2.7 Uniqueness of analytic continuation

The uniqueness of the analytic continuation of n Trρ to non-integer n may not seem obvious, especially if the field system in question is not relativistic so that there is no isometry in the polar angle ϕ, which would allow us, without any trouble, to glue together pieces of the Euclidean space to form a path integral over a conical space C α. However, some arguments can be given that the analytic continuation to non-integer n is in fact unique.

Consider a renormalized density matrix ρ ˆρ = Trρ-. The eigenvalues of ρˆ lie in the interval 0 < λ < 1. If this matrix were a finite matrix we could use the triangle inequality to show that

α α |Trˆρ | < |(Trˆρ) | = 1 if Re (α) > 1.

For infinite-size matrices the trace is usually infinite so that a regularization is needed. Suppose that 𝜖 is the regularization parameter and Tr 𝜖 is the regularized trace. Then

|Tr 𝜖ρˆα | < 1 if Re (α) > 1 . (17 )
Thus α Trˆρ is a bounded function in the complex half-plane, Re(α ) > 1. Now suppose that we know that α Tr𝜖ρ |α=n = Z0 (n) for integer values of α = n, n = 1,2,3,... Then, in the region Re (α) > 1, we can represent Z (α) = Tr𝜖ρα in the form
Z (α) = Z0 (α) + sin(π α)g(α) , (18 )
where the function g(α ) is analytic (for Re(α ) > 1). Since by condition (17View Equation) the function Z(α ) is bounded, we obtain that, in order to compensate for the growth of the sine in Eq. (18View Equation) for complex values of α, the function g (α ) should satisfy the condition
−π|y| |g(α = x + iy)| < e . (19 )
By Carlson’s theorem [36] an analytic function, which is bounded in the region Re (α) > 1 and which satisfies condition (19View Equation), vanishes identically. Thus, we conclude that g(α) ≡ 0 and there is only one analytic continuation to non-integer n, namely the one given by function Z0 (α).
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