Discussion of 339(1)

Agreed, the quantization scheme of the Schroedinger equation as you know, is H psi = E psi where for the hydrogen atom:

H psi = (T + U) psi = ( p squared / (2m)) – e squared / (4 psi epso r)) psi

The energy levels are actually <H>, denoted E in quantum mechanics. Most students still ask: Why can’t you cross the psi? The answer is that

p squared psi = – h bar squared del squared psi

is a differential operator. Many people have immense difficulty in understanding this. We also have

<H> = – h squared integral psi* del squared psi d tau

and del squared also operates on psi. One could quantize the Sommerfeld hamiltonian as

H psi = (gamma m c squared + U) psi

As recent papers show there are hundreds of way of proceeding once the Dirac approximation is discarded.

To: EMyrone@aol.com
Sent: 30/01/2016 10:38:12 GMT Standard Time
Subj: Re: Discussion of 339(1)

In quantum mechanics the expectation value of the Hamiltonian is often denoted as total energy, then E would be the kinetic total energy or so, but we know what we mean.

Am 30.01.2016 um 06:39 schrieb EMyrone:

Agreed, the hamiltonian H = E + U is always the starting point of the theory, where E = gamma m c squared is the total relativistic energy. For a free particle H = E, but any elementary particle is always interacting with vacuum, so U is always non zero.

To: EMyrone
Sent: 29/01/2016 12:48:19 GMT Standard Time
Subj: Re: 339(1): Development of the Correct Theory of Spin Orbit Structure

To recapitulate the concept of potential energy in general relativity: the “relativistic” energy
gamma m c^2
contains the kinetic and rest energy only. Any potential energy has to be introduced as an extra energy. I assume that we used this consistently in this way in all earlier papers on Dirac and similar theory.


Am 25.01.2016 um 13:31 schrieb EMyrone:

This note suggest the development of the correct theory of spin orbit fine structure before going on to a new theory of the Lamb shift. Some computer algebra is needed to evaluate Eq. (19) for the Coulomb potential.

Leave a Reply