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.
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| 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:
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| 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: Hω → 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:
UAΩω = AΩρ, A ∈ B(H).
But
Ωω =
1, Ωρ
= ρ½.
Therefore our formula reads
UA
= Aρ½.
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 Hω
=
B(H).
We
easily check that <UX,UY> =
(X,Y)ω.
(Do it!). Therefore U
is a bounded operator from Hω
to B2(H). Therefore it extends by
continuity to an isometry defined on the completion Hω
of
Hω.
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 en (n=1,2,....) consisting of eigenvectors of ρ in H such that
ρ = ∑n pn Pn,
where Pn are orthogonal projections on en.
Since ρ is positive of trace one, we have pn>0, ∑n pn = 1. The eigenvalues pn 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 ρ½ (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...
P.S. 15-03-25 9:35 I know that some Readers do not feel at home with all these infinite dimensions. But do not worry. Very soon I will return to the finite-dimensional home. As I have promised, we will move back to the Clifford algebras - the Clifford algebras for the space of spheres in 3D. Spheres are much more interesting than points - they have their boundaries and their interiors. They may serve as windows to other worlds.
"...some Readers do not feel at home with all these infinite dimensions".
ReplyDeleteI don't feel at home with functional analysis in general (we never studied it regularly, only some sprouts in the course of QM), so there is no much difference for me between treating finite and infinite cases :)
But i realized at last why people had to use all these Hilbert-Schmidt operators with finite norm - because they imply a descrete spectrum of eigenvalues! Just what was needed to describe quantum world instead of continuous classic. As soon as we bound or truncate a function, it gets obliged to have a discrete spectrum.
"The eigenvalues p_n are not necessarily all different, but each eigenvalue enters with a finite multiplicity. (Why?)"
Intuitively, because an infinite number of eigenvalues would sum up to an unbounded value, which is the trace of ρ, and it is assumed to be finite. Probably.
In order to define ρ½ and ρ-½, should we use some information from Fig.1?
Both are correct. Concerning Fig. 1, its use will become more clear in Part 50, which I am going to post today in a couple of hours.
DeleteConcerning functional analysis, I have studied it only from books during my 3 years of working on my PhD. But at that time I attended a course for mathematics students on harmonic analysis. But during my regular studies we had a seminar based von Neumann's book on foundations of quantum theory. It was a pretty good introduction to Hilbert spaces.
"It is easy to define ρ½ (How?)'
ReplyDeleteMay be, we can act similarly to as we did in Part 43 and take:
ρ(a) = , where L(a) is the restiriction of the left action to the space of left ideal, i.e. to B2(H) in this case. The operator which acts from Hω to B2(H) is U, so probably we can take it here to obtain
ρ(a) = , but i don't know whether this will do as a definition of ρ½ or not.