Secondary constraints leads to the value of lagrange multiplier From Lagrangian I got two primary constraint $\phi_i$ and $\phi$. And my Hamiltonian in presence of the constraints becomes- $$H_p=p\dot q-L+\lambda_i\phi_i+\lambda\phi$$ here the $\lambda_i$ and $\lambda$ are Lagrange undetermined multiplier. Now from $\dot \phi_i=[\phi_i,H_p]$ I got secondary constraint $\Sigma_k$ and from $\dot \phi=[\phi,H_p]$ I got another secondary constraint $\Sigma$ . To satisfy the consistency condition I calculated the $\dot \Sigma_k=[\Sigma_k,H_p]$ and $\dot\Sigma=[\Sigma,H_p]$.
$$$$ From the relation I have $\dot \Sigma\approx0$. But the $\dot \Sigma_k$ gives the value of of $\lambda_i$. Now can anyone help me how can I further analyze the constraints in this case? Do I have to put the value of $\lambda_i$ in the equation of $H_p$ and calculate the commutation again? An example would be lovely.
 A: With the validity of the Dirac's conjecture, all first-class constraints are generators of the gauge transformation, all the secondary constraints and the primary constraints should be added to the Hamiltonian $H_E$, called as an extended Hamiltonian. Otherwise, the answer given by Vahid is valid. It is worthy to notice that the calculational process of the time consistency of new constraints has to be continued except with the three results: (1) $0=0$, (2) a linear combination of other known constraints, and (3) the corresponding Lagrange undetermined multiplier is determined.
A: As I know, from the time consistency of $\Sigma_i$ you will not get another constraint. You have to substitute the Lagrange multiplier in the Hamiltonian by what you got. But I can not understand what you want to say by $\dot{\Sigma}=0$. If you mean 0=0, it is the end! But if it gives you an expression you have to continue and consider $\dot{\Sigma}=0$ as a new constraint. Check whether this new expression does not be a linear combination of other constraints. Generally It is not necessary to add secondary constraints to the complete Hamiltonian ($H_p$). 
