Dirac equation and clifford algebra

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?

ChrisVer
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..

vanhees71
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..
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.

ChrisVer
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 $E^2 = m^2 +p^2$. 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...

vanhees71