Time dependency in the Hamiltonian in Schwartz's book on Quantum Field Theory

Question

On page 258 of his book on Quantum Field Theory and the Standard Model (in the comment between formulas 14.25 and 14.26) , Schwartz writes:

"We will also write $\hat{H} (t) = \int d^3 x \hat{\mathscr{H}}$, with the $t$ dependence of $\hat{H}(t)$ coming from how the field operators change with time in the full interacting theory".

My problem is that while I do understand that the full interacting operators have a very complex dependency over time, I don't think the Hamiltonian should. After all even in the most complex theories, if time doesn't appear explicitly, $\hat{H}$ is a constant of motion, which means that it should remain the same at all times. I don't see why there should be a time dependency. Maybe Schwartz was just trying to stress how the operators that make up the Hamiltonian are time dependent?

Answer 1

I think you are correct. The Hamiltonian in the Heisenberg picture is (essentially by definition) time independent, since it obviously commutes with itself. I think that wether $\hat H$ is or is not time-independent seems to be irrelevant for the derivation of the path integral. Look at (14.27), where he uses $$ \langle \Pi_j | \hat{\mathcal H}(\vec x,t(?)) |\Phi_j \rangle = \langle \Pi_j | \left ( \frac{1}{2} \Pi_j(\vec x)^2 + \mathcal V(\Phi_j(\vec x)) \right) |\Phi_j \rangle $$ The RHS does clearly not depend on time, since $\Phi_j$ and $\Pi_j$ are time-independent classical fields.
In the path integral formalism one reinterprets $\Phi_j(\vec x) \equiv \Phi(t_j,\vec x)$ (and similarly for $\Pi$), where now $\Phi(t,\vec x)$ is the time dependent classical field. Similarly one can see $$ \mathcal H(\vec x, t) \equiv \frac{1}{2} \Pi(\vec x, t)^2 + \mathcal V(\Phi(\vec x, t)) $$ as time-dependent. Note that the $\Phi_j$ and $\Pi_j$ can be anything, they are integrated over in the functional integral. In particular their "time-evolution" must not be the classical time-evolution given by the Poisson bracket with the Hamiltonian, so $\mathcal H(\vec x, t)$ as defined above is somehow truly time-dependent. This is probably what Schwartz had in mind.

I may be wrong on this and I am happy to be corrected.