Schwinger-Dyson Equations and boundary conditions

Question

Schwinger-Dyson Equations (SDEs) can be understood in a myriad of different ways. In the Canonical formalism, they come from the non-commutation between time differentiation and the canonical time ordering (as opposed to covariant time-ordering in the path integral formalism). But in the Path Integral Formalism, there are two ways, at least that I'm aware of, to obtain them:

The first, is to integrate by parts, that is, given some functional of the field (we'll assume one scalar for simplicity) $F[\phi]$ $$ \frac{1}{Z[0]} \int\mathcal{D}\phi \frac{\delta F}{\delta{\phi}}e^{iS[\phi]}=\frac{1}{Z[0]} \int\mathcal{D}\phi\frac{\delta}{\delta{\phi}}\left(F[\phi]e^{iS[\phi]}\right)-\frac{i}{Z[0]}\int\mathcal{D}\phi\left(F[\phi]\frac{\delta S}{\delta{\phi}}\right)e^{iS[\phi]} $$

where $Z[0]$ is just the normalizing vacuum to vacuum amplitude. Then, assuming there are no non-trivial boundary conditions, the total functional derivative vanishes and we find the SDEs $$ \left\langle\frac{\delta F}{\delta{\phi}}\right\rangle=-i\left\langle F[\phi]\frac{\delta S}{\delta{\phi}}\right\rangle $$

where $\langle ...\rangle$ is shorthand for vacuum (covariantly) time-ordered expectation value.

Whereas the second method relies on redefining the field. That is, given $$ \frac{1}{Z[0]}\int\mathcal{D}\phi\, F[\phi]e^{iS[\phi]}. $$

One can take $$\phi(x)\to\phi(x)+\epsilon\, \eta(x),$$ and expand the path integral to first order in $\epsilon$. What one will find is that the zeroth order term saturates the equality, leaving us only with the first order term in the expansion $$ \epsilon\int d^4 x\,\eta (x)\left[\left(\frac{1}{Z[0]}\int \mathcal{D}\phi\frac{\delta F}{\delta \phi(x)}e^{iS[\phi]}\right) +i\left(\frac{1}{Z[0]}\int\mathcal{D}\phi \frac{\delta S}{\delta \phi(x)} e^{iS[\phi]}\right)\right]=0. $$

And because this should vanish for any choice of $\eta(x)$, the integrand must vanish identically, leading us to the SDEs again.

My question is about the assumptions necessary for these to hold. In particular, the first one requires trivial boundary conditions, whereas the second appears to not require this at all. I have no issue with the fact that in canonical quantization the SDEs come from an entirely different place, after all, the formalism is inherently different, but things should be consistent given a formalism. Maybe the trivial boundary condition requirement is hidden somewhere in the second method and I'm just not seeing it?

Answer 1

OP asks a very good question.

1. Consider for simplicity first a toy model where spacetime is just a single point, i.e. the spacetime dimension $d=0$ is zero. Then, the path integral is just an ordinary integral $$\begin{align} Z~:=~&\int_{\mathbb{R}}\!\mathrm{d}\phi~ B(\phi), \cr B(\phi)~:=~&\exp\left\{\frac{i}{\hbar}S(\phi)\right\}, \end{align}\tag{1}$$ and the expectation value is $$ \langle F(\phi)\rangle ~:=~\frac{1}{Z}\int_{\mathbb{R}}\!\mathrm{d}\phi~F(\phi)B(\phi).\tag{2}\label{eq:2}$$

2. Let us consider an infinitesimal field redefinition $$ \phi^{\prime}-\phi~=~\delta\phi ~=~\epsilon~Y(\phi),\tag{3}$$ where $\epsilon$ is an infinitesimal parameter.

