I would like to understand the following problem:

You have a classical fermion in one dimension. It has no mass, and no interactions. One can write its action as follows:


where $\theta(t)^{2}=0$.

This is just a $0+1$ dimensional analog of the Dirac spinor $\int d^{4}x\bar{\psi}i\gamma^{\mu}\partial_{\mu}\psi$.

This theory has some peculiar features:

First, since the Grassmann variable $\theta(t)$ transforms like a world-line scalar, the theory is of first-order, and is invariant under time-reparametrization, just like a relativistic bosonic particle $\int d\tau\sqrt{|\dot{x}|^{2}}$.

Second, because it has diffeomorphism invariance, its canonical Hamiltonian vanishes (both on-shell and off-shell). One can check:


and so

$$H=\pi\dot{\theta}-L=\pi\dot{\theta}-\pi\dot{\theta}\equiv 0.$$

Unlike in the bosonic case where one can use the einbein and apply Dirac's method for constrained systems, here there's no primary constraints to start with.

My question: How to eliminate this gauge redundancy of time-reparametrization in canonical quantization?

In the bosonic case, one may try the Polyakov action

$$S=\frac{1}{2}\int d\lambda\sqrt{g}g^{-1}\dot{x}\vphantom{x}^{\mu}\dot{x}\vphantom{x}_{\mu}.$$

The canonical momenta are given by

$$p_{\mu}=\frac{\dot{x}\vphantom{x}_{\mu}}{\sqrt{g}},\quad\quad \pi_{g}=\frac{\partial L}{\partial\dot{g}}=0.$$

So the primary constraint is given by $\phi_{1}=\pi_{g}=0$

The canonical Hamiltonian is


Adding the primary constraint, the primary Hamiltonian is given by


Then, the equation


produces a secondary constraint


One can check that $\phi_{1}$ and $\phi_{2}$ are first class constraints and there're no further constraints. Then, one can add gauge fixing conditions so that all constraints become second class constraints, and the time evolution in phase space would be


where in the last step one replaces the Poisson bracket by the Dirac bracekt.

In the fermionic case,


the only primary constraint one can find from the Lagrangian is


which is second class because the phase space is of Grassmann odd variables. The Poisson bracket in this case takes the form


The other first class constraint $H=0$ which generates time-reparametrization is trivial. There's nothing one can do about the equation $H=0$ following Dirac's algorithm because $H$ doesn't take any specific form.

Just in case:

I am not doing Berezin integrals

$$\int d\theta\theta=1.$$

Before you vote to close a question, make sure you understand what he is asking.

If any person is still confused by the odd differential form $\theta\dot{\theta}dt$, please read this answer by Qmechanics. It's just the odd symplectic one-form. I truly don't know what other details I can add.

If anyone is still confused by what I am asking, please click this link of "A Lagrangian formulation of the classical and quantum dynamics of spinning particles" by L. Brink, P. Di Vecchia, and P. Howe. On page 80, section 3, the author considered a massless spinning particle


This theory has both bosonic and fermionic degrees of freedom. It is invariant under time reparametrization, and is super-gauge invariant.

What I am doing here, is just considering the pure fermionic degrees of freedom. I missed a factor $\sqrt{-1}$ in my Lagrangian which makes it real, but it's not a big issue.

I just want to make it clear:

$$\int dtL$$

has nothing to do with Berezin integral. My question has nothing to do with path-integral quantization of fermions.

  • 1
    $\begingroup$ Does this help: physics.stackexchange.com/q/572231 $\endgroup$
    – hft
    Commented Oct 23, 2023 at 16:31
  • $\begingroup$ @hft Thanks for the new link. It's related, but not helpful. My question was about applying Dirac's methods for constrained systems for this action. $\endgroup$
    – Valac
    Commented Oct 23, 2023 at 16:34
  • 2
    $\begingroup$ Your new edit is still confusing: Why is there suddenly a $\theta$? Do you mean $\psi$? $\endgroup$
    – ACuriousMind
    Commented Oct 23, 2023 at 16:53
  • $\begingroup$ This answer on the massless Dirac field in 0+1 dimensions may help physics.stackexchange.com/a/572258/226902 $\endgroup$
    – Quillo
    Commented Oct 23, 2023 at 23:36

1 Answer 1

  1. OP's action $$S[\psi]~=~\int \!dt~L(t), \qquad L(t)~=~\frac{i}{2}\psi(t)\dot{\psi}(t)\tag{1}$$ of a real Grassmann-odd variable in 0+1D is already on first-order form, so we can use the Faddeev-Jackiw method rather than going through the whole Dirac-Bergmann analysis. Clearly the Hamiltonian $$H(t)~=~0\tag{2}$$ is zero, while the symplectic one-form potential is $$ \vartheta(t) ~=~ \frac{i}{2}\psi(t)~\mathrm{d}\psi(t),\tag{3}$$ leading to canonical equal-time Poisson brackets $$ \{\psi(t),\psi(t)\}~=~-i,\tag{4}$$ cf. my Phys.SE answer here.

  2. The action (1) is worldline (WL) reparametrization invariance, i.e. has a gauge symmetry. However since it only involves a non-dynamical variable (time $t$), there is no need to gauge-fix.

  • $\begingroup$ Thank you very much. $\endgroup$
    – Valac
    Commented Oct 24, 2023 at 10:55

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