Question

(a) A 2-fermion system has a wave function ψ(l, 2). What condition must it satisfy if the particles are identical?

(b) How does this imply the elementary statement of the Pauli exclusion principle that no two electrons in an atom can have identical quantum numbers?

(c) The first excited state of Mg has the configuration (3s,3p) for the valence electrons. In the LS-coupling limit, which values of L and S are possible? What is the form of the spatial part of their wave functions using the single-particle functions ψ_s(r) and \(\phi\)_p(r)? Which will have the lowest energy, and Why?

Answer 1

(a) ψ(1, 2) must satisfy the condition of antisymmetry for interchange of the two particles:

(b) In an atom, if there are two electrons having identical quantum numbers then ψ(1, 2) = ψ(2,1). The antisymmetry condition above then gives ψ(1, 2) = 0, implying that such a state does not exist.

(c) The electron configuration (3s, 3p) correspond to

The lowest energy state is \(\Psi_1^1\)(1, 2), i.e. the state of S = 1. Because the spatial part of the state S = 1 is antisymmetric for the interchange 1 \(\leftrightarrow\) 2, the probability that the two electrons get close together is small, and so the Coulomb repulsive energy is small, resulting in a lower total energy.