Exchange symmetry, spin pairs and Pauli exclusion
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| spin singlet | 自旋单态 | zì xuán dān tài |
| Slater determinant | 斯莱特行列式 | sī lái tè xíng liè shì |
A decision before an answer
- Two electrons can share one spatial orbital if their complete spatial-plus-spin state has the required antisymmetry.
- Your goal: Construct normalised symmetric and antisymmetric two-orbital spatial states.
Exchange complete particle labels
- For identical particles, exchanging every coordinate and spin label changes a fermionic total wavefunction by a minus sign and leaves a bosonic one unchanged. Mathematical slots 1 and 2 label arguments, not permanently distinguishable particles. For two distinct orthonormal orbitals a and b, spatial combinations are Ψ_±=[a(1)b(2)±b(1)a(2)]/√2.
- Orthogonality makes these combinations normalised and gives exchange eigenvalues ±1. If the two orbitals are identical, the antisymmetric combination is identically zero and the displayed symmetric formula is not correctly normalised; the double-occupation spatial state is simply a(1)a(2). Recheck normalization whenever orbitals or their overlaps change.
Two electrons share the same spatial orbital. Which spin state is allowed for their antisymmetric total state?
The same-orbital spatial product is symmetric; the spin factor must be antisymmetric.
Pair spatial and spin symmetry
- Two spin-1/2 particles have one spin singlet (↑↓−↓↑)/√2 with total spin S=0; it is antisymmetric under exchange. The three triplets ↑↑, (↑↓+↓↑)/√2 and ↓↓ have S=1 and are symmetric. Electrons require an antisymmetric total state, so a symmetric spatial state pairs with the singlet, while an antisymmetric spatial state pairs with a triplet.
- Two electrons in the same spatial orbital can form the singlet, but cannot form a triplet there in this simple two-electron state. Pauli exclusion prevents occupation of the same complete single-particle state, including spin. Opposite-spin electrons in one orbital are two different complete states; describing exclusion as one electron per orbital discards this distinction.
Two spinless identical bosons can use two spatial orbitals without an energy restriction. Occupation patterns number:
Allowed patterns are (2,0), (1,1) and (0,2); particle labels do not create separate versions of (1,1).
Count complete states
- For two spatial orbitals a,b and two spin states each, there are four complete single-particle states. Two identical electrons have choose(4,2)=6 occupation patterns: two double-occupation singlets, one different-orbital singlet, and three different-orbital triplets. For three spatial orbitals, there are six complete states and choose(6,2)=15 patterns.
- Counting ordered assignments overcounts identical particles. Spinless bosons in two orbitals instead have three occupations: both in a, one in each, both in b. These counts assume no additional energy restriction and the stated accessible single-particle states; restricting total energy or spin projection changes the allowed subset.
With two orthonormal orbitals a,b, the two-electron state Ψ_+χ_singlet is antisymmetric overall because (+1)(−1)=−1. Ψ_−χ_triplet is also allowed. If both electrons use a, the spatial product a(1)a(2) is symmetric, so only the singlet spin factor is allowed. In a well at positions L/4 and 3L/4, Ψ_−=−2/L gives density 4/L² while Ψ_+=0; each is normalised over the complete two-coordinate domain, rather than interpreted as a probability at one exact pair.
Two electrons access three spatial orbitals, each with two spin states. Complete-state occupation patterns number ____.
Six complete states give choose(6,2)=15.
Interpret the joint density
- An antisymmetric spatial state satisfies Ψ_−(x,x)=0, so its joint position density vanishes on the coincidence line. A symmetric spatial state need not vanish there. The difference arises from interference between exchanged amplitudes, even for noninteracting particles; it is not a separately imposed classical repulsive force.
- For a well of length L, choose a(x)=√(2/L)sin(πx/L), b(x)=√(2/L)sin(2πx/L). At x₁=L/4 and x₂=3L/4, the antisymmetric spatial amplitude is −2/L and its density is 4/L²; the symmetric amplitude is zero. These are joint probability densities per two lengths, not dimensionless probabilities at exact points. Physical spin compatibility still decides which total electronic state uses each spatial combination.
Exchange all spatial and spin arguments. Do not confuse antisymmetric space with antisymmetric total state, or use the distinct-orbital √2 factor for two identical orbitals.
Which answer fits this case?
Construct normalised symmetric and antisymmetric two-orbital spatial states
An antisymmetric two-orbital spatial state must have zero amplitude when the two spatial arguments coincide.
At x₁=x₂ the two products are equal and their difference vanishes.
Keep the distinctions
- spin singlet 自旋单态 — Antisymmetric two-spin-1/2 state with total spin S=0.
- Slater determinant 斯莱特行列式 — Antisymmetric fermionic construction from complete single-particle states; duplicate states make it vanish.
- Construct normalised symmetric and antisymmetric two-orbital spatial states.
- Combine spatial and spin symmetry to produce allowed two-electron states.
- Count complete-state occupations and interpret exchange-induced correlations.
Match each term with its precise meaning in this lesson.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.