## Impressions from FQMT 2017

I just came back from Vaxjo where I had a marvelous time. It does sounds cliche, but this year was the best conference organized by Professor Khrennikov and I got many pleasant and unexpected surprises.

The conference did had one drawback: everyday after the official talks we continue the discussions about quantum mechanics well past midnight at "The Bishops Arms" where we drank too many beers causing me to gained a few pounds :)

At the conference I had a chance to meet and talk with Phillipe Grangier (he worked with Aspect on the famous Bell experiment) and I witness him giving the best cogent comments on all talks: from experimental to theoretical. He even surprised me when he asked at the end of my presentation why I am using time to derive Leibniz identity, where any other symmetry will do? Indeed this is true, but the drawback is that any other symmetry lacks generality later on during composition arguments. Suppose we compose two physical systems: one with with a continuous symmetry and another without, then the composed system will lack that symmetry. The advantage of using time is that it works for all cases where energy is conserved.

Grangier presented his approach on quantum mechanics reconstruction using contextuality and continuity (like in Hardy's 5 reasonable axioms paper). The problem of continuity is that it lacks physical intuition/motivation. Why not impose right away the C* condition: $$||a^* a|| = {||a||}^2$$ and recover everything from it?

Bob Coecke and Aleks Kissinger book on the pictorial formalism: "Picturing Quantum Processes" was finally ready and was advertised at the conference. If you go to www.cambridge.org/pqp you can get with a 20% discount when you enter the code COECKE2017 at the checkout.

Coecke 's talk was about causal theories and his main idea was: "time reversal of any causal theory = eternal noise". This looks deep, but it is really a trivial observation: you can't get anything meaningful and you can't control signals which have an information starting point because the starting point corresponds to the notion of false and anything is derivable from false.

Robert Raussendorf from University of Vancouver had a nice talk about measurement based quantum computations where measurements are used to control the computation and he identified a cohomological framework.

One surprise talk for me was the one given by Marcus Appleby from University of Sydney who presented a framework of equivalence for Quantum Mechanics between finite and infinite dimensional cases. This is of particular importance to me as I recovered quantum mechanics in the finite dimensional case only and I am searching for an approach to handle the infinite dimensional case.

I made new friends there and I got very good advice and ideas - a big thank you. I also got to give many in person presentations of my quantum reconstruction program.

There was one person claiming he solved the puzzles of the many worlds interpretation. I sat next to him at the conference dinner and I invited him to have a guest post at this blog to present his solution. As a disclaimer, I think MWI lacks the proper notion of probability and I am yet to see a solution but I am open to listen to new arguments. What I would like to see is an explanation of how to reconcile the world split of 50-50% when the quantum probabilities are 80-20%? I did not see this explained in his presentation to my satisfaction, but maybe I was not understating the argument properly.

## Jordan-Banach, Jordan-Lie-Banach, C* algebras, and quantum mechanics reconstruction

This a short post written as waiting for my flight at Dulles Airport on my way to Vaxjo Sweden for a physics conference.

First some definitions. a Jordan-Banach algebra is a Jordan algebra with the usual norm properties of a Banach algebra. A Jordan-Lie-Banach algebra is a Jordan-Banach algebra which is a Lie algebra at the same time. A Jordan-Lie algebra is the composability two-product algebra which we obtained using category theory arguments.

Last time I hinted about this week's topic which is the final step in reconstructing quantum using category theory arguments. What we obtain from category theory is a Jordan-Lie algebra which in the finite dimensional case has the spectral properties for free because the spectrum in uniquely defined in an algebraic fashion (things gets very tricky in the infinite dimensional case). So in the finite dimensional case JL=JLB.

But how can we go from Jordan-Banach algebra to C*? In general it cannot be done. C* algebras correspond to quantum mechanics and on the Jordan side we have the octonionic algebra which is exceptional. Thus cannot be related to quantum mechanics because octonions are not associative. However we can define state spaces for both Jordan-Banach and C* algebras and we can investigate their geometry. The geometry is definable in terms if projector elements which obey: $$a*a = a$$. In turn this defines the pure states as the boundary of the state spaces. If the two geometries are identical, we are in luck.

