I am currently reading Schwartz's book on QFT and the Standard Model and I'm stuck on the beginning of the proof he gives for the LSZ reduction formula. Let the initial and final states of a scattering process be

$$|i\rangle= \sqrt{2\omega_1}\sqrt{2\omega_2} \, a_{p_1}^\dagger(-\infty)a_{p_2}^\dagger(-\infty)|\Omega\rangle\tag{6.4}$$ and $$|f\rangle= \sqrt{2\omega_3}\:...\sqrt{2\omega_n}\, a_{p_3}^\dagger(+\infty)\, ...\, a_{p_n}^\dagger(+\infty)|\Omega\rangle,\tag{6.5}$$ with $\Omega$ the vacuum of the interacting theory. In the next line, he writes:

$$\langle f | S | i \rangle=2^{n/2}\sqrt{\omega_1\omega_2...\omega_n} \langle \Omega | a_{p_3}(+\infty)\, ...\, a_{p_n}(+\infty)a_{p_1}^\dagger(-\infty)a_{p_2}^\dagger(-\infty)|\Omega\rangle.\tag{6.6}$$

Where is the $S$-matrix in this last equation? It seems to me like it is simply $\langle f |i \rangle$.


1 Answer 1


It's implicit because the operators $a_p$ are being treated as time-dependent. If instead we were in a picture where the operators did not have explicit time dependence, then we would need to explicitly include the $S$ matrix. The $S$ matrix allows us to figure out what the overlap is between the state at time $-\infty$ and the state at time $\infty$. In the formula you gave, the time dependence of the states is already taken into account by the time-dependent operators.


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