The Helium Atom

With two electrons the Schrödinger equation is no longer separable and no closed-form solution exists. Presented here in full is the hierarchy that replaces it — perturbative, variational, Hartree–Fock with the SCF algorithm that solves it, and the correlated expansions that define the exact answer — together with the complete converged wavefunction, the excited 1s·nl orbitals and their energies solved in the browser from the frozen-core potential, the term structure of para- and orthohelium, and a quantitative account of what the mean field omits.

Hartree 1928 · Hylleraas 1929 · Fock 1930 · Roothaan 1951 · Kato 1957 · Pekeris 1959 · Clementi & Roetti 1974 · SCF energy rebuilt in the browser on load
MODEL CLASS  SELF-CONSISTENT MEAN FIELD  Hartree–Fock; correlation not included
Z = 2 · 1s²
two electrons, closed shell
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EHF · recomputed live
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∫|φ|²dV · live
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⟨r⟩ vs HF lit · live
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Koopmans IE · live
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2¹P computed · live
A
active stage
Stage A · the Hartree–Fock 1s² density

The 1s² cloud

The point cloud is sampled from the self-consistent HF 1s density. The comparison shells place helium between hydrogen (Z = 1) and a bare Z = 2 nucleus: each electron partially screens the other, reducing the effective nuclear charge below 2.

Stage

Comparison shells
Cloud shows

Helium 1s² · Hartree–Fock

The orbital is of s type, so the density is exactly spherical. Drag to orbit; scroll to zoom.
Cloud samples16,000
Dot size1.0×
Slice contrast γ · Stages B, D0.45

1 · Formulation

Helium is the first atom for which the Schrödinger equation admits no closed-form solution. The repulsion term 1/r12 couples the two electronic coordinates and destroys the separability that makes hydrogen exact. What replaces the closed form is a hierarchy of constructions, each of which is stated here in full: first-order perturbation theory, the one-parameter variational solution, the Hartree–Fock equations with the algorithm by which this page solves them, and the correlated expansions that define the exact answer against which all of them are measured. Atomic units are used throughout (ℏ = me = e = 4πε₀ = 1; energies in hartree, 1 Ha = 27.211386 eV; lengths in a₀).

1.1Hamiltonian and constants of the motion

$$\hat H=\underbrace{-\tfrac12\nabla_1^2-\tfrac12\nabla_2^2-\frac{Z}{r_1}-\frac{Z}{r_2}}_{\hat H_0\ \text{— separable}}+\underbrace{\frac{1}{r_{12}}}_{\hat V\ \text{— couples them}},\qquad Z=2,\quad r_{12}=|\mathbf r_1-\mathbf r_2|$$
the infinite-nuclear-mass, non-relativistic, spin-free two-electron Hamiltonian; every term retained by this page appears here and nothing else does
$$\big[\hat H,\hat L^2\big]=\big[\hat H,\hat L_z\big]=\big[\hat H,\hat S^2\big]=\big[\hat H,\hat S_z\big]=\big[\hat H,\hat\Pi\big]=\big[\hat H,\hat P_{12}\big]=0,\qquad \hat{\mathbf L}=\hat{\mathbf L}_1+\hat{\mathbf L}_2,\quad \hat{\mathbf S}=\hat{\mathbf S}_1+\hat{\mathbf S}_2$$
1/r₁₂ depends only on the interelectronic distance, so total L, total S, parity Π and the exchange operator P̂₁₂ remain conserved — the states are labelled 2S+1LJ. Individual ℓ₁, ℓ₂ are not conserved; "1s2p" is a label for the dominant configuration, not an exact quantum number
$$\hat P_{12}\Psi(\mathbf x_1,\mathbf x_2)=-\Psi(\mathbf x_1,\mathbf x_2)\qquad\Longleftrightarrow\qquad \Psi(\mathbf x_2,\mathbf x_1)=-\Psi(\mathbf x_1,\mathbf x_2),\qquad \mathbf x=(\mathbf r,\sigma)$$
the Pauli antisymmetry requirement; it is a constraint on the admissible solutions, not a term in Ĥ

1.2Spin–space factorization: para- and orthohelium

Because Ĥ contains no spin operator, every eigenfunction factorizes into a spatial and a spin part. Antisymmetry of the product then forces the two to have opposite symmetry, which is what splits the helium spectrum into two nearly non-communicating systems.

$$\Psi=\underbrace{\Phi_{+}(\mathbf r_1,\mathbf r_2)}_{\text{symmetric}}\;\underbrace{\tfrac{1}{\sqrt2}\big(\alpha\beta-\beta\alpha\big)}_{S=0,\ \text{singlet · parahelium}} \qquad\text{or}\qquad \Psi=\underbrace{\Phi_{-}(\mathbf r_1,\mathbf r_2)}_{\text{antisymmetric}}\;\underbrace{\left\{\alpha\alpha,\ \tfrac{1}{\sqrt2}(\alpha\beta+\beta\alpha),\ \beta\beta\right\}}_{S=1,\ \text{triplet · orthohelium}}$$
the spin functions are exact; all the physics sits in Φ±. Φ− vanishes at r₁ = r₂, so triplet electrons avoid one another and lie lower — the Fermi hole, and the origin of Hund's first rule
$$\Phi_{-}=0\ \ \text{whenever}\ \ \mathbf r_1=\mathbf r_2\qquad\Longrightarrow\qquad E({}^3L)a purely statistical effect: no spin-dependent force appears in Ĥ. The measured 2³S–2¹S separation of 0.796 eV is exchange, not magnetism

