## 435(6): Rules for Quantization in m Space

435(6): Rules for Quantization in m Space

I arrived at Note 435(6) using the same transformation rules throughout. In all occurrences r is replaced by r / m(r) power half and t is replaced by m(r) power half t. It follows that psi (r, t) of Eq. (4) is replaced by psi(r, t) of Eq. (5). The quantization rules remain the same, because the frame (r, phi) is the same. The state of the particle is completely described by its wavefunction, so I decided to start the analysis with the wavefunction, and to replace r by r / m(r) power half and t by m(r) power half t inside the wavefunction. This gives Eq. (5), in which psi is a function of r and t. So the quantization rules are Eq. (2) and (3) because they are defined in frame (r, t) and psi is defined in terms of r and t. This procedure leads to the modified Planck quantization (10), which uses the expectation value of m power half computed self consistently with the wavefunction (5). This leads self consistently to Eq. (15), which is an equation for the quantized m(r) power half. This philosophy is based on the fundamental role of the wavefunction. The Born normalization remains the same. Since all this is completely new to physics, there is no precedent, so any self consistent procedure can be used. As usual, experiments must decide whether the results are acceptable or not.

comment-435(6).pdf

## 435(6): Rules for Quantization in m Space

435(6): Rules for Quantization in m Space

After experimenting with the notes for UFT435, I decided to adopt the rules (1) and (2) for quantization. They are applied for the free particle in this note, but can be applied to any wavefunction. They result in shifts and splittings given by Eq. (14). These shifts and splittings are due to m space, which can be thought of as the vacuum. So this gives an explanation of the radiative corrections in terms of m(r) functions, getting rid of all the obscurities of quantum electrodynamics. Eq. (14) shows that the Planck quantization is modified by the expectation values of m(r) power half. So the latter is quantized, meaning that the m space is quantized. The problem being considered defines the quantization. The free particle problem is the simplest case. This theory shows that general relativity and quantum mechanics have been unified, because m(r) of general relativity has been quantized. So I intend to base Sections 1 and 2 of UFT435 on this note.

a435thpapernotes6.pdf

## 435(5): Brief Review of Fundamentals Planck Quantization in m Space

435(5): Brief Review of Fundamentals Planck Quantization in m Space

This not checks fundamentals and demonstrates that the m theory is based on the Minkowski metric (13) in m space. This discovery was made in UFT416 and led to a number of important advances. In quantum mechanics the m theory leads to shifts and splittings due to m space itself, notably the Lamb shifts. The Planck quantization E = h bar omega is generalized in m space to Eq. (35), valid for any wavefunction psi and illustrated here by the free particle wave function. The general replacement rule is r goes to r / m(r) half and t goes to m(r) half t.

a435thpapernotes5.pdf

## 435(3): Splits and Shifts due to Frame Transformation

435(3) OK thanks, here it is. The complete interpretation of the Lamb shift must be based on these equations. The left hand side has the advantage of giving the number of energy levels without having to compute the expectation value of the hamiltonian (right hand side).

On Tue, Mar 26, 2019 at 6:52 PM Horst Eckardt <mail> wrote:

There was no document attached.
Horst

Am 26.03.2019 um 13:10 schrieb Myron Evans:

a435thpapernotes3.pdf

## PS: Re: Fwd: Fwd: Calculation of Lamb shift: sign problem

There is no doubt that the spectral effect found by Lamb and Retherford was a rise of about 1.0 GHz of the 2S sub 1/2 energy level above the 2P sub 1/2 level. Google "Lamb shift rise of 1 GHz", sites one and two that come up, and many other sites and books. The Lamb shift increases the energy of S states, but leaves the energy of P states unchanged. In the Dirac atom, 2S sub 1/2 and 2P sub 1/2 are degenerate. Also Google "Lamb shift of atomic H" first site that comes up. This contains Bethe’s original paper, "The Electromagnetic Shift of Energy levels" (Physical Review). The article mentions: "Bethe considered the Lamb Retherford discovery of the shift of the 2S state of H upward in energy. The easiest way to understand it is an increase in frequency of the 2S sub 1/2 absorption line with respect to the 2P sub 1/2 absorption line. This now known with great precision. I realize that it is important to understand the experimental result. Bethe in this very first paper introduced renormalization, I do not believe the claim of the standard model to incredible QED precision. We are doing very well in explaining it with m theory.

On Mon, Mar 25, 2019 at 3:33 PM Horst Eckardt <mail> wrote:

PS: To be precise, this diagram means that the absolute value of 2S1/2 is DECREASED:

This means that the correction integral of hbar omega must be smaller than one.

Horst

Am 25.03.2019 um 16:10 schrieb Horst Eckardt:

Does the absolute value of the 2S energy increase or decrease?
Horst

Am 25.03.2019 um 16:06 schrieb Myron Evans:

Calculation of Lamb shift

The frequency of the 2S state increases to about a gigahertz above the 2P state (Wiki, Lamb shift entry). So the energy of the 2S state increases with respect to that of the 2P state. The energy is h bar omega multiplied by a correction value integral psi* (1 / m(r) half psi dtau, so the correction value must increase, i..e become greater than 1.

