# Why does the Hamiltonian of a Lagrangian that consists of only a coupled term become zero?

Let's say our Lagrangian looks something like this:

$$L = \int dz\, Q\cdot \dot{A},\tag{1}$$

where $$Q$$ and $$A$$ are two generalized coordinates and $$\dot{Q}$$ and $$\dot{A}$$ would be the respective time-derivatives. If I wanted to Legendre-transform this, then considering the conjugate momenta $$P_Q = \frac{\partial L}{\partial \dot{Q}} = 0\tag{2}$$ and $$P_A = \frac{\partial L}{\partial \dot{A}} = Q\tag{3}$$ the Hamiltonian becomes:

$$H = P_Q\dot{A}-L= \int dz\,\, Q\cdot \dot{A} - Q\cdot \dot{A} = 0.\tag{4}$$

Is this correct? What does that even mean for the physical system?

• Jul 23 '19 at 2:00

Recall that the purpose of the Legendre transformation from Lagrangian to Hamiltonian formalism is to bring the equations of motion on first-order form. This is where the Faddeev-Jackiw method is so much simpler [than the traditional Dirac-Bergmann analysis which OP just performed]: OP's Lagrangian $$Q\dot{A}$$ is already on first-order form $$p\dot{q}-H$$ if we identify $$q~=~A,\qquad p~=~Q,\qquad H~=~0~!$$ A vanishing Hamiltonian means that all phase space variables are constants of motion. It reflects the world-line (WL) reparametrization invariance of the action, cf. e.g. this & this related Phys.SE posts.