Monday, May 15, 2017

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.  

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