Fractions and stochastic algebra

The atomic orbitals are not quantized to Shannon, meaning they do not make a two bit digit system, but make a digit system in the natural log. They do not have SNR greater than 1/5. So a wave function of the orbitals will have forms looking like:

e^(k) + e^(k-1)+  e^(k-1)....

A perfectly fine digit system, the k are integer, the quantum numbers of the orbitals.  And we can treat them like a digit system, add, subtract, multipy and so on.  But when we draw them, we convert to the twos system of our computers.  If we do that anyway, then just use Shannon with a high enough sample rate and find all the orbital quantum numbers.  Essentially what we would be doing is breaking the electron mass into fine granularity, applying a bit of special relativity.


Look what happens:
1) Convert e to log2(e) = 1.4427 = b then we get:

2^(l*k) + 2^(l*(k-1) + 2^(l * k-2)......

Nice, the the l*(k),l*(k-1)... are not integer, and we really do not have a nice digit system until we scale b up to an integer like 144.  Still works but we are dealing with a 144*k*j digit number. So, if we have some 20 quantum numbers total, we would apply Shannon across the spherical phase density using 20 * 144 integers; computing all the integers to quantize that phase density.

Why not? Good question, why not.  You end up with three variables, one for radius squared, one for theta angle and one for beta angle, and the pretty picture would be a series of binary numbers added up, each binary number ranging from about 1 to 1000 digits. So what? I dunno, why not.

All we are doing in atomic physics is minimizing the variance of {-1,0,1} within the proton.  The know wave/mass numbers that define the orbital boundaries, so we know the total phase. We should know the relative amount of phase in a unit of charge, so we initialize the proton  to that. We know the number of Nulls in a proton. We have chopped up the electron enough to accommodate relativity.   In finding the Shannon boundaries of the orbitals we have accommodated magnetism. Maximum entropy is minimum phase to the precision of 1000 digits over the sphere of the Proton. We ignore the quarks, they just give us an axis of symmetry.  The orbitals are simply the paths of uniform phase, so we map the proton phase function onto the orbitals

I simply cannot find fault.