- #1

victorvmotti

- 155

- 5

[QFT notes by Tong][1] [1]: http://www.damtp.cam.ac.uk/user/tong/qft/qft.pdfI am following the proof steps to obtain equation (3.95). But several intermediate steps of the proof are not clear.

**First question**

why we can write:

$$T\phi_{1I} \dots \phi_{nI}S=U_{I}(+\infty, t_{1})\phi_{1I}U(t_{1},t_{2})\phi_{2I}\dots \phi_{nI}U_{I}(t_{n},-\infty)$$

I mean after dropping the $$T$$ shouldn't we have?:

$$=\phi_{1I}\phi_{2I}\dots \phi_{nI}S$$$$=\phi_{1I}\phi_{2I}\dots \phi_{nI}U_{I}(+\infty,-\infty)$$

Does $$T$$ operate on the $$\phi_{1}\dots\phi_{n}$$ only or the $$\phi_{1}\dots \phi_{nI}S$$ and $$U_{I}(+\infty,-\infty)=U_{I}(+\infty, t_{1})U_{I}(t_{1},t_{2})\dots U_{I}(t_{n},-\infty)$$?**Second question**

How we convert each of the $$\phi_{I}$$ into $$\phi_{H}$$ using $$U_{I}(t_{k},t_{k+1})=Texp(-i\int_{t_{k}}^{t_{k+1}}H_{I})$$ to arrive at

$$T\phi_{1I} \dots \phi_{nI}S=U_{I}(+\infty, t_{0})\phi_{1H}\dots \phi_{nH}U_{I}(t_{0},-\infty)$$

**Third question**

Why we have

$$U_{I}(t, -\infty)=U(t,-\infty)$$