1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.