Now the key question is: under what circumstances can we complexify a Jordan-Banach algebra to get a C* algebra?

In nature, observables play a dual role as both observables and generators. In literature this is called dynamic correspondence. Dynamic correspondence is the essential ingredient: any JB algebra with dynamic correspondence is the self-adjoint part of a C* algebra. This result holds in general and can be established by comparing the geometry of the state spaces for JB and C* algebras.

Now for the punch line: a JL algebra comes with dynamic correspondence and I showed that in prior posts. The conclusion is therefore:

in the finite dimensional case: JL is a JLB algebra which gives rise to a C* algebra by complexification and by GNS construction we obtain the standard formulation of quantum mechanics.

Quantum mechanics is fully reconstructed in the finite dimensional case from physical principles using category theory arguments!

By the way this is what I'll present at the conference (the entire series on QM reconstruction).

## From composability two-product algebra to quantum mechanics

Last time we introduced the composability two-product algebra consisting of the Lie algebra $$\alpha$$ and the Jordan algebra $$\sigma$$ along with their compatibility relationship. This structure was obtained by categorical arguments using two natural principles of nature:

- laws of nature are invariant under time evolution
- laws of nature are invariant under system composition

What we did not obtain were spectral properties. However, in the finite dimensional case, we do not need spectral properties and we can fully recover quantum mechanics in this particular case. The trick is to classify all possible two-product algebras because there are only a handful of them. This is achieved with the help of the Artin-Weddenburn theorem

First some preliminary. We need to introduce a Lie-Jordan-Banach (JLB) algebra by augmenting the composability two-product algebra with spectral properties:
-a JLB-algebra is a composability two-product algebra with the following two additional properties:
• $$||x\sigma x|| = {||x||}^{2}$$
• $$||x\sigma x||\leq ||x\sigma x + y\sigma y||$$
Then we can define a C* algebra by compexification of a JLB algebra where the C* norm is:

$$||a+ib|| = \sqrt{{||a||}^{2}+{||b||}^{2}}$$

Conversely from a C* algebra we define a JLB algebra as the self-adjoint part and where the Jordan part is:

$$a\sigma b = \frac{1}{2}(ab+ba)$$

and the Lie part is:

$$a\alpha b = \frac{i}{\hbar}(ab-ba)$$

From C* algebra we recover the usual quantum mechanics formulation by GNS construction which gets for us:

- a Hilbert space H
- a distinguished vector $$\Omega$$ on H arising out of the identity of the C* algebra
- a representation $$\pi$$ of the algebra as linear operators on H
- a state $$\omega$$ on C* represented as $$\omega (A) = \langle \Omega, \pi (A)\Omega\rangle_{H}$$

Conversely, from quantum mechanics a C* algebra arises as bounded operators on the Hilbert space.

The infinite dimensional case is a much harder open problem. Jumping from the Jordan-Banach operator algebra side to the C* and von Neuman algebras is very tricky and this involves characterizing the state spaces of operator algebras. Fortunately all this is already settled by the works of Alfsen, Shultz, Stormer, Topping, Hanche-Olsen, Kadison, Connes.

## The algebraic structure of quantum and classical mechanics

Let's recap on what we derived so far. We started by considering time as a continous functor and we derived Leibniz identity from it. Then for a particular kind of time evolution which allows a representation as a product we were able to derive two products $$\alpha$$ and $$\sigma$$ for which we derived the fundamental bipartite relations.

Repeated applications of Leibniz identity resulted in proving $$\alpha$$ as a Lie algebra, and $$\sigma$$ as a Jordan algebra and an associator identity between them:

$$[A,B,C]_{\sigma} + \frac{J^2 \hbar^2}{4}[A,B,C]_{\alpha} = 0$$

where $$J$$ is a map between generators and observables encoding Noether's theorem.

