- #1

cazador970

## Main Question or Discussion Point

So, I am a newbie in quantum mechanics, took modern physics last fall for my physics minor.

I know that Schrodinger based his equation based on the equation K + V = E,

by using non-relativistic kinematic energy (P

p becoming the operator p= -iħ∇ for the wave equation eigenfunction when the wave function is introduced, and similarly, with energy E= iħ∂

For the time independent case:

-ħ

And this is used to solve the one electron atom and get the Azimuthal equation, the Polar equation, and the Radial equation when the Laplacian operator is used for spherical polar coordinates.

By Einstein equation:

E

He attempted to use V+√((mc

to make a relativistic wave equation.

But could not do it, given that the square root of operators do not "make sense".

There is the Klein-Gordon Equation relativistic wave equation that attempts to do it.

E

E

I know that Dirac solved this by using matrices unifiying quantum physics and special relativity agreeing similarly with klein-gordon equation for spinless particles. But looking into the Klein-Gordon Equation I noticed that it does not count for potential energy. E in the Klein-Gordon and Einstein equation is not total energy, but just the Rest mass Energy + Kinetic Energy

E

and

E

So E

And the equation becomes:

m

Given the way the equation is, getting the Angular and Azimuthal equations and solutions work just the same as in using the Schrodinger equation, the main difference would be in the radial equation. Any thoughts? I want to know if anyone has ever tried to do the equation this way, or why formulating the equation this way is a bad idea.

I know that Schrodinger based his equation based on the equation K + V = E,

by using non-relativistic kinematic energy (P

^{2}/2m + V = E)p becoming the operator p= -iħ∇ for the wave equation eigenfunction when the wave function is introduced, and similarly, with energy E= iħ∂

_{/∂t}.For the time independent case:

-ħ

^{2}∇^{2}ψ/2m + Vψ = Eψ.And this is used to solve the one electron atom and get the Azimuthal equation, the Polar equation, and the Radial equation when the Laplacian operator is used for spherical polar coordinates.

By Einstein equation:

E

^{2}= (mc^{2})^{2}+(pc)^{2}He attempted to use V+√((mc

^{2})^{2}+(pc)^{2}) = Eto make a relativistic wave equation.

But could not do it, given that the square root of operators do not "make sense".

There is the Klein-Gordon Equation relativistic wave equation that attempts to do it.

E

^{2}ψ= (mc^{2})^{2}ψ-ħ^{2}c^{2}∇^{2}ψE

^{2}ψ = m^{2}c^{2}ψ_{/ħ2}-∇^{2}ψ (time independent)I know that Dirac solved this by using matrices unifiying quantum physics and special relativity agreeing similarly with klein-gordon equation for spinless particles. But looking into the Klein-Gordon Equation I noticed that it does not count for potential energy. E in the Klein-Gordon and Einstein equation is not total energy, but just the Rest mass Energy + Kinetic Energy

E

_{Einstein}= M + Kand

E

_{total}= E_{Einstein}+ V_{Potential Energy}So E

_{Einstein}= E_{Total}- V_{Potential Energy}And the equation becomes:

m

^{2}c^{2}ψ_{/ħ2}-∇^{2}ψ = (E-V)^{2}ψGiven the way the equation is, getting the Angular and Azimuthal equations and solutions work just the same as in using the Schrodinger equation, the main difference would be in the radial equation. Any thoughts? I want to know if anyone has ever tried to do the equation this way, or why formulating the equation this way is a bad idea.