Quantum Mechanics, Volume 3. Claude Cohen-Tannoudji

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is already occupied, we must first permute the states to bring |ui〉 to the first position, which may eventually change the sign in front of the Fock space ket. The successive application on this ket of (A-25) and (A-19) shows that the action of the operator leaves this ket unchanged; we then move the state |ui〉 back to its initial position, which may introduce a second change in sign, canceling the first one. We finally obtain for fermions the same result as for bosons, except that the ni can only take the values 1 and 0. In both cases the Fock states are the eigenvectors of the operator with the occupation numbers as eigenvalues; consequently, this operator is named the “occupation number operator of the state |ui〉”. The operator associated with the total number of particles is simply the sum:

(A-29)

A-5. Commutation and anticommutation relations

      Creation and annihilation operators have very simple commutation (for the bosons) and anticommutation (for the fermions) properties, which make them easy tools for taking into account the symmetrization or antisymmetrization of the state vectors.

      To simplify the notation, each time the equations refer to a single basis of individual states |ui〉, we shall write ai instead of aui. If, however, it can lead 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

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