# Trying to Derive Dirac's equation following Dirac

1. Jan 27, 2013

### GravitatisVis

Hi everyone,

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.

$p^{\mu} p_{\mu} = m^2$ , so

$p^{\mu} p_{\mu} - m^2 = 0$,

Assuming that you can factor this, you get

$p^{\mu} p_{\mu} - m^2 = (\beta^{\kappa} p_{\kappa} - m)(\gamma^{\lambda} p_{\lambda}+m)$

Now we need to determine $\beta^{\kappa}$ and $\gamma^{\lambda}$

We can expand the righthand side so that

$\beta^{\kappa} \gamma^{\lambda} p_{\kappa}p_{\lambda} + (\beta^{\kappa}p_{\kappa} - \gamma^{\lambda}p_{\lambda}) - m^2$

and you pick up this cross term in the middle.

So now you can identify the $m^2$ on the RHS with the one on the LHS, and also identify the $\beta^{\kappa} \gamma^{\lambda} p_{\kappa}p_{\lambda}$ on the RHS with the $p^{\mu} p_{\mu}$ 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 $p^{\mu}$ four vectors and the $p_k$ factors. Is $p_k$ still a four vector? If not what is it?

Thanks for the help!

2. Jan 27, 2013

### samalkhaiat

you do not need to have two set of matrices. Just take $\gamma = \beta$. then symmetrize the first term
$$\gamma^{ \mu } \gamma^{ \nu } P_{ \mu } P_{ \nu } = \frac{ 1 }{ 2 } \{ \gamma^{ \mu } , \gamma^{ \nu } \} P_{ \mu } P_{ \nu }$$

Sam

3. Jan 28, 2013

### GravitatisVis

Hi Sam,
Thanks for the input.

I don't think it's fair game though at this point to symmetrize the first term, since you don't know the anti-commutation relation between the gamma matrices yet. That comes next when you expand everything out, compare terms on the RHS with the LHS, and realize that you need some kind of object with specific properties that happens to have the appropriate anti-commutation relation.

Also, I'm a little confused as to why you said "you do not need to have two set of matrices". Can you explain?

Thanks a lot!

4. Jan 28, 2013

### andrien

you can see here just above eqn 3 what sam means
http://www.mathpages.com/home/kmath654/kmath654.htm
However Sam's way is ingenious and you can get the usual way in above reference.Also by symmetrizing you get the anticommutation which is no artificial.

5. Jan 28, 2013

### samalkhaiat

Ok, do it your way. You want to factorize the expression
$$\eta^{ \mu \nu }p_{ \mu }p_{ \nu } - m^{ 2 }$$
as
$$( \gamma^{ \mu } p_{ \mu } - m ) ( \beta^{ \nu } p_{ \nu } + m )$$
and find the conditions under which such factorization is valid. So, you expand
$$\gamma^{ \mu } \beta^{ \nu } p_{ \mu } p_{ \nu } + ( \gamma^{ \mu } - \beta^{ \mu } ) p_{ \mu } m - m^{ 2 }$$
Now, you compare with original expression. Since there was no term linear in $m p_{ \mu }$, you must have $\gamma^{ \mu } = \beta^{ \mu }$. So you are left with
$$\eta^{ \mu \nu } p_{ \mu } p_{ \nu } = \gamma^{ \mu } \gamma^{ \nu } p_{ \mu } p_{ \nu }$$
In order to equate the coefficients on both sides that is valid for all values of the indices, you must make the right-hand side symmetric in $\mu$ and $\nu$ (because the left-hand side IS symmetric). This way the anti-commutator shows up naturally as a condition on your chosen matrices:
$$\{ \gamma^{ \mu } , \gamma^{ \nu } \} = 2 \eta^{ \mu \nu }$$

Sam