Spin Chronicles Part 50: Lurking infinity II

Sunday, March 16, 2025

Spin Chronicles Part 50: Lurking infinity II

We continue from Part 49.

Is all quantum? Or it is rather like this:

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 ....

P.S. 17-03-25 7:16 Changed the notation in the original post. Replaced P_m,nby e_m,n. P_m,nwas suggestive of "Projectors". But these are basis vectors of B₂(H), and not projectors. Notice that e_m,n* = e_n,m.

P.S. 17-03-25 10:46

Screenshot for Comment 10:01

P.S, 17-03-25 11:01 Reply to Anna's Comment 9:22:

"why do we need two different types of operators "trace-class" and "Hilbert-Schmidt" and what is the subspace B1 here: B1(H)⊂B2(H)⊂B(H)?"

Trace class operators - B₁(H) (sometimes denoted also L¹(H)) come naturally for the description of mixed states. There is Gleason's theorem from which it follows that any reasonable state on B(H) is described by a trace class density matrix. On the other hand we have a natural scalar product on operators given by Tr(A*B), with trace norm |||A|||² = Tr(A*A). Hilbert-Schmidt operators - B₂(H) - are those for which this norm is finite. Then one can prove that trace class is contained in Hilbert-Schmidt class, that both classes are two-sided ideals in B(H), and that B₂(H) is a completion of B₁(H) with respect to the trace norm. On the other hand B₂(H), while complete with respect to the trace norm, is incomplete with respect to the standard operator norm ||A|| in B(H). Its completion gives the ideal of compact operators K(H) (for Hilbert spaces these are the same as as completely continuous operators). Here we are not using this last two-sided ideal, the only two-sided ideal of B(H) closed with respect to the operator norm. The quotient B(H)/K(H) is a quite remarkable C*-algebra, the Calkin algebra, that does not have a faithful representation in any separable Hilbert space.

P.S. 19-03-25 15:16 Extract from J.P. McCarthy web page "Von Neumann Algebras: The Double Commutant Theorem"

If $C$ is a subset of an algebra $A$ , we define its commutant $C^\prime$ to be the set of all elements of $A$ that commute with all the elements of $C$ .

Observe that $C^\prime$ is a subalgebra of $A$ (straightforward).

The double commutant $C^{\prime\prime}$ of $C$ is $(C^\prime)^\prime$ .
Similarly, $C^{\prime\prime\prime}=(C^{\prime\prime})^\prime$ .

Always $C\subset C^{\prime\prime}$ and $C^{\prime}\subset C^{\prime\prime\prime}$

(let $c\in C$ . If $a\in C^\prime$ then $ac=ca$ . That is $c\in C^\prime$ '.

Now let $a\in C^\prime$ so that for all $c\in C$ , we have $ac=ca$ .
Now for any $b\in C^{\prime\prime}$ , $ab=ba$ so $a$ commutes with everything in $C^{\prime\prime}$ .
That is $a\in C^{\prime\prime\prime}$ .

The text claims that in fact $C^\prime=C^{\prime\prime\prime}$ .
Now let $a\in C'''$ . Now $ab=ba$ , for all $b\in C''$ .
However $C\subset C''$ so therefore $ab=ba$ for all $b\in C$ .
That is $a\in C'$ also.).
If $A$ is a normed algebra, then $C^\prime$ is closed (with more structure we could show this as follows.
Define $f_c:A\rightarrow A$ by $a\mapsto ac-ca$ . Now $f_c^{-1}(\{0\})$ is closed as is $\bigcap_{c\in C}f_c^{-1}(\{0\})=C^\prime$ .).
If $A$ is a *-algebra and $C$ is self-adjoint, then $C^\prime$ is a *-subalgebra of $A$ (straightforward).

51 comments:

SašaMarch 16, 2025 at 3:52 PM
Something seems to be eaten in Eq. (5).
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SašaMarch 16, 2025 at 8:53 PM
This nice post brought from memory one of my favorites during college days, statistical mechanics course and famous Boltzmann formula, S=k*ln(W). So now I have a desire and "need" to actually refresh my memory with those generator functions and three kinds of ensembles we talked about, if memory still serves me well.

