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  Dirac equation as canonical quantization?

+ 6 like - 0 dislike

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?

This post imported from StackExchange Physics at 2015-03-30 13:54 (UTC), posted by SE-user Sasha
asked Jul 7, 2012 in Theoretical Physics by Sasha (110 points) [ no revision ]
retagged Mar 30, 2015
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?

This post imported from StackExchange Physics at 2015-03-30 13:54 (UTC), posted by SE-user kηives
@kηives: Through relation $\bar{\psi} = (\gamma_0 \psi)^\dagger$.

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

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

2 Answers

+ 1 like - 0 dislike

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
answered Aug 9, 2012 by Ron Maimon (7,720 points) [ no revision ]
+ 0 like - 1 dislike

This is not answer to your question but a simpler analog of the question you asked : Consider 1 dimensional case. Take your phase space to be $T^*R$ (cotangent bundle of real line), and your 'velocity space' to be $TR$. In this case Dirac operator is $-i\partial/\partial x$ (I got this cool information from wikipedia:) which acts on space of complex valued functions on $R$. Classical analog of this operator would be "momentum function" on $T^*R$ (right?). Now apply rules of Legendre transformation to see what is the corresponding Lagrangian on $TR$. What you will find is that Legendre transformation is not well defined. I think the same will be the case in higher dimensions but I am not sure.

This post imported from StackExchange Physics at 2015-03-30 13:54 (UTC), posted by SE-user user10001
answered Jul 7, 2012 by user10001 (635 points) [ no revision ]

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