Assuming that the functional integral of a functional derivative is zero, so

$$ \int \mathcal{D}[\phi] \frac{i}{\hbar}\left\{ \frac{\delta S[\phi]}{\delta \phi}+J(x) \exp \left[ {i \over \hbar} \left( S[\phi]+\int \! \mathrm{d}^4 x \, J(x)\phi(x)\right)\right] \right\}$$

Apart from this trivial equation what else do I need to prove the Dyson-Schwinger equations?

  • $\begingroup$ In addition to be incorrect, I think the equation is not even readable... I though an equation had two members, one on the left (called in a highly imaginative way the left-hand-side) and one on the right (called the right-hand-side). The most important term in between is the symbol $=$ which is the parangon of truth in mathematics. So in short, $A$ is not an equation, but $A=0$ is ... Thanks for correcting your question. $\endgroup$ – FraSchelle Mar 5 '15 at 8:10

To formally show the Schwinger-Dyson equations, use the fact the

$$\int [d\phi]\frac{\delta}{\delta \phi^{\alpha}(x)} \exp\left[\frac{i}{\hbar}\left(S[\phi]+\int \!d^nx^{\prime}~J_{\alpha}(x^{\prime})\phi^{\alpha}(x^{\prime}) \right)\right] ~=~0,$$

cf. this Phys.SE post. Without specifying the action $S[\phi]$ and field content $\phi^{\alpha}$ further, it is impossible to provide a rigorous proof, since the path integral is not properly defined in the first place. See also this related Phys.SE post.

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  • $\begingroup$ yes but how sould i to replace $ \phi (X) $ by a linear differential operator $ \frac{\delta }{\delta J} $ $\endgroup$ – Jose Javier Garcia Apr 23 '14 at 18:59

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