The ground configuration 1s² admits only the symmetric spatial function, so the ground state is the singlet 11S0 and is unique. There is no 1s² triplet: that is the Pauli principle, stated as a theorem about this Hamiltonian rather than as a rule about boxes and arrows.

$$\Psi_{1s^2}(\mathbf x_1,\mathbf x_2)=\phi_{1s}(r_1)\,\phi_{1s}(r_2)\cdot\tfrac{1}{\sqrt2}\big(\alpha(1)\beta(2)-\beta(1)\alpha(2)\big) \;=\;\frac{1}{\sqrt2}\begin{vmatrix}\phi_{1s}\alpha(1)&\phi_{1s}\beta(1)\\[2pt]\phi_{1s}\alpha(2)&\phi_{1s}\beta(2)\end{vmatrix}$$
the closed-shell state as a single Slater determinant. This form — one spatial orbital, doubly occupied — is the entire content of the Hartree–Fock approximation for helium; §1.11 quantifies what it costs

1.3Why there is no closed form

Separation of variables fails because 1/r12 is not a sum of one-particle terms. Expanding it in the natural basis makes the obstruction explicit: the coupling connects every angular momentum on one coordinate to every other on the second, so no finite product ansatz can be exact.

$$\frac{1}{r_{12}}=\sum_{k=0}^{\infty}\frac{r_{<}^{\,k}}{r_{>}^{\,k+1}}\,P_k(\cos\theta_{12}) =\sum_{k=0}^{\infty}\sum_{q=-k}^{k}\frac{4\pi}{2k+1}\frac{r_{<}^{\,k}}{r_{>}^{\,k+1}}\,Y_k^{q*}(\hat r_1)\,Y_k^{q}(\hat r_2)$$
the multipole (Laplace) expansion; r< = min(r₁,r₂), r> = max(r₁,r₂). Only the k = 0 term survives for two s electrons, which is what makes the integrals of §1.4 and §1.8 elementary
$$\Psi\big(r_1,r_2,r_{12}\big)\;\underset{R\to0}{=}\;\sum_{k,m}R^{\,k}\big(\ln R\big)^{m}\,\psi_{km}(\alpha,\theta_{12}),\qquad R=\sqrt{r_1^2+r_2^2}$$
the Fock expansion (1954): the exact wavefunction is not analytic at the triple-coincidence point — logarithmic terms enter at R². No product of smooth one-electron functions can reproduce this, which is the structural reason orbital methods converge slowly
$$\left.\frac{\partial\langle\Psi\rangle}{\partial r_i}\right|_{r_i=0}=-Z\,\Psi\big|_{r_i=0}, \qquad \left.\frac{\partial\langle\Psi\rangle}{\partial r_{12}}\right|_{r_{12}=0}=+\tfrac12\,\Psi\big|_{r_{12}=0}$$
the Kato cusp conditions (1957) — exact constraints on any true solution. The second involves r₁₂ explicitly and cannot be satisfied by a determinant of one-electron orbitals; the first is violated by any Gaussian basis, as §5 records

1.4Solution I · first-order perturbation theory

Treat 1/r12 as a perturbation on the exactly solvable Ĥ₀. The zeroth-order state is a product of two hydrogenic 1s orbitals at Z = 2, and the first-order correction is a single closed-form integral.

$$\Phi^{(0)}=\phi^{Z}_{1s}(r_1)\,\phi^{Z}_{1s}(r_2),\qquad \phi^{Z}_{1s}(r)=\sqrt{\frac{Z^{3}}{\pi}}\,e^{-Zr},\qquad E^{(0)}=2\left(-\frac{Z^{2}}{2}\right)=-Z^{2}=-4.0000\ \text{Ha}$$
the independent-electron limit; it overbinds by 1.10 Ha because the electrons are not made to repel at all
$$E^{(1)}=\Big\langle \Phi^{(0)}\Big|\frac{1}{r_{12}}\Big|\Phi^{(0)}\Big\rangle=\frac{Z^{6}}{\pi^{2}}\!\int\!\!\int\frac{e^{-2Z(r_1+r_2)}}{r_{12}}\,d^{3}r_1 d^{3}r_2=\frac{5Z}{8}=1.2500\ \text{Ha}$$
evaluated with the k = 0 term of the multipole expansion; the classic 5Z/8 result
$$E\simeq E^{(0)}+E^{(1)}=-Z^{2}+\frac{5Z}{8}=-2.7500\ \text{Ha}\qquad\text{(exact: }-2.9037\text{ Ha; error }+0.1537\ \text{Ha}=4.18\ \text{eV})$$
5.3 % high. The full 1/Z expansion reads E = −Z² + 0.62500 Z − 0.157666 + 0.008699/Z − 0.000889/Z² …, giving −2.90354 Ha at Z = 2 through fourth order. It converges to the exact value, but it is not variational and is not used by this page

1.5Solution II · the one-parameter variational solution

The perturbative orbital is wrong in an obvious way: each electron sees a partially screened nucleus, not the bare Z = 2. Promote the exponent to a variational parameter Z′ and minimise. This is the cheapest quantitative theory of screening, and it produces the effective charge quoted in the live panel.

