Klein-Gordon equation: showing k_i k_i equals p^2/c^2

[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 can't 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 that's why he didnt show it, but I'm a bit stuck.

Any help would be appreciated.

edit

he also did something similar here:

 

[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.

 

[rwooduk]

I don't understand your questions (and I didn't see something similar in the notes you uploaded)
Momentum and frequencies are related via $p = \hbar k$

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!

 

[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.