# Time Evolution of Asymptotic Free States in QFT

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?

• Yes, the Hamiltonian of the free theory is time independent. – user178876 May 19 '18 at 3:15
• So the full hamiltonian is just the two free fields with some non-time dependent interaction term? – InertialObserver May 19 '18 at 3:19
• There are no interaction terms in the free Hamiltonian. – user178876 May 19 '18 at 3:25
• Then how can it be that there is anything nontrivial happening at $T\rightarrow\infty$, it's its just two non-interacting particles? – InertialObserver May 19 '18 at 5:11
• There is nothing nontrivial happening, that's the concept of asymptotically free states. – user178876 May 19 '18 at 14:14

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.
• i.e. when we write things like $H_I(t)$.. – InertialObserver May 20 '18 at 4:19
• 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 – tsufli May 20 '18 at 8:11