Let's define the interaction Hamiltonian as

$$\hat{H}(t) = \hat{H}_{\text{S}}+\hat{V}_{\text{S}}(t)\tag{1}$$

Where $\hat{V}_{\text{S}}\in \mathcal{L}(\mathcal{H})$ represents time-dependent perturbation to Schrödinger picture Hamiltonian $\hat{H}_{\text{S}}\in \mathcal{L}(\mathcal{H})$. Furthermore, $|\psi(t)\rangle_{\text{I}} = {\text{e}}^{\frac{i}{\hbar}\hat{H}_{\text{S}}t}|\psi(t)\rangle_{\text{S}}$ and $\hat{V}_{\text{I}}(t) = \mathrm {e}^{\frac{i}{\hbar}\hat{H}_{\text{S}}t}\hat{V}_{\text{S}}(t)\mathrm{e}^{-\frac{i}{\hbar}\hat{H}_{{\text{S}}}t}.$ From which we conclude that

$$\begin{align}i\hbar\frac{\partial |\psi(t)\rangle_{\text{S}}}{\partial t} = \hat{H}(t)|\psi(t)\rangle_{\text{S}}&\iff i\hbar \frac{\partial(e^{-\frac{i}{\hbar}\hat{H}_{\text{S}}t}|\psi(t)\rangle_{\text{I}})}{\partial t} = \hat{H}_{\text{S}}|\psi(t)\rangle_{\text{S}} + \hat{V}_{\text{S}}(t)|\psi(t)\rangle_{\text{S}} \\&\iff \hat{H}_{\text{S}}|\psi(t)\rangle_{\text{S}} + e^{-\frac{i}{\hbar}\hat{H}_{\text{S}}t}\frac{\partial |\psi(t)\rangle_{I}}{\partial t} = \hat{H}_{\text{S}}|\psi(t)\rangle_{\text{S}}+\hat{V}_{\text{S}}(t)|\psi(t)\rangle_{\text{S}}\\ &\iff i\hbar e^{-\frac{i}{\hbar}\hat{H}_{\text{S}}t}\frac{\partial |\psi(t)\rangle_{I}}{\partial t} = \hat{V}_S(t)|\psi(t)\rangle_{\text{S}} \\&\iff i\hbar\frac{\partial|\psi(t)\rangle_{I}}{\partial t} = e^{\frac{i}{\hbar}\hat{H}_{\text{S}}t}\hat{V}_{\text{S}}(t)e^{-\frac{i}{\hbar}\hat{H}_{\text{S}}t}|\psi(t)\rangle_{I}\\&\iff i\hbar\frac{\partial |\psi(t)\rangle_{I}}{\partial t} = \hat{V}_{\text{I}}(t)|\psi(t)\rangle_{\text{I}}.\end{align}\tag{2}$$

Dyson operator is then defined as $|\psi(t)\rangle_{I} = U_{I}(t, t_0)|\psi(t_0)\rangle_{I}$, where

$$\begin{align}U_{I}(t, t_0) &= 1-\frac{i}{\hbar}\int^{t}_{t_0}\hat{V}_{I}(t_1)dt_1+\biggr(\frac{i}{\hbar}\biggr)^2\int^{t}_{t_0}\int^{t_1}_{t_0}\hat{V}_{I}(t_1)\hat{V}_{I}(t_2)dt_{1}dt_{2}+\cdots \\ &= \mathcal{T}\biggr\{\exp\biggr(-\frac{i}{\hbar}\int^{t}_{t_0}\hat{V}_{I}(\tau)d\tau\biggr)\biggr\}.\end{align}\tag{3}$$

It seems that $S$-matrix of scattering process is $$\hat{S} = \lim_{t\rightarrow \infty, t_0\rightarrow -\infty}U_{I}(t, t_0)\tag{4}$$ However, what is the physical significance of taking the limit to very distant past and future?

  • $\begingroup$ Why do you include the whole discussion about the interaction picture? As far as I can see, you can remove everything before "Dyson operator [...]" and perhaps just give a link to the Wikipedia article. $\endgroup$ Oct 13, 2023 at 12:25
  • $\begingroup$ Hint: You consider a certain type of states as "in" and "out" states. For example, far in the past, the particles are supposed to be non-interacting, and thus evolve only under the free Hamiltonian... $\endgroup$ Oct 13, 2023 at 12:27

1 Answer 1


The basic idea behind the scattering matrix is that we have "free" states, which are easy to describe (or at least possible to write down). For example, we could accelerate two electrons towards each other. When the electrons are far apart, they are essentially propagating in free space: they are not interacting and so we can use the freely propagating solutions. When two electrons come near each other, they feel each others presence through the electromagnetic field and this is hard to describe mathematically.

So a scattering experiment is described as follows. We have some in-states in the far distant past, we let the particles collide with each other, and finally we look at the result in terms of a number of out-states in the distant future. The in- and out-states are solutions to the free Hamiltonian. They are often taken as planewaves, since we can decompose any state in terms of plane waves. Also, in particle accelerator experiments plane waves are often a good approximation.

To reiterate; we describe our problem in known free-propagating states and, because time evolution is linear in quantum mechanics (or QFT) these states should be related by a single linear operator: the scattering matrix.

Disclaimer: it has been a while since I did any QFT

  • $\begingroup$ From your answer, I understood that asymptotic limits are taken to assure that initial and final states evolve under free Hamiltonian, i.e they do not undergo interactions. Is that accurate? $\endgroup$
    – user375448
    Oct 13, 2023 at 15:55
  • $\begingroup$ I would say so yes $\endgroup$ Oct 13, 2023 at 20:44
  • 1
    $\begingroup$ I would put it the other way around : the need to compute transition amplitudes over long period of time (hence the limit $t\to \pm\infty$) is inherent to scattering experiments, as the detector and particle source are far away from the place of the collision. The formalism of the S-matrix, with its free asymptotic states, arises because it is the right way to deal with this limit (eg with the fact that for any 2 normalizable states, we have $\langle \psi|e^{-i\hat HT}|\phi\rangle \underset{T\to \infty}{\longrightarrow} 0$) $\endgroup$ Oct 15, 2023 at 13:17

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