The Helium Atom

Helium is the atom where the postulates about identical particles stop being abstract and start predicting spectra. Two electrons bound to a $Z=2$ nucleus is the smallest system that classical quantum mechanics cannot solve in closed form: the electron–electron term $e^2/4\pi\varepsilon_0 r_{12}$ couples the two coordinates and destroys separability. Yet almost everything interesting about atoms is already visible here — the ground state's spin, the splitting of excited levels, the meaning of "exchange energy," and the first hint of Hund's rules — so helium is the proving ground for the whole chapter. This article owns the structure of the two-electron problem. We write the non-separable Hamiltonian, show that Fermi antisymmetry forces the ground configuration $1s^2$ to be a spin singlet (parahelium), and compute the energy the honest way: first-order perturbation theory in the repulsion gives $\langle 1/r_{12}\rangle = \tfrac54 Z\,\mathrm{Ry}=34.0$ eV, lifting the crude $-108.8$ eV estimate to $-74.8$ eV. For excited configurations the direct and exchange integrals $J\pm K$ split each level into a lower triplet (orthohelium) and a higher singlet — the exchange energy in its first real appearance. Two boundaries are drawn deliberately. The variational refinement — the effective-charge trial that reaches $-77.5$ eV and the correlated calculation that reaches $-78.66$ eV toward the exact $-79.0$ eV — is owned by Variational Trial Wavefunctions and Variational Helium, and we recall their numbers by name rather than re-deriving them. The full mechanics of the exchange integral $K$ is owned by Exchange Forces and Exchange Energy, invoked here. What helium contributes is the punchline that makes the abstract rule visible: the lowest triplet state, orthohelium $2\,{}^3S_1$ , is metastable for about $7900$ seconds — nearly a million times the nanosecond life of an ordinary excited atom — because the singlet–triplet transition it would need is optically forbidden.

Key takeaways

Helium is the simplest non-separable atom: two electrons at charge $Z=2$ with a repulsion term $e^2/4\pi\varepsilon_0 r_{12}$ that couples the coordinates, so no product wavefunction is an exact eigenstate.
Fermi antisymmetry ties spatial symmetry to spin: the ground configuration $1s^2$ can only be spatially symmetric, so it must pair with the antisymmetric spin singlet — parahelium. The triplet is forbidden in the ground config because a spatially antisymmetric state from two identical $1s$ orbitals vanishes.
First-order perturbation theory gives the repulsion correction $\langle 1/r_{12}\rangle=\tfrac54 Z\,\mathrm{Ry}=34.0$ eV, lifting zeroth-order $-108.8$ eV to $-74.8$ eV versus the experimental $-79.0$ eV — a big but honest correction.
Excited $1s\,nl$ configurations split by the exchange integral $K$ : the triplet (orthohelium, spatially antisymmetric) always lies below the singlet (parahelium), a direct consequence of electrons avoiding each other. The mechanism $E_\pm=E_0+J\pm K$ is owned by Exchange Forces and Exchange Energy.
The owned punchline: the lowest triplet $2\,{}^3S_1$ is metastable for $\sim7900$ s, because the singlet–triplet transition to the ground state is optically forbidden — helium behaves almost like two separate gases.
The variational refinements ( $Z_{\text{eff}}=27/16\Rightarrow-77.5$ eV; correlated $-78.66$ eV) are owned by Variational Trial Wavefunctions and Variational Helium and recalled by name here.
Key idea: antisymmetry ties spatial symmetry to total spin, and Coulomb repulsion then turns that symmetry into an energy splitting — the spin triplet lies below the singlet not from any magnetic force but because a spatially antisymmetric wavefunction keeps the two electrons apart.

The core insight

For two electrons, total antisymmetry links the spatial and spin symmetries: a symmetric spatial state must carry the singlet spin, an antisymmetric spatial state the triplet. The electron–electron Coulomb repulsion, evaluated on those two spatial symmetries, then splits the energy — and the antisymmetric (triplet) case, in which the electrons cannot coincide, always lies lower. Helium turns "exchange force" from a phrase into a measured splitting.

