Spin Chronicles Part 50: Lurking infinity II

 We continue from Part 49.

Is all quantum? Or it is rather like this:

The Information Field

Let us recall the notation used there: H is a separable Hilbert space, with the scalar product (x,y), A = B(H) is the von Neumann algebra of all bounded operators on H, ρ is a (faithful) density matrix in the ideal of trace-class operators, B₂(H) is the ideal of Hilbert-Schmidt operators with the scalar product <X,Y> = Tr(X*Y), we have B₁(H)⊂B₂(H)⊂B(H), e_n (n=1,2,....) is an orthonormal basis consisting of eigenvectors of ρ in H, so that

ρ = ∑_n p_n P_n, p_n>0,  ∑_n p_n = 1,

where P_n are orthogonal projections on e_n. 

Notice that if H is infinite-dimensional, then lim_n p_n = 0 (Why?), so that, in this case 0 is an accumulation point in the spectrum of ρ.

It is useful to have an orthonormal basis in the Hilbert algebra B₂(H). To this end we define e_m,n∈B₂(H)

e_m,n ≐ |e_m)(e_n|

where

|e_m)(e_n| x = (e_n,x)e_m.

Exercise 1. Prove that e_m,n∈B₂(H).

Then they indeed form an orthonormal basis in B₂(H).

Exercise 2. Prove the last statement. Hint: to show that we have a basis consider the space of all finite linear combinations of E_mn. Use the fact that a subspace is dense in a Hilbert space if and only if the only vector orthogonal to all vectors of the subspace is the zero vector.

Exercise 3☺. Show that, with the notation as above, P_n = e_m,n.

In the following we shall employ a convenient notation used by Connes and Rovelli in their paper [1]. Given an element a∈A, it can be considered as a bounded operator in B(H). But if it is a Hilbert-Schmidt operator, it is an element of the Hilbert space B₂(H). In the later case we will write it as | a >. Thus, for example,

  Ω_ρ = | ρ^½ >.         (1)

The representations π and π' of A on B₂(H) become resp.

π(a)| b > = | ab >, π'(a)| b > = | ba >.

For the scalar product we can write

<a,b> = < a | b > = Tr(a*b).

The Tomita-Takesaki construction provides us with two (super-) operators: the anti-unitary involution J, and the unitary "Tomita flow" s ⟼ U_ρ(s), s ∈ R.

The involution J

We consider first the anti-unitary (super-) operator

J: B₂(H) → B₂(H),

defined by

J | a > = | a* >.                     (2)

We immediately get the anti-unitary property (How?):

<JS , JT> = cc(<T , S>),    S,T in B₂(H),

where cc stands for the complex conjugate.

We also have J² = 1 (the identity operator). Moreover, we have

JΩ_ρ = Ω_ρ , for any density matrix ρ.

Notice that, in our context, J does not depend on ρ - it is "universal".

We have

Jπ(A)J = π'(A).

Indeed, for any a∈A, b∈B₂(H), we have

Jπ(a)J| b > = Jπ(a)| b* > = J| ab* > = | ba*> = π'(a*) | b >.

Therefore Jπ(a)J = π'(a*), and the equality Jπ(A)J = π'(A) follows (Why?). But π'(A) = π(A)', therefore J transforms the von Neumann algebra π(A) into its commutant, and vice versa (as it follows from J²=1).

Tomita's thermal flow

Tomita's flow U_ρ(s) is defined by the formula:

U_ρ(s)| a > = | ρ^isaρ^-is >, a∈B₂(H), s∈R.      (3)

The formula above requires an explanation. Here it comes. First of all what is ρ^is? Here we use Fig. 1 of Part 49, with integrals replaced by infinite sums. Since
ρ = Σ_n p_n P_n, is a spectral resolution of ρ, ρ^is is defined as:

ρ^is = ∑_n p_n^is P_n.

But what is λ^is (here for λ>0)? It can be defined as

λ^is = e^{is log λ},

where log stands for the natural logarithm.

For λ>0 and s real, it is a complex number of modulus 1, thus nothing special. We thus have

ρ^is = ∑_n e^{is log p_n} P_n.          (4)

It follows then from the last statement in Fig 1 that ρ^is is a unitary operator in B(H) (How?). Moreover, denoting

U_ρ(s) = ρ^is  = e^{is log(ρ)},       (5)

we have (How?)

U_ρ(s) U_ρ(s') = U_ρ(s+s'),

so that we have a one-parameter group of unitary operators on H. Denoting

α_s(a) = U_ρ(s) a U_ρ(s)*,       a∈A,

we have a one-parameter group of *-automorphisms of A. It is called the group of modular automorphisms. It is this group that is called the Tomita modular flow. We notice that (Why?)

U_ρ(s) Ω_ρ = Ω_ρ,

so that the vector Ω_ρ, representing the state, is invariant under U_ρ(s). We can also write it as the invariance of the state ω under the modular flow

ω(α_s(a)) = ω(a),     s∈R.

References

[1] A. Connes, C. Rovelli, "Von Neumann algebra automorphisms and time-thermodynamics relation in generally covariant quantum theories", Class. Quantum Grav. 11 (1994) 2899 .

To be continued ....

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...