Why the lowest order of matrices in Dirac equation are 4x4 matrices?

Question

Why the lowest order of matrices in Dirac equation (Relativistic Quantums) are 4x4 matrices (and can not be 2x2 matrices)?

How to prove it?

Answer 1

It is not a proof, but at least some taste.

For a $3$ dimensional space (no time), a representation of the $3$ gamma matrices $\gamma^i$ ($i =1,2,3$) are simply the $2*2$ Pauli matrices $ \sigma^i$ verifying : {$\gamma^i, \gamma^i$} $= 2 \delta_{ij}$. So, for a space with $3$ spatial dimensions, a $2*2$ representation of the gamma matrices is possible.

Now, for a $3+1$ space-time, one could think to add a $4th$ $2*2$ gamma matrice $\gamma^0$, which must verify $(\gamma^0)^2=-2 ~\mathbb Id$ and {$\gamma^0, \gamma^i$} $= 0$.

Writing explicitely these equations for the $4$ components of $\gamma^0$, and you will find that $\gamma^0=0$, so it is a taste that there is not enough place in $2*2$ matrices, for the representation of the gamma matrices in $(3+1)$ dimensions.

Answer 2

The state space of a spin-$1/2$ particle is the two-dimensional complex Hilbert space $\mathbb{C}^2$. Any Hamiltonian acting on this state space is necessarily a $2\times 2$ matrix. The algebra of observables on the state space of a spin-$1/2$ particle is generated by the raising and the lowering operators (as well as the $2\times 2$ identity matrix) which in turn are generated by the Pauli operators.

Now, the $\gamma$ matrices in the Dirac equation can be written in terms of the block-diagonal matrices with blocks consisting of the Pauli operators. See here for some details. This then accounts for two facts simultaneously: (1) The order is necessarily even. (2) The lowest order is $4$.