Where this lives today

Helium is not a solved museum piece — it is a precision laboratory. Modern variational and correlated methods compute the helium ground-state energy to more than fifteen significant figures, and the comparison with equally precise spectroscopy is now sensitive to quantum-electrodynamic corrections, the finite nuclear size, and even proposed physics beyond the Standard Model; the "helium fine-structure" measurement is one route to the fine-structure constant. Metastable orthohelium, the $2\,{}^3S_1$ state whose long life this article computes, is a workhorse of atomic physics: it is laser-cooled and trapped, used in atom interferometry, and its Penning-ionization collisions probe ultracold chemistry. The two-electron problem also anchors quantum chemistry — the direct and exchange integrals introduced here are the atoms of Hartree–Fock theory, and the electron correlation that separates $-78.66$ eV from the exact $-79.0$ eV is the quantity every post-Hartree–Fock method is built to capture.

The physical picture

Put two electrons around a helium nucleus and ask a single question: how do they arrange their spins and their positions? The answer is dictated entirely by the requirement that the total wavefunction change sign when you swap the electrons. That antisymmetry cannot be imposed on the spins alone or the positions alone — it is a joint condition. A two-electron spin state is either the antisymmetric singlet ( $S=0$ ) or one of the three symmetric triplet states ( $S=1$ ), so to keep the total antisymmetric the spatial part must have the opposite symmetry: singlet spin with symmetric space, triplet spin with antisymmetric space. Now the Coulomb repulsion enters as the referee. A spatially symmetric wavefunction has a large amplitude for the two electrons to sit near each other, so it pays a high repulsion price. A spatially antisymmetric wavefunction vanishes when the electrons coincide — a statistical exclusion that keeps them apart and lowers the repulsion. So for any excited configuration where the two electrons occupy different orbitals, the triplet (antisymmetric space) lies below the singlet (symmetric space). This is the same "Fermi hole" that will later become Hund's first rule; helium is where you first see it.

Figure

Helium energy ladder: the honest estimates and the gap to experiment, with the orthohelium metastable outlier

The figure's message is the honesty of the perturbative step. The repulsion is not a small correction — at $+34.0$ eV it is a quarter of the total binding — so first-order theory overshoots. But it gets the qualitative energy right and, crucially, the spin structure exactly. Pushing past $-74.8$ eV toward the true $-79.0$ eV is the variational and correlated program owned downstream. What is bought cheaply here is the architecture.

The Hamiltonian and why it will not separate

The helium Hamiltonian is two hydrogen-like pieces plus the coupling that ruins them.

\hat H = -\frac{\hbar^2}{2m}\left(\nabla_1^2+\nabla_2^2\right) - \frac{Ze^2}{4\pi\varepsilon_0}\left(\frac{1}{r_1}+\frac{1}{r_2}\right) + \frac{e^2}{4\pi\varepsilon_0\,r_{12}} \qquad (1)

where: $Z=2$ is the nuclear charge; $r_1,r_2$ are the electron–nucleus distances; $r_{12}=|\mathbf r_1-\mathbf r_2|$ is the electron–electron separation. The physical read: drop the last term and the two electrons are independent hydrogenic problems, each with binding $-Z^2\,\mathrm{Ry}=-54.4$ eV, giving a total $-108.8$ eV. That last term is the villain of the piece — because $1/r_{12}$ depends on both positions at once and cannot be split into a piece for electron 1 plus a piece for electron 2, no product wavefunction $\psi(\mathbf r_1)\phi(\mathbf r_2)$ is an exact eigenstate. Helium is unsolvable in closed form for exactly this reason. The state carries a spin part and a spatial part, and their product must be antisymmetric.

\Psi(1,2)=\psi_{\text{space}}(\mathbf r_1,\mathbf r_2)\,\chi_{\text{spin}}(1,2),\qquad \Psi(2,1)=-\Psi(1,2) \qquad (2)

where: the spin space splits into the antisymmetric singlet $\chi_{0,0}=\tfrac{1}{\sqrt2}(\uparrow\downarrow-\downarrow\uparrow)$ and the three symmetric triplet states $\chi_{1,m}$ . The spin construction itself — the lowering operator, the singlet/triplet decomposition of $\tfrac12\otimes\tfrac12$ — is owned by Singlet and Triplet States; here we use its result. The physical read: to keep $\Psi$ antisymmetric, a symmetric spatial state must carry the singlet, an antisymmetric spatial state the triplet. That pairing is the whole engine of what follows.

The ground state must be parahelium

The lowest configuration puts both electrons in the $1s$ orbital. Try to build the triplet there and it evaporates.

Derivation AWhy the ground state is a spin singlet

Show that the ground configuration $1s^2$ of helium can only be a spin singlet (parahelium), and that this is exclusion at work, not a separate rule. The goal is to see that "the ground state is a singlet" is forced by the same antisymmetry that forbids two fermions from sharing a state — the triplet does not lose an energy competition here, it simply cannot be constructed. The payoff is that the spin of the ground state is fixed before any energy is computed.

