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

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


1 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 $\left|0\right>$.

  • $\begingroup$ I'd like to point out that your first and second equations are way non-trivial (unlike what you might seem to suggest). For one thing, the definition of $T$ for coinciding time arguments has many subtleties. Moreover, you cannot use $U(t_1,t_2)U(t_2,t_3)=U(t_1,t_3)$ inside the $T$ symbol (at least, not without the proper justification by analysing the different permutations and so on). $\endgroup$ Dec 13, 2016 at 11:20
  • $\begingroup$ Hmm... maybe I'm being too naive here. I tried to explain the ideas in the proof text in the reference (which says even less). I hope that my answer shows the path, but maybe it should've been much more detailed. I think I'll add a little paragraph a the beginning explaining this. $\endgroup$
    – coconut
    Dec 13, 2016 at 11:25
  • $\begingroup$ What you wrote is rather standard: you can find the very same (IMHO, naive) treatment in most introductory books on QFT. In this sense, your post is ok as far as Im concerned, but I just wanted to point out that, strictly speaking, matters are not so simple if one wants to formalise the theory. $\endgroup$ Dec 13, 2016 at 11:33
  • 1
    $\begingroup$ You're right. I'll leave it as it is and let the people know about those subtleties by reading your comment. $\endgroup$
    – coconut
    Dec 13, 2016 at 11:38
  • $\begingroup$ Thanks for your time. Are you discussing here how to time order $\phi_{1I}U_I(t_1,t_2)$ or $U_I(t_1,t_2)\phi_{1I}$? $\endgroup$
    – user56963
    Dec 13, 2016 at 13:14

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