I'm trying to follow chapter 4 about interacting fields in Peskin and Schröder. They define the S-matrix by $$_{\mathrm{out}}\langle p_1 p_2 | k_a k_b\rangle_{\mathrm {in}} = \langle p_1 p_2 | S | k_a k_b\rangle,$$ where $S = \lim_{T\rightarrow \infty}e^{-i2HT}$. The states on the right hand side 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?


2 Answers 2


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.

  • $\begingroup$ 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" $\endgroup$ Commented Dec 29, 2018 at 1:15
  • $\begingroup$ 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. $\endgroup$ Commented Dec 29, 2018 at 1:23
  • $\begingroup$ Also I don't think I can explain it better than this: physics.stackexchange.com/q/41439 $\endgroup$ Commented Dec 29, 2018 at 1:41
  • $\begingroup$ 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? $\endgroup$
    – tparker
    Commented Dec 29, 2018 at 1:58
  • 1
    $\begingroup$ 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. $\endgroup$
    – tparker
    Commented Dec 29, 2018 at 2:40

So I figured that the in and out states are indeed eigenstates of the full Hamiltonian. However, they can still have a non-trivial overlap since the energy-eigenstates of the full Hamiltonian obviously have a high degeneracy (eg with multiple particles you have a lot of possible momenta combinations that have the same total momentum and energy).

More precisely, the in- and out states are scattering states of the full Hamiltonian that fulfill certain boundary conditions such that they "match" free multi-particle states (but with physical mass due to self-interaction) long before and long after the scattering respectively, see Lippmann-Schwinger eq.

A nice way to gain some intuition is to look at potential scattering in non-relativistic quantum mechanics. For example chapter 32.1 Potentialstreuung illustrates nicely how in-states can be interpreted in terms of the scattering of an incoming wavepacket (the text only considers in-states so they're never explicitly called like that). Unfortunately, the text book is in German, but maybe anyone knows a similar text derivation in english..?

  • $\begingroup$ this is exactly what I'm trying to explain to arnold in link , but he won't listen to my arguments even if I've provided all the possible details. Please give a look at the discussion and tell me if I'm wrong or right, the two of us can't solve the issue as neither of us is backing up on their opinion. $\endgroup$
    – dallla
    Commented Feb 19 at 14:23

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