Let us consider Dirac equation
$$(i\gamma^\mu\partial_\mu -m)\psi ~=~0$$
as a classical field equation. Is it possible to introduce Poisson bracket on the space of spinors $\psi$ in such a way that Dirac equation becomes Hamiltonian equation
$$\dot{\psi}~=~\{ \psi,H\}_{PB}~?$$
Of course, such Poisson bracket would be graded (super Poisson bracket), but if it exists this would explain on classical level why $\frac{1}{2}$-spinors correspond to fermions.
Best Answer
Given the Dirac Lagrangian density
$$ \tag{1} {\cal L}~=~\overline{\psi}(i\sum_{\mu=0}^3\gamma^{\mu}\partial_{\mu}-m)\psi, \qquad \overline{\psi}~:=~\psi^{\dagger}\gamma^0, \qquad \{\gamma^{\mu},\gamma^{\nu}\}_{+} ~=~2\eta^{\mu\nu}{\bf 1}_{4\times 4}, $$
with Minkowski signature $(+,-,-,-)$, and $\psi$ is a Grassmann-odd Dirac-spinor, the question is How to find the corresponding Hamiltonian formalism?
The Legendre transformation of (1) is singular. The Dirac-Bergmann analysis of the theory (1) leads to constraints, cf. e.g. Ref. 1 or this Phys.SE post. Here we will instead take a shortcut using the Faddeev-Jackiw method.
I) Complex Grassmann-fields. We first identify the Hamiltonian density ${\cal H}$ as (minus) the terms in (1) that don't involve time derivatives:
$$ \tag{2} {\cal L}~=~i\psi^{\dagger}\dot{\psi}-{\cal H}, \qquad {\cal H}~=~ \overline{\psi} (-i\sum_{j=1}^3\gamma^{j}\partial_{j}+m)\psi. $$
The symplectic one-form potential can be transcribed from the kinetic term in (2):
$$ \tag{3} \vartheta(t) ~=~\int\! d^3x~ i\psi^{\dagger}({\bf x},t) ~\mathrm{d}\psi({\bf x},t), $$
where $\mathrm{d}$ denotes the exterior derivative$^1$ on the infinite-dimensional configuration space for the fermion field. The symplectic two-form is then
$$ \omega(t)~=~\mathrm{d}\vartheta(t) ~=~\int\! d^3x~ i\mathrm{d}\psi^{\dagger}({\bf x},t) \wedge \mathrm{d}\psi({\bf x},t) $$ $$ \tag{4} ~=~\int\! d^3x~d^3y~ i\mathrm{d}\psi^{\dagger}({\bf x},t) \wedge \delta^3({\bf x}-{\bf y}) ~\mathrm{d}\psi({\bf y},t). $$
The equal-time super-Poisson/Dirac bracket on fundamental fields is the inverse supermatrix of the supermatrix for the symplectic two-form (4):
$$ \tag{5} \{\psi_{\alpha}({\bf x},t), \psi^{\dagger}_{\beta}({\bf y},t)\}_{PB}~=~ -i \delta_{\alpha\beta}~\delta^3({\bf x}-{\bf y})~=~\{\psi^{\dagger}_{\alpha}({\bf x},t), \psi_{\beta}({\bf y},t)\}_{PB}, $$
and other fundamental super-Poisson brackets vanish. Due to the QM correspondence principle, the canonical anticommutation relations (CARs) are the super-Poisson brackets (5) multiplied with $i\hbar$:
$$ \tag{6} \{\hat{\psi}_{\alpha}({\bf x},t), \hat{\psi}^{\dagger}_{\beta}({\bf y},t)\}_{+} ~=~ \hbar\delta_{\alpha\beta}~\delta^3({\bf x}-{\bf y})\hat{\bf 1} ~=~\{\hat{\psi}^{\dagger}_{\alpha}({\bf x},t), \hat{\psi}_{\beta}({\bf y},t)\}_{+}, $$ and other CARs vanish.
I) Real Grassmann-fields. Alternatively, let us decompose the complex Dirac spinor
$$\tag{7}\psi_{\alpha}~\equiv~(\psi^1_{\alpha}+i\psi^2_{\alpha})/\sqrt{2} \quad\text{and}\quad \psi^{\dagger}_{\alpha}~\equiv~(\psi^1_{\alpha}-i\psi^2_{\alpha})/\sqrt{2}, $$
in real and imaginary parts. The Lagrangian density (2) reads up to total derivative terms$^2$
$$ \tag{2'} {\cal L}~=~\frac{i}{2}\left(\psi^{\dagger}\dot{\psi}- \dot{\psi}^{\dagger}\psi\right)-{\cal H} ~=~\frac{i}{2}\sum_{a=1}^2(\psi^a)^T\dot{\psi}^a-{\cal H}.$$
The corresponding symplectic one-form potential is
$$ \tag{3'} \vartheta(t) ~=~\sum_{a=1}^2\int\! d^3x~ \frac{i}{2}\psi^a({\bf x},t)^T ~\mathrm{d}\psi^a({\bf x},t). $$
The symplectic two-form is
$$ \omega(t)~=~\mathrm{d}\vartheta(t) ~=~\sum_{a=1}^2\int\! d^3x~ \frac{i}{2}\mathrm{d}\psi^a({\bf x},t)^T \wedge \mathrm{d}\psi^a({\bf x},t) $$ $$ \tag{4'} ~=~\sum_{a,b=1}^2\int\! d^3x~d^3y~ \frac{i}{2}\mathrm{d}\psi^a({\bf x},t)^T \wedge \delta_{ab}~\delta^3({\bf x}-{\bf y}) ~\mathrm{d}\psi^b({\bf y},t). $$
The equal-time super-Poisson is
$$ \tag{5'} \{\psi^a_{\alpha}({\bf x},t), \psi^b_{\beta}({\bf y},t)\}_{PB}~=~ -i \delta^{ab}~\delta_{\alpha\beta}~\delta^3({\bf x}-{\bf y}). $$
The CARs are
$$ \tag{6'} \{\hat{\psi}^a_{\alpha}({\bf x},t), \hat{\psi}^b_{\beta}({\bf y},t)\}_{+} ~=~ \hbar\delta^{ab}~\delta_{\alpha\beta}~\delta^3({\bf x}-{\bf y})\hat{\bf 1} . $$
References:
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$^1$ In our super-conventions, the exterior derivative $\mathrm{d}$ is Grassmann-even and carries form-degree +1.
$^2$ Note that adding a total time derivative
$$i\psi^{\dagger}\dot{\psi}~\longrightarrow~i\psi^{\dagger}\dot{\psi}+ \frac{d}{dt}(\alpha\psi^{\dagger}\psi)\tag{8} $$
to the kinetic term (2) corresponds to adding an exact term
$$ \vartheta(t)~\longrightarrow~\vartheta(t)+ \mathrm{d} \int\! d^3x~ \alpha\psi^{\dagger}({\bf x},t) \psi({\bf x},t) \tag{9} $$
to the symplectic one-form potential (3), which has no effect on the symplectic 2-form (4).