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Direct exchange is a quantum-mechanical spin-dependent energy arising from the Coulomb interaction and the required antisymmetry of a two-electron state. It requires spatial overlap of the relevant orbitals. It is not the classical magnetic dipole force between two small magnets, and it can be much stronger than that force.

For an explicit fixed-orbital illustration, take orthonormal spatial orbitals $a,b$. Orthonormality means $⟨a|b⟩=0$; it does not require their spatial densities to occupy disjoint regions. The normalized spatial states areThe symmetric spatial state combines with the antisymmetric spin-one-half singlet state, and the antisymmetric spatial state combines with a symmetric spin-one-half triplet state, so that the complete fermionic state is antisymmetric.

For Coulomb potential $V_{12}=e^2/(4\pi \epsilon |{\mathbf{r}}_1-{\mathbf{r}}_2|)$, distinguish the direct electrostatic integral and exchange integral:The Coulomb exchange splitting of two orbitals givesThe repulsive Coulomb potential makes $K\ge 0$ in this model. Thus the triplet is favored when the orbitals have nonzero exchange overlap. Its antisymmetric spatial wavefunction suppresses close electron encounters and reduces their repulsion. Since the Coulomb interaction contains no explicit spin operator, the spin dependence enters through the allowed spatial symmetry.

For dimensionless spin operators ${\mathbf{s}}_i={\mathbf{\sigma }}_i/2$, whose pair eigenvalues are $1/4$ in the triplet and $-3/4$ in the singlet, the same splitting is represented byThis explicitly states the sign convention: positive $K$ here favors parallel spins. A convention $H=J{\mathbf{s}}_1\cdot {\mathbf{s}}_2$ would instead have $J=-2K$.

In a crystal, direct exchange therefore depends sensitively on orbital extent, overlap and interatomic spacing. It often decreases rapidly as localized orbitals separate. The fixed-orbital example does not prove that every material must be ferromagnetic: orbital rearrangement and competing kinetic or virtual-hopping processes alter the singlet-triplet balance. In particular, antiferromagnetic superexchange through an intermediate ligand is a different mechanism and should not be confused with direct overlap exchange.

Spin isomers of hydrogen are two forms of molecular hydrogen, ${}^1{\mathrm{H}}_2$, distinguished by the combined nuclear spin of their two protons. They have the same chemical formula and electronic ground state, but different allowed molecular rotation states. Parahydrogen has total nuclear spin zero and even rotational quantum number $J$; orthohydrogen has total nuclear spin one and odd $J$. The difference makes heat capacity a probe of fermionic exchange symmetry, and makes cooling history matter for a tank of molecular hydrogen.

Assume the electronic ground state $X{}^1{\Sigma }_g^{+}$ and the vibrational ground state of a free molecule. The electronic factor is symmetric under exchanging the two protons, and the vibrational factor depends only on their separation, so it is also symmetric. These assumptions concern the exchange of protons; the electronic wavefunction separately obeys fermionic exchange symmetry under exchange of electrons. Neglect electronic excitation, vibrational excitation, external fields, and small interactions involving nuclear spin when calculating rotational energy eigenvalues.

The two protons are identical particles with spin $1/2$. Their complete wavefunction must change sign under their particle exchange operator. Interchanging the two ends reverses the internuclear direction $\hat{\mathbf{r}}$, and the spherical harmonic rotational factor satisfiesThe antisymmetric spin-one-half singlet state therefore requires symmetric, even-$J$ rotation; the symmetric spin-one-half triplet state requires antisymmetric, odd-$J$ rotation. This is exchange symmetry of two identical spin-one-half fermions applied inside a diatomic molecule, not an extra rotational force. A linear rigid rotor may omit spin-dependent splittings while still requiring this restriction on its allowed quantum states.

The two-spin singlet isSpin measurements along axes $\mathbf{a}$ and $\mathbf{b}$ have correlation $E(\mathbf{a},\mathbf{b})=-\mathbf{a}\cdot \mathbf{b}$, and equal axes give perfect anticorrelation.

This is the spin-one-half singlet state, an example of an angular momentum singlet state. The general singlet condition is zero total angular momentum, not the choice of a particular two-qubit Bell basis.

For two spin-one-half particles,has two eigenvalues $\alpha {\hslash }^2/4$ and two further eigenvaluesFor positive $\alpha$, the zero-field limit has a unique spin-one-half singlet state ground state and a triply degenerate spin-one-half triplet state excited level.