Now we can combine the Jordan and Lie algebra as:

$$\star = \sigma\pm \frac{J \hbar}{2}\alpha$$

and it is not hard to show that this product is associative (pick $$\hbar = 2$$ for convenience):

$$[f,g,h]_{\star} = (f\sigma g \pm J f\alpha g)\star h - f\star(g\sigma h \pm J g\alpha h)=$$
$$(f\sigma g)\sigma h \pm J(f\sigma g)\alpha h \pm J(f\alpha g)\sigma h + J^2 (f\alpha g)\alpha h$$
$$−f\sigma (g\sigma h) \mp J f\sigma (g\alpha h) \mp J f\alpha (g\sigma h) − J^2 f\alpha (g\alpha h) =$$
$$[f, g, h]_{\sigma} + J^2 [f, g, h]_{\alpha} ±J\{(f\sigma g)\alpha h + (f\alpha g)\sigma h − f\sigma (g\alpha h) − f\alpha (g\sigma h)\} = 0$$

because the first part is zero by associator identity and the second part is zero by applying Leibniz identity. In Hilbert space representation the star product is nothing but the complex number multiplication in ordinary quantum mechanics

Now we can introduce the algebraic structure of quantum (and classical) mechanics:

A composability two-product algebra is a real vector space equipped with two bilinear maps $$\sigma$$ and $$\alpha$$ such that the following conditions apply:

- $$\alpha$$ is a Lie algebra,
- $$\sigma$$ is a Jordan algebra,
- $$\alpha$$ is a derivation for $$\sigma$$ and $$\alpha$$,
- $$[A, B, C]_{\sigma} + \frac{J^2 \hbar^2}{4} [A, B, C]_{\alpha} = 0$$,
where $$J \rightarrow (−J)$$ is an involution mapping generators and observables, $$1\alpha A = A\alpha 1 = 0$$, $$1\sigma A = A\sigma 1 = A$$

For quantum mechanics $$J^2 = -1$$. In the finite dimensional case the composability two-product algebra is enough to fully recover the full formalism of quantum mechanics by using the Artin-Wedderburn theorem.

The same structure applies to classical mechanics with only one change: $$J^2 = 0$$.

In classical mechanics case, in phase space, the usual Poisson bracket representation for product $$\alpha$$ can be constructively derived from above:
$$f\alpha g = \{f,g\} = f \overset{\leftrightarrow}{\nabla} g = \sum_{i=1}^{n} \frac{\partial f}{\partial q^i} \frac{\partial g}{\partial p_i} - \frac{\partial f}{\partial p_i} \frac{\partial g}{\partial q^i}$$

and the product $$\sigma$$ is then the regular function multiplication.

In quantum mechanics case in the Hilbert space representation we have the commutator and the Jordan product:

$$A\alpha B = \frac{i}{\hbar} (AB − BA)$$
$$A\sigma B = \frac{1}{2} (AB + BA)$$

or in the phase space representation the Moyal and cosine brackets:

$$\alpha = \frac{2}{\hbar}\sin (\frac{\hbar}{2} \overset{\leftrightarrow}{\nabla})$$
$$\sigma = \cos (\frac{\hbar}{2} \overset{\leftrightarrow}{\nabla})$$

where the associative product is the star product.

Update: Memorial Day holiday interfered with this week's post. I was hoping to make it back home on time to write it today, but I got stuck on horrible traffic for many hours. I'll postpone the next post for a week.

## The Jordan algebra of observables

Last time, from concrete representations of the products $$\alpha$$ and $$\sigma$$ we derived this identity:

$$[A,B,C]_{\sigma} + \frac{i^2 \hbar^2}{4}[A,B,C]_{\alpha} = 0$$

Let's use this in a particular case when $$C = A\sigma A$$. What does the left hand side say?

$$[A,B,C]_{\sigma} = (A\sigma B) \sigma (A\sigma A)) - A\sigma (B \sigma (A \sigma A))$$

which if we drop $$\sigma$$ for convenience sake reads:

$$(AB)(AA) - A(B(AA))$$

If the right hand side is zero then we get the Jordan identity:

$$(xy)(xx) = x(y(xx))$$ where $$xy = yx$$

Now let's compute the right hand side and show it is indeed zero:

$$[A,B,A\sigma A]_{\alpha} = (A\alpha B) \alpha (A\sigma A)) - A\alpha (B \alpha (A \sigma A))$$

