Lagrangian to Hamiltonian in Quantum Field Theory

1. 
   
   

   While deriving Hamiltonian from Lagrangian density, we use the formula

   $$\mathcal{H} ~=~ \pi \dot{\phi} - \mathcal{L}.$$ But since we are considering space and time as parameters, why the formula $$\mathcal{H} ~=~ \pi_{\mu}\partial^{\mu} \phi - \mathcal{L}$$ is not used?
   
   


2. 
   

   Is there any particular book/lecture notes dealing with these kind of issues in theoretical physics, I would love to know them?

   
   

Answer

Vladimir's answer has the right essence but it is also misleading, so let me clarify.

The formula

$$H = \sum_i p_i\dot q_i - L$$ relating the Hamiltonian and the Lagrangian is completely general. It holds in all theories that admit both Lagrangians and Hamiltonians, whether they're relativistic or not, whether or not they have any other symmetry aside from the Lorentz symmetry.

When you have field theory, both the Hamiltonian and the Lagrangian may be written as spatial integrals of their densities.

$$H = \int d^3x \, {\mathcal H}, \quad L = \int d^3x\, {\mathcal L}$$ Combining that with the first formula, we get the relationship $$\mathcal{H} = \sum_i\pi_i \dot{\phi_i} - \mathcal{L}$$ Now, you proposed a different formula and I guess that the reason why you proposed it is that it looks more Lorentz-invariant to you, as appropriate for Lorentz-invariant field theories. That's a nice motivation.

However, what's wrong about your reasoning is the assumption that both the Hamiltonian density and the Lagrangian density are Lorentz-invariant. While the Lagrangian density is a nice scalar, so it is Lorentz-invariant (the density at the origin, at least), and it's because the integral of it is the Lorentz-invariant action which should be stationary, the same is not true for the Hamiltonian and its density.

The Hamiltonian is intrinsically linked to the time direction: it is the generator of the translations in time (the spatial counterparts of the Hamiltonian are the spatial components of the momentum); it is the energy, the 0th component of a 4-vector, $H\equiv p^0$. So the argument that this formula should be Lorentz-covariant is invalid, your proposed formula is wrong, and the right formula was justified at the beginning of my comment.