Quantum Dispersion Derivation

I got somewhat interested today in how the quantum dispersion relation for a photon was originally derived and the reason why was because I accidently derived it today, and through a series of simple gestures. But because the dispersion relation is quite an easy equation for photons, there is doubt in my mind my derivation was somewhat easier at all.

 

[math]\hbar c = Gm^2[/math]

 

Rearranging for the angular momentum component:

 

[math]\hbar = \frac{Gm^2}{c}[/math]

 

Then multiply [math]\frac{\partial}{\partial t}[/math] on both sides gives:

 

[math]\hbar \partial_t = \frac{GM^2}{c} \cdot \partial_t[/math]

 

Hit this with [latex]\psi[/latex] and it makes

 

[math]\hbar \cdot i\omega = \frac{Gm^2}{c} \cdot i \omega[/math]

 

Cancel out the [math]i[/math]'s this gives and replace the [math]\omega[/math] with [math]ck[/math] on the right

 

[math]\hbar \omega = \frac{Gm^2}{c} \cdot ck[/math]

 

clearly has units of energy. From now on, we will call [math]\frac{Gm^2}{c} = \beta[/math]. Since [math]\omega=ck[/math] then we can see this equation as freely exchanging the terms

 

[math]\hbar ck = \beta \cdot \omega[/math]

 

At first, I wasn't sure what to make of this equation, if indeed anything at all. So I carried on: turns out that a series of proportionalities could be measured from altering the equation:

 

[math]\frac{E}{(\beta \cdot \omega)} = k[/math]

 

because

 

[math]E= \hbar c k [/math]

 

so [math]k = \frac{E}{\hbar c}[/math]

 

It then struck me I'd seen the expression [math]\frac{E}{\hbar c}[/math] before. It is the case of electromagnetic radiation as a linear "dispersion relation" given as [math]E=\hbar ck[/math] with [math]k = \frac{E}{\hbar c}[/math]

 

This is obviously strictly a case only for a photon. I certainly derived a form of the photon dispersion relation but with a pesky mass sqaured term in there. As elegent it seemed deriving a dispersion relation by messing about with some terms, I wanted to have the equation suit a dispersion relation for a particle with mass... let us assume the simple case of an electron.

 

To derive the right one, I needed to go back to the derivations.

 

[math]\hbar ck = \beta \cdot \omega[/math]

 

Divide [math]c[/math] off the left hand side gives you a momentum expression

 

[math]p = \hbar k = \frac{\beta \cdot \omega}{c}[/math]

 

square

 

[math]\hbar^2 k^2 = (\frac{\beta \omega}{c})^2[/math]

 

Divide the left by [latex]2m[/latex] and you get the dispersion relation for an electron

 

[math]\frac{\hbar^2 k^2}{2m} = E[/math]

 

 

Has anyone got any links to how the dispersion formula was derived for a photon? I bet it was easier on second thoughts... I did a lot of fiddling about...

Edited October 13, 201114 yr by Mystery111