# Does one loose constraints with the total Hamiltonian?

Let me recall some of the discussion in M. Henneaux and C. Teitelboim, Quantization of Gauge Systems. Princeton University Press, 1992. for context. Assume we have a constrained Hamiltonian system with an action $$S(q,p,\lambda)=\int dt\left(p_i\dot{q}^i-H(p,q)-\lambda^m\phi_m\right).$$ The equations of motion are then $$\dot{F}=\{F,H\}+\{F,\phi_m\}\lambda^m.$$ The requirement that the constraints are preserved in the evolution of the system then leads us to the consistency conditions $$\{\phi_m,\phi_n\}\lambda^n=-\{\phi_m,H\}.$$ This is a set of linear equations for $$\lambda^m$$. However, it may not have solutions at every point on the constraint surface. Thus, the points at which it thus admit solutions define a new constrain surface obtained by adding new constraints to $$\phi_m$$. Let us denote the old constraints along with the new ones by $$\phi_j$$. We now have new consistency conditions, leading us to $$\{\phi_j,\phi_m\}\lambda^m=-\{\phi_j,H\}.$$ These may again not have solutions everywhere, leading to a new constraint surface and the need to add new constraints to the $$\phi_j$$'s.

Let us assume that the process has terminated and we are left with a set of consistency conditions with solutions everywhere on $$\phi_j=0$$. Then, the solutions for $$\lambda^m$$ are of the form $$\lambda^m=U^m+v^aV^m_a$$, where $$U^m$$ is a particular solution to the consistency conditions, $$V^m_a$$ is a basis of the kernel of $$\{\phi_j,\phi_m\}$$ and $$v^a$$ are completely undetermined. This allows us to define a first-class Hamiltonian $$H'=H+U^m\phi_m$$.

Here is where I am confused. On the initial constraint surface $$H$$ and $$H'$$ coincide. Since form the Lagrangian formulation one can only determine the Hamiltonian on this initial constraint surface, we could've decided to start with $$H'$$ instead of $$H$$, i.e. with the total action $$S_T(q,p,\lambda)=\int dt\left(p_i\dot{q}^i-H'-\lambda^m\phi_m\right).$$ However, from this action, the initial consistency conditions would've been $$\{\phi_m,\phi_n\}\lambda^n=0$$ since $$\{\phi_m,H'\}=0$$. These are clearly consistent, for they only require that $$\lambda^n$$ be in the kernel of the square matrix $$\{\phi_m,\phi_n\}$$. Then, how would've we recovered the rest of the constraints $$\phi_j$$?

If we didn't enforce the secondary, tertiary, etc, constraints, then OP's first-class Hamiltonian $$H^{\prime}$$ would in general correspond to a physically different theory than the original Hamiltonian $$H$$.
• Yeah! I just further realized that this is shown in formulas in the fact that $H'$ is only a first class Hamiltonian with respect to the final constraint surface, not the initial one. Thus, with respect to the total action the consistency conditions still read $$\{\phi_m,\phi_m\}\lambda^n=-\{\phi_m,H'\}$$ where the right hand side doesn't vanish because we are in the full initial constraint surface. Jun 28, 2021 at 11:01
• Then the right hand side actually is equivalent in the initial constraint surface to $-\{\phi_m,H\}$ due to the vanishing $\phi_m=0$ and the definition of $U^m$. Jun 28, 2021 at 11:07