Quantum Mechanics, Volume 3. Claude Cohen-Tannoudji

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choose a basis {|ui〉} for the individual states. The matrix elements fkl of the one-particle operator are given by:

      (B-3)

      They can be used to expand the operator itself as follows:

(B-4)

      B-2-a. Action of F^(N) on a ket with N particles

Using in (B-1) the expression (B-4) for leads to:

(B-5)

The action of on a symmetrized ket written as (A-9) therefore includes a sum over k and l of terms:

      (B-6)

      with coefficients fkl. Let us use (A-7) or (A-10) to compute this ket for given values of k and l. As the operator contained in the bracket is symmetric with respect to the exchange of particles, it commutes with the two operators SN and AN (§ C-4-a-β of Chapter XIV)), and the ket can be written as:

      (B-7)

In the summation over q, the only non-zero terms are those for which the individual state |ul〉 coincides with the individual state |um〉 occupied in the ket on the right by the particle labeled q; there are nl different values of q that obey this condition (i.e. none or one for fermions). For these nl terms, the operator |q : |uk〉 〈q : ul| transforms the state |um〉 into |ui〉, then SN (or AN) reconstructs a symmetrized (but not normalized) ket:

(B-8)

      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)

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