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In section 6.1, eqs(6.4) and eq(6.5), the $|i\rangle$ and $|f\rangle$ are defined as $$|i\rangle=\sqrt{2\omega_1}\sqrt{2\omega_2} a_{p_1}^{\dagger}(-\infty)a_{p_2}^{\dagger}(-\infty)|\Omega\rangle,\tag{6.4}$$ $$|f\rangle=\sqrt{2\omega_3}...\sqrt{2\omega_n}a_{p_3}^{\dagger}(\infty)...a_{p_n}^{\dagger}(\infty)|\Omega \rangle.\tag{6.5}$$

In the next, eq(6.6), the books says the $S$-Matrix is defined as $$\langle f|S|i\rangle=2^{n/2}\sqrt{\omega_1...\omega_n}\langle\Omega|a_{p_3}(\infty)...a_{p_n}(\infty)a_{p_1}^{\dagger}(-\infty)a_{p_2}^{\dagger}(-\infty)|\Omega \rangle .\tag{6.6}$$

But by the definitions, the righthand-side is nothing but $\langle f|i\rangle $, so we get $$ \langle f|S|i\rangle=\langle f|i\rangle~?$$ I think this is wrong. Or did I misunderstand something?

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Hint: The $S$-matrix $$\langle f|S|i\rangle={}_{\rm out}\langle f|i\rangle_{\rm in}$$ is defined as the unitary matrix connecting the in-Hilbert space and the out-Hilbert space.

References:

  1. M.E. Peskin & D.V. Schroeder, An Intro to QFT, 1995; eq. (4.71).
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    $\begingroup$ Are the kets $|i \rangle_{in}$ and $|f \rangle_{out}$ in the Heisenberg Picture? $\endgroup$
    – Sven2009
    Apr 3 '21 at 10:47

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