# Time dependency of interacting Hamiltonian in QFT

Suppose I have a Lagrangian of the form

$$\mathscr{L} = \frac{1}{2} \left( \partial_{\mu} \phi \partial^{\mu} \phi - m^2\right) - \frac{\lambda}{4!} \phi^4,$$

which is just an interacting real scalar field. The Hamiltonian of this field should be

$$H = \frac{1}{2} \int d^3\mathbf{x} \; \left( \dot{\phi}^2 + \nabla \phi \cdot \nabla \phi +m^2 \phi^2 +\frac{\lambda}{4!} \phi^4 \right) .$$ My lecturer said that in Heisenberg picture this Hamiltonian is time dependent and this makes it so the final vacuum and the initial vacuum are different. I understand how this Hamiltonian is time dependent in interaction picture but not how it can be time dependent in Heisenberg picture. After all operators in Heisenberg picture evolve as $$O_H (t) = e^{iH(t-t_0)} O_H(t_0) e^{-iH(t-t_0)}$$, and the Hamiltonian commutes with itself, so $$H_H(t) = H_H(t_0)$$. What am I missing?

• I think your teacher is mistaken. May 22 '21 at 17:09
• @mikestone it's entirely possible (and even more likely) that I might be mistaken and took my notes incorrectly even though I'm usually very careful May 22 '21 at 17:10
• Yep, in Heisenberg picture the Hamiltonian is time independent
– OON
May 22 '21 at 17:35
• Though if your teacher said this not about interacting but about interaction Hamiltonian he/she is correct
– OON
May 22 '21 at 17:37
• @OON yeah, that must be it, I misheard what he said. He must have been talking about the potential part of the Hamiltonian $\lambda/4! \phi^4$ and not the full Hamiltonian $H_0+V$. Thanks to you both May 22 '21 at 17:39