$$\Phi(Z')=\frac{Z'^{3}}{\pi}e^{-Z'(r_1+r_2)},\qquad E(Z')=\big\langle\hat T\big\rangle+\big\langle \hat V_{ne}\big\rangle+\big\langle \hat V_{ee}\big\rangle=Z'^{2}-2ZZ'+\frac{5}{8}Z'$$
⟨T̂⟩ = Z′², ⟨V̂ne⟩ = −2ZZ′, ⟨V̂ee⟩ = 5Z′/8; each term is elementary for 1s Slater functions
$$\frac{dE}{dZ'}=0\quad\Longrightarrow\quad Z'_{\min}=Z-\frac{5}{16}=\frac{27}{16}=1.6875, \qquad E_{\min}=-\Big(Z-\frac{5}{16}\Big)^{2}=-2.847656\ \text{Ha}$$
the screening constant 5/16 = 0.3125 is exactly Slater's empirical 0.30 for a 1s partner. Zeff = 1.6875 is the "Slater rule" figure reported in §4; the variational theorem guarantees Emin ≥ Eexact, and it is, by 0.056 Ha
$$\langle r\rangle_{Z'}=\frac{3}{2Z'}=0.8889\ a_0\qquad\text{versus}\qquad \langle r\rangle_{\rm HF}=0.9273\ a_0,\qquad \langle r\rangle_{\rm H\,1s}=1.5\ a_0$$
the single-exponent orbital is slightly too compact; relaxing the radial shape, not merely its scale, is exactly what Hartree–Fock does next

1.6Solution III · the Hartree–Fock equations

Hartree–Fock keeps the determinantal form of §1.2 but lets the radial function be arbitrary, minimising the energy over all normalised φ. Applying the variational principle to the determinant yields a nonlinear one-particle eigenvalue equation in which the potential depends on its own solution.

$$E_{\rm HF}[\phi]=2h+J,\qquad h=\int\!\phi^{*}\Big(-\tfrac12\nabla^{2}-\frac{Z}{r}\Big)\phi\,d^{3}r,\qquad J=\int\!\!\int\frac{|\phi(\mathbf r_1)|^{2}|\phi(\mathbf r_2)|^{2}}{r_{12}}\,d^{3}r_1d^{3}r_2$$
the closed-shell energy functional. For a single doubly occupied orbital the exchange integral equals the Coulomb integral, K = J, and the antisymmetry correction cancels exactly half of the double counting — hence one J, not two
$$\hat f\,\phi_{1s}=\varepsilon_{1s}\,\phi_{1s},\qquad \hat f=-\tfrac12\nabla^{2}-\frac{Z}{r}+\hat v_{\rm HF}[\phi],\qquad \hat v_{\rm HF}[\phi](\mathbf r_1)=\int\frac{|\phi(\mathbf r_2)|^{2}}{r_{12}}\,d^{3}r_2$$
the self-consistent field equation. The operator is built from its own eigenfunction, so it must be solved by iteration — the whole content of §1.9. For helium the exchange operator cancels one of the two Coulomb terms, leaving the single Hartree term above
$$\varepsilon_{1s}=h+J\qquad\Longrightarrow\qquad E_{\rm HF}=2\varepsilon_{1s}-J\ \neq\ 2\varepsilon_{1s}$$
orbital energies are not additive: summing them double-counts the repulsion. EHF = −2.86165 Ha, ε₁ₛ = −0.91796 Ha, so J = 2ε₁ₛ − EHF = 1.025752 Ha = 27.913 eV — the energetic cost of holding two electrons in the same orbital
$$\mathrm{IE}\;\simeq\;-\varepsilon_{1s}=0.91796\ \text{Ha}=24.98\ \text{eV} \qquad\text{(Koopmans);}\qquad \mathrm{IE}=E(\mathrm{He}^{+})-E(\mathrm{He})$$
Koopmans' theorem identifies −ε with the ionization energy by holding the remaining orbital frozen. Since He⁺ is hydrogenic and exact at −2 Ha, the ΔSCF value −2 − (−2.86165) = 0.86165 Ha = 23.45 eV; the measured 24.587 eV lies between the two, because relaxation lowers the ion and correlation lowers the neutral more

1.7Algebraic reduction · the Roothaan equations

The differential equation above is turned into linear algebra by expanding the unknown orbital in a fixed basis. This is the form actually solved for this page.

$$\phi_{1s}(r)=\sum_{\mu=1}^{M}C_{\mu}\,g_{\mu}(r),\qquad g_{\mu}(r)=e^{-\alpha_{\mu}r^{2}},\qquad M=11$$
a primitive s-type Gaussian basis, even-tempered from α = 0.09 to α = 1046.27. Gaussians are chosen because every integral in §1.8 is closed-form (Boys 1950)
$$\mathbf{F}\mathbf{C}=\mathbf{S}\mathbf{C}\,\varepsilon,\qquad F_{\mu\nu}=H^{\rm core}_{\mu\nu}+\sum_{\lambda\sigma}P_{\lambda\sigma}\Big[(\mu\nu|\lambda\sigma)-\tfrac12(\mu\lambda|\nu\sigma)\Big],\qquad P_{\lambda\sigma}=2\,C_{\lambda}C_{\sigma}$$
the Roothaan–Hall equations. The basis is non-orthogonal, so the overlap matrix S appears on the right: this is a generalized eigenvalue problem, not a plain one
$$E_{\rm HF}=\tfrac12\sum_{\mu\nu}P_{\mu\nu}\big(H^{\rm core}_{\mu\nu}+F_{\mu\nu}\big),\qquad \sum_{\mu\nu}C_{\mu}S_{\mu\nu}C_{\nu}=1$$
the total energy from the converged density matrix, and the normalization constraint carried through every iteration

