## 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.