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.
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.
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