Case Study 1: The Exchange Interaction — Why Helium Has a Singlet and Triplet

Overview

Helium is the simplest multi-electron atom — two electrons orbiting a nucleus with charge $Z=2$. It is also the first system where the exchange interaction matters. The existence of two distinct sets of helium spectral lines — so different that 19th-century spectroscopists believed they came from two different elements — is a direct consequence of the symmetrization postulate and the exchange integral.

This case study traces the history, develops the calculation, and connects the helium singlet-triplet splitting to the broader physics of exchange interactions.

Part 1: The Historical Puzzle — Parahelium and Orthohelium

Two Spectra, One Element

When spectroscopists first catalogued the emission lines of helium in the late 19th century, they found two completely separate sets of lines that never mixed. The two sets were called parahelium and orthohelium:

• Parahelium lines corresponded to singlet states ($S = 0$, antiparallel spins). The ground state of helium is a parahelium state.
• Orthohelium lines corresponded to triplet states ($S = 1$, parallel spins). There is no orthohelium ground state with both electrons in $1s$.

The two series had different term values, different fine-structure patterns, and — most puzzling — no transitions connecting them. Spectral lines connected parahelium states to other parahelium states, and orthohelium states to other orthohelium states, but never crossed between the two.

So complete was the separation that some physicists proposed helium was actually a mixture of two elements. (It is not.)

The Resolution

The explanation came from quantum mechanics. The two series correspond to the two possible spin configurations:

Transitions between singlet and triplet are forbidden because the electric dipole operator does not act on spin. A transition would require a simultaneous change in the spatial and spin parts of the wavefunction, which the spin-independent electromagnetic interaction cannot accomplish. (Magnetic dipole transitions and spin-orbit coupling weakly break this selection rule, but the resulting "intercombination" lines are extremely faint.)

📊 By the Numbers: The metastable $2^3S_1$ state of helium has a lifetime of approximately 7900 seconds (over two hours) — extraordinarily long for an excited atomic state (typical radiative lifetimes are $\sim 10^{-8}$ seconds). It decays primarily by two-photon emission or collisional de-excitation.

Part 2: Setting Up the Calculation

The Helium Hamiltonian

The Hamiltonian for the helium atom (neglecting spin-orbit and relativistic corrections) is:

$$\hat{H} = -\frac{\hbar^2}{2m_e}\nabla_1^2 - \frac{\hbar^2}{2m_e}\nabla_2^2 - \frac{Ze^2}{4\pi\epsilon_0 r_1} - \frac{Ze^2}{4\pi\epsilon_0 r_2} + \frac{e^2}{4\pi\epsilon_0 |\mathbf{r}_1 - \mathbf{r}_2|}$$

with $Z = 2$. The first four terms are single-particle (each electron interacting with the nucleus). The last term — the electron-electron repulsion — is the troublemaker that prevents exact solution.

Zeroth-Order Approximation

Ignoring the electron-electron repulsion, the Hamiltonian separates: $\hat{H}_0 = \hat{h}(\mathbf{r}_1) + \hat{h}(\mathbf{r}_2)$, where $\hat{h}$ is the hydrogen-like Hamiltonian with $Z=2$. The single-particle eigenstates are hydrogen-like wavefunctions with $Z=2$, and the energies are:

$$E_n^{(0)} = -\frac{Z^2 \times 13.6 \text{ eV}}{n^2} = -\frac{54.4 \text{ eV}}{n^2}$$

The Excited $(1s)(2s)$ Configuration

For the first excited configuration, one electron is in $1s$ and one in $2s$. The zeroth-order energy is:

$$E^{(0)} = E_1^{(0)} + E_2^{(0)} = -54.4 - 13.6 = -68.0 \text{ eV}$$

(measured from the fully ionized He$^{2+}$ state).

