# Understanding S-matrix elements in QFT

1. Apr 19, 2014

### "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!

2. Apr 19, 2014

### 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.

3. Apr 19, 2014

### "Don't panic!"

ok, thanks for providing the correct one.

Am I correct about the other bits though?

4. Apr 20, 2014

### 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

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

5. Apr 20, 2014

### "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?

6. Apr 21, 2014

### 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.

7. Apr 21, 2014

### "Don't panic!"

Ah, ok. Thank you for your help.