Читать книгу Quantum Mechanics, Volume 3 - Claude Cohen-Tannoudji - Страница 28

For fermions, let us first assume that the subscripts i and j are different. The successive action of and on an occupation number ket only yields a non-zero ket if ni = nj = 0; using twice (A-18) leads to:

(A-33)

but, if we change the order:

(A-34)

Consequently the sign change that goes with the permutation of the two individual states leads to:

(A-35)

If we define the anticommutator [A, B]₊ of two operators A and B by:

(A-36)

(A-35) may be written as:

(A-37)

Taking the Hermitian conjugate of (A-35), we get:

(A-38)

which can be written as:

(A-39)

Finally, we show by the same method that the anticommutator of ai and is zero except when it acts on a ket where ni = 1 and nj = 0; those two occupation numbers are then interchanged. The computation goes as follows:

(A-40)

and:

(A-41)

Adding those two equations yields zero, hence proving that the anticommutator is zero:

(A-42)

In the case where i = j, the limitation on the occupation numbers (0 or 1) leads to:

(A-43)

Equalities (A-37) and (A-39) are still valid if i and j are equal. We are now left with the computation of the anticommutator of ai and . Let us first examine the product ; it yields zero if applied to a ket having an occupation number ni = 1, but leaves unchanged any ket with ni = 0, since the particle created by is then annihilated by ai. We get the inverse result for the product where the order has been inverted: it yields zero if ni = 0, and leaves the ket unchanged if ni = 1. Finally, whatever the occupation number ket is, one of the terms of the anticommutator yields zero, the other 1 , and the net result is always 1 . Therefore:

(A-44)

All the previous results valid for fermions are summarized in the following three relations, which are for fermions the equivalent of relations (A-32) for bosons:

(A-45)