The two possible symmetrized spatial wavefunctions are:

$$\Phi_S(\mathbf{r}_1, \mathbf{r}_2) = \frac{1}{\sqrt{2}}[\phi_{1s}(\mathbf{r}_1)\phi_{2s}(\mathbf{r}_2) + \phi_{2s}(\mathbf{r}_1)\phi_{1s}(\mathbf{r}_2)]$$

$$\Phi_A(\mathbf{r}_1, \mathbf{r}_2) = \frac{1}{\sqrt{2}}[\phi_{1s}(\mathbf{r}_1)\phi_{2s}(\mathbf{r}_2) - \phi_{2s}(\mathbf{r}_1)\phi_{1s}(\mathbf{r}_2)]$$

Pairing with spin: - $\Phi_S$ (symmetric space) $\times$ singlet (antisymmetric spin) → parahelium $2^1S_0$ - $\Phi_A$ (antisymmetric space) $\times$ triplet (symmetric spin) → orthohelium $2^3S_1$

Part 3: Computing the Direct and Exchange Integrals

First-Order Perturbation Theory

Treating $\hat{V}_{12} = e^2/(4\pi\epsilon_0|\mathbf{r}_1 - \mathbf{r}_2|)$ as a perturbation, the first-order energy correction is:

$$E^{(1)}_\pm = \langle \Phi_{S/A} | \hat{V}_{12} | \Phi_{S/A} \rangle = J \pm K$$

where the $+$ corresponds to the symmetric spatial state (singlet) and the $-$ to the antisymmetric spatial state (triplet).

The Direct Integral

$$J = \int\!\!\int |\phi_{1s}(\mathbf{r}_1)|^2 \frac{e^2}{4\pi\epsilon_0|\mathbf{r}_1 - \mathbf{r}_2|} |\phi_{2s}(\mathbf{r}_2)|^2 \, d^3r_1 \, d^3r_2$$

This is the classical electrostatic interaction between the charge distributions $|\phi_{1s}|^2$ and $|\phi_{2s}|^2$. It is always positive (repulsive).

For the $(1s)(2s)$ configuration of helium:

$$J = \frac{17}{81} \times \frac{Z e^2}{4\pi\epsilon_0 a_0} = \frac{17}{81} \times Z \times 27.2 \text{ eV}$$

With $Z = 2$: $J = \frac{17}{81} \times 54.4 \text{ eV} \approx 11.4$ eV.

The Exchange Integral

$$K = \int\!\!\int \phi_{1s}^*(\mathbf{r}_1)\phi_{2s}^*(\mathbf{r}_2) \frac{e^2}{4\pi\epsilon_0|\mathbf{r}_1 - \mathbf{r}_2|} \phi_{2s}(\mathbf{r}_1)\phi_{1s}(\mathbf{r}_2) \, d^3r_1 \, d^3r_2$$

Notice the "exchange" of orbital labels: in the bra, electron 1 is in $1s$ and electron 2 in $2s$, but in the ket, electron 1 is in $2s$ and electron 2 in $1s$.

This integral has no classical interpretation. It involves the overlap of the two orbitals mediated by the Coulomb potential.

For the $(1s)(2s)$ configuration:

$$K = \frac{16}{729} \times \frac{Z e^2}{4\pi\epsilon_0 a_0} = \frac{16}{729} \times Z \times 27.2 \text{ eV}$$

With $Z = 2$: $K = \frac{16}{729} \times 54.4 \text{ eV} \approx 1.19$ eV.

Evaluating the Integral: Technique

The key technical step is the multipole expansion of the Coulomb potential:

$$\frac{1}{|\mathbf{r}_1 - \mathbf{r}_2|} = \sum_{\ell=0}^{\infty} \frac{4\pi}{2\ell+1} \frac{r_<^\ell}{r_>^{\ell+1}} \sum_{m=-\ell}^{\ell} Y_\ell^{m*}(\hat{\mathbf{r}}_1) Y_\ell^m(\hat{\mathbf{r}}_2)$$

where $r_< = \min(r_1, r_2)$ and $r_> = \max(r_1, r_2)$.

