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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?

Charlie
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user2224350
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2 Answers2

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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.

InertialObserver
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  • 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 01:15
  • 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 01:23
  • Also I don't think I can explain it better than this: https://physics.stackexchange.com/q/41439/ – InertialObserver Dec 29 '18 at 01:41
  • 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 01:58
  • All very good questions. I will my best to explain, but perhaps we should move this to chat. I think I can currently address your orthogonality concern though. While the Hamiltonian is hermitian, I believe that in a given Hilbert space they are orthonormal. But since $\mathcal{F} = \mathcal{H} \oplus \mathcal{H}^{\otimes 2} \oplus \cdots$ and hence includes many single particle hilbert spaces, the orthonormality conditions are different. – InertialObserver Dec 29 '18 at 02:06
  • And yes AN says that they are eigenstates of the free hamiltonian, but he doesn't stop there. He says that they are eigenstates of the free hamiltonian associated with the bound states of the theory.. now what exactly that means I'm unsure of.. – InertialObserver Dec 29 '18 at 02:11
  • I am not persuaded by your claim of a qualitative distinction between $\mathcal{F}$ and the individual $\mathcal{H}^{\otimes n}$'s. $\mathcal{F}$ is still a perfectly good Hilbert space, and I've never heard that "the orthonormality conditions are different" for it. – tparker Dec 29 '18 at 02:32
  • That's fair. I'm not entirely persuaded either. But please see my edit and link. It seems to be that the product states of the asymptotic states are not eigenstates of the full hamiltonian, and so we shouldn't expect them to be orthonormal. – InertialObserver Dec 29 '18 at 02:34
  • As for the "bound states" qualifier, I don't think that's too important - I think he just means that for an asymptotically free theory like QCD where interaction strengths between color-charged particles increase with distance, far-separated particles don't become noninteracting. So the whole scattering formalism doesn't apply, and you can only consider scattering between color-neutral bound states, i.e. hadrons, which do decouple at long distances. – tparker Dec 29 '18 at 02:35
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    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 02:40
  • @tparker But how do you define product states here? – user2224350 Dec 29 '18 at 16:40
  • @user2224350 It's just $|k_a , k_b \rangle := a^\dagger(k_a), a^\dagger(k_b) |0\rangle$. – tparker Jan 01 '19 at 14:28
  • @tparker i strongly believe that this answer by arnold is incorrect, see my comments and my answer at link. He is using the theory of scattering for non relativistic qft, while haag-ruelle scattering is done in a different way. We need external opinions since neither of us is backing up. Bjorkenn Drell says explicitly the asymptotic states are 4-momentum eigenstates in formula 16.11, ch. 16, and it's not wrong simply because it's old. – dallla Feb 19 '24 at 14:36
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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..?

user2224350
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  • 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. – dallla Feb 19 '24 at 14:23