The Jordan algebra of observables

In quantum mechanics the observables are represented by Hermitian operators: \(O^{\dagger} = O\). In general Hermitian operators need not commute and this corresponds to incompatible observables like position and momenta.  Given two observables A and B, can we generate another observable out of it?

Let's try the simplest idea: AB. Is this self-adjoint? Let's see:

\({(AB)}^{\dagger} = B^{\dagger} A^{\dagger} = BA \ne AB\)

So the next idea to try is a symmetrized product:  \(1/2 (AB + BA)\) .

This is called the Jordan product and the product \(\sigma \): \(A \sigma B = 1/2 (AB + BA)\) gives rise to the Jordan algebra of observables.

Pascual Jordan

This product has two property:

1) symmetry: \( A \sigma B = B \sigma A \)
2) Jordan identity: \((A \sigma B) \sigma (A \sigma A) = A \sigma (B \sigma (A \sigma A))\) 

Can we derive those properties from categorical arguments like we derived the properties of α last time? Drum-roll....Yes we can!!!

Last time we have proven that α is antisymmetric: \(A \alpha B = - B \alpha A \) and from the fundamental relationship:

\({\alpha}_{12} = \alpha \otimes \sigma  + \sigma \otimes \alpha \)    

it is trivial to show that σ must be symmetric thus respecting the first property of the Jordan algebra. Proving the second property is unfortunately much more involved project, about two orders of magnitude harder than what I left out last time as a simple exercise. Let me set the stage to where we want go. Both the products α and σ are not associative. To quantify the violation of associativity we introduce what is called the associator:

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

where "o" can be either \(\alpha \) or \(\sigma \). Then the Jordan and Lie algebras of quantum mechanics obey this identity:

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

where \(J \hbar /2 \) is a one-to-one map between observables and generators obeying \(J^2 = -1\): the imaginary unit which when multiplying a Hermitian operators generates an anti-Hermitean operator. 

You can convince yourself this is true using the usual realizations of the Jordan and Lie products:

\(A\sigma B = \frac{1}{2} (AB + BA)\)

\(A\alpha B = \frac{J}{\hbar} (AB - BA)\)

Now if this associator identity is true, proving the second Jordan identity is trivial because the identity can be rewritten as: \([A, B, A \sigma A {]}_{\sigma} = 0\)

What is the alpha associator: \([A, B, A \sigma A {]}_{\alpha}\) ?  I can leave this as an exercise to show this is indeed zero by expanding the associator and using the Leibniz identity for α.

Now our goal is to derive \([A, B, C {]}_{\sigma} + J^2 \hbar^2 / 4 [A, B, C {]}_{\alpha} = 0 \) from invariance under composability and time evolution.

But first where is the \(b_{11} = J^2 \hbar^2 / 4\) coming from? If you recall two posts ago I derived the following fundamental composition relationship as a coproduct:

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

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

\(b_{11}\) was a free parameter which can be normalized to -1 for quantum mechanics, 0 for classical mechanics and +1 for the unphysical "hyperbolic quantum mechanics". However I simply want to normalize it as:  \(b_{11} = J^2 \hbar^2 / 4\). Why? To recover the usual definition of the commutator and the Jordan product. In other words, convenience.

If last time we used \(\Delta (\alpha ) = \alpha \otimes \sigma  + \sigma \otimes \alpha \), now it is time to employ:\(\Delta (\sigma) = \sigma \otimes \sigma + \frac{J^2 \hbar^2}{4}\alpha \otimes \alpha\) as well.

The proof is rather long and was first obtained by Grgin and Petersen in 1976 and I won't present its gory bookkeeping detail, but I will only give the starting point:

\((A_1 \otimes A_2 ) \alpha_{12} ((B_1 \otimes B_2)\sigma_{12} (C_1 \otimes C_2))\)

Use the bipartite Leibniz identity along with the two delta coproducts and the Jacobi identity and you'll reach the associator identity (from Jacobi). The first time I double checked Grgin's paper it took me one full week to work out the result given the intermediate steps in the paper, but now I can do it in about an hour of careful and boring bookkeeping.

We now have all the ingredients to put together the C* algebras of quantum mechanics and arrive at the usual Hilbert space formulation. Please stay tuned.

Historical notes:

I am very grateful to Emile Grgin for introducing me to his approach which I managed to expand into a full blown reconstruction of quantum mechanics. The root idea came from Bohr himself who passed his intuition to Aage Petersen, his personal assistant. Later on Petersen developed those ideas in collaboration to Emile Grgin at Yeshiva University. Unfortunately Grgin left academia and his ideas were forgotten. I got in contact with him after he retired and later on I realized the categorical origin of the approach. Completely independent from me, Anton Kapustin of Caltech noticed the same 1976 paper and we both came out with almost identical papers, mine written from the physics point of view, and his from the math point of view. I uploaded my paper on the archive 3 weeks before him and I did not notice his paper. John Preskill made me aware of Kapustin's paper when we met at a conference. He was the very first person who understood my research.