Spin Chronicles Part 49: Lurking infinity I

 Tomita-Takesaki theory is usually presented in the environment of a von-Neumann algebra in its "standard form", that is with a cyclic and separating vector. Here we are dealing with a special case. Our von Neumann algebra is the algebra B(H) of all bounded operator on a complex separable Hilbert space H, and it is represented by left translations on the Hilbert space of B₂(H) of Hilbert-Schmidt operators. While for finite-dimensional Hilbert spaces the construction of the Tomita flow is straightforward (no unbounded operators, no need to discuss dense subspaces etc.), here I trying to introduce in a gentle way the necessary extra care for H of infinite dimensions.

John von Neumann

In Part 48, in order to prepare the discussion of the Tomita flow, we have reopened the GNS construction door to introduce the algebra B₂(H) of Hilbert-Schmidt operators on a separable Hilbert space H. Here we will use the notation introduced therein. So far we were dealing only with bounded operators. But here will meet some densely defined unbounded operators. I could have avoided them altogether, But, since these unbounded operators will be not of a dangerous type, there is no reason to be scared. So, first the necessary reminder from functional analysis. Here is the necessary for our purpose extract from the book P. Soltan, A Primer on Hilbert Space Operators, Birkhauser 2018:

Fig. 1. Spectral Integrals

Returning now to Part 48 let us start with solving Exercise 1 at the end of that post:

"Provide the explicit formula implementing the unitary isomorphism U: H_ω → B₂(H) mentioned in Remark 4. Show that U maps B(H) onto o proper subspace of B₂(H)."

This will prepare us for the following discussion. The solution can be, in fact, found the formula  Uρ(a)Ω = ρ'(a)Ω'  of Part 42. Except that there we were denoting the representation by ρ, while here ρ stands for the density matrix and the representation is denoted by π. So, let us rewrite this formula using our current notation. Both representations are just left actions:

UAΩ_ω = AΩ_ρ, A ∈ B(H).

But Ω_ω = 1, Ω_ρ = ρ^½. 

Therefore our formula reads

UA = Aρ^½. 

Remark 1. Here I have denoted U with a bold letter to stress the fact that (using the terminology borrowed from Ilya Prigogine) that we are dealing with a "superoperator", that is linear operator acting on operators.

This defines U on H_ω = B(H). We easily check that <UX,UY> = (X,Y)_ω. (Do it!). Therefore U is a bounded operator from H_ω  to B₂(H).  Therefore it extends by continuity to an isometry defined  on the completion H_ω  of  H_ω.

Now, what about U^-1? Here it will be useful to discuss a little bit the properties of ρ. It has a finite trace, so it is, in particular, a compact operator. As such it has discrete eigenvalues, and, since we assume that it defines a faithful state, zero is not an eigenvalue. Thus there exists an orthonormal basis e_n (n=1,2,....) consisting of eigenvectors of ρ in H such that

ρ = ∑_n p_n P_n,

where P_n are orthogonal projections on e_n.

Since ρ is positive of trace one, we have p_n>0, ∑_n p_n = 1. The eigenvalues p_n are not necessarily all different, but each eigenvalue enters with a finite multiplicity. (Why?)

Now, we will need also U^-1. But how to define it precisely? It is easy to define ρ^½  (How?), and it is a bounded operator (Why?). But what about ρ^-½? It is an unbounded self-adjoint operator (Why?), defined only on a dense domain (Which one? Write the spectral integral for it, Hint: ∫ becomes ∑ in our case. Why?).

And to define U^-1  it is useful to have an orthonormal basis in B₂(H). How to construct it?

To be continued...