First of all, I'm not a physicist, I'm mathematics phd student, but I have one elementary physical question and was not able to find answer in standard textbooks.

Motivation is quite simple: let me fix some finite dimensional vector space $V$. Then we can think about Clifford algebra $Cl(V \oplus V^*)$ as algebra of odd differential operators on $\bigwedge V$ i.e. as canonical quantization of the algebra of classical fermionic observables. ($\bigwedge V$ is an analog of functions on a space and $Cl(V \oplus V^*)$ is an analog of a Weil algebra.)

From the other hand, action of a Clifford algebra is used in the construction of the Dirac operator. Thus my question: can one write Lagrangian for classical fermion on a space-time $\mathbb{R}^{3,1}$ such that canonical quantization of such system gives Dirac operator as quantum Hamiltonian?

  • $\begingroup$ a related question: With the Dirac Hamiltonian $H=\int d^3 x (i\bar{\psi}\gamma^i \partial_i \psi +m \bar{\psi}\psi)$ how does one cast this into a form that clearly shows the constraints? I mean the theory is invariant under certain transformations, and the constraints should be the generators for these symmetries, how do I expose them in the Hamiltonaian formulation? $\endgroup$
    – kηives
    Aug 7 '12 at 5:20
  • $\begingroup$ @kηives: Through relation $\bar{\psi} = (\gamma_0 \psi)^\dagger$. $\endgroup$
    – Ron Maimon
    Aug 7 '12 at 22:22
  • $\begingroup$ A nitpick--- you shouldn't use "classical fermionic observables", they aren't "observables" because they are fermionic. You should say "classical fermionic variables" instead. The answer to your question is yes, but this point of view is not new, it is how people standardly do higher dimensional Dirac operators. $\endgroup$
    – Ron Maimon
    Aug 9 '12 at 8:18

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.


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