Since both $\phi_{1s}$ and $\phi_{2s}$ are spherically symmetric ($\ell = 0$), only the $\ell = 0$ term in the multipole expansion survives (the angular integrals of $Y_\ell^m$ with $\ell \neq 0$ against spherically symmetric wavefunctions vanish). This simplifies the computation enormously:

$$K = e^2 \int_0^\infty\!\!\int_0^\infty R_{1s}(r_1) R_{2s}(r_1) \frac{r_<^0}{r_>^1} R_{2s}(r_2) R_{1s}(r_2) \, r_1^2 r_2^2 \, dr_1 \, dr_2$$

where $R_{n\ell}(r)$ are the radial wavefunctions. The integral is doable analytically but tedious; the result is the $16/729$ factor given above.

Part 4: The Energy Splitting and Its Consequences

Results

The first-order corrected energies are:

$$E_{\text{singlet}} = E^{(0)} + J + K = -68.0 + 11.4 + 1.19 = -55.4 \text{ eV}$$

$$E_{\text{triplet}} = E^{(0)} + J - K = -68.0 + 11.4 - 1.19 = -57.8 \text{ eV}$$

The singlet-triplet splitting is:

$$\Delta E = E_{\text{singlet}} - E_{\text{triplet}} = 2K \approx 2.4 \text{ eV}$$

(The experimental splitting is about 0.80 eV. Our first-order calculation overestimates because using unshielded $Z=2$ wavefunctions is too crude. Better variational wavefunctions give much closer agreement.)

Physical Interpretation

Why is the triplet lower? Consider the spatial probability distributions:

Singlet (symmetric spatial): $|\Phi_S(\mathbf{r}_1, \mathbf{r}_2)|^2 \propto |\phi_{1s}(\mathbf{r}_1)\phi_{2s}(\mathbf{r}_2) + \phi_{2s}(\mathbf{r}_1)\phi_{1s}(\mathbf{r}_2)|^2$

When $\mathbf{r}_1 = \mathbf{r}_2$: the wavefunction is enhanced (bosonic bunching of the spatial part). The electrons are more likely to be close together.

Triplet (antisymmetric spatial): $|\Phi_A(\mathbf{r}_1, \mathbf{r}_2)|^2 \propto |\phi_{1s}(\mathbf{r}_1)\phi_{2s}(\mathbf{r}_2) - \phi_{2s}(\mathbf{r}_1)\phi_{1s}(\mathbf{r}_2)|^2$

When $\mathbf{r}_1 = \mathbf{r}_2$: the wavefunction vanishes. The electrons avoid each other.

Since the Coulomb repulsion $e^2/(4\pi\epsilon_0|\mathbf{r}_1 - \mathbf{r}_2|)$ increases when the electrons are close, the triplet state (electrons farther apart) has lower repulsion energy and thus lower total energy.

💡 Key Insight: The spin state indirectly controls the energy through exchange symmetry. The Hamiltonian contains no spin-dependent terms (we are neglecting spin-orbit coupling). Yet the energy depends on whether the spins form a singlet or triplet. The causal chain is:

$$\text{spin configuration} \to \text{spatial symmetry} \to \text{electron separation} \to \text{Coulomb energy}$$

This is the exchange interaction in action.

Part 5: Hund's First Rule and Ferromagnetism

Hund's First Rule

The helium singlet-triplet splitting is the simplest example of a general principle: Hund's first rule.

For a given electron configuration, the term with the greatest multiplicity ($2S+1$) has the lowest energy.

In other words, electrons in degenerate orbitals prefer to have parallel spins (maximizing total $S$). This is not because parallel spins attract — there is no magnetic spin-spin interaction strong enough to explain the effect. It is because parallel spins require an antisymmetric spatial wavefunction, which keeps the electrons farther apart and reduces the Coulomb repulsion.

