2.7 Uniqueness of analytic continuation
The uniqueness of the analytic continuation of
to non-integer
may not seem obvious,
especially if the field system in question is not relativistic so that there is no isometry in the polar angle
, which would allow us, without any trouble, to glue together pieces of the Euclidean space to form a
path integral over a conical space
. However, some arguments can be given that the analytic
continuation to non-integer
is in fact unique.
Consider a renormalized density matrix
. The eigenvalues of
lie in the interval
. If this matrix were a finite matrix we could use the triangle inequality to show
that
For infinite-size matrices the trace is usually infinite so that a regularization is needed. Suppose that
is
the regularization parameter and
is the regularized trace. Then
Thus
is a bounded function in the complex half-plane,
. Now suppose that we know
that
for integer values of
. Then, in the region
, we
can represent
in the form
where the function
is analytic (for
). Since by condition (17) the function
is
bounded, we obtain that, in order to compensate for the growth of the sine in Eq. (18) for complex values
of
, the function
should satisfy the condition
By Carlson’s theorem [36] an analytic function, which is bounded in the region
and
which satisfies condition (19), vanishes identically. Thus, we conclude that
and
there is only one analytic continuation to non-integer
, namely the one given by function
.