Quantum Mechanics, Volume 3. Claude Cohen-Tannoudji

      This ket is always the same for all the numbers q among the nl selected ones (for fermions, this term might be zero, if the state |uk〉 was already occupied in the initial ket). We shall then distinguish two cases:

(i) For k ≠ l, and for bosons, the ket written in (B-8) equals:

      (B-9)

      where the square root factor comes from the variation in the occupation numbers nk and nl, which thus change the numerical coefficients in the definition (A-7) of the Fock states. As this ket is obtained nl times, this factor becomes . This is exactly the factor obtained by the action on the same symmetrized ket of the operator , which also removes a particle from the state |ul〉 and creates a new one in the state |uk〉. Consequently, the operator reproduces exactly the same effect as the sum over q.

      For fermions, the result is zero except when, in the initial ket, the state |ul〉 was occupied by a particle, and the state |uk〉 empty, in which case no numerical factor appears; as before, this is exactly what the action of the operator would do.

      (ii) if k = l, for bosons the only numerical factor involved is nl, coming from the number of terms in the sum over q that yields the same symmetrized ket. For fermions, the only condition that yields a non-zero result is for the state |ul〉 to be occupied, which also leads to the factor nl. In both cases, the sum over q amounts to the action of the operator .

      We have shown that:

      (B-10)

The summation over k and l in (B-5) then yields:

(B-11)

      B-2-b. Expression valid in the entire Fock space

The right-hand side of (B-11) contains an expression completely independent of the space S(N) or A(N) in which we defined the action of the operator . Since we defined operator as acting as in each of these subspaces having fixed N, we can simply write:

(B-12)

This is the expression of one-particle symmetric operators we were looking for. Its form is valid for any value of N and the particles are no longer numbered; it contains equal numbers of creation and annihilation operators, which only act on occupation numbers.

       Comment:

      Choosing the proper basis {|ui〉} it is always possible to diagonalize the Hermitian operator and write:

      (B-13)

      Equality (B-11) is then simply written as:

      (B-14)

where is the occupation number operator in the state |uk〉 defined in (A-28).

B-3. Examples

A first very simple example is the operator , already described in (A-29), and corresponding to the total number of particles:

      (B-15)

      As expected, this operator does not depend on the basis {|ui〉} chosen to count the particles, as we now show. Using the unitary transformations of operators (A-51) and (A-52), and with the full notation for the creation and annihilation operators to avoid any ambiguity, we get:

      (B-16)

      which shows that:

      (B-17)

      For a spinless particle one can also define the operator corresponding to the probability density at point r₀:

      (B-18)

Relation (B-12) then leads to the “particle local density” (or “single density”) operator:

(B-19)

      The same procedure as above shows that this operator is independent of the basis {|ui〉} chosen in the individual states space.

Let us assume now that the chosen basis is formed by the eigenvectors |K_i〉 of a particle’s momentum ħk_i, and that the corresponding annihilation operators are noted a_ki. The operator associated with the total momentum of the system can be written as:

      (B-20)

      As for the kinetic energy of the particles, its associated operator is expressed as:

      (B-21)

B-4.