Calculation of Lamb shift: sign problem

To make it clear: If the energy level of 2S becomes less negative valued, the absolute value decreases. Therefore the correction factor of the 2S state becomes smaller than one. Do you agree to this argument?

Horst

Am 25.03.2019 um 15:11 schrieb Myron Evans:

Calculation of Lamb shift: sign problem

OK thanks, I would suggest experimenting to find the m(r) function that gives the experimental result. As long as m(r) becomes one in the Minkowski space, there is freedom of choice, so it can become greater than one. It is encouraging that an exponential m(r) gives the correct result of S being shifted and P not being shifted. There is a rise of about 1000 MHz of the 2S level. This means that the energy level of 2S becomes less negative valued, and the frequency increases. The absolute value of energy increases, so the absolute value of the theoretical result would match the experimental result.

Calculation of Lamb shift: sign problem

Accoring to the diagram of Fig. 1 in UFT429, the Lamb shift affects the
2S1/2 state, not the 2P1/2 state. This is verified by our exponential
m(r) functions. However the absolute value of the energy level shrinks
(because the levels in the diagram are negative). This is in conflict
with the expectation value

integral psi* 1/m(r)^(1/2) psi dtau

which gives values >1 because m(r)<1 near to the centre. The situation
could be remedied by using a m(r)>1, but this would change all our
I appended the diagram with relative energy change of the 2S1/2 state.
The experimental shift has been added. The parameter R is about 0.015
Bohr radii which is about twice the value obtained form s-o coupling in
UFT 429.

Horst

## PS: Re: Fwd: Fwd: Calculation of Lamb shift: sign problem

There is no doubt that the spectral effect found by Lamb and Retherford was a rise of about 1.0 GHz of the 2S sub 1/2 energy level above the 2P sub 1/2 level. Google "Lamb shift rise of 1 GHz", sites one and two that come up, and many other sites and books. The Lamb shift increases the energy of S states, but leaves the energy of P states unchanged. In the Dirac atom, 2S sub 1/2 and 2P sub 1/2 are degenerate. Also Google "Lamb shift of atomic H" first site that comes up. This contains Bethe’s original paper, "The Electromagnetic Shift of Energy levels" (Physical Review). The article mentions: "Bethe considered the Lamb Retherford discovery of the shift of the 2S state of H upward in energy. The easiest way to understand it is an increase in frequency of the 2S sub 1/2 absorption line with respect to the 2P sub 1/2 absorption line. This now known with great precision. I realize that it is important to understand the experimental result. Bethe in this very first paper introduced renormalization, I do not believe the claim of the standard model to incredible QED precision. We are doing very well in explaining it with m theory.

On Mon, Mar 25, 2019 at 3:33 PM Horst Eckardt <mail> wrote:

PS: To be precise, this diagram means that the absolute value of 2S1/2 is DECREASED:

This means that the correction integral of hbar omega must be smaller than one.

Horst

Am 25.03.2019 um 16:10 schrieb Horst Eckardt:

Does the absolute value of the 2S energy increase or decrease?
Horst

Am 25.03.2019 um 16:06 schrieb Myron Evans:

Calculation of Lamb shift

The frequency of the 2S state increases to about a gigahertz above the 2P state (Wiki, Lamb shift entry). So the energy of the 2S state increases with respect to that of the 2P state. The energy is h bar omega multiplied by a correction value integral psi* (1 / m(r) half psi dtau, so the correction value must increase, i..e become greater than 1.

Calculation of Lamb shift: sign problem

To make it clear: If the energy level of 2S becomes less negative valued, the absolute value decreases. Therefore the correction factor of the 2S state becomes smaller than one. Do you agree to this argument?

Horst

Am 25.03.2019 um 15:11 schrieb Myron Evans:

Calculation of Lamb shift: sign problem

OK thanks, I would suggest experimenting to find the m(r) function that gives the experimental result. As long as m(r) becomes one in the Minkowski space, there is freedom of choice, so it can become greater than one. It is encouraging that an exponential m(r) gives the correct result of S being shifted and P not being shifted. There is a rise of about 1000 MHz of the 2S level. This means that the energy level of 2S becomes less negative valued, and the frequency increases. The absolute value of energy increases, so the absolute value of the theoretical result would match the experimental result.

Calculation of Lamb shift: sign problem

Accoring to the diagram of Fig. 1 in UFT429, the Lamb shift affects the
2S1/2 state, not the 2P1/2 state. This is verified by our exponential
m(r) functions. However the absolute value of the energy level shrinks
(because the levels in the diagram are negative). This is in conflict
with the expectation value

integral psi* 1/m(r)^(1/2) psi dtau

which gives values >1 because m(r)<1 near to the centre. The situation
could be remedied by using a m(r)>1, but this would change all our