# Matrix order in Dirac equations

The trace of matrix is always sum of its eigen values , which can be seen if $\hat{U}$ transforms the matrix $\alpha_i$ into it's diagonal form . $$\begin{pmatrix} A_1 & 0 & \cdots & 0 \\ 0 & A_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & A_N \end{pmatrix}= \hat{U}\alpha_i\hat{U}^{-1} =tr\alpha_i\hat{U}\hat{U}^{-1} =tr\alpha_i$$ Therefore we can say that order of matrices have to be even.

We also have following requirements from dirac equations

$$(\alpha_i \alpha_j+ \alpha_j \alpha_i)= 2\delta_{i\,j} I$$

$$\left\{ \alpha_i , \alpha_i \right\}= 2\delta_{i\,j} I$$

$$(\alpha_i \beta+ \beta \alpha_i) =0$$

Now how can we prove that $\alpha$ and $\beta$ are 4*4 matrices? Please, don't skip too much.

• You begin your question by listing some facts about matrices, which don't really have anything to do with your question. Why? – user1504 Oct 20 '12 at 23:24
• @user1504: Because they have to do with the question. – Ron Maimon Oct 21 '12 at 0:56

## 1 Answer

You can't prove it, because it isn't true. Given any 4 by 4 matrices that satisfy the Dirac algebra, you can make 8 by 8 matrices $\gamma_i$ which are equal to the four by four matrices in their upper diagonal, and also equal to the same thing in the lower diagonal. The goal is to look for the lowest dimensional representation, the smallest possible matrices. In this stupid trick, the upper 4 components of the spinor transforms exactly the same as the lower 4 components, meaning that you can reduce the representation by setting the top and bottom components to be equal. You are looking for an irreducible representation, which just means you can't do that.

To understand the size of the lowest dimensional representation Dirac matrices, there is a nice trick described in an article of Scherk's which does this for arbitrary dimensions. In Euclidean signature, make complex dimensions out of pairs of real dimensions

$$z_1 = x_1 + i x_2$$ $$z_2 = x_3 + i x_4$$

and then linearly combine the Dirac matrices as indicated by this coordinate change:

$$\gamma'_1 = \gamma_1 + i \gamma_2$$ $$\gamma'_2 = \gamma_3 + i \gamma_4$$

Then the $\gamma$ algebra in terms of the new $\gamma'$ matrices and their conjugates turns into the commuting Fermionic raising and lowering operators. These have a minimal representation which begins by defining the state $|0>$ which is annihilated by all lowering operators, and then acting the raising operators at most once, to produce states. This produces $2^n$ different states, where n is the dimension divided by 2, and this is the basic starting point, ignoring three annoying complications.

These states produced by raising and lowering give you the dimension of the spinor (ignoring the two complications) This starting point tells you that the Dirac algebra should be represented by 4 by 4 matrices in 4d, by 8 by 8 matrices in 6d, by 16 by 16 in 8d, 32 by 32 in 10d. Two to the power of half the dimensions.

The three annoying complications are: odd dimensions, Weyl Fermions, and Majorana Fermions, which each are a separate not so long discussion, but you can see how big the Dirac matrices are supposed to be from the argument above, at least approximately, and you can figure out the useful forms in 2d,3d, and 4d from just piddling around, as Pauli, Dirac, Weyl and Majorana did. The generalization to higher dimensions for string theory is the only time you need to get systematic about it, so it isn't discussed well outside of string theory literature.