# Do physical results for spinors depend on the Clifford algebra representation?

As I understand it, the Dirac equation and its solutions depend on the representation of the $\gamma$ matrices one uses. So if I were to use the Dirac representation I would get different mathematical solutions that if I used the Majorana representation.

In particular, Majorana fermions are their own antiparticles (because the solutions are real), so I am pretty sure that solutions in the Majorana representation are not physically equivalent to the ones in Dirac representation, which are not their own antiparticles.

If the only difference between $\gamma_{dirac}$ and $\gamma_{majorana}$ is a change in representation (meaning a different way to express the same $\gamma$ object), why are the solutions physically not equivalent? Why is the physics of the Dirac equation dependant on the representation of the $\gamma$ matrices?

• Majorana fermions are solutions to the Majorana equation not a solution to the Dirac equation. – Y2H May 20 '18 at 22:20
• The basic confusion here is that the Majorana representation does not describe Majorana spinors. It's just that Majorana spinors look simpler in the Majorana representation. – knzhou May 20 '18 at 22:32
• Don't confuse the representation of the Clifford algebra, which is as physically meaningless as a choice of basis, with the Lorentz representation of the fields! – knzhou May 20 '18 at 22:33

$$\psi^* = \psi$$
The general Majorana condition (in a basis indepent form) is: $$C\psi = \psi ^*$$ where $C$ is the charge conjugation operator. The point is that $C$ has different coordinate (read: entries) in different basis. In particular, it's the $2\times 2$ identity in Majorana basis, hence the utility of this basis.
You can also find how a Majorana spinor looks in Dirac basis by using the fact that $C=-i\gamma^2$ in chiral basis.