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.

The fundamental bipartite relations

Continuing from where we left off last time, we introduced the most general composite products for a bipartite system:

\(\alpha_{12} = a_{11}\alpha \otimes \alpha + a_{12} \alpha\otimes\sigma + a_{21} \sigma\otimes \alpha + a_{22} \sigma\otimes\sigma\)

\(\sigma_{12} = b_{11}\alpha \otimes \alpha + b_{12} \alpha\otimes\sigma + b_{21} \sigma\otimes \alpha + b_{22} \sigma\otimes\sigma\)

The question now becomes: are the \(a\)'s and \(b\)'s parameters free, or can we say something abut them? To start let's normalize the products \(\sigma\) like this:

\(f\sigma I = I\sigma f = f\)

which can always be done. Now in:

\((f_1 \otimes f_2)\alpha_{12}(g_1\otimes g_2) = \)

\(=a_{11}(f_1 \alpha g_1)\otimes  (f_2 \alpha g_2) + a_{12}(f_1 \alpha g_1) \otimes (f_2 \sigma g_2 ) +\)

\(+a_{21}(f_1 \sigma g_1)\otimes  (f_2 \alpha g_2) + a_{22}(f_1 \sigma g_1) \otimes (f_2 \sigma g_2 )\)

if we pick \(f_1 = g_1 = I\) :

\((I \otimes f_2)\alpha_{12}(I\otimes g_2) = \)

\(=a_{11}(I \alpha I)\otimes  (f_2 \alpha g_2) + a_{12}(I \alpha I) \otimes (f_2 \sigma g_2 ) +\)

\(+a_{21}(I \sigma I)\otimes  (f_2 \alpha g_2) + a_{22}(I \sigma I) \otimes (f_2 \sigma g_2 )\)

and recalling from last time that \(I\alpha I = 0\) from Leibniz identity we get:

\(f_2 \alpha g_2 = a_{21} (f_2 \alpha g_2 ) + a_{22} (f_2 \sigma g_2)\)

which demands \(a_{21} = 1\) and \(a_{22} = 0\).

If we make the same substitution into:

 \((f_1 \otimes f_2)\sigma_{12}(g_1\otimes g_2) = \)

\(=b_{11}(f_1 \alpha g_1)\otimes  (f_2 \alpha g_2) + b_{12}(f_1 \alpha g_1) \otimes (f_2 \sigma g_2 ) +\)

\(+b_{21}(f_1 \sigma g_1)\otimes  (f_2 \alpha g_2) + b_{22}(f_1 \sigma g_1) \otimes (f_2 \sigma g_2 )\)

we get:

\(f_2 \sigma g_2 = b_{21} (f_2 \alpha g_2 ) + b_{22} (f_2 \sigma g_2)\)

which demands \(b_{21} = 0\) and \(b_{22} = 1\)

We can play the same game with \(f_2 = g_2 = I\) and (skipping the trivial details) we get two additional conditions: \(a_{12} = 1\) and \(b_{12} = 0\).

In coproduct notation what we get so far is:

\(\Delta (\alpha) = \alpha \otimes \sigma + \sigma \otimes \alpha + a_{11} \alpha \otimes \alpha\)
\(\Delta (\sigma) = \sigma \otimes \sigma + b_{11} \alpha \otimes \alpha\)

By applying Leibniz identity on a bipartite system, one can show after some tedious computations that \(a_{11} = 0\). The only remaining free parameters is \(b_{11}\) which can be normalized to be ether -1, 0, or 1 (or elliptic, parabolic, and hyperbolic). Each choice corresponds to a potential theory of nature. For example 0 corresponds to classical mechanics, and -1 to quantum mechanics.

Elliptic composability is quantum mechanics! The bipartite products obey:


\(\Delta (\alpha) = \alpha \otimes \sigma + \sigma \otimes \alpha \)
\(\Delta (\sigma) = \sigma \otimes \sigma - \alpha \otimes \alpha\)

Please notice the similarity with complex number multiplication. This is why complex numbers play a central role in quantum mechanics.

