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.