## 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:

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 \(f_1 = g_1 = I\) :

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:

\((f_1 \otimes f_2)\sigma_{12}(g_1\otimes g_2) = \)

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).

\(\Delta (\sigma) = \sigma \otimes \sigma - \alpha \otimes \alpha\)

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.**

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