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I have seen the transition amplitude was written in different ways, but I don't understand.

For example, in Mark Srednicki's textbook section $10,11$, it is written as $\langle f|i\rangle$. But in David Tong's notes, there is an insertion of $S$: $\langle f|S|i\rangle$, known as scattering matrix. Are they the same thing? Any help will be appreciated.

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    $\begingroup$ Are Srednicki's states time-dependent at all? $\endgroup$ Jan 25 at 13:44
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I don't know these books specifically, but what I think is happening is that you're missing information about $|i\rangle$. You can state the transition amplitude both ways with the adequate context, which is: to represent as $\langle f|i\rangle$ is implicitly that $|i\rangle$ is already an evolved state, more explicitly you could parametrize it with time and then you'd have a transition amplitude at a time $t$, $\langle f|i(t)\rangle=\langle f|(\textbf{U}|i(0)\rangle)$; on the other hand when you write $\langle f|\textbf{S}|i\rangle$ this means that the evolution of the initial state is being made by a scattering process, relating to the latter case, we are saying that the evolution of the initial state is done by $\textbf{U}=\textbf{S}$, or $\langle f|\textbf{S}|i\rangle$. Hope this helps you in some way!

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The notation is somewhat incomplete: In scattering theory, the in-states $|i\rangle_{in}$ and out-states $|f\rangle_{out}$ live on different Hilbert spaces. So $\langle f | i\rangle$ is really supposed to mean $_{out}\langle f | i \rangle_{in}$, with the meaning that an isomorphism is to be applied that translates $\mathcal H_{in}$ into $\mathcal H_{out}$. That isomorphism is the scattering matrix $S$, i.e. $_{out}\langle f | i \rangle_{in} := \langle f | S | i \rangle$ and on the right hand side $|i\rangle$ and $|f\rangle$ now live on the same Hilbert space.

As indicated by Joao, $S$ is essentially time evolution, i.e. a time evolution operator $U(t_{final}=+\infty ,\ t_{initial}=-\infty)$.

Usually in scattering theory, one considers momentum eigenstates and particular choices of particles. Then the shorthand notation $\langle f| i \rangle$ looks like it could only be non-zero if $|f\rangle = |i\rangle$. Of course, that is not true in scattering, and the illusion only arises because of lazy notation.

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  • $\begingroup$ Why do the in- and out-states live on different Hilbert spaces? $\endgroup$
    – Sven2009
    Jan 26 at 7:37

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