When canonical quantizing gauge theory, we find that the canonical momentum corresponding to $A_0$ vanish since the Lagrangian contains no $\dot{A_0}$ . Thus we need to choose a gauge, for example, $A_0=0$. However, this will impose a constraint. The equation of motion reads \begin{align} D_\mu F^{\mu\nu}=0 \end{align} here the $D_\mu$ take adjoint representation on $F$, which is just $\partial_\mu$ in Abelian case. If we set $A_0=0$, the zero component of this equation of motion is now not an equation for dynamic variables: generally, a EOM from Lagrangian could be get by using $\dot{q}=i[H,q],\dot{p}=i[H,p]$ when $p,q$ are canonical variables, but here we've set $A_0=0$ and the canonical momentum vanish. Thus this will be a constraint on the physical states (since its also easy to find that this operator itself is not zero) \begin{align} D_i F^{i0}|\psi\rangle=0 \end{align} However, it's also said that we will have $[D_iF^{i0},H]=0$ so it is enough to restrict the initial state in the physical space. I know how to check this here by directly calculate commutators, and it quite make sense too. However I wonder is it a general principle that any gauge fixing procedure by gauge redundancy will give constraints and these constraints will commute with Hamiltonian, and how we prove this commutation relation in a general way.


1 Answer 1

  1. In general, as part of the Dirac-Bergmann analysis, one introduces secondary constraints to ensure that the Hamiltonian commutes weakly$^1$ with the primary constraints, and so forth. As a result, the Hamiltonian commutes weakly with all constraints.

  2. In contrast, the Hamiltonian does not need to commute with the gauge-fixing conditions.


$^1$ Weak equality means equality modulo constraints.

  • $\begingroup$ Thanks!But I do not quite understand what you said. What does "weakly" mean?I know here the primary constraints is the canonical momentum is zero.So the $D_i F^{i0}|\psi\rangle=0$ is the secondary constraints and do not come from gauge fixing? $\endgroup$ Commented Apr 14, 2022 at 12:38
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Commented Apr 14, 2022 at 17:41

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