Quantum Mechanics, Volume 3. Claude Cohen-Tannoudji

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Quantum Mechanics, Volume 3 - Claude Cohen-Tannoudji


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been chosen to be the eigenvectors of image with eigenvalues image – see relation (26). We just replace in (56), Ĥ0 by image, and obtain:

      (59)image

      (60)image

      with:

      We now vary the eigenenergies image and eigenstates |θk〉 of image to find the value of the density operator image that minimizes the average value image of the potential. We start with the variations of the eigenstates, which induce no variation of image. The computation is actually very similar to that of Complement EXV, with the same steps: variation of the eigenvectors, followed by the demonstration that the stationarity condition is equivalent to a series of eigenvalue equations for a Hartree-Fock operator (a one-particle operator). Nevertheless, we will carry out this computation in detail, as there are some differences. In particular, and contrary to what happened in Complement EXV, the number of states |θi〉 to be varied is no longer fixed by the particle number N; these states form a complete basis of the individual state space, and their number can go to infinity. This means that we can no longer give to one (or several) state(s) a variation orthogonal to all the other |θj〉; this variation will necessarily be a linear combination of these states. In a second step, we shall vary the energies image.

       α. Variations of the eigenstates

      As the eigenstates |θi〉 vary, they must still obey the orthogonality relations:

      (62)image

      The simplest idea would be to vary only one of them, |θl〉 for example, and make the change:

      (63)image

      The orthogonality conditions would then require:

      (64)image

      It is actually more interesting to vary simultaneously two eigenvectors, which will be called |θl〉 and |θm〉, as it is now possible to give |θl〉 a component on |θm〉, and the reverse. This does not change the two-dimensional subspace spanned by these two states; hence their orthogonality with all the other basis vectors is automatically preserved. Let us give the two vectors the following infinitesimal variations (without changing their energies image and image):

      where da is an infinitesimal real number and χ an arbitrary but fixed real number. For any value of χ, we can check that the variation of 〈θl |θl〉 is indeed zero (it contains the scalar products 〈θl |θm〉 or 〈θm |θl〉 which are zero), as is the symmetrical variation of 〈θm |θm〉, and that we have:

      (66)image

      We now compute how they change the operator image defined in (40). In the sum over k, only the k = l and k = m terms will change. The k = l term yields a variation:

      (67)image

      whereas the k = m term yields a similar variation but where image is replaced by image. This leads to:

      (69)image

      As for image, it contains two contributions, one from image and one from image. These two contributions


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