The mathese in your question makes it difficult to understand, it is best to be more concrete rather than abstract. The answer is yes, this is how higher dimensional Dirac operators are standardly constructed.

If you have the Dirac algebra (clifford algebra) on a 2n-dimensional space

$$ \{ \gamma_\mu \gamma_\nu \} = 2 g_{\mu\nu}$$

say Euclidean, then you can split the space-coordinates into even and odd pairs, and define the raising and lowering operators:

$$ \sqrt{2}\gamma^-_{i} = \gamma_{2i} + i \gamma_{2i+1}$$
$$ \sqrt{2}\gamma^+_{i} = \gamma_{2i} - i \gamma_{2i-1}$$

These anticommute, and obey the usual fermionic raising and lowering operator algebra, you can define a 0 dimensional fermionic system for which the state space are the spin-states. The state space the gamma matrices act on can be labelled starting with the spin-state called |0>, which is annihilated by all the lowering operators, and the other states are found using raising operators applied to |0>.

Then the Dirac Hamiltonian is automatically a Hamiltonian defined on a system consisting of a particle at position x, and a fermionic variable going across the finite dimensional state space of spin-states.

This post imported from StackExchange Physics at 2015-03-30 13:54 (UTC), posted by SE-user Ron Maimon