Читать книгу Quantum Mechanics, Volume 3 - Claude Cohen-Tannoudji - Страница 165

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

(62)

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

(63)

The orthogonality conditions would then require:

(64)

preventing |dθl〉 from having a component on any ket |θi〉 other than |θl〉: in other words, |dθl〉 and |θl〉 would be colinear. As |θl〉 must remain normalized, the only possible variation would thus be a phase change, which does not affect either the density operator (1) or any average values computed with . This variation does not change anything and is therefore irrelevant.

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 and ):

(65)

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)

The variations (65) are therefore acceptable, for any real value of χ.

We now compute how they change the operator 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)

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

(68)

We now include these variations in the three terms of (61); as the distributions f are unchanged, only the terms and will vary. The infinitesimal variation of is written as:

(69)

As for , it contains two contributions, one from and one from . These two contributions are equal since the operator W₂ (1,2) is symmetric (particles 1 and 2 play an equivalent role). The factor 1/2 in disappears and we get:

(70)

We can regroup these two contributions, using the fact that for any operator O(12), it can be shown that:

(71)

This equality is simply demonstrated⁵ by using the definition of the partial trace Tr₂ {O(1,2)} of operator O(1, 2) with respect to particle 2. We then get:

(72)

Inserting now the expression (68) for , we get two terms, one proportional to eiχ, another one to e–iχ, whose value is:

(73)

Now, for any operator O(1), we can write:

(74)

so that the variation (73) can be expressed as:

(75)

The term in eiχ has a similar form, but it does not have to be computed for the following reason. The variation is the sum of a term in eiχ and another in e–iχ:

(76)

and the stationarity condition requires to be zero for any choice of Choosing χ = 0, yields c₁ + c₂ = 0; choosing χ = π/2, and multiplying by –i, we get c₁ – c₂ = 0. Adding and subtracting those two relations shows that both coefficients c₁ and c₂ must be zero. Consequently, it suffices to impose the terms in e±iχ, and hence expression (75), to be zero. When , the distribution functions fβ are not equal, and we get:

(77)

(if , however, we have not yet obtained any particular condition to be satisfied⁶).