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Distinguishable noninteracting particles. Interpret $V$ as the same external one-particle potential for each particle; the total potential is $V({\mathbf{r}}_1)+V({\mathbf{r}}_2)$. The Time-dependent Schrödinger equation isLet $\hat{h}{\varphi }_a={ϵ}_a{\varphi }_a$ and $\hat{h}{\varphi }_b={ϵ}_b{\varphi }_b$, where $\hat{h}=-{\hslash }^2{\nabla }^2/(2m)+V$. Applying ${\hat{h}}_1+{\hat{h}}_2$ to the product multiplies it by ${ϵ}_a+{ϵ}_b$. The time derivative of $e^{-iEt/\hslash }$ multiplies it by $E$. Thus the proposed stationary product is a solution providedThe tensor product separates the two independent one-particle dynamics.

Identical spin-one-half particles. Identical fermions require antisymmetry under exchange of the complete position-and-spin labels. First take $a,b$ to be distinct orthonormal spatial orbitals, and defineThe symmetric spatial state must multiply the antisymmetric spin-one-half singlet state; the antisymmetric spatial state must multiply one of the three symmetric spin-one-half triplet states. The four normalized possibilities, each with the common time factor $e^{-iEt/\hslash }$, areThese are the required product states under fermionic exchange symmetry. If $a=b$, the antisymmetric spatial combination vanishes. Only the product ${\varphi }_a({\mathbf{r}}_1){\varphi }_a({\mathbf{r}}_2)$ times the singlet remains, as required by the Pauli exclusion principle. A nonorthogonal pair would require different normalization factors; the displayed $1/\sqrt{2}$ assumes an orthonormal stationary-state basis.

Weak repulsive Coulomb interaction. For fixed orbitals $a,b$, first-order perturbation theory gives spatial energy shifts $J\pm K$. Here $J$ is the direct interaction energy, while the exchange integral for repulsion strength $g>0$ isThe positivity follows from the positive Fourier transform of the Coulomb kernel. Consequently Coulomb exchange splitting of two orbitals givesAll three triplet spin components remain degenerate for a spin-independent Coulomb interaction. Their antisymmetric spatial factor vanishes at coincidence and reduces close encounters. The inequality is strict when $K>0$; for negligible orbital overlap, the splitting may vanish. This comparison fixes one particle in each of the specified orbitals and does not rank singlets with a different orbital occupation.

Spin-zero bosons. For one particle in each of the same distinct orthonormal orbitals, the spin factor is symmetric and bosonic exchange symmetry requires ${\Phi }_{+}$. At coincident positions,Thus the bosonic coincidence enhancement in two orthogonal orbitals isThese are joint probability densities, or small-detector-volume limits; an exactly specified point has zero probability in a continuous distribution. If both particles occupy the same orbital, the properly normalized product is already symmetric and the corresponding enhancement relative to distinguishable particles in that same orbital is one. The paper's usual factor-two answer presumes distinct orthogonal input orbitals.

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.