The Dirac Lagrangian $$\mathcal{L} = i\bar{\psi}\gamma^{\mu}\partial_\mu \psi - m \bar{\psi}\psi$$ gives a Hamiltonian $$\mathcal{H}(\Pi,\bar{\Pi},\psi,\bar{\psi})=\Pi \dot{\psi}-\mathcal{L}=-\bar{\psi}\gamma^{i}\partial_i \psi + m \bar{\psi}\psi,$$ $$\text{where}\quad \Pi=i\bar{\psi}\gamma^0.$$ So the Hamiltonian is actually independent of $\Pi,\bar{\Pi}$. Suppose I wanted to start from this Hamiltonian and tried to find the equations of motion via Hamilton's equations $\dot{\psi} = \partial\mathcal{H}/\partial \Pi$ and $\dot{\Pi} = -\partial\mathcal{H}/\partial \psi$, plus equivalent ones for conjugate field.

Clearly something is wrong as the first of Hamilton's equations gives $\dot{\psi}=0$, while the second gives $\dot{\Pi}=i \partial_i\bar{\psi}\gamma^i +m\bar{\psi}$ and I don't see the Dirac equation popping out...

  • $\begingroup$ in the transformation from lagrangian to hamiltonian you change variables $(\psi,\dot{\psi})\rightarrow(\psi,\Pi)$. I still see $\dot{\psi}$ in your hamiltonian instead of $\Pi $. Perhaps changing the variables would solve the issue? $\endgroup$
    – Umaxo
    Jun 3, 2019 at 19:30
  • $\begingroup$ I don't think there's a $\dot{\psi}$ in my Hamiltonian. $\partial_i$ is a spatial derivative, which sstays as is in the Hamiltonian formulation. $\endgroup$
    – Rudyard
    Jun 3, 2019 at 19:37
  • 1
    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/43502/2451 and links therein. $\endgroup$
    – Qmechanic
    Jun 3, 2019 at 19:54