What is the Hilbert dimension of a Fock space?

Quantum field theory in curved spacetimes is often described in the algebraic approach, which consists of describing observables as elements of a certain $$*$$-algebra. To recover the notion of a Hilbert space, one represents this algebra as operators acting on said Hilbert space. Given a state on the algebra, the GNS construction allows one to obtain a particular representation of this algebra. Given two different states, such as the Minkowski and Rindler vacua in Minkowski spacetime, it might happen that the representations are not unitarily equivalent.

What I find curious, though, is the existence of the following theorem in Functional Analysis (see Kreyszig's Introductory functional analysis with applications Theorem 3.6-5)

Two Hilbert spaces $$H$$ and $$\tilde{H}$$, both real or both complex, are isomorphic if and only if they have the same Hilbert dimension.

The Hilbert dimension is the cardinality of an orthonormal basis of the Hilbert space.

Now my question is: how to make sense of this? For example, both the Minkowski and Rindler vacua lead to representations in Fock spaces. Isn't a Fock space always separable, and hence has Hilbert dimension $$\aleph_0$$? Shouldn't then any two Fock space representations be unitarily equivalent? In particular, shouldn't the Minkowski and Rindler vacua lead to unitarily equivalent representations? Why don't they?

• I might be saying something stupid, but could it be that the isomorphism guaranteed by the theorem is not necessarily unitary? Commented Jan 3 at 13:03
• @LucasBaldo That's a good first guess, but Kreyszig defines an isomorphism between Hilbert spaces as always being unitary. Another Functional Analysis book I consulted (in Portuguese) is explicit about the isomorphism being unitary Commented Jan 3 at 13:07
• Does this answer your question? What is the importance of unitary (in-)equivalent representations? Commented Jan 3 at 13:08
• @LucasBaldo Any separable, infinite-dimensional complex Hilbert spaces $H_1$ and $H_2$ are isomorphic; here, "isomorphic" means exactly that there exists a unitary operator such that for all $\psi\in H_2$ it holds that there exists a unique $\varphi\in H_1$ with $\psi=U\varphi$. Similarly, any two finite-dimensional complex Hilbert spaces are isomorphic in the same sense. Commented Jan 3 at 13:09
• @NíckolasAlves The Hilbert space dimension of a (bosonic or fermionic) Fock space $F(\mathfrak h)$ -- corresponding to/constructed by some single-particle Hilbert space $\mathfrak h$-- is always infinite-dimensional if $\mathfrak h$ is infinite-dimensional. If $\mathfrak h$ is finite-dimensional, then the bosonic Fock space is still infinite-dimensional; for fermions, the dimension is $2^{\mathrm{dim}\mathfrak h}$. Commented Jan 3 at 13:12

In addition to an isomorphism of Hilbert spaces $$U : H_1 \to H_2$$, for unitary equivalence of representations we have two representations $$\pi_i : A\to \mathfrak{gl}(H_i)$$ (where $$A$$ is the algebra of fields or any other equivalent presentation of the CCR), and $$U$$ is required to be an intertwiner between these, i.e. $$\pi_2(a)U = U\pi_1(a)$$ for all $$a\in A$$.
The statement that certain spaces like the Minkowksi and Rindler Fock spaces are not unitarily equivalent is not about the non-existence of an isomorphism of Hilbert spaces $$U$$, it's about no such $$U$$ fulfilling the intertwiner condition.