# Transition amplitude, $\langle f|i\rangle$ or $\langle f|S|i\rangle$?

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.

• Are Srednicki's states time-dependent at all? Jan 25, 2021 at 13:44

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!
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.