# Time ordering and time derivative in path integral formalism and operator formalism

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In operator formalism, for example a 2-point time-ordered Green's function is defined as

$\langle\mathcal{T}\phi(x_1)\phi(x_2)\rangle_{op}=\theta(x_1-x_2)\phi(x_1)\phi(x_2)+\theta(x_2-x_1)\phi(x_2)\phi(x_1),$

where the subscript "op" refers to operator formalism. Now if one is to take a time derivative of it, the result will be $\frac{\partial}{\partial x_1^0}\langle\mathcal{T}\phi(x_1)\phi(x_2)\rangle_{op}=\langle\mathcal{T}{\frac{\partial \phi(x_1)}{\partial x_1^0}}\phi(x_2)\rangle_{op}+\delta(x_1^0-x_2^0)[\phi(x_1),\phi(x_2)]$, the delta function comes from differentiating the theta functions. This means time derivative does not commute with time ordering.

If we consider path integral formalism, the time-ordered Green's function is defined as

$\langle\mathcal{T}\phi(x_1)\phi(x_2)\rangle_{pi}=\int\mathcal{D}\phi\phi(x_1)\phi(x_2)e^{iS(\phi)}$.

Of course $\langle\mathcal{T}\phi(x_1)\phi(x_2)\rangle_{op}=\langle\mathcal{T}\phi(x_1)\phi(x_2)\rangle_{pi},$ as is proved in any QFT textbook. However in path integral case time derivative commutes with time ordering, because we don't have anything like a theta function thus $\frac{\partial}{\partial x_1^0}\int\mathcal{D}\phi\phi(x_1)\phi(x_2)e^{iS(\phi)}=\int\mathcal{D}\phi\frac{\partial}{\partial x_1^0}\phi(x_1)\phi(x_2)e^{iS(\phi)}$. I did a bit googling and found out that for the path integral case the time-ordered product is called "$\mathcal{T^*}$ product" and operator case just "$\mathcal{T}$ product".

I am not that interested in what is causing the difference(still explanations on this are welcomed), because I can already vaguely see it's due to some sort of ambiguity in defining the product of fields at equal time. The question that interests me is, which is the right one to use when calculating Feynman diagrams?

I did find a case where both give the same result, i.e. scalar QED(c.f. Itzykson & Zuber, section 6-1-4), but is it always the case? If these two formulations are not effectively equivalent, then it seems every time we write down something like $\langle\partial_0\phi\cdots\rangle$, we have to specify whether it's in the sense of the path integral definition or operator definition.

EDIT: As much as I enjoy user1504's answer, after thinking and reading a bit more I don't think analytic continuation is all the mystery. In Peskin&Schroeder chap 9.6 they manage to use path integral to get a result equivalent to operator approach, without any reference to analytic continuation. It goes like this : Consider a T-product for free KG field $\langle T\{\phi(x)\phi(x_1)\}\rangle=\int\mathcal{D}\phi\phi(x)\phi(x_1)e^{iS(\phi)}$. Apply Dyson-Schwinger equation, we get $\int\mathcal{D}\phi(\partial^2+m^2)\phi(x)\phi(x_1)e^{iS}=-i\delta^4(x-x_1)$, then they just assume the $\partial^2$ commute with path integration(which is already weird according to our discussion) and they conclude

$(\partial^2+m^2)\int\mathcal{D}\phi\phi(x)\phi(x_1)e^{iS}=(\partial^2+m^2)\langle T\{\phi(x)\phi(x_1)\}\rangle=-i\delta^4(x-x_1)$.

This is just the right result given by operator approach, in which $\delta(x^0-x_1^0)$ comes from $\theta$ function. Given my limited knowledge on the issue, this consistency looks almost a miracle to me. What is so wicked behind these maths?

