---
title: "Helium: the Prototype Two-Electron Atom"
module: Many-Electron Atoms
moduleNumber: 5
lessonNumber: 4
order: 504
summary: >
  Helium is the smallest atom the Schrödinger equation cannot solve exactly, and
  the smallest that shows every many-electron effect. Ignoring the electron
  repulsion overbinds the ground state by 30 eV; first-order perturbation theory
  and a one-parameter variational calculation with an effective charge close most
  of the gap. The excited configurations split into para (singlet) and ortho
  (triplet) states separated by the exchange integral, with the triplet lower —
  and the absence of a 1s² triplet is the Pauli principle in its plainest form.
topics: [Many-Electron Atoms]
sources:
  - book: Bransden & Joachain
    ref: "Ch. 6 — Two-Electron Atoms; §6.1–6.4 Perturbation and Variational Treatments, Exchange and the Para/Ortho Split"
  - book: Griffiths & Schroeter
    ref: "Ch. 5 — Identical Particles; §5.2 Atoms; Ch. 7 §7.2 The Variational Principle Applied to Helium"
  - book: Foot
    ref: "Ch. 3 — Helium; §3.1–3.2 The Ground State and the Excited States"
draft: false
---

Helium has two electrons and $Z=2$, and its Hamiltonian is one repulsion term
away from two independent hydrogenic problems. That single term makes the equation
non-separable and, at the same time, makes helium the cleanest laboratory for
every many-electron idea: the mean field, the variational method, exchange, and
the singlet-triplet split. The
[central field](/atomic-physics/many-electron-atoms/central-field-self-consistent)
and [Hartree-Fock](/atomic-physics/many-electron-atoms/identical-particles-hartree-fock)
constructions are heavy machinery; on helium they can be checked against numbers
by hand.

## The helium Hamiltonian

With the nucleus fixed and both electrons measured from it, the non-relativistic
Hamiltonian is[^bj-61]

$$
H = -\frac{\hbar^2}{2m}\nabla_1^2 - \frac{\hbar^2}{2m}\nabla_2^2
- \frac{Ze^2}{4\pi\varepsilon_0}\!\left(\frac1{r_1}+\frac1{r_2}\right)
+ \frac{e^2}{4\pi\varepsilon_0\,r_{12}},
$$

with $r_{12} = |\vec r_1 - \vec r_2|$ and $Z=2$. Drop the last term and the
equation separates into two hydrogenic problems of charge $Z$; the ground state is
the product $1s(1)\,1s(2)$ with both electrons in the $Z=2$ ground orbital. Its
energy is twice the hydrogenic ground energy at $Z=2$,

$$
E^{(0)} = 2\left(-\frac{Z^2}{2}\right)\text{Ha} = -Z^2\,\text{Ha} = -4\,\text{Ha} = -108.8~\text{eV},
$$

writing energies in Hartree ($1~\text{Ha} = 27.211$ eV). The measured value is
$-79.0$ eV.[^nist-he] The zeroth-order estimate overbinds by nearly $30$ eV,
exactly the mean repulsion the product ignored. Helium is not a small perturbation
away from independent electrons, and any honest treatment must handle the
repulsion.

$$
% caption: Helium coordinates. Both electrons attract to the nucleus (r₁, r₂) and
% repel each other across r₁₂; the repulsion is the only non-separable term.
\begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth]
\definecolor{acc}{HTML}{4A6FA5}
\fill[black!70] (0,0) circle (3.2pt);
\node[black!70, anchor=north] at (0,-0.14) {nucleus $Z=2$};
\coordinate (e1) at (-2.1,1.3);
\coordinate (e2) at (2.2,0.9);
\fill[black] (e1) circle (2.8pt); \node[black, anchor=south east] at (e1) {$e_1$};
\fill[black] (e2) circle (2.8pt); \node[black, anchor=south west] at (e2) {$e_2$};
\draw[black, thick] (0,0) -- (e1); \node[black, anchor=south] at (-1.05,0.75) {$r_1$};
\draw[black, thick] (0,0) -- (e2); \node[black, anchor=north] at (1.1,0.55) {$r_2$};
\draw[acc, thick, dashed] (e1) -- (e2); \node[acc, anchor=south] at (0.05,1.25) {$r_{12}$};
\end{tikzpicture}
$$

## First-order perturbation theory

Treat the repulsion as a perturbation on the product ground state. The first-order
shift is its expectation value in the unperturbed state,[^bj-61]

$$
E^{(1)} = \left\langle 1s\,1s\left|\frac{e^2}{4\pi\varepsilon_0\,r_{12}}\right|1s\,1s\right\rangle
= \frac{5}{8}Z\,\text{Ha}.
$$

