Proof of quantum correlation functions

Question

I'm reading through David Tong's lecture notes on QFT.

On pages 76-77, he gives a proof about correlation functions. See the below link:

QFT notes by Tong

I'm following the proof steps to obtain equation (3.95). But several intermediate steps of the proof are not clear.

First question

Why can we 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 $T$ relate to the $\phi_{1}\dots\phi_{n}$ only, or to 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 do 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 do we have

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

Answer 1

First question

Using that $S=U_I(+\infty,-\infty)=U_I(+\infty, t_1)U_I(t_1,t_2)\cdots U_I(t_n,-\infty)$, as you state, you have that \begin{align} T\phi_{1I}\phi_{2I}\cdots\phi_{nI}S &= T\phi_{1I}\phi_{2I}\cdots\phi_{nI} U_I(+\infty, t_1)U_I(t_1,t_2)\cdots U_I(t_n,-\infty) \\ & = U_I(+\infty, t_1)\phi_{1I}U_I(t_1,t_2)\phi_{2I}\cdots \phi_{nI}U_I(t_n,-\infty), \end{align} where the second equality is given by the definition of time ordering.

Second question

Choosing the operators in the interaction picture and the Heisenberg picture to be equal at some time $t_0$, we have that $\phi_{kI}=U(t_0,t_k)^{-1}\phi_{kH}U_I(t_0,t_k)$. Subtituting into the result for the previous question: \begin{align} T\phi_{1I}\phi_{2I}\cdots\phi_{nI}S =& U_I(+\infty, t_1)U(t_0,t_1)^{-1}\phi_{1H}U_I(t_0,t_1) U_I(t_1,t_2) U(t_0,t_2)^{-1}\\ & \phi_{2H}U_I(t_0,t_2) \cdots U(t_0,t_n)^{-1}\phi_{nH}U_I(t_0,t_n)U_I(t_n,-\infty) \\ =& U_I(+\infty,t_0)\phi_{1H}\phi_{2H}\cdots\phi_{nH}U_I(t_0,-\infty) \end{align}

Third question

Notice that Tong is not saying that $U_I(t,-\infty)=U(t,-\infty)$, but that for any $\left|\Psi\right>$, we have $\left<\Psi\right| U_I(t,-\infty)\left|0\right>=\left<\Psi\right|U(t,-\infty)\left|0\right>$. This statement is equivalent to \begin{equation} U_I(t,-\infty)\left|0\right>=U(t,-\infty)\left|0\right> \end{equation} By definition $\left|0\right>$ is an eigenvector of $H_0$ with eigenvalue $0$, so \begin{equation} H_I\left|0\right>=H_Ie^{iH_0t}\left|0\right>=H_I\left|0\right>_I= i\frac{d}{dt}\left|0\right>_I= i\frac{d}{dt}\left(e^{iH_0t}\left|0\right>\right)= i\frac{d}{dt}\left|0\right>=H\left|0\right>. \end{equation} Thus, the interaction picture time evolution $U_I(t,-\infty)$ (obtained by exponentiating the integral of $H_I$) and the Schrödinger picture time evolution $U(t,-\infty)$ (the exponential of the integral of $H$) are the same when applied to $\left|0\right>$.