This question comes out of my other question "Time ordering and time derivative in path integral formalism and operator formalism", especially from the discussion with drake. The original post is somewhat badly composed because it contains too many questions, and till today I finally get energetic enough to compose another question that hopefully clarifies what was asked in that post.
I have no problem with all the textbook derivations of DSE, but after changing a perspective I found it very curious that DSE actually works, I'll take the equation of motion(EOM) of time-ordered Green's function of free Klein-Gordon(KG) field as an example and explain what I actually mean.
The EOM of free KG T-ordered Greens function is
$$(\partial^2+m^2)\langle T\{\phi(x)\phi(x')\}\rangle=-i\delta^4(x-x')\cdots\cdots(1).$$
The delta function comes from the fact that $\partial^2$contains time derivatives and it doesn't commute with T-ordering symbol. In general for Bosonic operators
$$\partial_t \langle T\{A(t)\,B(t')\}\rangle=\langle T \{ \dot A(t)B(t')\rangle+\delta (t-t')\,\langle [A(t),B(t')]\rangle \cdots\cdots(2),$$
$(1)$ can be derived from $(2)$ and the equal time canonical commutation relation of the fields.
However $(1)$ isn't very obvious from path integral approach:
$$(\partial^2+m^2)\langle T\{\phi(x)\phi(x')\}\rangle=(\partial^2+m^2)\int\mathcal{D}\phi e^{iS}\phi(x)\phi(x')\\\quad \quad \quad \qquad \qquad \qquad \qquad =\int\mathcal{D}\phi e^{iS}[(\partial^2+m^2)\phi(x)]\phi(x')\cdots\cdots(3).$$
Now if we formally and naively think
$$\int\mathcal{D}\phi e^{iS}[(\partial^2+m^2)\phi(x)]\phi(x')=\langle T\{[(\partial^2+m^2)\phi(x)]\phi(x')\}\rangle\cdots\cdots(4),$$
with the notation $\langle\cdots\rangle$ always denoting expectation value in operator approach. Then the final result will be 0(due to the field equation) instead of $-i\delta^4(x-x')$, and a result like equation(2) cannot be obtained.
As drake has pointed out, this is because of the ambiguity in the definition of the equal time operator product when time derivative is present in the path integral, it's very important(e.g. CCR in path integral) to define clearly the time derivative on time lattice, that is, the discretization. There are 3 possible definitions of $\dot \phi$(omitting the spatial variables):
(a)forward derivative $$\dot \phi(t)=\frac{\phi(t+\epsilon^+)-\phi(t)}{\epsilon^+}$$;
(b)backward derivative $$\dot \phi(t)=\frac{\phi(t)-\phi(t-\epsilon^+)}{\epsilon^+}$$
(c)centered derivative $$\dot \phi(t)=\frac{\phi(t+\epsilon^+)-\phi(t-\epsilon^+)}{2\epsilon^+}=\frac{1}{2}(\text{forward}+\text{backward})$$
These different time discretizations will lead to different equal-time operator orderings(see Ron Maimon's answer in this post ), respectively they are:
(a)$$\int\mathcal{D}\phi e^{iS}\dot \phi(t)\phi(t)=\langle \dot \phi(t)\phi(t)\rangle$$
(b)$$\int\mathcal{D}\phi e^{iS}\dot \phi(t)\phi(t)=\langle \phi(t)\dot \phi(t)\rangle$$
(c)$$\int\mathcal{D}\phi e^{iS}\dot \phi(t)\phi(t)=\frac{1}{2}[\langle \dot \phi(t)\phi(t)\rangle+\langle \phi(t)\dot \phi(t)\rangle]$$
With these in mind, we can now get equation $(1)$ from path integral(I'll just show it for the point of equal time, because for $t\neq t'$ there isn't any inconsistency): first I take definition (c) for $\dot \phi$, but define $\ddot\phi$ using forward derivative(which I agree is contrived), so we have
$\int\mathcal{D}\phi e^{iS}\ddot \phi(t)\phi(t)\equiv\int\mathcal{D}\phi e^{iS}\frac{1}{\epsilon^+}[\dot \phi(t+\epsilon^+)-\dot \phi(t)]\phi(t) =\frac{1}{\epsilon^+}\{\langle\dot \phi(t+\epsilon^+)\phi(t)\rangle-\frac{1}{2}\langle \dot \phi(t)\phi(t)\rangle-\frac{1}{2}\langle \phi(t)\dot \phi(t)\rangle\}\\ =\frac{1}{\epsilon^+}\{\langle \dot \phi(t+\epsilon^+)\phi(t)\rangle-\langle \dot \phi(t)\phi(t)\rangle+\frac{1}{2}\langle [\dot \phi(t),\phi(t)]\rangle\}\\ =\langle \ddot \phi(t)\phi(t)\rangle+\frac{1}{2\epsilon^+ }\langle[\dot \phi(t),\phi(t)]\rangle=\langle \ddot \phi(t)\phi(t)\rangle+\frac{1}{2\epsilon^+ }\delta^3(\mathbf{x}-\mathbf{x'})\cdots\cdots(5)$
Now we can formally think $\lim_{\epsilon^+\to 0}\frac{1}{2\epsilon^+}=\delta(0)$ because $\delta (t)= \lim_{\epsilon^+\to 0}\,{1\over 2\epsilon^+}\,e^{-\pi\,t^2/(4\,{\epsilon^+}^2)}$. So $(5)$ becomes
$$\int\mathcal{D}\phi e^{iS}\ddot \phi(t)\phi(t)=\langle \ddot \phi(t)\phi(t)\rangle+\delta(0)\delta^3(\mathbf{x}-\mathbf{x'})\cdots\cdots(6)$$
The rest is trivial, just apply the spatial derivatives, add it to $(6)$ and apply the field equation, then $(1)$ will be reproduced. The above derivation is mostly due to drake, in a more organized form.
Now it's clear that carefully defining time-derivative discretization is crucial to get the correct result, a wrong discretization won't give us $(1)$. However the derivation of DSE makes absolutely no reference to any discretization scheme, but it always gives a consistent result with the operator approach, why does it work so well?
Many thanks to who are patient enough to read the whole post!
UPDATE: Recently I had a lucky chance to communicate with Professor Dyson about this problem. His opinion is that neither these manipulations nor DSE is true math, because of the lack of mathematical rigor of the underlying theory, so there could be situations where DSE might just fail too, but unfortunately he couldn't immediately provide such an example. Although not very convinced(in the sense that even there's such an example, I still think the "degrees of naivety" of different approaches can be discerned, DSE is clearly more sophisticated and powerful than a direct application of $\partial^2+m^2$ ), I'd be partially satisfied if someone can provide a situation where DSE fails .
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