The exchange interaction and its mark on spectra

Nineteenth-century spectroscopists saw something odd in the helium spectrum: the lines sorted themselves neatly into two sets, with transitions between the sets almost never occurring — as if the bottle held two different gases. For a while people genuinely believed it did, naming them “orthohelium” and “parahelium” and treating them as two substances.

Only later did it become clear: the bottle holds one kind of helium. The two sets of lines correspond to the two values of the two electrons’ total spin — singlet ($S=0$, parahelium) and triplet ($S=1$, orthohelium). And with that came something odder still:

For the same configuration (say one electron in $1s$ and one in $2s$), the singlet and triplet differ in energy by about 0.80 eV.

What is odd about that? Helium’s Hamiltonian

$$\hat{H}=\frac{\hat p_1^2}{2m}+\frac{\hat p_2^2}{2m} -\frac{2e^2}{4\pi\epsilon_0 r_1}-\frac{2e^2}{4\pi\epsilon_0 r_2} +\frac{e^2}{4\pi\epsilon_0 r_{12}}\tag{8.4.1}$$

contains no spin from start to finish (spin-field couplings and spin-orbit coupling are small enough to neglect for now; chapter 6’s fine structure is only of order $10^{-4}$ eV). Spin appears nowhere in the energy, yet the energies sort themselves by spin, differing by nearly 1 eV — four orders of magnitude too much. Where does this energy come from?

The perturbation tools of chapter 7 are ready to hand, and the last missing puzzle piece was fitted in the previous section: the total spin, via the antisymmetry requirement, dictates the symmetry of the spatial wavefunction. And the shape of the spatial wavefunction is something the Coulomb energy can see.

Settling the account: helium’s 1s2s configuration

The singlet-triplet energy splittingadvanced~12 minexpand

Step 0: set up the zeroth-order states. Neglecting the electron-electron repulsion $e^2/4\pi\epsilon_0 r_{12}$, each electron lives in a hydrogen-like orbital (nuclear charge $Z=2$). Consider the excited configuration: one electron in $\phi_{1s}$, one in $\phi_{2s}$. By section 8.2, the spatial part has only two legal combinations:

$$\psi_\pm(\vec{r}_1,\vec{r}_2)=\frac{1}{\sqrt2}\bigl[\phi_{1s}(\vec{r}_1)\phi_{2s}(\vec{r}_2)\pm\phi_{2s}(\vec{r}_1)\phi_{1s}(\vec{r}_2)\bigr]\tag{8.4.2}$$

$\psi_+$ (symmetric) must pair with the spin singlet; $\psi_-$ (antisymmetric) with the spin triplet. The zeroth-order energies are equal: $E^{(0)}=E_{1s}+E_{2s}$.

Step 1: treat the electron repulsion as a perturbation and find the first-order correction. Write $\hat V=\dfrac{e^2}{4\pi\epsilon_0 r_{12}}$. The first-order energy correction is the expectation value (the spin part is normalised and contracts away):

$$E^{(1)}_\pm=\braket{\psi_\pm}{\hat V\,\psi_\pm} =\frac12\Bigl[\braket{ab}{\hat V\,ab}+\braket{ba}{\hat V\,ba} \pm\braket{ab}{\hat V\,ba}\pm\braket{ba}{\hat V\,ab}\Bigr]\tag{8.4.3}$$

Here $\ket{ab}$ abbreviates $\phi_{1s}(\vec{r}_1)\phi_{2s}(\vec{r}_2)$. Since $\hat V$ is symmetric under $1\leftrightarrow2$, the four terms are equal in pairs, giving

$$E^{(1)}_\pm=J\pm K\tag{8.4.4}$$

Step 2: meet the two integrals.