1.8The integrals, in closed form

For s-type Gaussians on a common centre every required integral reduces to a Gamma function. Nothing here is tabulated or fitted; with these four expressions and the exponents of §1.10 the calculation is reproducible from scratch.

$$S_{\mu\nu}=\int g_\mu g_\nu\,d^3r=\left(\frac{\pi}{p}\right)^{3/2},\qquad T_{\mu\nu}=-\tfrac12\!\int g_\mu\nabla^{2}g_\nu\,d^3r=\frac{3\alpha_\mu\alpha_\nu}{p}\left(\frac{\pi}{p}\right)^{3/2},\qquad p=\alpha_\mu+\alpha_\nu$$
overlap and kinetic energy
$$V_{\mu\nu}=-Z\!\int\frac{g_\mu g_\nu}{r}\,d^3r=-\frac{2\pi Z}{p},\qquad H^{\rm core}_{\mu\nu}=T_{\mu\nu}+V_{\mu\nu}$$
nuclear attraction; the 1/r singularity is integrable and the result is elementary
$$(\mu\nu|\lambda\sigma)=\int\!\!\int\frac{g_\mu(\mathbf r_1)g_\nu(\mathbf r_1)\,g_\lambda(\mathbf r_2)g_\sigma(\mathbf r_2)}{r_{12}}\,d^3r_1d^3r_2 =\frac{2\pi^{5/2}}{p\,q\,\sqrt{p+q}},\qquad p=\alpha_\mu+\alpha_\nu,\ \ q=\alpha_\lambda+\alpha_\sigma$$
the two-electron repulsion integral — the reason Gaussians displaced Slater functions in electronic-structure practice. M = 11 gives 11⁴ = 14 641 raw integrals, reduced to 2 211 by the eightfold permutational symmetry (μν|λσ) = (νμ|λσ) = (λσ|μν)…

1.9The algorithm · self-consistent field

The following is the complete procedure by which the orbital rendered on the stage was obtained. It is deterministic, and converges for helium from a core-Hamiltonian guess to better than 10−9 Ha in nine iterations, and is stationary to 10−12 Ha thereafter.

  1. Input. Z = 2; exponents {αμ}, μ = 1…11; thresholds τE = 10−10 Ha, τD = 10−8.
  2. One-electron integrals. Build S, T, V from §1.8; set Hcore = T + V. — O(M²)
  3. Two-electron integrals. Build (μν|λσ) once, exploiting eightfold symmetry; hold in memory. — O(M⁴), 2 211 unique values
  4. Orthogonalization. Diagonalize S = UsU†; form X = S−1/2 = Us−1/2U†. Discard eigenvectors with si < 10−6 to suppress near-linear dependence of the basis.
  5. Initial guess. F(0) = Hcore, i.e. neglect the repulsion entirely; equivalently P(0) = 0.
  6. Transform. F′ = X†FX. — converts the generalized problem to a standard one
  7. Diagonalize. F′C′ = C′ε; back-transform C = XC′; take the eigenvector of lowest ε as the occupied 1s orbital (the aufbau selection).
  8. Density. P = 2CC† over the one occupied orbital.
  9. New Fock matrix. F = Hcore + G[P] with Gμν = Σ Pλσ[(μν|λσ) − ½(μλ|νσ)].
  10. Energy. E = ½ Σ Pμν(Hcoreμν + Fμν).
  11. Convergence test. Stop when |E(k) − E(k−1)| < τE and RMS(FPS − SPF) < τD — the commutator residual, which vanishes identically at self-consistency. Otherwise return to step 6.
  12. Acceleration (optional). Extrapolate F by DIIS over the last 6 error vectors e(k) = X†(FPS − SPF)X; for helium plain iteration already converges monotonically and DIIS is not required.
  13. Output. {Cμ}, ε1s, EHF; the coefficients are the numbers tabulated in §1.10 and evaluated live by this page.

Convergence trace, hartree: E(1) = −2.7499726 · E(2) = −2.8602713 · E(3) = −2.8616144 · E(4) = −2.8616472 · E(5) = −2.8616485 · E(6…10) = −2.86164852 (stationary to 10−12 Ha); ε1s = −0.917948 Ha. Two features of the trace are checks rather than coincidences. The first diagonalization of Hcore returns ε = −1.999975 Ha, the hydrogenic −Z²/2 of a bare Z = 2 nucleus reproduced to 1×10−5 by the finite basis; and the first energy built from that density, −2.74997 Ha, is precisely the first-order perturbation result −Z² + 5Z/8 of §1.4, which the same initial orbital defines. The remaining 0.112 Ha is relaxation of the orbital shape under the screened potential.

1.10The converged wavefunction, in full

The two-electron state is the determinant of §1.2 built from the single spatial function below. It is spherical (ℓ = 0), nodeless, and everywhere of one sign; the point cloud on the stage is sampled from |φ1s|² and the radial curve from P(r).

