Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Dirac equation and clifford algebra

  1. Jan 28, 2015 #1
    Is it a must to know clifford algebra in order to derive the dirac equation?
    I recently watch drphysics video on deriving dirac equation and he use two waves moving in opposite directions to derive it, without touching clifford algebra. If this possible, what is the intuition behind it?
     
  2. jcsd
  3. Jan 29, 2015 #2

    ChrisVer

    User Avatar
    Gold Member

    No you don't "need to" know Clifford Algebra... you derive it in the process in order to get your equation working [and reproducing the relativistic relationship between energy and mass/momenta] . At least by the time one adds the gamma matrices..
     
  4. Jan 29, 2015 #3
    Do you have any sources that will allow me to derive the dirac equation without clifford algebra? Cause I have been searching on the Internet and all of it involves clifford algebra. Besides the one by drphysics he didnt really derive it fully such as the dirac matrix.
     
  5. Jan 29, 2015 #4

    ChrisVer

    User Avatar
    Gold Member

    I haven't seen what you are talking about, I don't know which is that drphysics book... Maybe this can help? http://arxiv.org/vc/quant-ph/papers/0607/0607001v1.pdf (sec.4)

    However, I don't understand this "clifford algebra" thing. In every introductory book, they never used the Clifford algebra to derive the Dirac equation.
    Most of them are starting by a linear Hamiltonian operator, which they act to reproduce the result of the relativistic energy-momentum-mass relation [itex]E^2 = m^2 +p^2 [/itex]. By doing so, you see that the objects that you used in your operator were not just complex numbers, but they should satisfy certain relations, one of them is the Clifford Algebra. You don't "use" it, you "result" in it...
     
  6. Jan 30, 2015 #5

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Well, you can call it as you like. The Clifford algebra is basically the algebra realized in the usual treatment by the Dirac matrices in the one or the other representation. At some time ago in history also quaternions and octonions were en vogue. In the 19th century at least; Maxwell formulated his equations first in terms of quaternions; the usual vector notation was introduced into physics by Heaviside and became widely used in the beginning of the 20th century. It was not the least spread by the famous Handbook "Enzykloädie der mathematischen Wissenschaften", edited by Felix Klein around 1900. You can find a derivation of the Dirac equation, using quaternions in the even more famous book by Sommerfeld "Atombau und Spektrallinien" (freely translated: Atomic Structure and Spectral Lines). As stressed before: No matter, how you derive it you end up with somehow with the Dirac formalism. This is no surprise, because it's a very natural representation of the proper orthochronous Lorentz group augmented by spatial reflections to be able to describe the parity-conserving interactions, especially the electromagnetic interaction, which was the first application of the Dirac formalism of course.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Dirac equation and clifford algebra
Loading...