Hund's first rule determines the ground-state configuration of atoms across the periodic table:

• Carbon ($1s^2 2s^2 2p^2$): The two $2p$ electrons occupy different $m_\ell$ orbitals with parallel spins ($S=1$, $^3P$), not the same orbital with antiparallel spins.
• Nitrogen ($1s^2 2s^2 2p^3$): All three $2p$ electrons have parallel spins ($S=3/2$, $^4S$), filling each $m_\ell$ orbital once before pairing.
• Iron ($[Ar]3d^6 4s^2$): The six $3d$ electrons maximize spin with 4 unpaired electrons ($S=2$), giving iron its magnetic moment.

Connection to Ferromagnetism

In ferromagnetic materials (iron, cobalt, nickel), the exchange interaction between unpaired $d$-electrons on neighboring atoms is strong enough to align spins across macroscopic domains. The exchange energy favoring parallel alignment overcomes the thermal energy ($k_BT$) below the Curie temperature ($T_C = 1043$ K for iron).

The Heisenberg exchange Hamiltonian for a lattice of spins is:

$$\hat{H} = -\sum_{\langle i,j \rangle} J_{ij} \, \hat{\mathbf{S}}_i \cdot \hat{\mathbf{S}}_j$$

where $J_{ij}$ is the exchange coupling constant between sites $i$ and $j$, and the sum is over nearest-neighbor pairs. When $J_{ij} > 0$ (ferromagnetic), parallel alignment is energetically favored. When $J_{ij} < 0$ (antiferromagnetic), antiparallel alignment is preferred.

The entire field of magnetism — from compass needles to hard drives to MRI machines — traces its quantum mechanical origin to the exchange interaction.

Part 6: Beyond First Order — Improving the Helium Calculation

Variational Approaches

The first-order perturbation theory result for helium is only qualitatively correct. Quantitative accuracy requires:

1. Effective nuclear charge: Rather than using $Z=2$ for both electrons, one can use a variational parameter $Z_{\text{eff}} < 2$ to account for screening. This is the simplest variational treatment (Chapter 19 preview).

2. Configuration interaction: The true ground state is not a single Slater determinant but a superposition of many configurations: $(1s)^2 + c_1(2s)^2 + c_2(2p)^2 + \ldots$. Optimizing the coefficients $c_i$ systematically improves the energy.

3. Hylleraas coordinates: Using coordinates that explicitly include the inter-electron distance $r_{12} = |\mathbf{r}_1 - \mathbf{r}_2|$ allows extremely accurate calculations. Hylleraas (1929) achieved six significant figures for the helium ground-state energy with just a few variational parameters.

📊 By the Numbers: The best theoretical value for the helium ground-state energy is $E = -2.903\,724\,377\,034\,119\,598\,31$ hartrees (Drake, 2006), computed with a 2358-term variational wavefunction. This agrees with experiment to 15 significant figures — making helium one of the most precisely tested systems in all of physics.

Discussion Questions

1. The helium singlet-triplet splitting arises from the Coulomb interaction, not from any magnetic or spin-dependent interaction. Yet the resulting physics looks exactly like a spin-spin interaction (the Heisenberg model). Is the Heisenberg model "real" or is it an effective description? What is the relationship between effective models and fundamental physics?

2. Hund's first rule says "maximize spin." But the ground state of helium ($1s^2$, $S=0$) has the minimum spin. Is this a violation of Hund's rule? (Careful: Hund's rule applies to electrons in degenerate orbitals within the same subshell.)

3. The singlet-triplet structure of helium was observed spectroscopically decades before quantum mechanics provided the explanation. What does this tell us about the relationship between experimental observation and theoretical understanding in physics?

4. Ferromagnetism arises from the exchange interaction, which in turn arises from the Pauli exclusion principle. Follow the chain of logic backward: what would happen to magnets if electrons were bosons instead of fermions?