Hi everyone,(adsbygoogle = window.adsbygoogle || []).push({});

I'm trying to follow how Dirac went about deriving his wave equation. I know there's a couple of different ways to "derive" this, but I'm trying to follow the original method that Dirac used. There's a lot of good stuff online for this, but there's a step that I get stuck at and was hoping you could help.

Just as Schrodinger stars with the classical Hamiltonian and make's the quantum substitution, Dirac does the same thing, only starting with the energy-momentum relation from special relativity. The problem is that this is quadratic, so he assumes you can factor it in order to get a linear expression for the energy.

[itex] p^{\mu} p_{\mu} = m^2 [/itex] , so

[itex] p^{\mu} p_{\mu} - m^2 = 0 [/itex],

Assuming that you can factor this, you get

[itex] p^{\mu} p_{\mu} - m^2 = (\beta^{\kappa} p_{\kappa} - m)(\gamma^{\lambda} p_{\lambda}+m) [/itex]

Now we need to determine [itex] \beta^{\kappa} [/itex] and [itex] \gamma^{\lambda} [/itex]

We can expand the righthand side so that

[itex] \beta^{\kappa} \gamma^{\lambda} p_{\kappa}p_{\lambda} + (\beta^{\kappa}p_{\kappa} - \gamma^{\lambda}p_{\lambda}) - m^2 [/itex]

and you pick up this cross term in the middle.

So now you can identify the [itex] m^2 [/itex] on the RHS with the one on the LHS, and also identify the [itex] \beta^{\kappa} \gamma^{\lambda} p_{\kappa}p_{\lambda} [/itex] on the RHS with the [itex] p^{\mu} p_{\mu} [/itex] on the LHS. This seems to suggest that the cross terms in the middle should be zero.

The problem is that I'm just now sure how to get rid of them. Also I'm a little confused about the relationship between the [itex] p^{\mu} [/itex] four vectors and the [itex] p_k [/itex] factors. Is [itex] p_k [/itex] still a four vector? If not what is it?

Thanks for the help!

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# Trying to Derive Dirac's equation following Dirac

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