Simplifying units

I am looking at atomic orbitals, and the necessity of dealing with the units of physics that keep coming up. I need to simplify, because everything is really, just phase and null and how much of those we have with respect to two quantization ratios.  So I am dumping the units of physics and replacing them with units of things counted, as follows:

Primary units:

m is the quantization ratio of Nulls
f is the quantization ratio of Phase

Secondary units:

j*m is m multiplied by an integer
k*f is f multiplied by an integer

m^j that is m taken to an integer power
f^k  and f taken to an integer power

There, that is simpler, and eliminates charge and magnetism and particles and mass and time and energy and power and force and space.  As long as I can write equations with these, in some symmetrical frame, I can count them up and let physicists add their semantics later.

I add some secondary conjectures.

2^(j*m) and 2^(k*f)  define a digit system when operations on m and f are Shannon separable, and
e^(j*m) and e^(k*f) define the digits system when operations are not Shannon separable.

A symmetric frame is one in which either conjecture always holds. Symmetry is determined by the density of phase and nulls samples. To obtain density I construct two operators:

D1 = 1/2(m^k - f^k) and D2 = 1/2(m^k + f^k) k integer

When the operators are Shannon equal to one, we have a binary system, otherwise a natural log system.

I suspect my symmetric frames will be defined by integers:  
{k, j, j * k}; k, j   integer.

The trick to the orbitals is to find the set j,k that define non-Shannon frames. I can find them in two ways; 1) Cheat and look at the centuries of work done by physicists, or 2) Use some methods of group theory. Once the integer set is found, I count them out and draw the natural log results.

Phase theory predicts the Hamiltonian is always the sum of variance in Phase + variance in Nulls, Vp +  Vn Physics is the art of finding these two quantities. In the Shannon equation we have the term: log2(1+ S/N) which now gets converted to:
log2((Vp+Vn)/Vp).

Lets call that the Shannon operator. I think we can fix the Shannon sample rate to the value 2. So then the Shannon operator will be a function of (j*k,j,k). The Shannon condition not met when the predicted quant value some j,k is < 1. The trick then becomes finding the form of the Shannon operator.

We know why we have the j,k axis of symmetry, because the proton is made of a threesome of quarks which would thus have separable axis of symmetry.  Group theory likely tells us then, we have a third axis, j*k. I think that is how it is going to work.  We make a simplification by treating the proton as a sphere along the three axis, and the value Vp and Vn will be mappings of their values inside the proton along the three axis. We are damn near done! We know the orbitals are constructed in, our spectral charge, between two points, one where the m,f for magnetic is Shannon and one where m,f for the electron is Shanon. We know how much proton precision is available, so we just take our spread the precision among the density operators along the three axis. We scale up the precision to the point where Shannon is almost met, take the 'int' of j an k and presto.

At any given point in the proton, Vn is  (D1+D2)^2, Vp = (D1-D2)^2 Computed over the three axis, up to the precision of the proton.  Precision is:
sum(j*k*p + j*p + k*p); where p is a scale factor and must scale each term evenly. We need only find the integer set where the manifold over the three axis converge uniformly, the manifold curvature being the limit of proton precision. That manifold curvature makes the Shannon Uncondition in the Shannon operator.

Charge falls out because that curve component prevents EM light from forming. Spin falls out because two adjacent mass in our spectral chart.  The j*k axis defines null density that prevents mass quantization,  The magnetic is really the phase angle of the phase imbalance in the orbital, and when that is to straight, the electron flies away.