$$\Psi_{1s^2}(\mathbf r_1,\mathbf r_2)=\phi_{1s}(r_1)\phi_{1s}(r_2)\cdot\tfrac{1}{\sqrt2}(\alpha\beta-\beta\alpha), \qquad \phi_{1s}(r)=\sum_{\mu=1}^{11}d_{\mu}\,e^{-\alpha_{\mu}r^{2}}$$
the complete solution rendered by this page — every figure is evaluated from these 22 numbers and nothing else
The converged Hartree–Fock 1s orbital · exponents αμ (a₀−2) and contraction coefficients dμ, normalised so ∫|φ|²dV = 1
μαμdμμαμdμ
10.0900000−0.00376985724.744800−0.1372320
20.2295000−0.06452000863.099200−0.0836482
30.5852250−0.186417009160.90300−0.0629014
41.4923200−0.2595040010410.30200−0.0207522
53.8054300−0.25713200111046.2700−0.0426115
69.7038400−0.19731200
$$\rho(r)=2\big|\phi_{1s}(r)\big|^{2},\qquad P(r)=4\pi r^{2}\big|\phi_{1s}(r)\big|^{2},\qquad \int_0^\infty P(r)\,dr=1$$
the total electron density and the one-electron radial distribution; P(r) is the curve plotted in Stage B against the Z = 1 and Z = 2 hydrogenic limits
$$\big\langle r^{k}\big\rangle=4\pi\sum_{\mu\nu}d_{\mu}d_{\nu}\int_{0}^{\infty}\! r^{k+2}e^{-p r^{2}}dr =2\pi\sum_{\mu\nu}\frac{d_{\mu}d_{\nu}}{p^{(k+3)/2}}\,\Gamma\!\left(\frac{k+3}{2}\right),\qquad p=\alpha_\mu+\alpha_\nu$$
all radial moments in closed form. This gives ⟨r⟩ = 0.92733 a₀, ⟨r²⟩ = 1.18487 a₀², ⟨1/r⟩ = 1.68716 a₀−1; the page recomputes ⟨r⟩ by quadrature at run time as a check (§4)
$$Z_{\rm eff}=\frac{3}{2\langle r\rangle}=1.618,\qquad \frac{\langle r\rangle_{\rm H}}{\langle r\rangle_{\rm He}}=\frac{1.500}{0.9273}=1.618,\qquad \left.\frac{d\phi_{1s}}{dr}\right|_{r=0}=0\ \neq\ -Z\phi_{1s}(0)$$
the screening result, and the one structural defect of the basis: a sum of Gaussians is flat at the origin and cannot reproduce the Kato cusp of §1.3. The virial theorem, exact for any Coulomb eigenstate, holds to 3×10⁻⁵: ⟨T⟩ = 2.861562 Ha, ⟨V⟩ = −5.723211 Ha, −⟨V⟩/⟨T⟩ = 2.000030

1.11What remains · correlation and the exact solution

Hartree–Fock is variationally optimal within the determinantal form, so its residual error is definite and is called the correlation energy. Recovering it requires functions that depend on r12 explicitly, or an expansion in many determinants.

$$E_{\rm corr}\equiv E_{\rm exact}-E_{\rm HF}=-2.903724-(-2.861680)=-0.042044\ \text{Ha}=-1.1441\ \text{eV}$$
1.4 % of the total energy, but 4.7 % of the ionization energy and larger than most chemical bond-energy differences — which is why mean-field accuracy is insufficient for thermochemistry
$$\Psi_{\rm Hyl}=e^{-\zeta s}\sum_{l,m,n}c_{lmn}\,s^{\,l}t^{\,m}u^{\,n},\qquad s=r_1+r_2,\quad t=r_1-r_2,\quad u=r_{12}\ \ (m\ \text{even})$$
the Hylleraas expansion (1929): u = r₁₂ enters as a coordinate, so the electron–electron cusp is representable. Three terms already give −2.9024 Ha — better than Hartree–Fock at the HF limit
$$\Psi_{\rm CI}=\sum_{i}c_i\,\Phi_i,\qquad E_{\rm CI}\xrightarrow[\ \text{complete basis}\ ]{\ }E_{\rm exact},\qquad \Delta E\sim (L+1)^{-3}$$
configuration interaction converges as the inverse cube of the highest angular momentum retained — the slow partial-wave convergence caused by the cusp; explicitly correlated (R12/F12) methods restore rapid convergence
The ladder of solutions · non-relativistic, infinite nuclear mass, ground-state total energy
methodwavefunctionE (Ha)error vs exact
independent electronsφZ=2φZ=2, §1.4−4.000000−1.0963
1st-order perturbationsame, ⟨1/r₁₂⟩ added−2.750000+0.1537
variational Z′ = 27/16screened 1s, §1.5−2.847656+0.0561
Hartree–Fock (this page)11-Gaussian SCF, §1.10−2.861650+0.0421
Hartree–Fock limitnumerical, complete basis−2.861680+0.0420
Hylleraas, 3 termsexplicit r₁₂, §1.11−2.902400+0.0013
Pekeris 1959, 1078 termsexplicit r₁₂−2.9037244< 10−7
Drake 2006 (reference)doubled basis, 2 358 terms−2.90372437710−19 Ha
experiment—−2.903572relativity + QED + finite mass

The last two rows differ by 1.5×10−4 Ha (4.1 meV). That residue is not an error in the solution of the Schrödinger equation; it is the part of the physics the Schrödinger equation does not contain — relativistic kinematics, the Lamb shift, and the finite mass of the nucleus. §5 itemises it.

2 · Excited states and the energy levels

Above the ground configuration the spectrum is that of a screened one-electron atom: promoting one electron to nl leaves the other in 1s, so the excited electron sees a nucleus of charge 2 screened almost completely to 1 at large r. The levels therefore lie close to a hydrogenic ladder converging on He⁺, shifted by a quantum defect and split into singlet and triplet by exchange.

