In his Harvard lectures on QFT, Sidney Coleman defines the $S$-matrix as,

$$ S \equiv U_{I}(\infty, -\infty) $$

Where $U_{I}(-\infty, \infty)$ is the time evolution operator in the interaction picture. The $S$-matrix, according to the text, is evaluated between states in the free theory evaluated at $t = 0$ ($0$ being a matter of convention). This seems perfectly reasonable to me given that he defines the $S$-matrix as a connection between the transition probabilities in the interacting theory and the free theory, $$ \langle \phi(0)|S|\psi(0)\rangle \ = \ ^{out}\langle\Phi|\Psi\rangle^{in} \equiv \text{Transition probability in interacting theory}$$ Where the states on the left side are the ones in the free theory and the ones on the right are in the complete (or interacting) theory.

Peskin's text, on the other hand, says,

We can now relate the probability of scattering in a real experiment to an idealized set of transition amplitudes between the asymptotically defined in and out states of definite momentum, $$ _{out}\langle \textbf{p}_{1}\textbf{p}_{2}...|\textbf{k}_{A}\textbf{k}_{B}\rangle_{in} = \lim_{T\rightarrow \infty} \langle \underbrace{\textbf{p}_{1}\textbf{p}_{2}...}_{T}|\underbrace{\textbf{k}_{A}\textbf{k}_{B}}_{-T}\rangle = \lim_{T\rightarrow \infty} \langle \textbf{p}_{1}\textbf{p}_{2}...|e^{-2iHT}|\textbf{k}_{A}\textbf{k}_{B}\rangle \Rightarrow S \equiv e^{-2iHT}$$

I am confused about the kets here. The paragraph quoted above seems to the say that the states on the extreme left are idealized and asymptotically defined - which means they are the free theory states.

My question is this: I think Coleman is saying that we use the asymptotic states (defined in the free theory) with the $S$-matrix to evaluate the probabilities in the interacting theory - we already know the free theory states and we can evaluate the matrix as well, so instead of looking at what the states in the complete theory are, we calculate the overlap by exploiting the connections in the far past and far future through the $S$-matrix. Is my understanding correct?

And, if it is, why is Peskin calculating what seems to be a free theory overlap and evaluating the $S$-matrix between the interacting theory states, as the quoted paragraph seems to imply?

  • 4
    $\begingroup$ As a side note in case you're interested, the first 50 or so pages of John Taylor's Scattering Theory textbook do a really good job of removing some of the ambiguity necessarily introduced in QFT texts have have to do a (usually very brief) crash course on scattering theory. In particular making the definition of the in and out states clear and their connection to the formal definition of the scattering matrix. $\endgroup$
    – Charlie
    Nov 5, 2023 at 16:09
  • $\begingroup$ @Charlie Thanks for the heads up, I will definitely take a look. $\endgroup$
    – ShKol
    Nov 5, 2023 at 16:19

1 Answer 1


Assuming I am properly interpreting it, the summary you gave is correct.

Regarding Peskin's treatment here, he actually does not refer to any free theory states. The states $|\mathbf{k}_A\mathbf{k}_B\rangle_{\rm in}$ and $|\mathbf{p}_1\mathbf{p}_2\cdots\rangle_{\rm out}$ are products of eigenstates of the momentum operator in the interacting theory at asymptotic times (which are the only times at which single-particle momentum eigenstates can be constructed btw; see e.g. Srednicki ch. 5). So they correspond to a particular choice of Coleman's $|\Psi\rangle^{\rm in}$ and $|\Phi\rangle^{\rm out}$ states.

  • $\begingroup$ That makes sense! The word "idealised" kept confusing me, since I thought he was talking about the free theory. He probably means that single particle states can't be constructed at any other time, so that's why it's an idealised situation. Am I right? $\endgroup$
    – ShKol
    Nov 6, 2023 at 4:40
  • 1
    $\begingroup$ @ShKol looking at the context, it seems he is using the word "idealized" to distinguish them from the wavepackets, which are superpositions of the asymptotic one-particle eigenstates. As far as I can tell, no reason is provided for why we have to construct these states in the far past/future, which is an unfortunate shortcoming. But you're right that the use of one-particle asymptotic eigenstates is in itself another level of idealization, since real experiments are of course finite duration and never involve exact eigenstates. $\endgroup$ Nov 6, 2023 at 5:54

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