But, before going down that lane, I owe you a geometric, or as close to possible to a geometric explanation of a spinor. So, here's the current understanding of that concept in geometrical terms.
Take a normal to a plane, or any vector representing a direction orthogonal to some plane, rotate it in any plane, which would be perpendicular to the plane of origin, while simultaneously rotating the original plane.
To get back in the same position it would take a 2pi rotation in the plane where the normal actually rotated, and a 2pi rotation of the plane of origin that the normal stands for. After half of a full rotation of the normal vector and the plane it stands for, the vector would be pointing in the opposite to its starting original direction, i.e. a "coupled" rotation for in total 2pi gives spatial reflection, which would explain so called two valuedness of a spinor.
In terms of geometrical bodies, this can be represented by a coupled orthogonal rotations of a simple ball, one as angle phi and the other as angle theta in spherical coordinates, or in terms of geography, along the longitudinal and latitudinal lines or directions.

Would that be satisfactory enough for your taste?
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AnnaMarch 17, 2025 at 9:22 AM
From the very beginning of this post, realized that i have an awful mess of subspaces, ideals, and operators...((
Ark, could you explain once again why do we need two different types of operators "trace-class" and "Hilbert-Schmidt" and what is the subspace B1 here: B1(H)⊂B2(H)⊂B(H)?

One consolation is that it dawned on me why ideals are crucial elements of our consideration! Ideals are like boudaries - they have zero norm, i.e., zero size in the metrics of the host space: like 2d surface has zero 3d volume, etc.
Ideals capture the idea of the 'edge of world', the infinity, where one will never get (without adding an extra dimension).
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AnnaMarch 17, 2025 at 9:50 AM
The first "Why?" is not hard
"Notice that if H is infinite-dimensional, then lim_n p_n = 0 (Why?)"

Since ∑p_n = 1, we need a finite sum of infinite number of terms, therefore, starting from some n, they need to be all zeros.

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AnnaMarch 17, 2025 at 10:09 AM
This comment has been removed by the author.
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AnnaMarch 18, 2025 at 10:04 AM
"Exercise 1. Prove that e_m,n∈B2(H)"
First attempt to do it:
We want to show that Tr ((e_mn)*e_mn)<∞, so we assess the trace:
Tr((e_mn)* e_mn) = Sum_i(e_i, (e_mn)* e_mn e_i) =
Sum_i(e_i, e_nm e_mn e_i) = Sum_i(e_i, |e_n)(e_m||e_m)(e_n| e_i) =
Sum_i(e_i, |e_n)(e_n| e_i) = Sum_i(e_i, |e_n)δ_ni = (e_n|e_n) = 1
A strange result...
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AnnaMarch 19, 2025 at 12:28 PM
"get the anti-unitary property (How?):
= cc(), S,T in B2(H)"

i think we can just write formally
(JS , JT) = (S*, T*) = (T, S)* = cc(T, S)
Will that do?
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AnnaMarch 19, 2025 at 9:34 PM
"Therefore, Jπ(a)J = π'(a*), and the equality Jπ(A)J = π'(A) follows (Why?)"

I would say that it is because algebra A is a *-algebra with involution and every element 'a' has its conjugate pair a*. Thus, by taking one by one all elements 'a' of algebra A on the left-hand side and looking for the corresponding a* on the right-hand side, we obtain the algebra A of a*s, which is again A but "in another order". So the whole algebra A is translated into itself, and one can write 'A' instead of 'a' in the equality considered above.

Such an informal explanation, but i don't know how to formalize it.
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AnnaMarch 24, 2025 at 10:01 AM
@Ark, here are my belated attempts to grasp something else.
"It follows then from the last statement in Fig. 1 that ρ^is is a unitary operator in B(H) (How?)"
Does the last statement at Fig.1 (x_f)* = x_f mean that operator x_f is unitary? The bar over f confuses me a little...
we have (How?)

The statement "Uρ(s) Uρ(s') = Uρ(s+s')" is simply addition of the exponent powers.

The statement "Uρ(s) Ωρ = Ωρ" is the consequence of (3):
Uρ(s)| a > = | ρ^is a ρ^-is > =| ρ^is ρ^1/2 ρ^-is > = |all factors commutate| => ρ^1/2
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Thank you for your comment..

Arkadiusz Jadczyk - Open system

Sunday, March 16, 2025