The integral is done with the multipole expansion
$1/r_{12} = \sum_\ell (r_<^\ell/r_>^{\ell+1})P_\ell(\cos\theta_{12})$; only the
$\ell=0$ term survives against the spherical $1s$ densities, and the remaining
radial integral gives the clean fraction $5/8$. With $Z=2$,

$$
E^{(1)} = \frac{5}{8}(2)\,\text{Ha} = \frac{5}{4}\,\text{Ha} = 34.0~\text{eV},
$$
$$
E \approx E^{(0)} + E^{(1)} = -108.8 + 34.0 = -74.8~\text{eV}.
$$

First order lands within $5$ eV of experiment, an error of about $5\%$. The
perturbation is not really small — $E^{(1)}$ is a third of $E^{(0)}$ — so the
agreement is better than the method has any right to give, and the higher orders
do not converge quickly. A variational estimate does better with less work.

## The variational estimate

The physical flaw in the product $1s(Z{=}2)\,1s(Z{=}2)$ is that each electron
screens the nucleus from the other, so neither sees the full charge $Z=2$. Promote
the orbital charge to a variational parameter $Z'$ and minimize the energy over
it.[^gs-72] For a $1s$ orbital of charge $Z'$ in the field of the true nucleus $Z$,
the pieces are

$$
\langle T\rangle = \frac{Z'^2}{2},
\qquad
\left\langle -\frac{Z}{r}\right\rangle = -Z Z',
\qquad
\left\langle\frac{1}{r_{12}}\right\rangle = \frac{5}{8}Z',
$$

each in Hartree. The trial energy is the sum over both electrons plus the
repulsion,

$$
E(Z') = 2\!\left(\frac{Z'^2}{2}\right) - 2ZZ' + \frac58 Z'
= Z'^2 - \left(2Z - \frac58\right)Z'.
$$

Setting $\d E/\d Z' = 0$ gives $2Z' = 2Z - 5/8$, so the optimal effective charge is

$$
Z' = Z - \frac{5}{16} = 2 - \frac{5}{16} = \frac{27}{16} = 1.6875,
$$

and back-substitution collapses to $E_{\min} = -Z'^2$:

$$
E_{\min} = -\left(\frac{27}{16}\right)^2\text{Ha} = -2.848\,\text{Ha} = -77.5~\text{eV}.
$$

The number $Z' = 1.69$ is the physics: each electron sees not the full $+2$ but a
partially screened $+1.69$, the other electron's cloud cancelling about a third of
one unit of charge. The variational energy, guaranteed to lie above the true
ground energy, sits at $-77.5$ eV against the measured $-79.0$ eV — an error of
$2\%$ from a single parameter.[^gs-72]

$$
% caption: The variational energy E(Z') for helium in Hartree. It is a parabola in
% the effective charge with a minimum at Z' = 27/16 = 1.69, below the full nuclear
% charge because each electron screens the other.
\begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (5.6,0) node[right, black!70] {trial charge};
\draw[->, black] (0.0,0) -- (0.0,3.4) node[above, black!70] {trial energy};
% parabola E = Z'^2 - 3.375 Z', min at Z'=1.6875; plot shifted up for display
% map Z' in [1.0,2.4] to x in [0.6,5.2]; energy inverted (up = higher)
\draw[black, very thick]
  (0.6,2.9) .. controls (1.7,1.4) and (2.5,0.62) .. (3.05,0.5)
  .. controls (3.6,0.62) and (4.4,1.4) .. (5.2,2.9);
\fill[acc] (3.05,0.5) circle (2.2pt);
\draw[black, dashed] (3.05,0.5) -- (3.05,0);
\node[acc, anchor=north] at (3.05,-0.05) {$\tfrac{27}{16}$};
\node[acc, anchor=west] at (3.2,0.55) {minimum};
\node[black, anchor=west] at (0.55,3.05) {full charge not optimal};
\end{tikzpicture}
$$

