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?

  • $\begingroup$ Is $H$ here the Hamiltonian for the "full interacting theory" or for some other (presumably non-interacting) theory? $\endgroup$ Oct 6, 2021 at 14:27
  • $\begingroup$ @BySymmetry I am 99.9% sure he's using the full Hamiltonian here, because he's using it in the construction of a path integral $\endgroup$ Oct 6, 2021 at 14:28

1 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.


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