Deductions of Formulas for Energy

  • #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 (P2/2m + 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:
22ψ/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:
E2 = (mc2)2+(pc)2
He attempted to use V+√((mc2)2+(pc)2) = E
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.
E2ψ= (mc2)2ψ-ħ2c22ψ
E2ψ = m2c2ψ2-∇2ψ (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
EEinstein = M + K
and
Etotal = EEinstein + VPotential Energy
So EEinstein = ETotal - VPotential Energy
And the equation becomes:
m2c2ψ2-∇2ψ = (E-V)2ψ

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.
 

Answers and Replies

  • #2
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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 don't know if anyone has tried your suggestons but I don't think it would be fruitful.

First we know how to take that square-root you allude to nowadays:
http://cds.cern.ch/record/944002/files/0604169.pdf

If you read chapter 3 of Ballentine you discover that the potential term comes from symmetry considerations if the Galilean transforms are assumed ie in the non relativistic limit. You apply the same symmetry ideas to relativistic regions and yoi get QFT. There is a reason for that - the proof can be found here:
https://www.amazon.com/dp/019969933X/?tag=pfamazon01-20

It inevitably leads to not single particle systems - but to particles not being conserved. This was the solution to the negative probabilities that appeared when the Klein-Gordon equation was used - instead of negative probabilities it was positive probabilities of antiparticles. Since the potential term comes from symmetry, it was natural to apply the same to relativistic QM and that indeed proved to work well. Not that it was that sort of logic is how it happened historically, but from our vantage it looks rather obvious.

Thanks
Bill
 
Last edited:
  • #3
Demystifier
Science Advisor
Insights Author
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And the equation becomes:
m2c2ψ2-∇2ψ = (E-V)2ψ

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.
This is essentially the Klein-Gordon equation in electric potential. You can find it's solution for the hydrogen atom in Schiff's Quantum Mechanics.
 

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