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In equation (4.70) of Peskin, he states that

$$_{out}\langle \mathbf{p_1, p_2, \cdots} | \mathbf{k_A,k_B}\rangle_{in} = \lim_{T\rightarrow \infty}\langle \mathbf{p_1, p_2, \cdots} | e^{-iH(2T)} |\mathbf{k_A,k_B}\rangle \tag{4.70}$$

where $H$ is the hamiltonian of the full interacting theory. This seems to imply that the hamiltonian of the full interacting theory is time independent. Why would we assume this? Shouldn't this be a time-ordered exponential?

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  • $\begingroup$ Yes, the Hamiltonian of the free theory is time independent. $\endgroup$ – user178876 May 19 '18 at 3:15
  • $\begingroup$ So the full hamiltonian is just the two free fields with some non-time dependent interaction term? $\endgroup$ – InertialObserver May 19 '18 at 3:19
  • $\begingroup$ There are no interaction terms in the free Hamiltonian. $\endgroup$ – user178876 May 19 '18 at 3:25
  • $\begingroup$ Then how can it be that there is anything nontrivial happening at $T\rightarrow\infty$, it's its just two non-interacting particles? $\endgroup$ – InertialObserver May 19 '18 at 5:11
  • $\begingroup$ There is nothing nontrivial happening, that's the concept of asymptotically free states. $\endgroup$ – user178876 May 19 '18 at 14:14
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Usually in QFT, we assume Poincare invaraiance - Invariance to both boosts and translations, In particular time translations. The full interacting Hamiltonian is assumed to be time independent - we do not expect strength of interactions to vary in time. A typical interacting Hamiltonian is $H(\phi)=\int d^{3}x\frac{1}{2}(\partial_t\phi)^{2} +\frac{1}{2}(\nabla\phi)^{2} +V(\phi) $, and the states of the theory (superpositions of field configurations at a certain time) evolve according to $e^{iHt}$ as there is no explicit time dependence in H.

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  • $\begingroup$ It looks like equation 4.89 of Peskin seems to imply that the Hamiltonian does depend on time? $\endgroup$ – InertialObserver May 20 '18 at 3:59
  • $\begingroup$ i.e. when we write things like $H_I(t)$.. $\endgroup$ – InertialObserver May 20 '18 at 4:19
  • $\begingroup$ They are switching from Schroedinger picture where it is ok to write $e^{iHt}$ to the Heisenberg picture where H is evolving. See the discussion on section 4.2, in particular eq. 4.19 $\endgroup$ – tsufli May 20 '18 at 8:11

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