The direct integral $J$:

$$J=\int|\phi_{1s}(\vec{r}_1)|^2\,\frac{e^2}{4\pi\epsilon_0 r_{12}}\,|\phi_{2s}(\vec{r}_2)|^2\,\dd^3r_1\dd^3r_2\tag{8.4.5}$$

This is just the classical Coulomb repulsion energy of the two charge clouds $|\phi_{1s}|^2$ and $|\phi_{2s}|^2$ — no quantum strangeness whatsoever.

The exchange integral $K$:

$$K=\int\phi_{1s}^*(\vec{r}_1)\phi_{2s}^*(\vec{r}_2)\,\frac{e^2}{4\pi\epsilon_0 r_{12}}\,\phi_{2s}(\vec{r}_1)\phi_{1s}(\vec{r}_2)\,\dd^3r_1\dd^3r_2\tag{8.4.6}$$

Notice that the orbitals on the two sides of the kernel have swapped places — this has no classical counterpart. It is an interference term: the interference between the two histories “particle 1 in $1s$, particle 2 in $2s$” and “the other way round”. For nodeless overlapping real orbitals like $1s,2s$, $K>0$ (one can prove it is strictly positive here).

Step 3: read off the physics.

$$E_{\text{singlet}}=E^{(0)}+J+K,\qquad E_{\text{triplet}}=E^{(0)}+J-K,\qquad \Delta E=2K\tag{8.4.7}$$

The triplet is lower. For helium 1s2s, experiment gives: triplet $2^3S_1$ at excitation energy 19.82 eV, singlet $2^1S_0$ at 20.62 eV — a splitting of 0.80 eV, i.e. $K\approx0.40$ eV. Evaluating the integral directly with hydrogen-like orbitals gives the same order of magnitude (somewhat too large, because screening is neglected).

Why is the triplet lower? — the Fermi hole

The formula has answered; intuition is still owed. Look at $\psi_-$ at $\vec{r}_1=\vec{r}_2$:

$$\psi_-(\vec{r},\vec{r})=\frac{1}{\sqrt2}\bigl[\phi_{1s}(\vec{r})\phi_{2s}(\vec{r})-\phi_{2s}(\vec{r})\phi_{1s}(\vec{r})\bigr]=0\tag{8.4.8}$$

In the triplet, the probability density for the two electrons to meet at the same point is identically zero, and it is suppressed throughout each other’s neighbourhood — every electron is surrounded by a “moat” its companion refuses to enter, called the Fermi hole. The electrons stay far apart on average, so the Coulomb repulsion $\propto1/r_{12}$ costs less. The singlet is the opposite: $\psi_+$ is enhanced at coincidence (a “Fermi heap”), and the repulsion bill runs high.

From helium to iron: the quantum origin of ferromagnetism

Abstract the conclusion one step. Since the energy depends only on whether the total spin is singlet or triplet, we can rewrite it as a function of spin operators. Using the fact that $\hat{\vec S}_1\cdot\hat{\vec S}_2$ takes the value $-\frac{3}{4}\hbar^2$ in the singlet and $+\frac{1}{4}\hbar^2$ in the triplet (the addition results of chapter 5), the two levels can be packaged into a single effective Hamiltonian:

$$\hat{H}_{\text{eff}}=\text{const}-\frac{2K}{\hbar^2}\,\hat{\vec S}_1\cdot\hat{\vec S}_2\tag{8.4.10}$$

Check by substitution: the triplet gets a relative shift of $-K/2$, the singlet $+3K/2$, and the gap is exactly $2K$. ✔

The significance of this form goes far beyond the helium atom: it says that pure Coulomb + antisymmetry is equivalent to a spin-spin coupling. Generalised to the countless atoms of a crystal lattice, it is the starting point of the theory of magnetism — the Heisenberg model $\hat H=-\sum_{ij}J_{ij}\,\hat{\vec S}_i\cdot\hat{\vec S}_j/\hbar^2$. For $J_{ij}>0$ the spins prefer to align in parallel: ferromagnetism.