3. Using integration by substitution (IBS) the infinitesimal change of integration variable yields $$\begin{align} 0~=~&\frac{1}{\epsilon}\int_{\mathbb{R}}\! \mathrm{d}\phi^{\prime}~ B(\phi^{\prime})F(\phi^{\prime}) -\frac{1}{\epsilon}\int_{\mathbb{R}}\!\mathrm{d}\phi~ F(\phi)B(\phi)\cr ~=~&\frac{1}{\epsilon}\int_{\mathbb{R}}\!(\mathrm{d}\phi^{\prime} -d\phi)~F(\phi)B(\phi)\cr ~+~&\frac{1}{\epsilon}\int_{\mathbb{R}}\!\mathrm{d}\phi~ \{F(\phi^{\prime})B(\phi^{\prime}) -F(\phi)B(\phi)\}\cr ~=~&\int_{\mathbb{R}}\!\mathrm{d}\phi~\left\{\frac{\partial Y(\phi)}{\partial\phi}F(\phi)B(\phi) +Y(\phi)\frac{\partial\{F(\phi)B(\phi)\}}{\partial\phi}\right\}\cr ~=~&\int_{\mathbb{R}}\!\mathrm{d}\phi~\frac{\partial \{Y(\phi)F(\phi)B(\phi)\}}{\partial\phi}\cr ~=~&Z\left< B(\phi)^{-1}\frac{\partial \{Y(\phi)F(\phi)B(\phi)\}}{\partial\phi}\right>\cr ~=~&\left[Y(\phi)F(\phi)B(\phi) \right]_{{\phi=-\infty}}^{\phi=\infty}. \end{align}\tag{4}\label{eq:4}$$ Note that the 2nd-last expression of eq. \eqref{eq:4} is (one of the sides of) the Schwinger-Dyson (SD) equations $$\begin{align} \left< \frac{\partial \{Y(\phi)F(\phi)\}}{\partial\phi}\right>&\cr+\frac{i}{\hbar}\left< Y(\phi)F(\phi)\frac{\partial S(\phi)}{\partial\phi}\right>&~=~0,\end{align}\tag{5} $$ while the 3rd-last expression of eq. \eqref{eq:4} is the starting point for the integration by parts (IBP) proof.

4. Now, let us address OP's question. The last expression of eq. \eqref{eq:4} is the boundary terms. On one hand, in the IBP proof of the SD equations, we assume that the last expression of eq. \eqref{eq:4} vanishes. This implies that the limits $$ \lim_{\phi\to-\infty}Y(\phi)F(\phi)B(\phi) \quad\text{and}\quad \lim_{\phi\to\infty}Y(\phi)F(\phi)B(\phi)\tag{6}$$ exist (and are equal). On the other hand, in the IBS proof of the SD equations, for the first equality of eq. \eqref{eq:4} to make sense, we need the integral \eqref{eq:2} to be convergent. This, in turn, restricts$^1$ the behavior of the integrand $F(\phi)B(\phi)$ as $\phi\to\pm\infty$. Hence, there is a lot less (if practically any) daylight between the mathematically valid applications of the IBS and IBP proofs than one would perhaps initially think. This is our main answer. $\Box$

5. In the remainder of this answer, we sketch for completeness the case with a $d$-dimensional spacetime. This is technically more challenging, but the main idea remains the same.

6. Define the Boltzmann factor $$\begin{align}\exp\left\{\frac{i}{\hbar}S[\phi]\right\} ~:=~&B[\phi]~:=~\prod_{x\in\mathbb{R}^d}b(\phi(x),x),\cr b(\phi(x),x)~:=~&\exp\left\{\frac{i}{\hbar}d^dx~ {\cal L}(\phi(x),x)\right\}, \end{align}\tag{7}$$ where the product $\prod\limits_{x\in\mathbb{R}^d}$ is a sufficiently fine discretization of spacetime $\mathbb{R}^d$.

7. Let us assume that the $F$-functional is of the form $$ F[\phi]~=~\prod_{x\in\mathbb{R}^d}f(\phi(x),x). \tag{8}$$

8. Let us define the path integral $$Z~:=~\int \!{\cal D}\phi~B[\phi]~=~\prod_{x\in\mathbb{R}^d} \left[\int_{\mathbb{R}} \!\mathrm{d}\phi(x)~b(\phi(x),x) \right]\tag{9}$$ and the expectation value $$\begin{align} Z\langle F[\phi] \rangle ~:=~&\int \!{\cal D}\phi~F[\phi]~B[\phi]\cr ~=~&\prod_{x\in\mathbb{R}^d} \left[\int_{\mathbb{R}} \!\mathrm{d}\phi(x)~f(\phi(x),x)~b(\phi(x),x) \right].\end{align} \tag{10} $$

9. Let us, for simplicity, only consider ultra-local infinitesimal field redefinitions $$ \phi^{\prime}(x)-\phi(x)~=~\delta\phi(x) ~=~\epsilon~Y(\phi(x),x).\tag{11}$$