2.1Configuration energies and the exchange splitting

$$\Phi_{\pm}=\frac{1}{\sqrt2}\Big[\phi_{1s}(\mathbf r_1)\phi_{nl}(\mathbf r_2)\pm\phi_{nl}(\mathbf r_1)\phi_{1s}(\mathbf r_2)\Big], \qquad E_{\pm}=\varepsilon_{1s}+\varepsilon_{nl}+J_{1s,nl}\pm K_{1s,nl}$$
+ for the singlet (parahelium), − for the triplet (orthohelium). J is the classical repulsion of the two charge clouds; K has no classical counterpart
$$J_{1s,nl}=\!\int\!\!\int\frac{|\phi_{1s}(1)|^{2}|\phi_{nl}(2)|^{2}}{r_{12}},\qquad K_{1s,nl}=\!\int\!\!\int\frac{\phi_{1s}^{*}(1)\phi_{nl}^{*}(2)\,\phi_{nl}(1)\phi_{1s}(2)}{r_{12}},\qquad \Delta E_{\rm ST}=2K$$
the singlet–triplet separation is twice the exchange integral. Measured: 2K(1s2s) = 0.796 eV → K = 0.398 eV; 2K(1s2p) = 0.254 eV → K = 0.127 eV, smaller because 2p has less overlap with 1s
$$E_{nl}\simeq-2-\frac{1}{2\big(n-\delta_{l}\big)^{2}}\ \text{Ha},\qquad \delta:\ {}^1S\ 0.140,\ \ {}^3S\ 0.297,\ \ {}^1P\ -0.012,\ \ {}^3P\ 0.068,\ \ {}^{1,3}D\ 0.002$$
the quantum-defect form. δ measures penetration of the 1s core: s orbitals penetrate most and are depressed most; by ℓ = 2 the defect is negligible and the levels are hydrogenic to four figures

2.2The level table

Helium levels, energy above the 11S0 ground state (NIST ASD; fine structure averaged where unresolved)
configurationtermE (eV)Etotal (Ha)note
1s²1 ¹S₀0.00000−2.903572ground state; unique, no triplet partner
1s2s2 ³S₁19.81961−2.175215metastable, τ = 7 870 s (M1); the lowest triplet
1s2s2 ¹S₀20.61577−2.145956metastable, τ = 19.7 ms (two-photon)
1s2p2 ³P20.96409−2.1331562³S→2³P at 1083.0 nm — the helium laser line
1s2p2 ¹P₁21.21802−2.123824resonance line, 58.43 nm; τ = 0.56 ns
1s3s3 ³S₁22.71847−2.068684
1s3s3 ¹S₀22.92032−2.061266
1s3p3 ³P23.00707−2.058078
1s3d3 ³D23.07365−2.055631δ ≈ 0.003 — hydrogenic to four figures
1s3d3 ¹D₂23.07407−2.0556162K = 0.4 meV; exchange nearly extinct at ℓ = 2
1s3p3 ¹P₁23.08702−2.05514053.70 nm
1s∞limit24.58739−2.000000He⁺ 1s, exact hydrogenic −Z²/2 = −2 Ha
2s2p¹P°60.15−0.693100doubly excited, above the limit: autoionizes, Γ = 38 meV (τ = 17 fs)
He²⁺ + 2e⁻—79.005170.000000second ionization 54.418 eV = 2 Ha exactly

2.3Selection rules and the two spectra

$$\Delta S=0,\qquad \Delta L=\pm1,\qquad \Delta l=\pm1,\qquad \text{parity must change}$$
electric-dipole rules for LS coupling. ΔS = 0 forbids singlet↔triplet radiation, so parahelium and orthohelium form two almost disjoint spectra — which is why they were taken for two distinct gases until 1926

The prohibition is not absolute: spin–orbit coupling mixes the two systems weakly, and the 2³S₁ → 1¹S₀ magnetic-dipole decay proceeds at 1.27×10⁻⁴ s⁻¹, a lifetime of 7 870 s. That is the longest-lived neutral excited state known in any atom, and it is long precisely because the transition violates a rule that the Hamiltonian of §1.1 enforces exactly.

Two further consequences of the level structure are visible in the table. First, the doubly excited 2s2p configuration lies above the single-ionization limit: it is a bound state of the two-electron problem embedded in a continuum, and it decays by autoionization in 17 fs rather than by radiation. Such resonances are outside any single-configuration description entirely. Second, the ionization energy 24.587 eV is the largest of any element — the consequence of a filled 1s shell in a Z = 2 field with only 5/16 of a unit of screening to work against.

2.4The frozen-core construction solved by this page

Stages C, D and E do not read the table above. They solve for the excited orbital directly, in the one approximation that follows naturally from §1: hold the 1s core at the converged Hartree–Fock function of §1.10, and let the outer electron move in the field of the nucleus plus that frozen charge cloud. This is the frozen-core Hartree problem. It is a genuine one-electron eigenvalue problem in a non-Coulomb central potential, and it is what produces a quantum defect from first principles rather than as a fitted parameter.

