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N (α 2 ) =√ ... ... N! ϕ1 (α N ) ϕ2 (α N ) . . ϕ N (α N ) (3-19) Note that the properties of determinant guarantee that the Pauli exclusion principle is obeyed. Indeed, if two columns in (3-19) are identical, which means that two electrons are in the same state, then the function vanishes giving zero probability for such a situation to happen. In addition to a requirement of an appropriate behavior with respect to the permutations, a many electron wave function has to be also a solution of the eigenvalue problems of all the operators that commute with the Hamiltonian.

This means that the state is described by the quantum numbers S = 1, L = 5, and J = 6. The state is (2J + 1) = 13fold degenerate, and the degenerate states are distinguished by the quantum number M. These states could be described by the kets |S L J M in the so-called RussellSaunders or L S-coupling, where S is coupled to L to give a total angular momentum J . Alternatively the states could be described by the kets defined in an uncoupled momenta scheme, |S M S L M L . These two sets of states correspond to two different bases that are linked by the Clebsch-Gordan coupling coefficients as in (2-8).

However, Hmag does not preserve S O3 symmetry, since a particular direction in 3−space is chosen. The total angular momentum J is no longer a good quantum number. Therefore there are non-zero matrix elements of Hmag that couple the states with J = ±1. We first note that Jz = L z + Sz , and hence L z + gs Sz = Jz + (gs − 1)Sz . But, at the same time the matrix elements of Jz are diagonal in J , and therefore to calculate the off-diagonal matrix elements of Hmag we need only to calculate the off-diagonal matrix element of Sz , αS L J M|S0(1) |αS L J + 1M = (−1) J −M J −M 1 0 J +1 M αS L J S (1) αS L J + 1 (2-55) Explicit evaluation of the 3 j−symbol gives (−1) J −M J −M 1 0 J +1 M =− (J + M + 1)(J − M + 1) (2J + 1)(J + 1)(2J + 3) (2-56) P1: Binod February 15, 2007 5:22 7264 7264˙Book Notes on the Quantum Theory of Angular Momentum 21 Using subsequently (2-45) and (2-37b) the reduced matrix element in (2-55) has the form αS L J S (1) αS L J + 1 = (−1) S+L+J = − J S (2J + 1)(2J + 3) 1 L J +1 S S S (1) S (S + L + J + 2)(S + J + 1 − L)(J + 1 + L − S)(S − J + L) (2-57) 4(J + 1) and finally the off-diagonal matrix element is determined by the following expression αS L J M|Hmag |αS L J + 1M = Bz µ0 (gs − 1) × (J + 1)2 − M 2 (S + L + J + 2)(S + J + 1 − L)(J + 1 + L − S)(S − J + L) (2-58) 4(J + 1)2 (2J + 1)(2J + 3) In a particular case of a 3 P term, with S = 1 and L = 1, we can have J = 0, 1 and 2.

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