Step 1. In the ground configuration both electrons occupy the same spatial orbital, $\psi_a=\psi_b=\psi_{1s}$ . The two-electron spatial state comes in symmetric or antisymmetric flavors, $\psi_\pm\propto\psi_{1s}(\mathbf r_1)\psi_{1s}(\mathbf r_2)\pm\psi_{1s}(\mathbf r_2)\psi_{1s}(\mathbf r_1)$ .

Step 2. The antisymmetric combination vanishes identically: $\psi_{1s}(\mathbf r_1)\psi_{1s}(\mathbf r_2)-\psi_{1s}(\mathbf r_1)\psi_{1s}(\mathbf r_2)=0$ . There is no spatially antisymmetric state built from two copies of the same orbital — this is the Slater-determinant vanishing (two equal rows) recalled from Slater Determinants, and the double-occupancy collapse recalled from Pauli Exclusion Principle.

Step 3. To keep the total state antisymmetric, the surviving symmetric spatial part $\psi_+$ must pair with an antisymmetric spin part — the singlet. Therefore the ground state is $\psi_{1s}(\mathbf r_1)\psi_{1s}(\mathbf r_2)\times\chi_{0,0}$ : parahelium, $S=0$ , the $1\,{}^1S_0$ term. There is simply no $1s^2$ triplet to compete. Why this matters: the ground-state spin is settled by symmetry alone, before energies enter. The spin triplet only becomes available for excited configurations where the two electrons occupy different orbitals, and that is where the singlet–triplet energy competition begins. Verified: `verify_aq-qm-helium-atom.py` — the antisymmetric spatial state from two identical $1s$ orbitals is identically zero (sympy).

The energy the honest way: first-order perturbation theory

Treat the repulsion as a perturbation on the independent-electron zeroth order. The correction is the expectation of $1/r_{12}$ in the product of two $1s$ orbitals — a standard Slater integral with a clean closed form.

Go deeper · Derivation B · 1 minFirst-order repulsion energy and the ground-state estimate

Derivation BFirst-order repulsion energy and the ground-state estimate

Compute the first-order energy shift from the electron–electron repulsion in the helium ground state and produce a real number for the binding energy. The goal is to show that the correction $\langle 1/r_{12}\rangle$ evaluates to exactly $\tfrac54 Z\,\mathrm{Ry}$ — a specific, verifiable value — and that adding it to the crude independent-electron estimate lands within a few eV of experiment. The payoff is a controlled, honest number: not exact, but right in spirit and structure.

Step 1. The zeroth-order energy treats the electrons as independent hydrogenic $1s$ states at charge $Z$ : $E^{(0)}=-Z^2\,\mathrm{Ry}\,(1/n_1^2+1/n_2^2)$ . For $n_1=n_2=1$ , $E^{(0)}=-2Z^2\,\mathrm{Ry}=-2(4)(13.6)=-108.8$ eV. This is far too deep — it ignores that the electrons repel each other.

Step 2. The first-order correction is $\langle V_{ee}\rangle=\dfrac{e^2}{4\pi\varepsilon_0}\displaystyle\int\dfrac{|\psi_{1s}(\mathbf r_1)|^2|\psi_{1s}(\mathbf r_2)|^2}{r_{12}}\,\mathrm d^3r_1\,\mathrm d^3r_2$ . Expanding $1/r_{12}$ in Legendre multipoles, only the monopole survives the spherically symmetric $1s$ densities, and the radial Slater integral evaluates in closed form to $\langle 1/r_{12}\rangle=\tfrac{5Z}{8}$ in atomic units.

Step 3. Converting, $\langle V_{ee}\rangle=\tfrac54 Z\,\mathrm{Ry}=\tfrac54(2)(13.6)=34.0$ eV. This is enormous — one quarter of the binding — so perturbation theory is not "small," but it is meaningful.

Step 4. Add the pieces: $E\approx E^{(0)}+\langle V_{ee}\rangle=-108.8+34.0=-74.8$ eV, against the experimental $-79.0$ eV. First order overshoots by about $4$ eV; the variational effective-charge trial (owned by Variational Trial Wavefunctions) does better, reaching $-77.5$ eV with $Z_{\text{eff}}=27/16$ , and the correlated calculation (owned by Variational Helium) reaches $-78.66$ eV. Verified: `verify_aq-qm-helium-atom.py` — $\langle 1/r_{12}\rangle=5Z/8$ a.u. by direct numerical Slater integral, $\langle V_{ee}\rangle=34.0$ eV, $E^{(0)}=-108.8$ eV, first-order $E=-74.8$ eV, all to the stated precision; the recalled variational $Z_{\text{eff}}=27/16$ and $E_{\min}=-77.5$ eV cross-checked.