$$V_{\rm dir}(r)=\int\frac{\big|\phi_{1s}(\mathbf r')\big|^{2}}{|\mathbf r-\mathbf r'|}\,d^{3}r' =\frac{1}{r}\int_{0}^{r}\!P_{1s}(r')\,dr'+\int_{r}^{\infty}\!\frac{P_{1s}(r')}{r'}\,dr', \qquad P_{1s}=4\pi r^{2}\big|\phi_{1s}\big|^{2}$$
the Hartree potential of one core electron, obtained from the multipole expansion of §1.3 with only k = 0 surviving because the core is an s orbital. It is built at run time from the 22 numbers of §1.10
$$-\frac12\frac{d^{2}u_{nl}}{dr^{2}}+\underbrace{\left[-\frac{Z}{r}+V_{\rm dir}(r)+\frac{l(l+1)}{2r^{2}}\right]}_{V_{\rm eff}(r)}u_{nl}=\varepsilon_{nl}\,u_{nl}, \qquad u_{nl}=r\,R_{nl},\qquad \int_{0}^{\infty}\!u_{nl}^{2}\,dr=1$$
the radial equation for the outer electron. Veff → −2/r as r → 0 and → −1/r as r → ∞: the outer electron sees the bare nucleus when it penetrates the core and a singly charged ion when it does not. That crossover is the quantum defect
$$\varepsilon_{nl}=-\frac{1}{2\nu_{nl}^{2}},\qquad \nu_{nl}=n-\delta_{l}, \qquad \delta_{l}\ \text{large for}\ l=0,\ \ \delta_{l}\to0\ \text{for}\ l\ge2$$
the Rydberg–Ritz form. δ is not fitted here: ν is read off the computed eigenvalue, and δ = n − ν follows. The live panel reports both
$$E\big({}^{1,3}L\big)=E_{\rm core}+\varepsilon_{nl}\pm K_{1s,nl},\qquad K_{1s,nl}=\frac{1}{2l+1}\int_{0}^{\infty}\!\!\int_{0}^{\infty}\frac{r_{<}^{\,l}}{r_{>}^{\,l+1}}\, u_{1s}u_{nl}(r_1)\,u_{1s}u_{nl}(r_2)\,dr_1dr_2$$
the Slater exchange integral Gl(1s,nl)/(2l+1), evaluated by the same quadrature. Only k = l contributes, because the core carries l = 0. Ecore = −2 Ha, the exact He⁺ energy to which the whole ladder converges

The equation is integrated by the Numerov method on a linear grid to r = 90 a₀, with the outward and inward solutions matched at the outermost classical turning point; eigenvalues are the energies at which the logarithmic derivatives agree. The node count n − ℓ − 1 identifies which root is which, so no level is assigned by hand. Every number in the panels of Stages C, D and E comes out of this procedure.

2.5What the frozen core gets right, and where it fails

Computed term energies above the 11S0 ground state, against NIST ASD. Computed values are those the page produces on load; the reference row for the ground state is experimental and is used as the zero
configurationδ computed³L computed³L measured¹L computed¹L measured
1s2s0.22319.6619.82020.8920.616
1s2p0.01920.9020.96421.3421.218
1s3s0.21922.6822.71822.9822.920
1s3p0.02222.9923.00723.1223.087
1s3d0.000323.07423.073723.07723.0741
1s4d0.000523.73623.736123.73823.7363
1s4f0.000023.73723.736123.73723.7361

Three things follow from the table, and they are the reason the stage is worth solving rather than tabulating. The quantum defect emerges with the right magnitude and the right ℓ dependence — 0.22 for s, 0.02 for p, 3×10−4 for d — from nothing but the shape of the core charge; by ℓ = 2 the outer electron no longer reaches the core and the level is hydrogenic to four figures, exactly as §2.1 asserts. The ordering E(³L) < E(¹L) comes out right for every configuration, and it comes out of the sign of a single integral rather than from any spin-dependent term in Ĥ.

The failure is equally definite and is concentrated in the exchange splitting of the low states: 2K(1s2s) computes to 1.23 eV against a measured 0.796 eV, and 2K(1s2p) to 0.44 eV against 0.254 eV. Freezing the core is the reason. The real core relaxes outward when an electron is promoted, and it polarizes in the field of the outer electron; both effects reduce the 1s–nl overlap that K measures, and neither is available to a fixed φ1s. The error is largest for 2s, whose orbital still overlaps the core substantially, and vanishes with the overlap itself: by 1s3d the computed splitting is below a millielectronvolt, as measured. §5 records this alongside the other omissions.

3 · Stages

Stagewhat it showscaveat
A · The 1s² clouda point cloud sampled from the converged HF density of §1.10, with optional ⟨r⟩ comparison shells at Z = 1, Zeff = 1.618 and Z = 2points are position-measurement outcomes, not trajectories; the two electrons occupy the same spatial orbital, so the cloud is their common one-electron density and shows no correlation hole
B · Screening and the cuspa section through φ1s in the x–z plane and P(r) = 4πr²φ² against the two hydrogenic limitsthe orbital is of one sign and nodeless; the centre is smooth because a Gaussian basis has zero derivative at r = 0 and cannot satisfy the Kato condition of §1.3
C · Excited orbitals|ψ|² of the outer electron in any 1s·nl configuration to n = 4, sampled as a three-dimensional point cloud and coloured by the sign of ψ, with the 1s core available as an overlay. n, ℓ, m, the term (¹L or ³L) and the angular basis are all selectablethe orbital is the frozen-core solution of §2.4, not a two-electron wavefunction: the picture is of one electron in the field of the nucleus and a fixed core. The lobed forms are one choice of real basis; the complex m eigenstate is azimuthally symmetric
D · Nodes and P(r)a signed section through ψ in the x–z plane for the selected 1s·nl, with the radial distribution P(r) = u²(r) drawn against the 1s core on the same scalen − ℓ − 1 radial and ℓ angular nodes, counted at run time from the integrated solution; the outer orbital is one to two orders of magnitude more extended than the core, which is what the screened picture of §2 rests on
E · Term diagramthe computed singlet and triplet ladders arranged by ℓ, with the ionization limit, the 58.43 nm resonance line and the 1083 nm helium laser line markedevery level is solved on load by §2.4, so the exchange splittings of the low terms are systematically too large — see §2.5. Fine structure is not resolved and J is not shown
F · Energy and correlationthe 1s binding energies of H, He (HF) and He⁺ with their ionization arrows, and the mean-field energy deficitε1s is a Koopmans estimate with the remaining orbital frozen; the 1.14 eV correlation energy of §1.11 is not represented in the rendered density, only quantified in the panel

Select a configuration:

4 · Verification computed live

Only the 11 exponents and 11 coefficients of §1.10 are stored. Every quantity below — including the total energy, which requires the full 11⁴ sum of two-electron integrals — is rebuilt from them in the browser on load, using the closed forms of §1.8. Nothing in this table is a transcribed literature value.

