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


where the subscript "op" and "pi" refer to operator and path integral formalism, respectively. 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


Of course


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 and this post), 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


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 positive 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.

  • 1
    $\begingroup$ 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 : $\int d\phi ~\phi(x)~\phi(y) ~e^{i\int_0^T dt L(\phi,t)} =\langle 0|T[\hat \phi(x),\hat \phi(y)]e^{-i \hat HT}|0\rangle$ $\endgroup$
    – Trimok
    Commented Jul 23, 2013 at 15:13
  • $\begingroup$ @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/… $\endgroup$
    – Jia Yiyang
    Commented Jul 24, 2013 at 1:50
  • $\begingroup$ This may seem odd, but I believe this is actually the same ambiguity as in physics.stackexchange.com/q/69828. Its not very transparent, but see Sec II of doi.org/10.1103/PhysRevD.3.2486 for an example where the derivative should not commute with path integral. In terms of practical operations, the best advice seems to be "work the Hamiltonian formulation". There are no time derivatives in this case and then you can integrate out the momenta at the end. $\endgroup$ Commented Jul 24, 2013 at 3:06
  • $\begingroup$ @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. $\endgroup$
    – Trimok
    Commented Jul 24, 2013 at 9:39
  • $\begingroup$ Hi @JiaYiyang . No, it's not like that. The first line should be $1/a$ times $A(t+a)B(t)- \theta(0) (A(t)B(t)+B(t)A(t))$ and then use or define $\theta(0)=1/2$, what in my opinion is totally right. As you don't like to take this value for $\theta (0)$, I'll write a derivation without using it. $\endgroup$ Commented Jul 30, 2013 at 4:27

2 Answers 2


Good question! It made me think. Here it's the answer:

  • Strictly speaking, it is not possible to compute $\theta(t-t')\langle \dot\phi(t)\phi(t')\rangle+\theta (t'-t)\langle \phi (t')\dot\phi (t)\rangle$ ( shorthand notation: $\langle\,\equiv \langle 0 |\,$. Also note that I'm omitting the spatial arguments of the fields ) using the Lagrangian version of the path integral, because to derive this, we are to assume that the insertions (factors multiplying $e^{iS}$ in the integrand of the path integral) are functionals of the fields at a given time (i.e., they are independent of momenta $\pi$). Thus,

$$\partial_t \,\left[\theta(t-t')\langle \phi(t)\phi(t')\rangle+\theta (t'-t)\langle \phi (t')\phi (t)\rangle\right]=\lim_{a\to 0}{1\over a}\left[\theta(t+a-t')\langle \phi(t+a)\phi(t')\rangle+\theta (t'-t-a)\langle \phi (t')\phi (t+a)\rangle - \theta(t-t')\langle \phi(t)\phi(t')\rangle+\theta (t'-t)\langle \phi (t')\phi (t)\rangle\right]=\lim_{a\to 0}{1\over a}\int D\phi\, (\phi(t+a)\phi(t')-\phi(t)\phi(t'))\,e^{iS}=\int D\phi\, \dot\phi(t)\phi(t')\,e^{iS}$$

which agrees with $\theta(t-t')\langle \dot\phi(t)\phi(t')\rangle+\theta (t'-t)\langle \phi (t')\dot\phi (t)\rangle$ only if $t\neq t'$.$^1$

  • Example. Let $A(t)$ and $B(t)$ be two functional at a given time, then you can check that

$$\int \dot A(t) B(t) \,e^{iS}=\langle\dot A(t)B(t)\rangle+\lim_{a\to 0}{1\over 2a}\langle [A(t),B(t)]\rangle$$

The last term represents the Dirac delta you found after deriving the step function.

