Understanding S-matrix elements in QFT

["Don't panic!"]

Hi,

I was wondering if I could test my understanding on the S-matrix and its role in evolving initial states of systems to final states (after some scattering process has occurred).

Would it be correct to say the following:

Given a system in an initial state $\vert i \rangle$, the final state, $\vert f \rangle$ of the system, at a sufficiently long time after some scattering process can be mapped to by the so-called S-operator, $S=\lim_{t\rightarrow\infty,\,t_{0}\rightarrow\infty} U\left(t,t_{0}\right)$ (where $U$ is the unitary time-evolution operator) i.e.

$\vert f \rangle=S\vert i \rangle$​

Is it then correct to say that the S-operator annihilates the initial state $\vert i \rangle$ and creates the final state $\vert f \rangle$?
Also, would it then be correct to say that the S-matrix element $\langle f\vert S\vert i \rangle$ corresponding to the given scattering process gives the vacuum expectation values (v.e.v) for the appropriate annihilation and creation operators involved?

Sorry for any inaccuracies, hoping to gain a more in depth understanding of the concept. Thanks in advance!

 

[meopemuk]

The formula you wrote

$S=\lim_{t\rightarrow\infty,\,t_{0}\rightarrow -\infty} U\left(t,t_{0}\right)$

doesn't have a fixed limit. The correct formula for the S-operator is

$S= \lim_{t\rightarrow\infty,\,t_{0}\rightarrow -\infty} U^{-1}_0\left(t,t_{0}\right) U\left(t,t_{0}\right) = \lim_{t\rightarrow\infty,\,t_{0}\rightarrow -\infty} U_0\left(t_{0}, t\right) U\left(t,t_{0}\right)$

where $U_0\left(t_{0}, t\right)$ is the free (non-interacting) time evolution operator.

 

["Don't panic!"]

ok, thanks for providing the correct one.

Am I correct about the other bits though?

 

[meopemuk]

Assuming that we are working in the Schroedinger picture and that the initial state $\vert i \rangle$ is at time $t_0 = -\infty$ and the final state $\vert f \rangle$ is at time $t = +\infty$, then correct formulas connecting these states are

$\vert f \rangle=S U_0\left(t,t_{0}\right) \vert i \rangle = U_0\left(t,t_{0}\right) S \vert i \rangle$​

 

["Don't panic!"]

In my course we've been working in the interaction picture.
Would my heuristic explanation of what the S-operator physically does be correct though?

 

[meopemuk]

In my opinion, the interaction picture is rather confusing and not easy to visualize. But the Schroedinger picture is very transparent. For example, when there is no interaction, which means that $S=1$, then the time evolution of states is governed by the free time evolution operator

$\vert f \rangle=U_0\left(t,t_{0}\right) \vert i \rangle$​

When there is interaction, the time evolution is governed by the full interacting time evolution operator

$\vert f \rangle=U\left(t,t_{0}\right) \vert i \rangle$​

This formula is very difficult for calculations, but the whole idea of the scattering theory is that this formula can be simplified without any loss of accuracy

$\vert f \rangle=U\left(t,t_{0}\right) \vert i \rangle = S U_0\left(t,t_{0}\right) \vert i \rangle$​

This means that you can evolve your state by the (very simple) free evolution operator, and then multiply the result by the S-operator. So, basically, the S-operator tells you by how much the interacting time evolution differs from the free time evolution.

 

["Don't panic!"]

Ah, ok. Thank you for your help.