The spectrum puzzle, revisited

Finally, close the case we opened with. The two sets of lines rarely cross-transition because the operator for optical transitions (electric dipole) contains no spin and cannot change the total spin: the singlet-to-triplet matrix element is approximately zero (the “spin selection rule” $\Delta S=0$). So parahelium and orthohelium each transition internally, forming two independent line systems; $2^3S_1$ cannot drop back to the $1^1S_0$ ground state and becomes a metastable state living over two hours — one of the longest-lived metastable states in atomic physics. The “two heliums” are one helium’s two spin families, driven apart by exchange symmetry.

∎Key formulas

• First-order energies

  $$E_\pm=E^{(0)}+J\pm K$$

  + pairs with symmetric space (spin singlet), − with antisymmetric space (triplet)

• Direct integral

  $$J=\int|\phi_a(\vec{r}_1)|^2\,\frac{e^2}{4\pi\epsilon_0 r_{12}}\,|\phi_b(\vec{r}_2)|^2\,\dd^3r_1\dd^3r_2$$

  The classical Coulomb energy of two charge clouds

• Exchange integral

  $$K=\int\phi_a^*(\vec{r}_1)\phi_b^*(\vec{r}_2)\,\frac{e^2}{4\pi\epsilon_0 r_{12}}\,\phi_b(\vec{r}_1)\phi_a(\vec{r}_2)\,\dd^3r_1\dd^3r_2$$

  No classical counterpart; set by orbital overlap, short-ranged

• Effective spin coupling

  $$\hat{H}_{\text{eff}}=\text{const}-\frac{2K}{\hbar^2}\,\hat{\vec S}_1\cdot\hat{\vec S}_2$$

  Prototype of the Heisenberg model; K>0 → parallel spins more stable, ferromagnetism

?Self-check4 questions

1. 1.

   Helium's Hamiltonian contains no spin, yet the singlet and triplet differ by about 0.8 eV. The essence of this energy difference is:

   • Coulomb repulsion energy — the spin, via the antisymmetry requirement, sets the symmetry of the spatial wavefunction and hence the average electron separation
   • The magnetic interaction between the two electrons' spin magnetic moments
   • A fine-structure effect of spin-orbit coupling
   • A new type of fundamental interaction appearing only between identical particles

   Submit
2. 2.

   In helium's 1s2s configuration the triplet 2³S lies at excitation energy 19.82 eV and the singlet 2¹S at 20.62 eV. Using E± = E⁽⁰⁾ + J ± K, find the exchange integral K in eV.

   eV2% relative tolerance

   Submit
3. 3.

   Which statements about the exchange integral K are correct? (Select all that apply.)

   Select all that apply

   • It arises from interference between the "two ways of occupying the orbitals" and has no counterpart in classical electromagnetism
   • It decays as the overlap of the two orbitals shrinks, and vanishes when the orbitals do not overlap
   • It always lowers the energy of the parallel-spin configuration
   • In the effective spin Hamiltonian it plays the role of the spin coupling constant

   Submit
4. 4.

   Iron's Curie temperature is as high as 1043 K. Why can we rule out the magnetic dipole interaction as what sustains ferromagnetism?

   • The dipole energy converts to about 0.04 K, over four orders of magnitude below 1043 K — thermal motion would have destroyed it long ago
   • The magnetic dipole interaction exists only in vacuum
   • Iron atoms have no magnetic moment
   • The dipole interaction always drives spins antiparallel

   Submit

What comes next

Helium has just two electrons, and we have already filled pages with symmetrized combinations. Iron has 26, whose antisymmetrized expansion runs to $26!\approx4\times10^{26}$ terms; a mole of metal does not bear thinking about. Clearly, the road of “writing out every term” has come to its end.

The next section switches bookkeeping systems: stop asking “which particle is in which state” (a question that should never have been asked — particles carry no labels) and ask instead “how many particles are in each state”. A new language — second quantization — will shrink the $N!$-term determinant to a single line.