  • While your question is very interesting, I think that your example is unfortunate since $\delta (t-t')[\phi(x),\phi(x')]=\delta (t-t')[\phi(t,\vec x),\phi(t,\vec x')]=0$. Let me choose an example in which this last commutator is different from zero to show how the time derivative of a correlation function splits in the two terms you mention: $\partial_t \langle \, T\, \phi(t)\, A(t')\,\rangle$, where $A$ is a functional of fields and momenta at a given time. Let's make a general variation (which doesn't modify the path integral measure) of the Hamiltonian or phase-space path integral ($S=\int dt\, \dot \phi\,\pi-H \, $ ). Since the momentum is an integration variable, the integral may not change:

$$0={\delta\over \delta \pi (t)}\int D\phi D\pi \, A(t')\,e^{iS}\,\delta \pi=\\ \int D\phi D\pi\, \left( {\delta A(t')\over \delta \pi (t)}+A(t')i\dot\phi (t)-A(t')(-i){\delta H (t)\over \delta \pi (t)}\right)\,e^{iS}\,\delta \pi$$

As $\delta \pi$ is a general variation and:

$${\delta A(t')\over \delta \pi (t)}=\delta(t-t')(-i)\,[\phi(t),A(t')]$$ $${\delta H (t)\over \delta \pi (t)}=-i\,[\phi(t), H]$$

we obtain:

$$\partial_t \langle \, T\, \phi(t)\,A(t')\,\rangle=i\langle \, T\, [\phi(t),H]\,A(t')\,\rangle+\delta (t-t')\,\langle \, [\phi(t),A(t')]\,\rangle \\ =\theta(t-t')\langle \, \dot\phi(t)\,A(t')\,\rangle+\theta(t'-t)\langle \, A(t')\,\dot\phi(t)\,\rangle+\delta (t-t')\,\langle \, [\phi(t),A(t')]\,\rangle$$

If, for example, $A(t')=\pi (t',\vec x')$, the last term gives $i\,\delta ^4 (x-x')$.


Just in case it is not clear enough, let me remark that derivatives do commute with the path integral measure. The key point is that

$$\partial_t \,\left[\theta(t-t')\langle \phi(t)\phi(t')\rangle+\theta (t'-t)\langle \phi (t')\phi (t)\rangle\right]=\int D\phi\, {\phi(t+\epsilon^+)-\phi (t)\over \epsilon ^+}\phi(t')\,e^{iS}\neq\theta(t-t')\langle \dot\phi(t)\phi(t')\rangle+\theta (t'-t)\langle \phi (t')\dot\phi (t)\rangle$$

In addition, I would like to emphasize that ${\delta H (t)\over \delta \pi (t)}$ is a functional at a given time, while $\dot\phi (t)$ in the integrand of a path integral must be interpreted as ${\phi(t+\epsilon^+)-\phi (t)\over \epsilon ^+}$, that is, as a difference between fields evaluated at different times.

$^1$ Note that the derivative can be defined in different ways giving rise to different operator orderings, see Maimon's excellent answer Path integral formulation of quantum mechanics, be careful with some typo in the expressions: where says $x(t)p(t)$ should say $p(t)x(t)$.

EDIT: To derive some of the results above, one needs to take $\theta (0)=1/2$. However, one can proceed in a slightly different manner to avoid such a choice (which, in my opinion, is totally right). For example,

$$\int \dot A(t) B(t) \,e^{iS}=\int {A(t+a)-A(t)\over a } B(t+a/2) \,e^{iS}=\langle\dot A(t)B(t)\rangle+\lim_{a\to 0}{1\over a}\langle [A(t),B(t)]\rangle$$

  • $\begingroup$ @JiaYiyang On the right track? Directly to the top of the mountain ;). Have you tried to derive it? It's easy. Take the definition of derivative I give in the answer and $\theta (0)=1/2$. Note that $\delta (t)= \lim_{a\to 0}\,{1\over 2a}\,e^{-\pi\,t^2/(4\,a^2)}$ and evaluate it in $t=0$ $\endgroup$ Commented Jul 28, 2013 at 19:06
  • $\begingroup$ Well, I tried your definition $\int D\phi\, {\phi(t+\epsilon^+)-\phi (t)\over \epsilon ^+}\phi(t)\,e^{iS}$, but according to Ron Maimon's thread isn't it just $\langle\dot{\phi}(t)\phi(t)\rangle$? $\endgroup$
    – Jia Yiyang
    Commented Jul 29, 2013 at 4:55
  • $\begingroup$ @JiaYiyang If $A=\phi$ and $B=\phi$, the last term (commutator term) vanishes. But $A\neq B$ in general. Apply the standard relation between path integrals and time ordered product of fields. It's straightforward. $\endgroup$ Commented Jul 29, 2013 at 16:46
  • $\begingroup$ Two questions:1. Even if it's $\int D\phi\, {\dot{\phi}(t+\epsilon^+)-\dot{\phi (t)}\over \epsilon ^+}\phi(t)\,e^{iS}$, according to my understanding of Ron Maimon's thread, will be equal to $\langle\ddot{\phi}(t)\phi(t)\rangle$(as long as we also use forward discretization to define $\dot{\phi}(t)$. 2. Again, the only way I could get a $\delta(0)\langle[A(t),B(t)]\rangle$ is to first get back to the operator definition where we have step functions, then differentiate, but this would be a circular logic. I want a calculation without any reference to operator definition, and compare the result $\endgroup$
    – Jia Yiyang
    Commented Jul 29, 2013 at 17:44
  • $\begingroup$ with the operator one, see if they are consistent. $\endgroup$
    – Jia Yiyang
    Commented Jul 29, 2013 at 17:45

