# Why any expectation value can be computed by this path integral, and not just the time-ordered ones?

This is quite a basic question about the path integral. In Polchinki's String Theory book, Chapter 2, he says:

Expectation values are defined by the path integral

$$\langle \mathscr{F}[X]\rangle=\int[dX] \exp(-S)\mathscr{F}[X],\tag{2.1.14}$$

where $$\mathscr{F}[X]$$ is any functional of $$X$$, such as a product of local operators.

Now I believe I have gotten something wrong. My issue is with the any functional part. If I recall what the path integral gives are time-ordered mean values, so that it would not give the mean of "any functional of $$X$$".

In fact, in Appendix A, Polchinski reviews the path integral. He derives this result, and in fact in Eq. (A.1.17) we see:

$$\int[dq]_{q_i,0}^{q_f,T}\exp (iS)q(t)q(t')=\langle q_f,T|\mathrm{T}[\hat{q}(t)\hat{q}(t')]|q_i,0\rangle\tag{A.1.17}.$$

So I confess I am a bit lost, but that's probably something very basic that I'm missing.

How to reconcille Polchinski's statement, Eq. (2.1.14), that we may get the expectation value of any functional of $$X$$ by that path integral, with the fact that the path integral actually computes time-ordered expectation values? Is there some way in which the path integral may, in fact, compute the expectation value of any functional?

• I don't understand the question. In the text quoted, Polchinski uses (2.1.14) as the definition of the expectation value of any functional, i.e. he is defining the expectation value of "any functional" to be the same as the expectation value of its time-ordered version. What exactly is the problem here? – ACuriousMind Jul 28 '19 at 15:40
• $\langle F\rangle$ means $\langle 0|T[F]|0\rangle$. – AccidentalFourierTransform Jul 28 '19 at 15:45
• @ACuriousMind if that is the case, then I fail to grasp the motivation for said definition. Is it because in the end we only need the time-ordered ones (e.g., in computing the $S$-matrix), so we simply don't care with the non-time-ordered ones? I mean, in the operator formalism we have a distinction between the expectation value of $\mathscr{F}[X]$ and $\mathrm{T} \mathscr{F}[X]$. Why would it be reasonable to define the expectation value of a functional to be the same as the expectation value of its time-ordered version? Wouldn't this even clash with the operator formalism? – Gold Jul 28 '19 at 15:56
• Frankly, I've never seen anyone keep writing the $T$ for time-ordered expectation values beyond elementary intro texts. There is so little use for "non-time-ordered expectation values" that you'll see n-point functions in many places simply written as $\langle \phi(x_1)\dots\phi(x_n)\rangle$, and the time-order is implicit. I don't know what sources you're reading that always carefully keep mentioning the time order, but they are not representative of the majority of theoretical QFT. – ACuriousMind Jul 28 '19 at 16:02
• Ok, I see that it is then a matter of conventions and wording. The whole point in the end seems to be that since the correlation functions that are useful are the time-ordered ones, we focus on these and assume time-ordering by default. Anyway, thanks for pointing this out @ACuriousMind. – Gold Jul 28 '19 at 20:09