# Stuck on last part of class derivation (proof) K-G Eqn

1. Apr 26, 2015

### rwooduk

Mentor note: fixed formulas so they get displayed properly

If we take the K-G eqn and the following term for the wave function

$$( \partial^2 + \frac{m^{2}c^{2}}{\hbar^{2}})\Psi =0 \\\\\Psi = Re^{-i\omega t + i k_{i}x_{i}}$$

We worked through to this $\hbar \omega = \pm \sqrt{\hbar^{2} c^{2}k_{i}k_{i}+ m^{2}c^{4}}$ which is fine and recognisable, but I cant get the $\hbar^{2} c^{2}k_{i}k_{i}$ term to equal the familiar $\rho^{2}c^{2}$. I'm assuming it's obvious and thats why he didnt show it, but I'm a bit stuck.

Any help would be appreciated.

edit

he also did something similar here:

Last edited: Apr 26, 2015
2. Apr 26, 2015

### ChrisVer

I don't understand your questions (and I didn't see something similar in the notes you uploaded)
Momentum and frequencies are related via $p = \hbar k$...you can see this fast with some dimensional analysis.

3. Apr 26, 2015

### rwooduk

Sorry, I'm trying to show that this $$\hbar \omega = \pm \sqrt{\hbar^{2} c^{2}k_{i}k_{i}+ m^{2}c^{4}}$$ is the same as this:

edit which you have just solved, sorry and thanks!!

4. Apr 26, 2015

### ChrisVer

Yup it is... use the DeBroglie relations: $E= \hbar \omega$ and $p= \hbar k$
since your wave function is expressed in terms of frequency/wavenumber instead of energy/momenta.

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