For the free scalar theory we can derive the Hamiltonian in terms of the creation and annihilation operators as: $$H_{S/H}= \int \frac{d^3 p}{(2\pi)^3} \omega_{\vec p}( a_{\vec p}^\dagger a_{\vec p}+\frac{1}{2} [ a_{\vec p} , a_{\vec p}^\dagger])$$ I know this to hold in both the Schrödinger and Heisenberg pictures but the derivation is much simpler in the former case. Is it valid to say that since in the Schrödinger picture $H_S$ is time independent then $H_S=H_H$? If not what is a valid way of determining that $H_S=H_H$ without having to explicitly derive $H_H$ in this case.

In fact what does it even mean to be time-independent in QFT where $t$ is treated on a level playing field with $x$ is. e.g. a Hamiltonian of the form $H=\phi(\vec x,t)$ explicitly time independent?

  • $\begingroup$ I don't understand the question. By definition, $H_H(t) = U(t) H_S U(t)^\dagger$, but since $U$ is simply an exponential of $H_S$ - as $H_S$ is not explicitly time-dependent - and therefore commutes with it, this is just $H_H(t) = H_S$ for all $t$. What does it matter whether we are doing QFT or not? Also, given that you are doing a Hamiltonian approach, $t$ is very obviously not treated the same as $x$ in this part of QFT. That the results of the Hamiltonian approach to QFT are covariant in the end is a non-trivial result. $\endgroup$
    – ACuriousMind
    Dec 31 '17 at 13:20
  • $\begingroup$ @ACuriousMind This is what I thought, but then I read this answer physics.stackexchange.com/a/333421/70392 to another question which seems to say otherwise. I agree with your last comment about $t$, but my confusion of weather $H=\phi(\vec x, t)$ would be counted as time dependent or independent still holds. $\endgroup$ Dec 31 '17 at 13:25

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