# QFT Why do in and out states have a non-trivial overlap?

Im trying to follow chapter 4 about interacting fields in Peskin and Schröder. They define the S matrix by $$_{out}_{in} = $$, where $$S = \lim_{T\rightarrow \infty}e^{-i2HT}$$. The states on the right hand site of the equation are eigenstate of the momentum operator. Furthermore, they are said to be eigenstates of H as well (in 4.6 below eq. 4.87).

But if the states are eigenstates of H, the above scalar product becomes very trivial right? So what's going on here?

This answer is essentially a citation of this answer, given by Arnold Neumaier.

The resolution is essentially that while the asymptotic single particle states in the full hamiltonian $$|p \rangle$$ (which is what is what I'm guessing is what Peskin meant) are eigenstates of the full hamiltonian, the product states of those asymptotic states (the only ones that have non-trivial scattering) $$|p_1, p_2 \rangle$$ are not eigenstates of the full hamiltonian $$H$$, and so we would expect

$$_{in}\langle p_1, p_2 \cdots | k_A, k_B \rangle_{out} = \lim_{t\to \infty} \langle p_1, p_2 \cdots| e^{-iH2t} | k_A, k_B \rangle$$

to have a nontrivial overlap.

• This is what is written in the book, below equation 4.87: "To compute this quantity we would like to replace the external plane-wave states in (4.87) (the eq. in my question), which are eigenstates of H, with their counterparts in the unperturbed theory, which are eigenstates of H0" – user2224350 Dec 29 '18 at 1:15
• It's because the eigenstates of the full hamiltonian don't have a trivial overlap. If they did then we wouldn't have to go through all this mess. So perhaps what I said in my answer is somewhat misleading. Now that I understand your question I'll clarify. – InertialObserver Dec 29 '18 at 1:23
• Also I don't think I can explain it better than this: physics.stackexchange.com/q/41439 – InertialObserver Dec 29 '18 at 1:41
• I'm confused - P&S and Arnold Neumaier seem to be saying two different things. AN seems to be saying that the in/out states are eigenstates of the free Hamiltonian, while P&S say they're eigenstates of the interacting Hamiltonian. Which are you claiming is correct? And anyway, both Hamiltonians (free and interacting) are Hermitian, so shouldn't their eigenstates be orthonormal? – tparker Dec 29 '18 at 1:58
• Ah, I think you have it with your edit. When P&S says "the external plane-wave states", they mean the single-plane-wave states, but not the "tensor product of multiple plane wave" states. – tparker Dec 29 '18 at 2:40