# The constraint commute with Hamiltonian in Gauge theory

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. 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.
$$^1$$ Weak equality means equality modulo constraints.
• 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? Apr 14, 2022 at 12:38