Using Leibniz identity in the second term we get:

$$(A\alpha B) \alpha (A\sigma A)) - (A\alpha B) \alpha (A\sigma A) - B \alpha (A\alpha (A\sigma A))) = - B \alpha (A\alpha (A\sigma A))$$

But $$A\alpha (A\sigma A) = 0$$ because

$$A\alpha (A\sigma A) = (A\alpha A) \sigma A + A\sigma (A\alpha A)$$

and $$A\alpha A = -A\alpha A = 0$$ by skew symmetry.

Therefore due to the associator identity, the product $$\sigma$$ is a Jordan algebra. Now we need to arrive at the associator identity using only the ingredients derived so far. This is tedious but it can be done using only Jacobi and Leibniz identity. Grgin and Petersen derived it in 1976 and you can see the proof here

The associator identity is better written as:

$$[A,B,C]_{\sigma} + \frac{J^2 \hbar^2}{4}[A,B,C]_{\alpha} = 0$$

where $$J$$ is a map from the the product $$\alpha$$ to the product $$\sigma$$. The existence of this map is equivalent with Noether's theorem. It just happens that in quantum mechanics case $$J^2 = -1$$ and the imaginary unit maps anti-Hermitean generators to Hermitean observables.

In classical physics case, $$J^2 = 0$$ and this means that the product $$\sigma$$ is associative (in fact it is the ordinary function multiplication) and the product $$\alpha$$ can be proven to be the Poisson bracket, but that is a topic for another day as we will continue to derive the mathematical structure of quantum mechanics. Please stay tuned.

## Lie, Jordan algebras and the associator identity

Before I continue the quantum mechanics algebraic series, I want to first state my happiness for the defeat of the far (alt)-right candidate in France despite Putin's financial and hacking support. Europe has much better antibodies against the scums like Trump than US. In US the diseases caused by the inoculation of hate perpetuated over many years by Fox News has to run its course before things will get better.

Back to physics, first I will show that the product $$\alpha$$ is indeed a Lie algebra. This is utterly trivial because we need to show antisymmetry and the Jacobi identity:

$$a\alpha b = -b\alpha a$$
$$a\alpha (b\alpha c) + c\alpha (a\alpha b) + b\alpha (c\alpha a) = 0$$

We already know that  the product $$\alpha$$ is antisymmetric and we know that the it obeys Leibniz identity:

$$a\alpha (b\circ c) = (a\alpha b) \circ c + b\circ (a\alpha c)$$

where $$\circ$$ can stand for either $$\alpha$$ or $$\sigma$$. When $$\circ = \alpha$$ we get:

$$a\alpha (b\alpha c) = (a\alpha b) \alpha c + b\alpha (a\alpha c)$$

which by antisymmetry becomes

$$a\alpha (b\alpha c) = - c \alpha (a\alpha b) - b\alpha (c\alpha a)$$

In other words, the Jacobi identity.

Therefore the product $$\alpha$$ is in fact a Lie algebra. Now we want to prove that the product $$\sigma$$ is a Jordan algebra.

This is not as simple as proving the Lie algebra, and we will do it with the help of a new concept: the associator. Let us first define it. The associator of an arbitrary product $$\circ$$ is defined as follows:

$$[a,b,c]_{\circ} = (a\circ b)\circ c - a\circ (b\circ c)$$

as such it measures the lack of associativity.

