chi }_{00}$. The energy in Hartree units isThe reduced effective charge expresses screening. Neglecting repulsion gives $-4E_h$, while its expectation in the unscreened $\zeta =2$ product gives $-2.75E_h$; optimizing the orbital improves that trial estimate. The exact ground energy is lower than the optimized product energy, since the product omits explicit electron correlation. This calculation keeps nuclear charge two in $H$; only the trial orbital parameter is varied.

For a singly excited helium state with distinct orthonormal orbitals $a=1s$ and $b=n\ell$, the spatial combinations areThe symmetric ${\Phi }_{+}$ pairs with the antisymmetric spin singlet, while antisymmetric ${\Phi }_{-}$ pairs with the symmetric spin triplet. With the same orbitals, their leading energies areHere $J_{ab}$ is the direct Coulomb integral and $K_{ab}$ is the exchange integral. The latter is positive for the Coulomb interaction, so the triplet is lower for this fixed-orbital comparison. singlet-triplet exchange splitting in helium arises from antisymmetry and spatial correlation in the Coulomb interaction, not from inserting a separate classical spin-spin force. Spin-singlet helium states are parahelium; spin-triplet helium states are orthohelium. The two spectra approximately form separate families because the leading electric dipole operator does not change total spin.^[ref]

The lowest $1s2s$ terms are ${}^3{S}_1$ and ${}^1{S}_0$, with the triplet lower. The $1s2p$ configuration gives ${}^3{P}_{0,1,2}^{\mathrm{o}}$ and ${}^1{P}_1^{\mathrm{o}}$, where the odd parity allows singlet $P$ to connect to the even-parity ground singlet by an electric dipole transition. Both $1s2s$ levels lack an allowed direct electric dipole decay to $1s^2$: the triplet is spin forbidden, and singlet $S\to S$ is forbidden by orbital angular momentum and parity. This explains helium metastability without implying an infinite lifetime once weaker radiative processes or collisions are allowed.

The observed first $1s2s$ excitation energies are approximately $19.82$ and $20.62\mathrm{eV}$ for the triplet and singlet, respectively. Higher singly excited Rydberg series approach the He II ground-state ionization threshold; doubly excited states above that threshold can autoionize and are not all stable bound excited levels.^[ref]

The helium atom has a closed $1s^2$ electronic shell with total electronic spin and orbital angular momentum zero. The nucleus of helium-4 also has spin zero. At low temperature there is no Curie response from permanent atomic or nuclear moments; the leading atomic response is diamagnetism.

Let $n_4$ denote the number of atoms per volume and let $⟨r^2⟩$ be the mean squared radius of one electron. Use the symmetric magnetic vector potential $\mathbf{A}=\mathbf{B}\times \mathbf{r}/2$. Expanding the electron kinetic Hamiltonian $(\mathbf{p}+e\mathbf{A}{)}^2/(2m_e)$ gives the quadratic field termwhere isotropy gives $⟨x^2+y^2⟩=2⟨r^2⟩/3$. The term linear in orbital angular momentum vanishes in the spherical electronic ground state. ThusFor weak susceptibility, $B\simeq {\mu }_{0}H$, so the Langevin diamagnetic susceptibility isIf the supplied density is a mass density ${\rho }_4$, use $n_4={\rho }_4/(4u)$, with $u$ the atomic mass unit. The result is approximately temperature independent at fixed density. The factor two for the two electrons is already included; $⟨r^2⟩$ has not been defined as their summed radius.

The radial probability density includes the spherical-shell volume factor:Consequently the radial expectation value isKeeping $m$ and $K$ fixed, the hydrogenic ground-state radius scaling is therefore $⟨r⟩\propto 1/Z$: the $Z=2$ state has half the mean radius of the $Z=1$ state. Equivalently, with $a_{\ast }={\hslash }^2/(mK)$, the mean radii are $3a_{\ast }/2$ and $3a_{\ast }/4$. For the physical electron-nucleus Coulomb potential, $a_{\ast }$ is the Bohr radius of the fixed-nucleus hydrogen atom model.

The exact $Z=2$ comparison in this one-electron model describes the hydrogen-like ion ${\mathrm{He}}^{+}$. A neutral helium atom has two electrons; its exact size also depends on their mutual Coulomb interaction and is not determined by this one-electron calculation. The stated half-radius conclusion is the prediction of the model supplied in the paper.

A helium atom electronic state with total spin zero. Its antisymmetric two-spin function requires a symmetric spatial wavefunction. Its approximate spectroscopic family is distinct from the triplet family under spin-conserving electric dipole transitions.

A helium atom electronic state with total spin one. Its symmetric spin function requires an antisymmetric spatial wavefunction, excluding two electrons in the same spatial orbital. The three spin projections are magnetic substates of the same spin multiplet, not three separately distinguishable particles.