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 41: density matrix

The proof of the pudding is in the eating.

In Part 40, we embarked on a journey through the GNS construction—a profound bridge between abstract algebra and the tangible world of Hilbert spaces. Starting with a finite-dimensional *-algebra 

$A$ and a positive functional $f$ (a state), we constructed a Hilbert space $H$, a *-representation $\rho$ of $A$ by linear operators acting on $H$, and a unit cyclic vector $\mathrm{\Omega }$ in $H$. This elegant framework satisfied the condition:

$$(\mathrm{\Omega },\rho (a)\mathrm{\Omega })=f(a)$$

for all $a\in A$.

Yet, in our exploration, we treated these states as distant, almost alien entities—exotic creatures whose inner lives remained a mystery. They performed their roles impeccably, but we never paused to ask: What animates them? What is their essence? Do they hum with the quiet resonance of mathematical truth, or do they roar with the intensity of physical reality? Where do they come from, and what do they yearn to reveal? In this note, we will rectify this oversight. We will cultivate a deeper connection with these states, learning to appreciate their beauty and, perhaps, even growing to cherish them.

Friedrich Engels once invoked the English proverb: The proof of the pudding is in the eating. In our case, the states are the ingredients with which we prepare the pudding—a rich, nourishing dish of mathematical insight. But what is pudding without chocolate? To make our exploration more palatable, more resonant with the human spirit, we will sweeten it with a familiar tool: the algebra of $2\times 2$ complex matrices. Though our *-algebra may seem abstract, even when rooted in the geometric Clifford algebra of space, its isomorphism to this well-known matrix algebra brings it closer to our hearts. This matrix algebra, wielded with care, will be the chocolate that enriches our pudding—a tool, yes, but one that transforms the unfamiliar into the delightful.

As we proceed, let us remember that mathematics is not merely a cold, mechanical exercise. It is a dance of ideas, a symphony of structures, and a journey of discovery. By developing a deeper connection with these states, we not only illuminate their mathematical significance but also uncover the poetry hidden within their formalism. Let us eat the pudding, savor the chocolate, and celebrate the beauty of the journey.

So, what is state? First of all it is a linear functional on A. But our A is not only an algebra. It is also a Hilbert space. In fact, it is even a Hilbert algebra (see Part 26), with its scalar product satisfying:

<ba,c> = <b,ca*>.           (0)

So, if f is a state, it is, first of all, a linear functional on a Hilbert space. I assume that the Reader has an elementary knowledge of Hilbert spaces. When learning about Hilbert space, one of the early things about them that we learn is that every continuous linear functional on a Hilbert space is given by a scalar product with a certain vector, in our case say Φ_f. Here this vector is an element of A: Φ_f∈A. Thus

f(a) = <Φ_f,a>  for all a∈A.                           (*)

Well, why call it Φ_f? Perhaps we can use the same symbol f, now f denoting the element of the algebra representing the functional f? This would be eco-friendly, while  the meaning would be clear from the context.

That was my first idea, but that was a bad idea. Here is why: The left-hand side of (*) is linear in f, but on the right-hand side we have the scalar product, which is anti-linear in the first argument, so writing f(a) = <f,a> would be inconsistent, it would make a bad pudding. Yet a simple change fixes that: we call f the element of the algebra that accomplishes the following:

f(a) = <f*,a>.             (1)

Now it is consistent, and we can continue. The functional f should be a state, that is it should be positive:

f(a*a) ≥ 0 ⩝a∈A.

This implies, as we have seen before, that f has the Hermitian property

f(a*)=f(a)*.

How is this property reflected in the properties of the functional f represented through (1)  as an element of the algebra denoted by the same symbol?

At first I tried to use the Hilbert algebra identity (0), but, unfortunately, this identity is written in a form that is not quite suitable for this purpose. So, it is better to reach for chocolate: we realize A as the algebra Mat(2,C), where the scalar product is given by

<a,b> = ½Tr(a*b).

Then

f(a) = <f*,a> =  ½Tr(f**a) = ½Tr(fa),

and using the trace property Tr(uv)=Tr(vu):

f(a*) =  ½Tr(fa*) = ½Tr(a*f) = <a,f>,   (2)

while

f(a)* = <f*,a>* = <a,f*> .                   (3)

For f to have the Hermitian property the difference of (2) and (3) should be 0, so we get

<a, (f-f*)> = 0 for all a.

This implies f=f*. 

Thus the algebra element f representing the functional f must be Hermitian.

What about the positivity condition? Let's use the chocolate again. The matrix f representing the functional f is a Hermitian matrix. So it has real eingenvalues. Suppose one of its eigenvalues  λ is negative. Let p=p*=pp  be the orthogonal projection on the corresponding subspace. Then fp=λp. So

f(p*p)=<f*,pp>= ½Tr(fp) = ½Tr(λp) = λ/2

if p projects on 1-dimensional subspace, and it is λ if p projects on 2-dimensional subspace. This would be negative. Therefore the matrix representing f must have only positive (we say "positive" to denote non-negative) eigenvalues. 

Thus positive matrix f^½ exists, so that f=(f^½)². Then

f(a) = ½Tr(fa) = ½Tr(f^½ f^½ a) = ½Tr(f^½ af^½ ) =<f^½ ,a f^½ >.

We still have the condition f(1)=1. That means that <f^½ , f^½ > =1. (Why?) 

Thus f^½ is a unit vector, and  thus Tr(f) = 2. (Can you see it?)

The formula

f(a) = =<f^½ ,a f^½ >

looks almost exactly the same as the formula

f(a) = (Ω,ρ(a)Ω)

from Part 40: GNS construction. It will look even better if instead of "a f^½" we write L(a)f^½, where L denotes the left regular representation of A on A:

f(a) = <f^½ ,L(a)f^½ >.

For it to look exactly the same we must ensure that f^½ is a cyclic vector. But is it?

We will go into these little  details (where the devil is hiding) in the next post.

By the way: in quantum theory mixed states are represented by "density matrices": positive matrices of trace 1. Thus our  ½f is the standard density matrix of quantum theory.