EDIT: I'm leaving this up as background reading to @drake's answer. (The point of the following is that the path integral does indeed give the correct time ordering, so it is producing the correct $\theta$-function weighted, time-ordered sums, which must be accounted for when differentiating its output.)

The two formalisms are equivalent; if they don't give the same result, something is wrong in the calculation. To see this you have to understand a subtlety which is not usually well-explained in textbooks, namely that the path integral is not defined merely by taking the limit of a bunch of integrals of the form $\int_{\mbox{lattice fields}} e^{iS(\phi)} d\phi$.

The problem is that these finite-dimensional integrals are not absolutely convergent, because $|e^{iS(\phi)}| = 1$. To define even the lattice path integral in Minkowski signature, you have to specify some additional information, to say exactly what is meant by the integral.

In QFT, the additional information you want is that the path integral should be calculating the kernel of the time evolution operator $e^{iH\delta t}$, which is an analytic function of $\delta t$. This fact is usually expressed by saying that the Minkowski signature path integral is the analytic continuation of a Euclidean signature path integral: The Euclidean $n$-point functions $E(y_1,...,y_n)$ defined by

$E(y_1,...,y_n) = \int \phi(y_1)...\phi(y_n) e^{-S_E(\phi)} d\phi$

are analytic functions of the Euclidean points $y_i \in \mathbb{R}^d$. This function $E$ can be continued to a function $A(z_1,...,z_n)$ of $n$ complex variables $z_i \in \mathbb{C}^d$. This analytic function $A$ does not extend to the entire plane; it has singularities, and several different branches. Each branch corresponds to a different choice of time-ordering. One branch is the correct choice, another choice is the 'wrong sign' time-ordering. Other choices have wrong signs on only some subsets of the points. If you restrict $A$ to the set $B$ of boundary points of the correct branch, you'll get the Minkowski-signature $n$-point functions $A|_B = M$, where $M(x_1,...,x_n) = \langle \hat{\phi}(x_1)...\hat{\phi}(x_n)\rangle_{op}$ and the $x_i$ are points in Minkowski space.

In perturbation theory, most of this detail is hidden, and the only thing you need to remember is that the $+i\epsilon$ prescription selects out the correct time-ordering.

  • $\begingroup$ +1. I didn't realize this subtlety. So I guess in path integral the issue is $\partial_0$ does not commute with analytic continuation? And it implies when differentiating w.r.t time it's safer to use operator definition of T-ordering, while in path integral one has to take the trouble first differentiating in Euclidean signature then analytically continue it? It would be nicer if you can talk more about the time derivative issue. $\endgroup$
    – Jia Yiyang
    Commented Jul 23, 2013 at 15:46
  • $\begingroup$ I don't have much to add on the time derivatives. Take them on the operator side, and be aware that naively passing them through the path integral is hazardous. $\endgroup$
    – user1504
    Commented Jul 23, 2013 at 17:42
  • 2
    $\begingroup$ Hi. I upvoted your question because I think everything you say is correct, however I think it doesn't answer the question. Do you have any reason why passing time derivatives through the path integral is so dangerous? Can you have a look at my answer? $\endgroup$ Commented Jul 27, 2013 at 7:00

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