It is helpful now to look at the concrete realizations of the products $$\alpha$$ and $$\sigma$$ in quantum mechanics to know where we want to arrive. In quantum mechanics the product alpha is the commutator, and the product sigma is the anticommutator:

$$A \alpha B = \frac{i}{\hbar}[A,B] = \frac{i}{\hbar}(AB - BA)$$
$$A\sigma B = \frac{1}{2}\{A, B\} = \frac{1}{2}(AB+BA)$$

Let's compute alpha and sigma associators:

$$[A,B,C]_{\alpha} = \frac{-1}{\hbar^2}([AB-BA, C] - [A, BC-CB]) =$$
$$=\frac{-1}{\hbar^2}(ABC-BAC-CAB+CBA - ABC+ACB+BCA-CBA)$$
$$= \frac{-1}{\hbar^2}(-BAC-CAB +ACB+BCA)$$

$$[A,B,C]_{\sigma} = \frac{1}{4}(\{AB+BA, C\} - \{A, BC+CB\}) =$$
$$=\frac{1}{4}(ABC+BAC+CAB+CBA - ABC-ACB-BCA-CBA) =$$
$$=\frac{1}{4}(BAC+CAB -ACB-BCA)$$

and so we have the remarkable relationship:

$$[A,B,C]_{\sigma} + \frac{i^2 \hbar^2}{4}[A,B,C]_{\alpha} = 0$$

What is remarkable about this is that the Jordan and Lie algebras lack associativity in precisely the same way and because of this they can be later combined into a single operation. The identity above also holds the key for proving the Jordan identity.

Next time I'll show how to derive the identity above using only the ingredients we proved so far and then I'll show how Jordan identity arises out of it. Please stay tuned.

## The origin of the symmetries of the quantum products

Quantum mechanics has three quantum products:
• the Jordan product of observables
• the commutator product used for time evolution
• the complex number multiplication of operators
The last product is a composite construction of the first two and it is enough to study the Jordan product and the commutator. In the prior posts notation, the Jordan product is called $$\sigma$$, and the commutator is called $$\alpha$$. We will derive their full properties using category theory arguments and the Leibniz identity. Bur before doing this, I want to review a bit the two products. The commutator is well known and I will not spend time on it. Instead I will give the motovation for the Jordan product.

In quantum mechanics the observables are represented as self-adjoint operators: $$O = O^{\dagger}$$ If we want to create another self-adjoint operator out of two self-adjoint operators A and B, the simple multiplication won't work because $$(AB)^{\dagger} = B^{\dagger} A^{\dagger} = BA \ne AB$$. The solution is to have a symmetrized product: $$A\sigma B = (AB+BA)/2$$. A lot of the quantum mechanics formalism transfers to the Jordan algebra of observables, but this is a relatively forgotten approach because it is rather cumbersome (the Jordan product is not associative but power associative) and (as it is expected) it does not produce any different predictions than the standard formalism based on complex numbers.

Now back to obtaining the symmetry properties of the Jordan product $$\sigma$$ and commutator $$\alpha$$, at first we cannot say anything about the symmetry of the product $$\sigma$$. However we do know that the product $$\alpha$$ obeys the Leibniz identity. We have already use it to derive the fundamental composition relationships, so what else can we do? We can apply it to a bipartite system:

$$f_{12}\alpha_{12}(g_{12}\alpha_{12}h_{12}) = g_{12}\alpha_{12}(f_{12}\alpha_{12}h_{12}) + (f_{12}\alpha_{12}g_{12})\alpha_{12} h_{12}$$

where

$$\alpha_{12} = \alpha\otimes \sigma + \sigma\otimes\alpha$$

Now the key observation is that in the right hand side, $$f$$ and $$g$$ appear in reverse order. Remember that the functions involved in the relationship above are free of constraints, by judicious picks of their value can lead to great simplifications because $$1 \alpha f = f\alpha 1 = 0$$. The computation is tedious and I will skip it, but what you get in the end is this:

$$f_1\alpha h_1 \otimes [f_2 \alpha g_2 + g_2 \alpha f_2 ] = 0$$

which means that the product alpha is anti-symmetric $$f\alpha g = -g\alpha f$$

If we use this property in the fundamental bypartite relationship we obtain in turn that the product sigma is symmetric: $$f\sigma g = g\sigma f$$

Next time we will prove that $$\alpha$$ is a Lie algebra and that $$\sigma$$ is a Jordan algebra. Please stay tuned.