Further S matrix clarifications

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Hello! I attached a SS of the part of my book that I am confused about. So there they write the initial and final states in term of creation and annihilation operators, acting on the (not free) vacuum i.e. ##|\Omega>##. So first thing, the value of the creation (annihilation) operators at ##\pm\infty## are not the same, right? They evolve in time with the full interaction hamiltonian, so differently from the free case they have different values in time. Second thing, the last line, if I understand it well, and plugging in the values for the initial and final states, would be $$<f|S|i>=<f|i>$$ Is this right? It looks like the S matrix does nothing here. What am I doing wrong? Thank you!

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A few sentences down Schwartz says the creation/annihilation operators are rotated by the interacting hamiltonian and so are different at different times t and t'.

The use of the ##S## in ##<f|i>## here is a bit confusing, compare to Srednicki.

A few sentences down Schwartz says the creation/annihilation operators are rotated by the interacting hamiltonian and so are different at different times t and t'.

The use of the ##S## in ##<f|i>## here is a bit confusing, compare to Srednicki.
Thank you for this! I am sorry I honestly didn't have time to go further yet. However, the creation/annihilation operators, don't they evolve with the full interaction hamiltonian? Shouldn't we have some terms of the form ##e^{iHt}## in between them, to evolve them between the 2 times? Why do we use the S-matrix and not the usual time evolution operator? Edit: I read the notes from the link you mentioned. He doesn't even mention the S matrix there... Also even if he uses the interaction theory he uses for the vacuum ##|0>## and not ##|\Omega>##. Is his ##|0>## the ground state of the interaction theory? And what happens to the S matrix?

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Schwartz mentioned the dependence of the ##a(t)##'s on ##e^{iHt}## on the same page and goes over the whole derivation in the next few chapters, e.g. I think he brings up ##|\Omega >## more in the next chapter, though I do not see where he explained why ##S## does not appear, he should have mentioned something like ##S|in > = |out>##, ##<out|in > = <in|S|in>## somewhere, something analogous to slide 18 here.

George Jones
Staff Emeritus