The Pauli Exclusion Principle — Question 8

Question 8

A fixed electron number gives a combinatorial many-particle state count

A truncated one-electron space contains three spatial orbitals, each with two spin states. Place two identical electrons in this six-spin-orbital space.

1. Count the independent two-electron Slater determinants.

2. Count the determinants with both spins up, both down, and one of each, and verify the total.

Question 8 - Solution

A fixed electron number gives a combinatorial many-particle state count

(a) Method and calculation

$$\boxed{d=\binom62=6(5)/2=15}.$$$$\text{The factor }2!\text{ removes ordering of occupied spin orbitals}.$$

(b) Interpretation and further calculation

$$\boxed{d_{\uparrow\uparrow}=\binom32=3,\quad d_{\downarrow\downarrow}=3,\quad d_{\uparrow\downarrow}=3\times3=9}.$$$$\boxed{3+3+9=15}.$$

Physical check

These are basis determinants, not necessarily distinct energy levels. Opposite spins may occupy the same spatial orbital, which is included in the nine mixed-spin choices.