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?

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.