How is the conjugate momentum defined in the Schrödinger picture of QFT?

[alphaone]

Hi,
I have been wondering why we can consider d(phi)/dt when we are in Schrödinger picture (phi is just the usual scalar field here). Isn't this 0 as operators do not depend on time in this picture? However then how does it make sense to talk about the conjugate momentum in this picture which is
d(L)/d(d(phi)/dt)?

 

[Demystifier]

To answer it, I would suggest to make it simpler.
First, replace QFT by single-particle QM, the problem is essentially the same.
Second, instead of quantum physics in the Schrodinger picture, study classical physics in the Hamiltonian formulation. The problem is still essentially the same.
Now I hope that you will be able to answer your question by yourself.