Читать книгу Quantum Mechanics, Volume 3 - Claude Cohen-Tannoudji - Страница 163

We now have to compute the grand potential written in (30). As the exponential form (28) for the trial operator makes it easy to compute ln, we see that the terms in μ cancel out, and we get:

(53)

We now have to compute the average energy, with the density operator , of the difference between the Hamiltonians Ĥ and respectively defined by (1) and (25).

We first compute the trace:

(54)

starting with the kinetic energy contribution Ĥ₀ in (1). We call K₀ the individual kinetic energy operator:

(55)

(m is the particle mass). Equality (43) applied to Ĥ₀ yields the average kinetic energy when the system is described by :

(56)

This result is easily interpreted; each individual state contributes its average kinetic energy, multiplied by its population.

The computation of the average value follows the same steps:

(57)

(as in Complement E_XV, operator V₁ is the one-particle external potential operator).

To complete the calculation of the average value of Ĥ, we now have to compute the trace , the average value of the interaction energy when the system is described by . Using relation (51) we can write this average value as a double trace:

(58)

We now turn to the average value of . The calculation is simplified since is, like Ĥ₀, a one-particle operator; furthermore, the |θi〉 have been chosen to be the eigenvectors of with eigenvalues – see relation (26). We just replace in (56), Ĥ₀ by , and obtain:

(59)

Regrouping all these results and using relation (36), we can write the variational grand potential as the sum of three terms:

(60)

with:

(61)