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:
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:
and thus leads to variations of expressions (61) of and . Their sum is:
(79)
where the factor 1/2 in has been canceled since the variations induced by and double each other. Inserting (78) in this relation and using again (74), we get:
(80)
As for , its variation is the sum of a term in coming from the explicit presence of the energies in its definition (61), and a term in . If we let only the energy vary (not taking into account the variations of the distribution function), we get a zero result, since:
(81)
Consequently, we just have to vary by the distribution function, and we get:
(82)
Finally, after simplification by (which, by hypothesis, is different from zero), imposing the variation to be zero leads to the condition:
