Put one particle in orbital $\phi_{a}$ and one in $\phi_{b}$. The product $\phi_{a}(x_{1})\phi_{b}(x_{2})$ will not do for identical particles; its symmetric and antisymmetric combinations will. For electrons the spin enters too, and the whole state must be antisymmetric: a symmetric spatial part goes with the antisymmetric spin singlet, an antisymmetric spatial part with the symmetric spin triplet — the singlet and triplet of the chapter on adding angular momenta, now tied to where the electrons can be. The difference shows where the particles meet. The antisymmetric combination vanishes when $x_{1}=x_{2}$: two electrons of the same spin avoid each other, an exchange hole around each. The symmetric one is twice as likely there as for distinguishable particles: bosons bunch.

Two dancers who must mirror each other can never stand on the same spot, while two who must copy each other are drawn together. Antisymmetry makes fermions keep their distance; symmetry makes bosons crowd.

The probability density of the separation $x_{2}-x_{1}$ of two particles in a box of unit width, one in its lowest level and one in the next: for identical bosons, for distinguishable particles (dashed), and for identical fermions of the same spin. At zero separation the bosons’ density is twice the distinguishable one and the fermions’ vanishes — an exchange hole.

For orthonormal orbitals $\phi_{a}\neq\phi_{b}$, the two-particle states of definite symmetry are

• For two electrons the whole state is antisymmetric: $\psi_{+}(\mathbf{r}_{1},\mathbf{r}_{2})$ times the spin singlet, or times one of the three triplet states.

At coinciding arguments the two terms of $\psi_{\pm}$ are the same product, so they cancel in the antisymmetric state and add in the symmetric one. The normalisation uses the orthogonality of the orbitals, which kills the cross terms.

The cross terms are $|\braket{\phi_{a}|\phi_{b}}|^{2}$, zero for orthogonal orbitals.

The two terms coincide.

The two terms add.

Distinguishable particles, one in each orbital: bosons are twice as likely to meet, fermions of one spin never.

For orthonormal orbitals, the symmetric and antisymmetric combinations of the two products each have norm 1 after dividing by $\sqrt{2}$.

The whole state of two electrons is antisymmetric: a symmetric spatial part goes with the spin singlet, an antisymmetric one with the triplet.

An antisymmetric spatial state vanishes when the two positions coincide: electrons with the same spin keep apart.

Two electrons in the same orbital have a symmetric spatial part, $\phi(x_{1})\phi(x_{2})$, so their spins must form the singlet.

Great work! You now know how to:

• build symmetric and antisymmetric states of two particles
• pair the spatial states of two electrons with singlet and triplet
• show the exchange hole of fermions and the bunching of bosons