Thanks Julio and Barry, for this offers me more clarity on Schrodinger Equation of a cubic set solution versus general set. The Elements Beyond Uranium, Seaborg & Loveland, 1990, pages 72-73.
Now I take as the math equations that govern all of Physics as the calculus on New Ohm's Law Voltage = current*magnetic field*electric field, and that calculus delivers these EM laws that govern all of physics-- and governs all of mathematics also, for math is a subset of physics.
Here I use Angular Momentum L as electric field E, both are interchangeable.
0) domain law as Atomic Theory
1) Magnetic primal unit law Magnetic Field B = kg /A*s^2
2) V = C*B*L New Ohm's law, law of electricity
3) V' = (C*B*L)' Capacitor law
4) (V/C*L)' = B' Ampere-Maxwell law
5) (V/(B*L))' = C' Faraday law
6) (V/(C*B))' = L' the new law of Coulomb force with EM gravity force
All the calculus permutations possible are V', B', C', and E' (same as L').
Notice that in all the calculus, only one is Addition the Voltage derivative. While the other three are derivatives of division which ends up as Subtraction. For example, the quotient rule of calculus is (V/i*L)' = B' = (V'*i*L - V*i' *L - V*i*L') / (i*L)^2. Do you see that in every division derivative, we have two subtractions over one number.
Algebra of 3D Calculus, for remember we did the algebra of
V' = (iBL)'
i' = (V/BL)'
B' = (V/iL)'
L' = (V/iB)'
--- quoting 1st year calculus from Teaching True ---
Using the Product Rule which is (fgh)' = (f'gh + fg'h + fgh')
Capacitor Law (i*B*L)' = i'*B*L + i*B'L + i*B*L'
V' = (iBL)' = i'*B*L + i*B'*L + i*B*L' here we have three terms explaining capacitors
Using the Quotient Rule, which is (f/gh)' = (f'gh - fg'h - fgh')/(gh)^2
(V/i*L)' = B' = (V'*i*L - V*i' *L - V*i*L') / (i*L)^2
Maxwell had two terms in the Ampere-Maxwell law-- the produced magnetic field and a displacement current, but above we see we have also a third new term.
(V/B*L)' = i' = (V'*B*L - V*B' *L - V*B*L') / (B*L)^2
V' = (iBL)' = i'*B*L + i*B'*L + i*B*L' reduces to
= iBL + iVL + iBL'
i' = V'*B*L/ (B*L)^2 - V*B' *L/ (B*L)^2 - V*B*L' / (B*L)^2 reduces to
i' = B^2*L/ (B*L)^2 - V^2 *L/ (B*L)^2 - V*B*L' / (B*L)^2 further reduces
= 1/L - V^2/B^2*L - VL'/BL^2
B' = V'*i*L/ (i*L)^2 - V*i' *L/ (i*L)^2 - V*i*L' / (i*L)^2 reduces to
B' = B*i*L/ (i*L)^2 - V*i *L/ (i*L)^2 - V*i*L' / (i*L)^2 further reduces to
= B/iL - V/iL - VL'/iL^2
L' = (V/i*B)' = (V'*i*B - V*i' *B - V*i*B') / (i*B)^2 reduces to
L' = i*B^2 / (i*B)^2 - V*i *B / (i*B)^2 - V^2*i / (i*B)^2 further reduces to
= 1/i - V/iB - V^2/iB^2
(1) V' = iBL + iVL + iBL'
(2) i' = 1/L - V^2/B^2*L - VL'/BL^2
(3) B' = B/iL - V/iL - VL'/iL^2
(4) L' = 1/i - V/iB - V^2/iB^2
Alright, so I replace L' in (1) with 1/i - V/iB - V^2/iB^2
I get V' = iBL + iVL + iB*(1/i - V/iB - V^2/iB^2 )
= iBL + iVL + B - V - V^2/ B
Doing the replacement in (2)
i' = 1/L - V^2/B^2*L - VL'/BL^2
= 1/L - V^2/B^2*L - V*(1/i - V/iB - V^2/iB^2) /BL^2
= 1/L - V^2/B^2*L - (V/iBL^2) - (V^2/iB^2L^2) - (V^3/(iB^3L^2))
Doing the replacement in (3)
B' = B/iL - V/iL - VL'/iL^2
= B/iL - V/iL - V(1/i - V/iB - V^2/iB^2)/iL^2
= B/iL - V/iL - (V/i^2L^2) - (V^2/i^2*B*L^2) - (V^3/( i^2B^2L^2))
Julio likes geometry, and that is my preference also, I prefer to solve problems of both science, and math, best solved by geometry, for I believe in Darwin Evolution that what evolved us from apes was the throwing of rocks and stones to garner advantage, and throwing requires the brain to advance in geometry, more than advance in quantity = algebra. To hit a target needs geometry.
So in the above, I replace Voltage = current*magnetic field*electric field (note, the * symbol is generalized multiplication and can be dot vector product or cross vector product or even scalar product).
So, well in math we can convert Voltage = i*B*E, into volume and make it easy on ourselves, as Volume = length*width*height.
Now that Volume decomposes into calculus of four differential equations of Volume', length', width', height'.
The length, width, height differential equations will end up with two subtraction terms. The Volume differential equation will be pure addition of terms.
And now, this leads me to the clarity of Cubic versus General Set in Schrodinger Equation.
If we have volume of rectangular box such as Length = 10, Width = 4, Height = 3.
But what if we have the volume of Length = 10, Width = 10, Height = 10, then we are faced with two possibilities, for our volume can be the volume of a cube in 3D or the volume of a sphere with a radius of 5.
This is why Schrodinger Equation has a General Set along with a Cubic Set.