I'm trying to derive the BZ Lagrangian (density) from the Dirac Hamiltonian density and some questions popped up.
BZ Lagrangian is $$\mathcal{L} = \frac{i}{2}(\dot{\bar{\psi}}\psi - \bar{\psi}\dot{\psi}) + p_\mu(\dot{x}^\mu - \bar{\psi}\gamma^\mu\psi)\tag{1}$$ where the four-potential $A_\mu$ is set to zero for simplicity.
The first question is: how can I derive it? As it can be seen, the first right member is the time derivative inside the Dirac Lagrangian (which corresponds to the conjugate momenta) and if I set $p_\mu\gamma^\mu = m$, then second member in brackets is the second term of Dirac Lagrangian.
I though that it can be derived from the Legendre transform starting with the Dirac Hamiltonian and the conjugate momenta (that I know) and inserting the other conjugate momenta linked to the dynamical variables $(x_\mu, p_\mu)$, i.e. $p_\mu\dot{x}^\mu$ .
The problem is that there's the divergence term of the Dirac Hamiltonian that doesn't appear. Any advice?
The second question is the following: I saw that the Lagrangian density depends on fields and their derivatives: in this case we have $\psi$,$\bar{\psi}$ as fields and their derivative there aren't. Although, in this case we also have a dependence from the dynamical variables $(x_\mu, p_\mu)$. How can I justify this?


1 Answer 1

  1. The Barut-Zanghi (BZ) Lagrangian $$\begin{align} L ~=~& \frac{i}{2}(\dot{\bar{\psi}}\psi - \bar{\psi}\dot{\psi}) + p_{\mu}\dot{x}^{\mu} \cr &-(p_{\mu}-qA_{\mu}) \bar{\psi}\gamma^{\mu}\psi, \cr \bar{\psi} ~=~&\psi^{\dagger}\gamma^0 ,\end{align}\tag{1} $$ is a point-mechanical (rather than a field-theoretic) model, i.e. all the variables depend only on time. In contrast the spinor in the Dirac equation is a field.

  2. The BZ Hamiltonian $$H~=~ (p_{\mu}-qA_{\mu}) \bar{\psi}\gamma^{\mu}\psi\tag{2}$$ is minus the terms without time derivatives in the BZ Lagrangian (1), cf. e.g. the Faddeev-Jackiw method.

  • $\begingroup$ This seems relevant: arxiv.org/abs/hep-th/9607212 $\endgroup$
    – Qmechanic
    Apr 15, 2023 at 13:56
  • $\begingroup$ Thanks, I wrongly thought that was CFT. Anyways, from where can I start deriving this Lagrangian? Because I’m assuming that only the Dirac Hamiltonian is known. $\endgroup$
    – Gyro
    Apr 15, 2023 at 14:24
  • $\begingroup$ Perhaps this is explained/motivated in the original BZ paper? $\endgroup$
    – Qmechanic
    Apr 15, 2023 at 14:36
  • $\begingroup$ Unfortunately it is not explained anywhere $\endgroup$
    – Gyro
    Apr 15, 2023 at 14:39

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