Response to @drake:If $a$ is a positive infinitesimal, then $\int \dot A(t) B(t) \,e^{iS}\equiv\int D\phi\, {A(t+a)-A(t)\over a}B(t)\,e^{iS}=\frac{1}{a}\langle T\{A(t+a)B(t)\}\rangle-\frac{1}{a}\langle A(t)B(t)\rangle$, notice the second term has an ordering ambiguity from path integral(say $A=\dot{\phi},B=\phi$), and we can make it in any order we want by choosing an appropriate time discretization, c.f. Ron Maimon's post cited by drake. Keeping this in mind we proceed:

$\frac{1}{a}\langle T\{A(t+a)B(t)\}\rangle-\frac{1}{a}\langle A(t)B(t)\rangle\\=\frac{1}{a}\theta(a)\langle A(t+a)B(t)\rangle+\frac{1}{a}\theta(-a)\langle B(t)A(t+a)\rangle-\frac{1}{a}\langle A(t)B(t)\rangle\\=\frac{1}{a}\theta(a)\langle A(t+a)B(t)\rangle+\frac{1}{a}[1-\theta(a)]\langle B(t)A(t+a)\rangle-\frac{1}{a}\langle A(t)B(t)\rangle\\=\frac{\theta(a)}{a}\langle [A(t+a),B(t)]\rangle+\frac{1}{a}[\langle B(t)A(t+a)\rangle-\langle A(t)B(t)\rangle]$

Now taking advantage of ordering ambiguity of the last term to make it $\langle B(t)A(t)\rangle$(this amounts to defining A using backward discretization, say $A=\dot{\phi}(t)=\frac{\phi(t+\epsilon^-)-\phi(t)}{\epsilon^-}$), then the finally:

$\frac{\theta(a)}{a}\langle [A(t+a),B(t)]\rangle+\frac{1}{a}\langle B(t)[A(t+a)-A(t)\rangle]\to \frac{1}{2a}\langle [A(t),B(t)]\rangle+\langle B(t)\dot{A}(t)\rangle$.(Here again a very dubious step, to get $\frac{1}{2a}$ we need to assume $\theta(a\to 0^+)=\theta(0)=\frac{1}{2}$, but this is really not true because $\theta$ is discontinuous)

However on the other hand, since $a$ was defined to be a postive infinitesimal, at the very beginning we could've written $\frac{1}{a}\langle T\{A(t+a)B(t)\}\rangle-\frac{1}{a}\langle A(t)B(t)\rangle=\frac{1}{a}\langle A(t+a)B(t)\rangle-\frac{1}{a}\langle A(t)B(t)\rangle$, then all the above derivation doesn't work. I'm sure there are more paradoxes if we keep doing these manipulations.

This post imported from StackExchange Physics at 2014-03-31 22:24 (UCT), posted by SE-user Jia Yiyang
I think that the problem is that you cannot represent a commutator of 2 operators at equal time, by a path integral representation (it would be zero), but this does not mean that the derivative and the integral commute in the path integral formalism. These are equivalent formalism : $\int d\phi ~\phi(x)~\phi(y) ~e^{i\int_0^T dt L(\phi,t)} =\langle 0|T[\hat \phi(x),\hat \phi(y)]e^{-i \hat HT}|0\rangle$
@Trimok: Maybe my example is not that good because $[\phi(x_1),\phi(x_2)]$ is indeed 0 at equal time. It's better to consider $[\phi(x_1),\dot{\phi}(x_2)]$, in this case path integral can also give non-zero result by this argument: en.wikipedia.org/wiki/…
@JiaYiyang : When you derive a time-ordered product, you have always the last term which looks like $\langle0|\delta(x_0-y_0)[A(x), B(y)]e^{iS}|0\rangle$. And this term, no matter $A$ and $B$ are, cannot be represented by a path integral. But this term is not null, even if it cannot be represented by a path integral. So derivative and integral (in the path integral formalism) do not commute, even if the (not null) difference, cannot be represented by a path integral.