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.

 

[denian]

In the Schrödinger picture of quantum field theory, the fields are treated as operators that do not depend on time, while the states of the system evolve in time. This means that the time evolution of the fields is given by the Heisenberg equation of motion, which is d(phi)/dt = [H, phi], where H is the Hamiltonian operator.

Now, when we talk about the conjugate momentum in this picture, we are actually referring to the momentum operator, which is defined as the derivative of the Lagrangian with respect to the time derivative of the field, i.e. d(L)/d(d(phi)/dt). This is different from the classical conjugate momentum, which is defined as the derivative of the Hamiltonian with respect to the canonical coordinate.

In the Schrödinger picture, the Lagrangian does not depend on time, so the momentum operator is well-defined. It represents the momentum of the field at a particular time, and it is an important quantity in quantum field theory as it is used to calculate quantities such as the energy density and the Hamiltonian.

So, while it may seem counterintuitive that we can consider d(phi)/dt in the Schrödinger picture, it is important to remember that we are talking about the time derivative of the field operator, not the field itself. And this quantity is well-defined and plays a crucial role in the theory.

I hope this helps to clarify any confusion about the Schrödinger picture in QFT.