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It is well-known how to canonically quantize the Lagrangian

$$L = i \bar{\psi} \dot{\psi} - \omega \bar\psi \psi$$

I now wonder how one quantizes the Lagrangian with one real fermion

$$L = i \psi \dot\psi$$

Obviously there can be no mass term since it is anticommuting so $\psi \psi = 0$.

I find no obstacles when going through Dirac's procedure. First I get the conjugate momentum as

$$\pi = i \psi$$

which I view as a constraint since there is no time derivative in this relation,

$$\Phi = \pi - i \psi = 0$$

Next I construct the Dirac bracket (DB) from the Poisson brackets (PB)

$$\{\psi,\pi\}_{PB} = 1$$

$$\{\psi,\psi\}_{PB} = 0$$

$$\{\pi,\pi\}_{PB} = 0$$

by following the standard procedure. First I define

$$C = \{\Phi,\Phi\}_{PB} = - 2 i$$

and then

$$\{\psi,\psi\}_{DB} = \{\psi,\pi\}_{PB} C^{-1} \{\pi,\psi\}_{PB} = 1\cdot(-2i)^{-1}\cdot 1 = i/2$$

Quantizing amounts to replacing DB by anticommutator as

$$[\psi,\chi]_+ = i\hbar\{\psi,\chi\}_{DB}$$

In this case this gives

$$[\psi,\psi]_+ = i\hbar(i/2) = - \hbar/2$$

This amounts to

$$\psi \psi = - \hbar/4$$

contradicting the fact that $\psi \psi = 0$.

Is there no way out of this? Is it impossible to canonically quantize this theory? The path integral seems to exist and make sense.

MarianD
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Andreas
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1 Answers1

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Comments to the post (v2):

  1. Concretely, a Grassmann-odd operator $\hat{\psi}$ does not have to square to zero, cf. e.g. this Phys.SE post, even though it is true that a Grassmann-odd number $\psi$ always squares to zero: $\psi^2=0$.

  2. It seems OP is interested in Grassmann-odd point mechanics rather than field theory.

  3. The quantization of fermions are discussed in various Phys.SE posts, e.g. here, here and links therein.

Qmechanic
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