The excited-state story is where the exchange integral appears. For a configuration $1s\,nl$ with the electrons in different orbitals $\psi_a,\psi_b$ , the diagonal energy is $E_\pm=E_0+J\pm K$ , with a direct integral $J$ (the classical repulsion of the two charge clouds) and an exchange integral $K$ (a purely quantum overlap term). The mechanics of $J$ and $K$ — their integral definitions, the proof $K>0$ , the "exchange force" that is no force at all — is owned by Exchange Forces and Exchange Energy. What matters here is the sign: because $K>0$ , the antisymmetric-space state (triplet, $E_0+J-K$ ) lies below the symmetric-space state (singlet, $E_0+J+K$ ).

Figure

Ortho below para: the exchange splitting of an excited configuration, and the metastable outlier

Limiting cases

Case	Result	Meaning
Drop $1/r_{12}$ (independent electrons)	$E=-2Z^2\,\mathrm{Ry}=-108.8$ eV	separable, exactly solvable, but far too deep — repulsion ignored
First-order PT	$E=-108.8+34.0=-74.8$ eV	one-quarter correction; overshoots experiment by $\sim4$ eV
Variational $Z_{\text{eff}}=27/16$	$E=-77.5$ eV	screening each electron sees; owned by Variational Trial Wavefunctions
Correlated trial (explicit $r_{12}$ )	$E=-78.66$ eV toward $-79.0$ eV exact	electron correlation; owned by Variational Helium
Excited $1s\,nl$ , $l\neq0$	triplet below singlet by $2K$	exchange splitting; magnitude owned by Exchange Forces and Exchange Energy
$Z\to\infty$ (heliumlike ions)	repulsion becomes a small fraction of binding	PT becomes accurate; He $^0$ is the hardest case, not the easiest

Common trapThe singlet–triplet splitting is electrostatic, not magnetic — and "exchange force" is not a force

Two traps sit here. First, there is no magnetic interaction between the two spins driving the splitting. The gap between orthohelium and parahelium comes entirely from the electrostatic Coulomb repulsion evaluated on spatial wavefunctions of different symmetry; the spin merely bookkeeps which spatial symmetry antisymmetry allows. If you look for a spin–spin magnetic term to explain the $\sim0.80$ eV gap you will find nothing remotely that large — the electron magnetic dipole interaction is a million times too weak. Second, the "exchange force" that pushes triplet below singlet is not a new interaction; it is a statistical consequence of indistinguishability, the same one owned quantitatively by Exchange Forces and Exchange Energy. Naming it a force is a historical convenience, not a fifth fundamental interaction.

Worked examples

Example 1 — the ground-state energy, step by step. Estimate the helium ground-state energy in first-order perturbation theory and compare with experiment.

Step 1. Zeroth order. Treat the two electrons as independent hydrogenic $1s$ states at $Z=2$ : each binds by $-Z^2\,\mathrm{Ry}=-4(13.6)=-54.4$ eV, so $E^{(0)}=2\times(-54.4)=\mathbf{-108.8}$ eV.

Step 2. Apply the repulsion. The first-order correction is $\langle V_{ee}\rangle=\tfrac54 Z\,\mathrm{Ry}=\tfrac54(2)(13.6)=\mathbf{+34.0}$ eV — the Slater integral $\langle 1/r_{12}\rangle=5Z/8$ a.u. converted to eV.

Step 3. Boundary — the total and the honest gap. Adding, $E\approx-108.8+34.0=\mathbf{-74.8}$ eV, against the experimental $\mathbf{-79.0}$ eV: first order overshoots by $4.2$ eV. The effective-charge variational trial (owned by Variational Trial Wavefunctions) reaches $-77.5$ eV with $Z_{\text{eff}}=27/16=1.6875$ , and the correlated calculation (owned by Variational Helium) reaches $-78.66$ eV. The perturbative number is not exact, but it is a genuine prediction that captures three-quarters of the effect from a one-line integral.

Example 2 — the ionization energy and the metastable outlier. Deduce helium's first ionization energy from the total energies, and explain why orthohelium lives so long.

