Is Hamiltonian being a time evolution operator a consequence of minimizing action?

Question

When I write down the Lagrangian for a quantum field, I can derive the equation of motion for the field. Therefore, Lagrangian specifies how field evolves with time completely. Can I derive Schrödinger's equation $$\phi(x, t) = e^{iHt}\phi(x, 0)e^{-iHt}\tag{1}$$ from Euler-Lagrange equation $$\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)}\right) - \frac{\partial\mathcal{L}}{\partial\phi}~?\tag{2}$$

I got this question while reading Peskin's "An introduction to quantum field theory". It first re-derives all the classical Lagrangian and Hamiltonian field mechanics, and then quantizes the classical fields with commutators, using quantum harmonic oscillators as inspiration. So far so good, but right after that it switches to Heisenberg picture where the equation (1) appears. Sure, I've seen that equation in non-relativistic QM courses, but that comes from Schordinger's equation without using any Lagrangian mechanics. This got me thinking if eq. (1) is a consequence of minimizing action, or is Schrödinger's equation something new that we impose on the system on top of Lagrangian mechanics?

Answer 1

1. At the classical level, a field theory with a Hamiltonian density ${\cal H}(\phi(x),\pi(x),x)$ does satisfy a stationary action principle for the Hamiltonian action $$ S_H[\phi,\pi]~=~\int\! d^4x~\left(\pi(x) \dot{\phi}(x)-{\cal H}(\phi(x),\pi(x),x)\right). $$ The corresponding Euler-Lagrange (EL) equations imply Hamilton's equations. The Hamiltonian functional $$H[\phi(\cdot,t),\pi(\cdot,t),t]~=~\int\! d^3{\bf x}~{\cal H}(\phi({\bf x},t),\pi({\bf x},t),{\bf x},t)$$ becomes the generator of time-evolution $$\frac{dF}{dt}~=~\{F,H\} + \frac{\partial F}{\partial t}.$$

2. Upon quantization (which is a non-trivial procedure) functions are replaced by operators and Poisson brackets are replaced by commutators, cf. e.g. this Phys.SE post.

   Similarly, the Hamilton's equations are replaced wirh the Heisenberg EOM, or equivalently, the TDSE, cf. OP's question.