Читать книгу Quantum Mechanics, Volume 3 - Claude Cohen-Tannoudji - Страница 131

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).