Читать книгу Quantum Mechanics, Volume 3 - Claude Cohen-Tannoudji - Страница 155

Consider two density operators ρ and ρ′, both having a trace equal to 1:

(7)

As we now show, the following relation is always true:

(8)

We first note that the function x lnx, defined for x ≥ 0, is always larger than the function x – 1, which is the equation of its tangent at x = 1 (Fig. 1). For positive values of x and y we therefore always have:

(9)

or, after multiplying by y:

(10)

the equality occurring only if x = y.

Figure 1: Plot of the function x lnx. At x = 1, this curve is tangent to the line y = x – 1 (dashed line) but always remains above it; the function value is thus always larger than x – 1.

Let us call pn the eigenvalues of ρ corresponding to the normalized eigenvectors |un〉, and the eigenvalues of ρ′ corresponding to the normalized eigenvectors |vm〉. Used for the positive numbers pn and , relation (10) yields:

(11)

We now multiply this relation by the square of the modulus of the scalar product:

(12)

and sum over m and n. For the term in pn lnpn of (11), the summation over m yields in (12) the identity operator expanded on the basis {|v_m〉}; we then get 〈un|un〉 = 1, and are left with the sum over n of pn lnpn, that is the trace Tr{ρ lnρ}. As for the term in pn ln, the summation over m introduces:

(13)

and we get:

(14)

As for the terms on the right-hand side of inequality (11), the term in pn yields:

(15)

and the one in also yields 1 for the same reasons, and both terms cancel out. We finally obtain the inequality:

(16)

which proves (8).

Comment:

One may wonder under which conditions the above relation becomes an equality. This requires the inequality (11) to become an equality, which means whenever the scalar product (12) is non-zero; consequently all the eigenvalues of the two operators ρ and ρ′ must be equal. In addition, the eigenvectors of each operator corresponding to different eigenvalues must be orthogonal (their scalar product must be zero). In other words, the eigenvalues and the subspaces spanned by their eigenvectors are identical, which amounts to saying that ρ = ρ′.