## 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.