# The True Quantum

1. Apr 12, 2004

### Antonio Lao

What is the true quantum? The quantization of matter, energy, charge, angular momentum (spin), scalar and vector field and many other forms of quantization to date can all be approximation to the true quantum.

The true quantum is the square of the energy. This is a quantum of time (with two directions) and quantum of double-spin. The true quantum is a scalar field like the Higgs field.

2. Apr 13, 2004

### Antonio Lao

In a way, the complete understanding of quantum entanglement is intimately related to the true quantum.

This understanding of entanglement is presently beyond the reach of current physics. For us (creatures of reality), like what Einstein's demanded, our understanding depends on some element of reality. But John Bell and all the experiments on entanglement have shown, these elements of reality simply did not materialize at this time.

3. Apr 13, 2004

### ahrkron

Staff Emeritus
There's no such a thing as "the true quantum". There seems to be a "quantum principle" in all of physics, in the sense that the measurement process, by which we acquire information from experimental setups, behaves according to QM principles. However, the "quantum" in QM is a property, not a noun.

4. Apr 13, 2004

### Antonio Lao

It is not a noun because it is still not found. Once located, it's ghostly appearance will eventually disappear. At the moment, the quantum is intimately linked to the wave function of QM. And it's measuement requires the probability amplitude which is the square absolute value of the wave function.

My independent research strongly indicated to me that the "true" quantum of reality is the square of energy and the internal structure is that of two Hopf rings.

5. Apr 13, 2004

### ahrkron

Staff Emeritus
In many circumstances, QM systems do allow for continuous spectra. For instance, the energy of a free electron can have any value. If there was such a thing as "the true quantum" as a fundamental element of reality, every system would be quantized, instead of only bound states (as is the case).

6. Apr 13, 2004

### ahrkron

Staff Emeritus
Energy is a number. As such, it can be squared, and the result of such operation is also a number; hence, you are saying that the "true" quantum is only a number, not a physical entity.

7. Apr 13, 2004

### Antonio Lao

In a simplistic way, you are correct to say that the "true' quantum is a number. But in reality, it's a matrix. But once the operations of addition and multiplication are done on the matrices, numbers are produced. Addition produces numbers for values of electric charge and multiplication produces values for mass.

8. Apr 13, 2004

### Antonio Lao

A first hint that leads me to suspect the "trueness" of quantum of energy squared is th following relativistic energy equation:

$$E^2 = c^2 p^2 + m^2c^4$$

Dirac used this to propose the existence of antimatter. Further, he used the same square root form of the equation for the concept of spin. But the quantization of spin is only an approximation to the "trueness" of the quantum. A deeper and more subtle symmetry can be found in the double spins of two Hopf rings. But this implies reality as being one-dimensional instead of 4-dim (3 of space and one of time) as we are used to believe.

Last edited: Apr 13, 2004
9. Apr 13, 2004

### ahrkron

Staff Emeritus
Just in a very loose sense. He used the operation version of the equation, and defined a type of "number" (that turned out to be representable as a matrix) that was able to solve it.

10. Apr 13, 2004

### Antonio Lao

Dirac's matrices contain 0,1, -1, and i as elements. The matrices I use only contains 1, and -1. I was able to calculate the mass ratio of proton to electron within a percent of the experimental value.

11. Apr 13, 2004

### ahrkron

Staff Emeritus
A number, or a matrix, are representations of physical quantities, not physical entities themselves. The "true quantum" would only be a concept, and not able to be the building block of anything else.

Sure

This is an absurd generalization. Addition of electrical charges produces an electrical charge. Addition of masses produces a value of mass. As of multiplication, you need to specify what magnitudes you are multiplying.

12. Apr 13, 2004

### ahrkron

Staff Emeritus
1. Are "your matrices" supposed to have the same function as that of Dirac matrices? what commutation relations do your matrices satisfy? what is their relation to Dirac's eqn? If they have only real entries, they cannot possibly have the same role, which makes the comparison pointless.

2. What elements go into the calculation? I'm not asking for the full thing; only for a summary of the main ideas.

13. Apr 13, 2004

### Antonio Lao

The matrices always commute. The matrices are symmetrical and square with alternate elements of 1 and -1. They are basically Hadamard matrices. What is done to these matrices as operators operating on themselves to generate numerical values for charge and mass.

14. Apr 13, 2004

### ahrkron

Staff Emeritus
If they always commute, they cannot be used to solve Dirac's eqn. comparing them with Dirac matrices is then mixing apples and oranges.

15. Apr 13, 2004

### Antonio Lao

They are not used to solve any equation. it's only the elements that are different.
The Pauli's and Dirac's matrices contains 1, -1, 0, and i. Hadamard matrices contain only 1 and -1.

These matrices are all singular. Their determinants are zero. At first, I tried to fit these matrices into a group. But the multiplication operation does not satisfy the group property of possessing an inverse and also successive matrix addition or matrix multiplication produced scalar factors that cannot be part of the group. These matrices more or less formed an algebraic ring that are Abelian group of matrix addition and semi-group of matrix multiplication.

16. Apr 13, 2004

### Antonio Lao

If one can assume the existence of infinitesimal distances and forces r1, F1, and r2, F2 then the square of energy E is given by:

$$E^2 = r_i \times F_i \cdot r_j \times F_j$$

where i=1 and j=2.

Expanded by Langrange's identity give

$$E^2 = (r_i \cdot r_j)(F_i \cdot F_j) - (r_i \cdot F_j)(r_j \cdot F_i)$$

17. Apr 13, 2004

### Antonio Lao

The square of energy can also be equally given by

$$E^2 = r_i \times F_i \cdot F_j \times r_j$$

and when expanded give

$$E^2 = (r_i \cdot F_j)(r_j \cdot F_i) - (r_i \cdot r_j)(F_i \cdot F_j)$$

which one of these two forms of E^2 happens more often in reality can be determined by probability theory. One form represents the kinetic energy (or kinetic mass) and the other represents the potential energy (or potential mass).