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Understanding S-matrix elements in QFT

  1. Apr 19, 2014 #1
    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 [itex] \vert i \rangle[/itex], the final state, [itex] \vert f \rangle[/itex] of the system, at a sufficiently long time after some scattering process can be mapped to by the so-called S-operator, [itex]S=\lim_{t\rightarrow\infty,\,t_{0}\rightarrow\infty} U\left(t,t_{0}\right)[/itex] (where [itex]U[/itex] is the unitary time-evolution operator) i.e.

    [itex]\vert f \rangle=S\vert i \rangle[/itex]​

    Is it then correct to say that the S-operator annihilates the initial state [itex] \vert i \rangle[/itex] and creates the final state [itex] \vert f \rangle[/itex]?
    Also, would it then be correct to say that the S-matrix element [itex] \langle f\vert S\vert i \rangle[/itex] 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. jcsd
  3. Apr 19, 2014 #2
    The formula you wrote

    [itex]S=\lim_{t\rightarrow\infty,\,t_{0}\rightarrow -\infty} U\left(t,t_{0}\right)[/itex]

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

    [itex]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) [/itex]

    where [itex] U_0\left(t_{0}, t\right) [/itex] is the free (non-interacting) time evolution operator.
     
  4. Apr 19, 2014 #3
    ok, thanks for providing the correct one.

    Am I correct about the other bits though?
     
  5. Apr 20, 2014 #4
    Assuming that we are working in the Schroedinger picture and that the initial state [itex] \vert i \rangle[/itex] is at time [itex] t_0 = -\infty [/itex] and the final state [itex] \vert f \rangle[/itex] is at time [itex] t = +\infty [/itex], then correct formulas connecting these states are

    [itex]\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[/itex] ​
     
  6. Apr 20, 2014 #5
    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?
     
  7. Apr 21, 2014 #6
    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 [itex]S=1 [/itex], then the time evolution of states is governed by the free time evolution operator

    [itex]\vert f \rangle=U_0\left(t,t_{0}\right) \vert i \rangle [/itex] ​

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

    [itex]\vert f \rangle=U\left(t,t_{0}\right) \vert i \rangle [/itex] ​

    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


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

    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.
     
  8. Apr 21, 2014 #7
    Ah, ok. Thank you for your help.
     
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