Quantum Mechanics, Volume 3. Claude Cohen-Tannoudji

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rel="nofollow" href="#fb3_img_img_fae1fb00-869c-57d7-a514-d9c26efaa34e.jpg" alt="image"/> on |θ〉 must have zero components on all the vectors orthogonal to |θ〉; its only non-zero component must be on the ket |θ〉 itself, which means it is necessarily proportional to |θ〉. In other words, |θ〉 must be an eigenvector of that operator, with eigenvalue μ (real since the operator is Hermitian):

      (43)image

      We have just shown that the optimal value |φ〉 of |θ〉 is the solution of the Gross-Pitaevskii equation:

       Comment:

      The Gross-Pitaevskii operator image is simply a partial trace over the second particle:

      where (2) is the projection operator (2) of the state of particle 2 onto |θ〉:

      (47)image

      (48)image

      which is simply the initial definition (40) of image. Relation (45) is therefore another possible definition for the Gross-Pitaevskii potential.

      We have established which conditions the variational wave function must obey to make the energy stationary, but we have yet to study the actual value of this energy. This will allow us to show that the parameter μ is in fact the chemical potential associated with the system of interacting bosons. We shall then introduce the concept of a relaxation (or “healing”) length, and discuss the effect, on the final energy, of the fragmentation of a single condensate into several condensates, associated with distinct individual quantum states.

      (49)image

      (50)image

      which is simply twice the potential interaction energy given in (33) when |θ1〉 = |φ〉. This leads to:

      To find the energy image, note that N μ/2 is the sum of image and of half the kinetic and external potential energies. Adding the missing halves, we finally get for image:

      (52)image

      An advantage of this formula is to involve only one- (and not two-) particle operators, which simplifies the computations. The interaction energy is implicitly contained in the factor μ.

      (53)image

      For large N, one can safely replace in this equation (N — 1/2) by (N — 1); after multiplication by N, we obtain a sum of average energies:

      (54)image

      (55)image

      We know (Appendix VI, § 2-b) that in the grand canonical ensemble, and at zero temperature, the derivative of the energy with respect to the particle number (for a fixed volume) is equal to the chemical potential. The quantity μ introduced mathematically as a Lagrange multiplier, can therefore be simply

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