Step 1. Ionization energy. Removing one electron leaves He $^+$ , a hydrogenic ion at $Z=2$ with $E(\text{He}^+)=-Z^2\,\mathrm{Ry}=-54.4$ eV. The first ionization energy is $I_1=E(\text{He}^+)-E(\text{He})=-54.4-(-79.0)=\mathbf{+24.6}$ eV, in excellent agreement with the NIST value $24.587$ eV.

Step 2. Orthohelium's grace period. The lowest triplet $2\,{}^3S_1$ cannot decay to the $1\,{}^1S_0$ ground state by an ordinary electric-dipole transition: that transition would require $\Delta S\neq0$ and is doubly forbidden (spin change plus $S\to S$ ). The only channels left are heavily suppressed higher-order processes.

Step 3. Boundary — the lifetime. The result is a lifetime of about $\mathbf{7900}$ seconds — over two hours — versus the $\sim10^{-9}$ s of an allowed atomic decay, a factor near $10^{12}$ . This single number is why nineteenth-century spectroscopists thought there were two heliums: parahelium and orthohelium barely interconvert, so they behave like two nearly independent gases.

Origins · 1 minhelium atom and how this was discovered

helium atom

the two-electron testbed of quantum mechanics · Kirchhoff–Bunsen 1868 · Heisenberg 1926 · Hylleraas 1929

The helium atom is the smallest system where the quantum theory of identical particles becomes quantitative, and its history tracks the theory's own maturation. Helium was discovered spectroscopically in the Sun in 1868 (the name comes from helios) before it was found on Earth. The puzzle that made it famous among theorists was its spectrum: two apparently separate sets of lines, as if two different gases were present, which spectroscopists labeled parahelium and orthohelium. In 1926 Werner Heisenberg resolved the mystery by recognizing that the two "gases" were the spin-singlet and spin-triplet sectors of a single two-electron atom, split by what he identified as an exchange interaction — the first physical appearance of exchange energy. Making helium quantitative fell to Egil Hylleraas, who in 1929 built trial wavefunctions containing the interelectron distance $r_{12}$ explicitly and computed a ground-state energy agreeing with experiment to a fraction of a percent — a landmark that established the variational method as a precision tool and whose descendants now compute helium to many significant figures. Between them, Heisenberg supplied the structure and Hylleraas the numbers.

History

1868

— helium discovered as an unexplained yellow line in the solar spectrum (Janssen, Lockyer)

1895

— helium first isolated on Earth from uranium ore (Ramsay)

1926

— Heisenberg explains the para/ortho "two heliums" as spin singlet and triplet split by exchange

1927

— Kellner and Unsöld give early perturbative and variational helium calculations

1929

— Hylleraas computes the helium ground state with explicit

r_{12}

correlation, matching experiment

1930

s — Bethe and Salpeter systematize two-electron atomic structure

Going further

References. Griffiths & Schroeter, Introduction to Quantum Mechanics, 3rd ed., §5.2 and §8.2 · Sakurai & Napolitano, Modern Quantum Mechanics, 2nd ed., §7.4 · Shankar, Principles of Quantum Mechanics, 2nd ed., Ch. 10 · Cohen-Tannoudji, Diu & Laloë, Quantum Mechanics, Vol. II, Ch. XIV · Bransden & Joachain, Physics of Atoms and Molecules, Ch. 6

Provenance. The Hamiltonian, non-separability, spin-space pairing, and the direct/exchange structure $E_\pm=E_0+J\pm K$ are standard across Griffiths §5.2, Sakurai §7.4, and Cohen-Tannoudji Vol. II Ch. XIV. Every number added here is recomputed in the committed verification script: the first-order Slater integral $\langle 1/r_{12}\rangle=5Z/8$ a.u. (direct numerical integration), $\langle V_{ee}\rangle=34.0$ eV, $E^{(0)}=-108.8$ eV, first-order $E=-74.8$ eV, the ionization energy $I_1=24.6$ eV against NIST $24.587$ eV, and the sign $K>0$ (contact-toy and positive-definite-kernel arguments). Editorial note for this rebuild: the variational effective-charge result $Z_{\text{eff}}=27/16\Rightarrow-77.5$ eV and the correlated $-78.66$ eV are owned by Variational Trial Wavefunctions and Variational Helium — this article recalls them by name (their served content computes these values; here they are cross-checked but not re-derived) and picks up the "full $r_{12}$ chase" exactly where those articles hand it off. The exchange-integral mechanics ( $J$ , $K$ , $K>0$ ) is owned by Exchange Forces and Exchange Energy. The orthohelium $\sim7900$ s lifetime and $\sim0.80$ eV $1s2s$ gap are standard atomic-data values.