10. The infinitesimal change of integration variables yields $$\begin{align} 0~=~&\frac{1}{\epsilon}\int \!{\cal D}\phi^{\prime}~F[\phi^{\prime}]B[\phi^{\prime}] -\frac{1}{\epsilon}\int\!{\cal D}\phi~F[\phi]B[\phi]\cr ~=~&\frac{1}{\epsilon}\int \!({\cal D}\phi^{\prime} -{\cal D}\phi)~F[\phi]B[\phi]\cr ~+~&\frac{1}{\epsilon}\int\!{\cal D}\phi~(F[\phi^{\prime}]B[\phi^{\prime}] -F[\phi]B[\phi])\cr ~=~&\int \!{\cal D}\phi~F[\phi]B[\phi] \sum_{x\in\mathbb{R}^d}\cr &\quad\left[\frac{\partial Y(\phi(x),x) }{\partial\phi(x)} +Y(\phi(x),x)\frac{\partial \ln\{f(\phi(x),x)b(\phi(x),x)\}}{\partial\phi(x)}\right]\cr ~=~&\int \!{\cal D}\phi~F[\phi]~B[\phi] \int_{\mathbb{R}^d}\!d^dx\cr &\quad\left[\delta^d(0)\frac{\partial Y(\phi(x),x) }{\partial\phi(x)} +Y(\phi(x),x)\frac{\delta\ln\{F[\phi]B[\phi]\} }{\delta\phi(x)}\right]\cr ~=~&\int \!{\cal D}\phi~ \sum_{x\in\mathbb{R}^d} \underbrace{\frac{F[\phi]B[\phi]}{f(\phi(x),x)b(\phi(x),x)} }_{\text{indep. of }\phi(x)} \frac{\partial\{Y(\phi(x),x)f(\phi(x),x)b(\phi(x),x)\} }{\partial\phi(x)}\cr ~=~& \sum_{x\in\mathbb{R}^d} \left[\prod_{y\in\mathbb{R}^d\backslash\{x\}} \int_{\mathbb{R}}\!\mathrm{d}\phi(y)~f(\phi(y),y)b(\phi(y),y)\right]\cr &\quad \int_{\mathbb{R}}\!\mathrm{d}\phi(x) \frac{\partial\{Y(\phi(x),x)f(\phi(x),x)b(\phi(x),x)\} }{\partial\phi(x)}\cr ~=~& \sum_{x\in\mathbb{R}^d} \left[\prod_{y\in\mathbb{R}^d\backslash\{x\}} \int_{\mathbb{R}}\! \mathrm{d}\phi(y)~f(\phi(y),y)b(\phi(y),y)\right]\cr &\quad \left[Y(\phi(x),x)f(\phi(x),x)b(\phi(x),x)\} \right]_{{\phi(x)=-\infty}}^{\phi(x)=\infty}\cr ~=~&\int \!{\cal D}\phi~\sum_{x\in\mathbb{R}^d} \frac{\delta\{Y(\phi(x),x)F[\phi]B[\phi]\} }{\delta\phi(x)}\cr ~=~&Z\left< \sum_{x\in\mathbb{R}^d}B[\phi]^{-1} \frac{\delta\{Y(\phi(x),x)F[\phi]B[\phi]\} }{\delta\phi(x)}\right>.\tag{12}\label{eq:12} \end{align}$$ Note that the last expression of eq. \eqref{eq:12} is (one of the sides of) the SD equations $$\begin{align} \left< \sum_{x\in\mathbb{R}^d} \frac{\delta\{Y(\phi(x),x)F[\phi]\} }{\delta\phi(x)}\right>&\cr +\frac{i}{\hbar}\left< \sum_{x\in\mathbb{R}^d}Y(\phi(x),x)F[\phi] \frac{\delta S[\phi]}{\delta\phi(x)}\right>&~=~0, \end{align}\tag{13}$$ while the 2nd-last expression of eq. \eqref{eq:12} is the starting point for the IBP proof. This is similar to the $d=0$ case \eqref{eq:4}. $\Box$

TL;DR: The IPB and the IBS proofs of the SD equations are practically equivalent.

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$^1$ Be aware of pathological counterexamples a la this Phys.SE post. A semi-classical argument is that boundary contributions are deeply off-shell, and hence do not contribute significantly to the path integral.