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This question already has an answer here:

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

How to prove it?

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marked as duplicate by ACuriousMind quantum-mechanics Jul 14 '16 at 17:46

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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 $4$.

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    $\begingroup$ Perhaps I'm missing something obvious here, but it's not clear to me how (2), that the lowest dimension is $4$ follows from what you've outlined. Perhaps you could include a bit more logical detail? $\endgroup$ – joshphysics Feb 4 '14 at 18:34
  • $\begingroup$ @joshphysics: Please see the Wikipedia link in the answer. In the Dirac equation, we have $\alpha_i = \gamma^0\gamma^k$ (with $\gamma^j$ as in the Wiki link). Notice that $\gamma^k$ can be viewed as an action by $\sigma^k$ (where $\sigma^k$, $k=1,2,3$ are the Pauli spin operators) simultaneously on two decoupled spins, say $s_1\oplus s_2$ (notice the direct sum here instead of the tensor product). Since Pauli spin operators are $2\times 2$, it follows that $\alpha_k$ is $4\times 4$. $\endgroup$ – user3657 Feb 4 '14 at 20:22
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    $\begingroup$ I understand that the standard construction of the representation of the Dirac matrix algebra in four spacetime dimensions leads to $4\times 4$ matrices, but as far as I can tell, the question is asking about whether it's possible to construct lower-dimensional matrices that satisfy the same Clifford algebra. It's not clear to me how you've demonstrated this. $\endgroup$ – joshphysics Feb 4 '14 at 20:57
  • $\begingroup$ Agreed. In this case I'm answering the question as I understood it (differently from what you claim the question was). $\endgroup$ – user3657 Feb 4 '14 at 21:01
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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.

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