Builds on: Indistinguishability of Identical Particles · Symmetric and Antisymmetric States · Slater Determinants Leads to: Exchange Forces and Exchange Energy · Singlet and Triplet States in Two-Electron Systems · Hund's Rules · Aufbau Principle and Electron Configurations

Why can't the helium Schrödinger equation be solved exactly in closed form?

Because the electron–electron term $e^2/4\pi\varepsilon_0 r_{12}$ depends on both electron coordinates simultaneously and does not factor into a piece for electron 1 plus a piece for electron 2. Separation of variables requires the potential to split that way; the $1/r_{12}$ coupling forbids it, so no product wavefunction $\psi(\mathbf r_1)\phi(\mathbf r_2)$ is an exact eigenstate. This is why helium is treated perturbatively or variationally, and why "electron correlation" — the part of the true wavefunction that cannot be written as any product — exists at all.

Helium's ground state is a singlet, not a triplet. Why doesn't the triplet win as it does for excited states?

In the ground configuration both electrons occupy the same $1s$ orbital. The spatially antisymmetric state built from two identical orbitals is identically zero (two equal rows in a Slater determinant), so there is no ground-state triplet to compete — the singlet is the only option. The triplet only becomes available for excited configurations $1s\,nl$ where the electrons occupy different orbitals; there, the spatially antisymmetric (triplet) state exists and, keeping the electrons apart, lies below the singlet. So the rule "triplet is lower" applies to excited states; the ground state has no triplet at all.

Is the "exchange force" that lowers the triplet a real force between the electrons?

No. There is no new interaction and no magnetic coupling doing the work. The splitting comes entirely from the ordinary electrostatic Coulomb repulsion, evaluated on two spatial wavefunctions of different symmetry: the antisymmetric one vanishes when the electrons coincide, so it accumulates less repulsion energy. The spin state only decides, through antisymmetry, which spatial symmetry is allowed. "Exchange force" is a historical name for this statistical effect; the quantitative machinery lives in Exchange Forces and Exchange Energy.

How large is the repulsion correction, and does perturbation theory even make sense here?

The first-order correction is $\tfrac54 Z\,\mathrm{Ry}=34.0$ eV — roughly a quarter of the total binding, so it is not "small" in the usual perturbative sense. Strictly, the expansion parameter is order one and higher orders are not negligible. But first order still captures the physics: it gets the spin structure exactly right and lands the energy at $-74.8$ eV, within a few eV of the experimental $-79.0$ eV. It is a genuine, if rough, prediction; the variational and correlated methods (owned downstream) do the precision work.

Why did helium look like two different elements, "parahelium" and "orthohelium"?

Because the singlet ( $S=0$ , para) and triplet ( $S=1$ , ortho) sectors barely interconvert. Optical transitions that change total spin ( $\Delta S\neq0$ ) are forbidden to leading order, so a helium atom prepared in a triplet state cannot easily radiate down to the singlet ground state — the lowest triplet $2\,{}^3S_1$ is metastable for about $7900$ s. Spectroscopically the two sets of lines look like two nearly independent gases, which is exactly how nineteenth-century observers described them, before Heisenberg identified them in 1926 as the two spin sectors of one atom.

What is helium's first ionization energy, and how does it come out of these energies?

The first ionization energy is the work to remove one electron: $I_1=E(\text{He}^+)-E(\text{He})$ . He $^+$ is a hydrogenic ion at $Z=2$ with $E=-Z^2\,\mathrm{Ry}=-54.4$ eV, and the neutral atom has $E(\text{He})=-79.0$ eV, so $I_1=-54.4-(-79.0)=24.6$ eV, matching the NIST value $24.587$ eV. This is the largest first ionization energy of any element — helium's closed $1s^2$ shell is exceptionally tightly bound, which is the quantum-mechanical root of its chemical inertness.

Where does the exact -79.0 eV come from if perturbation theory only reaches -74.8 eV?

From going beyond the independent-electron picture. The effective-charge variational trial, which lets each electron see a screened charge $Z_{\text{eff}}=27/16$ , reaches $-77.5$ eV — that calculation is owned by Variational Trial Wavefunctions. Adding explicit dependence on the interelectron distance $r_{12}$ (Hylleraas-type correlated trials) reaches $-78.66$ eV and, with many parameters, the full $-79.0$ eV — owned by Variational Helium. The gap between the product estimate and the exact value is precisely electron correlation, the quantity that no single-configuration method can capture.