Testthis pagereferencestatus
EHF rebuilt from the 22 tabulated numbers (live)…−2.861649 Ha; HF limit −2.861680 (deviation 3×10⁻⁵)…
ε₁ₛ = h + J, same coefficients (live)…−0.917948 Ha, §1.6…
Virial ratio −⟨V⟩/⟨T⟩ (live)…2 exactly, for any Coulomb eigenstate…
⟨r⟩: Γ-function closed form vs quadrature (live)…independent routes to §1.10, must agree…
Normalization ∫|φ₁ₛ|²dV (live)…unity (Born normalization)…
⟨r⟩ of the 1s orbital (live)…0.9273 a₀ (Clementi & Roetti 1974)…
Koopmans ionization energy −ε₁ₛ (live)…24.98 eV (HF); 24.587 eV (measured)…
Effective charge from ⟨r⟩ = 1.5/Zeff (live)…1.618; Slater rule gives 1.688…
Frozen-core eigenvalue ε(2p), §2.4 (live)…ν ≈ 1.98, i.e. δ(p) ≈ 0.02 — a nearly hydrogenic p level…
Quantum defect δ(ns) − δ(nd), computed (live)…≈ 0.22 against ≈ 0.0003: penetration is an ℓ effect…
Radial node count for 2s / 3p / 4f (live)…1 / 1 / 0, from n − ℓ − 1…
1s3d term energy above ground, computed (live)…23.074 eV (NIST); hydrogenic to four figures…
Exchange splitting 2K(1s2s), computed (live)…0.796 eV measured — frozen core overshoots, §2.5…
Orthogonality ⟨1s|2s⟩ of core and outer orbital (live)…the two are eigenfunctions of the same Veff only approximately; the residual overlap measures the frozen-core error…

5 · Domain of validity

Terms retained in the mean-field treatment and terms omitted

Model class: self-consistent mean field. The rendered density is that of the converged Hartree–Fock orbital, which is itself an approximation to the exact two-electron state. The nature of the approximation and the magnitude of its error are stated below.

Orbital shape and size

The HF 1s radial form and ⟨r⟩ = 0.927 a₀ agree with the exact one-electron density to better than one percent. Screening is reproduced quantitatively.

valid · density, size, screening
Correlation energy

Hartree–Fock yields −2.8617 Ha against an exact non-relativistic value of −2.9037 Ha. The difference, 0.0421 Ha or 1.14 eV, is the correlation energy: the instantaneous interelectronic avoidance that a mean field cannot represent.

omitted · 1.14 eV; requires CI or QMC
Koopmans' theorem

−ε₁ₛ = 24.98 eV against a measured 24.587 eV. Holding the remaining orbital fixed during ionization neglects relaxation, a systematic bias of the theorem.

approximate · relaxation neglected
Frozen core (Stages C–E)

The excited orbitals of §2.4 are solved with φ1s held fixed. The real core relaxes and polarizes when an electron is promoted, which reduces the 1s–nl overlap. The computed exchange splitting is consequently too large: 1.23 eV against 0.796 eV for 1s2s, and 0.44 eV against 0.254 eV for 1s2p.

omitted · core relaxation and polarization
Quantum defect

δ = 0.22 (s), 0.02 (p), 3×10⁻⁴ (d) emerge from the core charge distribution alone, with the correct ℓ dependence and the correct near-independence of n. The ℓ ≥ 2 levels are reproduced to four figures.

valid · penetration and the level ordering
Configuration mixing

"1s2p" labels a dominant configuration, not an exact state (§1.1). Doubly excited states such as 2s2p lie above the ionization limit and autoionize; they have no single-configuration description and are absent from Stage E entirely.

omitted · CI and continuum coupling
Fine structure and J

Spin–orbit coupling splits the ³P terms by ~10⁻⁵ eV and is what makes the 2³S₁ → 1¹S₀ decay possible at all. No spin-dependent term appears in Ĥ, so J is not a label anywhere on this page.

omitted · requires the Breit–Pauli Hamiltonian
Nuclear cusp

The exact wavefunction satisfies the Kato cusp condition, ∂φ/∂r|₀ = −Zφ(0). A sum of Gaussians has zero derivative at the origin. The discrepancy is a basis-set artefact, negligible for the density plotted here but significant for nuclear-region properties.

artefact · Kato cusp not satisfied

Summary. The spherical 1s density, its radial extent and the screening it exhibits are accurate at the Hartree–Fock level. The 1.14 eV correlation energy and the Koopmans bias are the systematic cost of treating the two electrons as independent. Above the ground configuration the frozen core reproduces the quantum defects and the ordering of the two spectral systems, and overstates the exchange splitting of the low terms by roughly a factor of 1.5 — a definite error with a definite cause, quantified in §2.5.

References