The three estimates and the datum, side by side, show each correction pulling the
binding energy toward the measured value:

$$
% caption: Helium ground-state binding energy |E|. Zeroth order overbinds;
% first-order perturbation and the variational estimate bracket the measured
% 79.0 eV from below in binding (above in energy).
\begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (0,0) -- (0,3.6) node[above, black!70] {binding $|E|$ (eV)};
\draw[black, dashed] (0,2.72) -- (6.4,2.72);
\node[black, anchor=east] at (-0.1,2.72) {79.0};
% bars: zeroth 108.8 -> off top; scale so 79 -> 2.72 => factor 0.03443; 108.8 -> 3.75 (cap)
% first-order 74.8 -> 2.575; variational 77.5 -> 2.668; exp 79.0 -> 2.72
\draw[black] (0.6,0) rectangle (1.7,3.55);
\node[black, anchor=south, rotate=90] at (1.15,0.3) {108.8};
\node[black, anchor=north] at (1.15,-0.05) {zeroth};
\draw[black] (2.2,0) rectangle (3.3,2.575);
\node[black, anchor=south, rotate=90] at (2.75,0.3) {74.8};
\node[black, anchor=north] at (2.75,-0.05) {first order};
\fill[acc!12] (3.8,0) rectangle (4.9,2.668); \draw[acc] (3.8,0) rectangle (4.9,2.668);
\node[acc, anchor=south, rotate=90] at (4.35,0.3) {77.5};
\node[black, anchor=north] at (4.35,-0.05) {variational};
\draw[black] (5.4,0) rectangle (6.3,2.72);
\node[black, anchor=south, rotate=90] at (5.85,0.3) {79.0};
\node[black, anchor=north] at (5.85,-0.05) {experiment};
\end{tikzpicture}
$$

## The first ionization energy

Removing one electron leaves the one-electron ion $\text{He}^+$, a hydrogenic
system with $Z=2$ and energy $-Z^2/2 = -2\,\text{Ha} = -54.4$ eV. The first
ionization energy is the difference,

$$
\text{IE} = E(\text{He}^+) - E(\text{He}) = -54.4 - (-79.0) = 24.6~\text{eV},
$$

the largest first ionization energy of any element and the reason helium is inert.
The closed $1s^2$ shell, tightly bound and with no low-lying vacancy, resists both
losing and sharing an electron.

## Para- and ortho-helium

The excited configurations put one electron in $1s$ and the other in a higher
orbital $n\ell$ (the doubly excited states lie above the first ionization limit and
autoionize). Two electrons in different spatial orbitals can form a spatially
symmetric or antisymmetric combination, and antisymmetry of the total state pairs
each with the opposite spin symmetry:[^bj-63]

- **para-helium** — spatially symmetric, spin **singlet** ($S=0$), terms $^1L_J$;
- **ortho-helium** — spatially antisymmetric, spin **triplet** ($S=1$), terms $^3L_J$.

The energy of a $1s\,n\ell$ configuration, to first order, is the sum of the two
orbital energies plus the direct integral, shifted by the exchange integral with a
sign set by the spin:

$$
E_\pm = I_{1s} + I_{n\ell} + J_{1s,n\ell} \pm K_{1s,n\ell},
$$

with $+$ for the singlet (symmetric space) and $-$ for the triplet. Since the
exchange integral $K > 0$, the **triplet lies below the singlet** of the same
configuration by $2K$. The physical cause is the
[exchange hole](/atomic-physics/many-electron-atoms/identical-particles-hartree-fock):
the antisymmetric spatial state keeps the electrons apart, lowering their Coulomb
repulsion, and the parallel-spin triplet is the one forced into that arrangement.

$$
% caption: A 1s n-ell configuration splits by exchange. The direct integral J
% raises both; the exchange integral K then pushes the triplet down and the
% singlet up by an equal amount, leaving the triplet 2K below the singlet.
\begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth]
\definecolor{acc}{HTML}{4A6FA5}
% bare configuration energy
\draw[black, thick] (0,1.6) -- (1.6,1.6);
\node[black!70, anchor=east] at (0,1.6) {no repulsion};
% after direct J (raised)
\draw[black, dashed] (2.0,2.5) -- (3.2,2.5);
\node[black, anchor=west] at (3.25,2.5) {$+\,J$};
% singlet (up by K)
\draw[acc, thick] (4.0,3.1) -- (5.6,3.1);
\node[acc, anchor=west] at (5.65,3.1) {singlet $^1L$};
% triplet (down by K)
\draw[acc, thick, dashed] (4.0,1.9) -- (5.6,1.9);
\node[acc, anchor=west] at (5.65,1.9) {triplet $^3L$};
% split marker
\draw[black, <->] (4.8,3.05) -- (4.8,1.95);
\node[black!70, anchor=west] at (4.85,2.5) {$2K$};
\draw[black] (3.2,2.5) -- (4.0,3.1);
\draw[black] (3.2,2.5) -- (4.0,1.9);
\end{tikzpicture}
$$

