## 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.