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

  1. Apr 26, 2015 #1
    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:

    MB063El.jpg
     
    Last edited: Apr 26, 2015
  2. jcsd
  3. Apr 26, 2015 #2

    ChrisVer

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    Gold Member

    I don't understand your questions (and I didn't see something similar in the notes you uploaded)
    Momentum and frequencies are related via [itex]p = \hbar k[/itex]...you can see this fast with some dimensional analysis.
     
  4. Apr 26, 2015 #3
    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:

    emcpc.gif

    edit which you have just solved, sorry and thanks!!
     
  5. Apr 26, 2015 #4

    ChrisVer

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    Gold Member

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