Now at the moment the two products do not respect any other properties. But we can continue this line of argument and prove their symmetry/anti-symmetry. And from there we can derive their complete properties arriving constructively at the standard formulation of quantum mechanics. Please stay tuned.

Time evolution for a composite system

Continuing where we left off last time, let me first point out one thing which I glossed over too fast: the representation of \(D\) as a product \(\alpha\): \(Dg = f\alpha g\). This is highly nontrivial and not all time evolutions respect it. In fact, the statement above is nothing but a reformulation of Noether's theorem in the Hamiltonian formalism. I did not build up the proper mathematical machinery to easily show this, so take my word on it for now. I might revisit this at a later time.

Now what I want to do is explore what happens to the product \(\alpha\) when we consider two physical systems 1 and 2. First, let's introduce the unit element of our category, and let's call it "I":

\(f\otimes I = I\otimes f = f\)

for all \(f \in C\)

Then we have \((f_1\otimes I) \alpha_{12} (g_1\otimes I) = f \alpha g\)

On the other hand suppose in nature there exists only the product \(\alpha\). Then the only way we can construct a composite product \(\alpha_{12}\) out of \(\alpha_1\) and \(\alpha_2\) is:

\((f_1\otimes f_2) \alpha_{12} (g_1 \otimes g_2) = a(f_1 \alpha_1 g_1)\otimes (f_2\alpha_2 g_2)\)

where \(a\) is a constant. 

Now if we pick \(f_2 = g_2 = I\) we get:

\((f_1\otimes I) \alpha_{12} (g_1 \otimes I) = a(f_1 \alpha_1 g_1)\otimes (I \alpha_2 I)  \)

which is the same as \(f \alpha g\) by above. 

But what is \(I\alpha I\)? Here we use the Leibniz identity and prove it is equal with zero:

\(I \alpha (I\alpha A) = (I \alpha I) \alpha A + I \alpha (I \alpha A)\)

for all \(A\) and hence \(I\alpha I = 0\)

But this means that a single product alpha by itself is not enough! Therefore we need a second product \(\sigma\)! Alpha will turn out to be the commutator, and sigma the Jordan product of observables, but we will derive this in a constructive fashion.

Now that we have two products in our theory of nature, let's see how can we build the composite products out of individual systems. Basically we try all possible combinations:

\(\alpha_{12} = a_{11}\alpha \otimes \alpha + a_{12} \alpha\otimes\sigma + a_{21} \sigma\otimes \alpha + a_{22} \sigma\otimes\sigma\)

\(\sigma_{12} = b_{11}\alpha \otimes \alpha + b_{12} \alpha\otimes\sigma + b_{21} \sigma\otimes \alpha + b_{22} \sigma\otimes\sigma\)

which is shorthand for (I am spelling out only the first case):

\((f_1 \otimes f_2)\alpha_{12}(g_1\otimes g_2) = \)

\(=a_{11}(f_1 \alpha g_1)\otimes  (f_2 \alpha g_2) + a_{12}(f_1 \alpha g_1) \otimes (f_2 \sigma g_2 ) +\)

\(+a_{21}(f_1 \sigma g_1)\otimes  (f_2 \alpha g_2) + a_{22}(f_1 \sigma g_1) \otimes (f_2 \sigma g_2 )\)

For the mathematically inclined reader we have constructed what it is called a coalgebra where the operation is called a coproduct: \(\Delta : C \rightarrow C\otimes C\). In category theory a coproduct is obtained from a product by reversing the arrows.

Now the task is to see if we can say something about the coproduct parameters: \(a_{11},..., b_{22}\). In general nothing can constrain their values, but in our case we do have an additional relation: Leibniz identity which arises out the functoriality of time evolution. This will be enough to fully determine the products \(\alpha\) and \(\sigma\), and from them the formalism of quantum mechanics. Please stay tuned.