Quantum Mechanics, Volume 3. Claude Cohen-Tannoudji

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alt="image"/> (j = 1, 2, ..N) we will find products of N operators . Relation (A-43) of Chapter XV however indicates that the squares of any creation operators are zero, which means that the only non-zero products are those including once and only once each of the N different operators . Each term is then proportional to the ket built from the . Consequently, the two variational kets built from the two bases are necessarily proportional. As definition (1) ensures they are also normalized, they can only differ by a phase factor, which means they are equivalent from a physical point of view. It is thus the operator that best embodies the trial ket .

2-b. Optimization of the one-particle density operator

      We now vary to look for the stationary conditions for the total energy:

      (62)

      We therefore consider the variation:

      (63)

      which leads to the following variations for the average values of the one-particle operators:

      (64)

      As for the interaction energy, we get two terms:

      (65)

      which are actually equal since:

      (66)

      and we recognize in the right-hand side of this expression the trace:

      (67)

      As we can change the label of the particle from 2 to 1 without changing the trace, the two terms of the interaction energy are equal. As a result, we end up with the energy variation:

(68)

      To vary the projector PN, we choose a value j₀ of j and make the change:

      (69)

where |δθ〉 is any ket from the space of individual states, and χ any real number; no other individual state vector varies except for |θ_j0〉. The variation of PN is then written as:

(70)

      We assume |δθ〉 has no components on any |θi〉, that is no components in , since this would change neither εN, nor the corresponding projector PN. We therefore impose:

(71)

which also implies that the norm of |θ_j0〉 remains constant⁶ to first order in |δθ〉. Inserting (70) into (68), we obtain:

      (72)

      For the energy to be stationary, this variation must remain zero whatever the choice of the arbitrary number χ. Now the linear combination of two exponentials e^iχ and e^–iχ will remain zero for any value of χ only if the two factors in front of the exponentials are zero themselves. As each term can be made equal to zero separately, we obtain:

      (73)

      This relation must be satisfied for any ket |δθ〉 orthogonal to the subspace . This means that if we define the one-particle Hartree-Fock operator as:

(74)

      the stationary condition for the total energy is simply that the ket HHF |θ_j0〉 must belong to :

(75)

As this relation must hold for any |θ_j0〉 chosen among the |θ₁〉, |θ₂〉, ….|θN〉, it follows that the subspace is stable under the action of the operator (74).

2-c. Mean field operator

      We can then restrict the operator HHF to that subspace:

      (76)

This operator, acting in the subspace spanned by the N kets |θj〉, is a Hermitian linear operator, hence it can be diagonalized. We call |φn〉 its eigenvectors (n = 1, 2, ..N), which are linear combinations of the kets |θj〉. The stationary condition for the energy

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