Quantum Mechanics, Volume 3. Claude Cohen-Tannoudji

to ambiguity, we will return to the full notation.

      A-5-a. Bosons: commutation relations

      Consider, for bosons, the two operators and . If both subscripts i and j are different, they correspond to orthogonal states |ui〉 and |uj〉. Using twice (A-16) then yields:

      (A-30)

      Changing the order of the operators yields the same result. As the Fock states form a basis, we can deduce that the commutator of and is zero if i ≠ j. In the same way, it is easy to show that both operator products aiaj and ajai acting on the same ket yield the same result (a ket having two occupation numbers lowered by 1); ai and aj thus commute if i ≠ j. Finally the same procedure allows showing that ai and commute if i ≠ j. Now, if i = j, we must evaluate the commutator of ai and . Let us apply (A-16) and (A-22) successively, first in that order, and then in the reverse order:

      (A-31)

      The commutator of ai and is therefore equal to 1 for all the values of the subscript i. All the previous results are summarized in three equalities valid for bosons:

(A-32)

      A-5-b. Fermions: anticommutation relations

      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)

      A-5-c. Common relations for bosons and fermions

      To regroup the results valid for bosons