Exchange effects

Intuition

The Hamiltonian of two electrons contains no spin at all, only their kinetic energies, the nucleus’s pull and their repulsion. Yet the spin decides the energy, through the symmetry of the spatial part. Put one electron in orbital $a$ and one in $b$ . The average repulsion has two parts: the direct energy $J$ of the two charge clouds, which any two charges would have, and the exchange energy $K$ , which exists only because the spatial state is symmetric or antisymmetric. The singlet, with symmetric space, gets $J+K$ ; the triplet, with antisymmetric space, gets $J-K$ . $K$ is positive, so the triplet lies lower: electrons with parallel spins keep out of each other’s way and repel less. The resulting splitting looks like a strong magnetic coupling of the spins, but it is electrostatics and antisymmetry.

Two commuters who happen to take the same train sit close together half the time; two who have agreed never to be in the same carriage bump into each other less. The triplet’s antisymmetry is such an agreement, and it saves repulsion.

The energy of two electrons in different orbitals: without their repulsion on the left, raised by the direct energy $J$ in the middle, and split by the exchange energy on the right — the singlet at $J+K$ above, the triplet at $J-K$ below, $2K$ apart.

Two electrons in orthonormal orbitals $\phi_{a},\phi_{b}$ , with the repulsion $\hat{V}=e^{2}/4\pi\varepsilon_{0}|\mathbf{r}_{1}-\mathbf{r}_{2}|$ , have the average energies

\begin{aligned}E_{\pm}&=E_{a}+E_{b}+J\pm K\\ J&=\braket{ab|\hat{V}|ab},\qquad K=\braket{ab|\hat{V}|ba}\end{aligned}

Properties

$\ket{ab}=\phi_{a}(\mathbf{r}_{1})\phi_{b}(\mathbf{r}_{2})$ ; the upper sign, symmetric space, is the singlet, the lower sign the triplet.

Direct and exchange energies

\braket{\psi_{\pm}|\hat{V}|\psi_{\pm}}=J\pm K,\qquad K>0

Expand the average repulsion in the symmetrised state into four terms. The repulsion is unchanged by swapping the electrons, so the four are two equal pairs: the direct energy and the exchange energy. The exchange energy is the Coulomb energy of a complex charge density with itself, which an integration by parts turns into the positive integral of the square of its field.

Proof steps

\psi_{\pm}=\tfrac{1}{\sqrt2}\bigl(\ket{ab}\pm\ket{ba}\bigr)

The spatial states of the singlet, $+$ , and the triplet, $-$ .

\braket{\psi_{\pm}|\hat{V}|\psi_{\pm}}=\tfrac12\bigl(V_{ab,ab}+V_{ba,ba}\pm V_{ab,ba}\pm V_{ba,ab}\bigr)

Four terms, with $V_{ab,ba}=\braket{ab|\hat{V}|ba}$ .

V_{ba,ba}=V_{ab,ab}=J,\qquad V_{ba,ab}=V_{ab,ba}=K

$\hat{V}$ is symmetric in $\mathbf{r}_{1},\mathbf{r}_{2}$ : renaming the variables of integration makes each pair equal, and real.

E_{\pm}=E_{a}+E_{b}+J\pm K

Add the energies of the two orbitals.

K=\varepsilon_{0}\int|\nabla\varphi|^{2}\,d^{3}r>0

$K$ is the Coulomb energy of the complex density $\rho=e\,\phi_{a}^{*}\phi_{b}$ with itself; with its potential, Poisson’s equation and an integration by parts give this integral.

Applications

Practice

Direct and Exchange

The average repulsion of two electrons is $J+K$ in the singlet and $J-K$ in the triplet.

E_{\pm}=E_{a}+E_{b}+J\pm K

Try it

Two electrons have $J=10$ eV and $K=1.5$ eV. How far above the triplet does the singlet lie, in eV?