The ground configuration $1s^2$ has no such partner. Both electrons occupy the
same $1s$ spatial orbital, which is necessarily symmetric, so the spin **must** be
the antisymmetric singlet. A $1s^2$ triplet would need a symmetric spin state on a
symmetric spatial state — a fully symmetric total wave function, forbidden for
fermions. There is therefore **no** $1s^2$ triplet, and the helium ground term is
$1\,^1S_0$ with no low-lying triplet beneath it.

$$
% caption: The helium term diagram. Para (singlet) levels on the left, ortho
% (triplet) on the right, each configuration's triplet below its singlet. The
% ground state 1 singlet-S has no triplet partner.
\begin{tikzpicture}[scale=1.0, font=\footnotesize, >=stealth]
\definecolor{acc}{HTML}{4A6FA5}
% columns
\node[black] at (1.4,4.7) {para (singlet)};
\node[black] at (5.2,4.7) {ortho (triplet)};
% ground singlet only
\draw[acc, thick] (0.6,0.4) -- (2.2,0.4);
\node[acc, anchor=east] at (0.5,0.4) {$1\,^1S_0$};
% 1s2s
\draw[black, thick] (0.6,3.0) -- (2.2,3.0);
\node[black, anchor=east] at (0.5,3.0) {$2\,^1S_0$};
\draw[black, thick, dashed] (4.4,2.55) -- (6.0,2.55);
\node[black, anchor=west] at (6.05,2.55) {$2\,^3S_1$};
% 1s2p
\draw[black, thick] (0.6,3.7) -- (2.2,3.7);
\node[black, anchor=east] at (0.5,3.7) {$2\,^1P_1$};
\draw[black, thick, dashed] (4.4,3.35) -- (6.0,3.35);
\node[black, anchor=west] at (6.05,3.35) {$2\,^3P$};
% ionization limit
\draw[black, dashed] (0.4,4.15) -- (6.2,4.15);
\node[black, anchor=west] at (4.4,4.05) {ionization limit};
% metastable note
\draw[black, ->] (4.4,2.55) .. controls (3.2,1.4) .. (2.2,0.5);
\node[black, anchor=north] at (3.3,1.55) {intercombination forbidden};
\end{tikzpicture}
$$

Because the electric-dipole operator does not touch spin, transitions between
singlet and triplet — **intercombination** lines — are strongly forbidden by the
$\Delta S = 0$ rule. Para- and ortho-helium behave almost like two separate gases:
the nineteenth century catalogued their spectra as two elements. The lowest
triplet $2\,^3S_1$ cannot decay to the $1\,^1S_0$ ground state by any allowed
route and is **metastable**, with a lifetime of about $7900$ s, an eternity on
atomic timescales.[^bj-63]

The ordering "triplet below singlet" and its origin in the exchange integral is
the physical content of the [first Hund
rule](/atomic-physics/many-electron-atoms/hund-rules-ground-terms), which the next
lessons extend from helium's two electrons to open shells of any size, once the
[coupling schemes and term
symbols](/atomic-physics/many-electron-atoms/ls-jj-coupling-term-symbols) are in
place.

[^bj-61]: **Bransden & Joachain**, _Physics of Atoms and Molecules_, 2nd ed., §6.1 — the helium Hamiltonian, the zeroth-order product ground state $E^{(0)}=-Z^2\,\text{Ha}$, and the first-order repulsion shift $E^{(1)}=\tfrac58 Z\,\text{Ha}$ from the multipole expansion of $1/r_{12}$. <https://www.pearson.com/en-gb/subject-catalog/p/physics-of-atoms-and-molecules/P200000005386>
[^gs-72]: **Griffiths & Schroeter**, _Introduction to Quantum Mechanics_, 3rd ed., §7.2 — the variational treatment of helium with an effective nuclear charge, the minimum at $Z' = Z - 5/16 = 27/16$, and the resulting ground-state energy $-2.85$ Ha $= -77.5$ eV.
[^nist-he]: NIST Atomic Spectra Database — helium ground-state energy and ionization energy ($24.587$ eV). <https://www.nist.gov/pml/atomic-spectra-database>
[^bj-63]: **Bransden & Joachain**, _Physics of Atoms and Molecules_, 2nd ed., §6.3–6.4 — para- and ortho-helium, the exchange splitting $E_\pm = I_{1s}+I_{n\ell}+J\pm K$ with the triplet $2K$ below the singlet, the absence of a $1s^2$ triplet, the $\Delta S=0$ intercombination rule, and the metastable $2\,^3S_1$ level.