# Deriving Schwinger-Dyson equation in the book Gauge/Gravity Duality: Foundations and Applications

In the book Gauge/Gravity Duality: Foundations and Applications by M. Ammon and J. Erdmenger they derive the Schwinger-Dyson equation by considering the generating functional $$J[Z]$$ and a change of variables $$\phi(x) \rightarrow \widetilde{\phi}(x)= \phi(x)+\delta \phi(x)$$, with $$\delta\phi(x)$$ being an arbitrary infinitesimal shift. This leads to the identity $$0=\delta Z[J]=i\int D\phi e^{i(S+\int d^dJ(x)\phi(x)}\int d^dx \left(\frac{\delta S}{\delta\phi(x)}+J(x) \right)\delta \phi(x)$$ The next step is then to take functional derivatives with respect to $$J(x_i)$$ and setting $$J$$ to zero. The book notes the result in eq. 1.234 as $$0=i\left\langle \frac{\delta S}{\delta\phi(x)}\phi(x_1)...\phi(x_n)\right\rangle +\sum_{j=1}^n\left\langle \phi(x_1)...\phi(x_{j-1})\delta(x-x_j)\phi(x_{j+1})...\phi(x_n) \right\rangle$$ However when I try calculate the functional derivatives I find (and I now go to the specific case of $$n=2$$) $$0=\frac{\delta^2}{\delta J(x_1) \delta J(x_2)} \left[i\int D\phi e^{i(S+\int d^dJ(x)\phi(x)}\int d^dx \left(\frac{\delta S}{\delta\phi(x)}+J(x) \right)\delta \phi(x) \right]_{J=0}$$

$$=\frac{\delta}{\delta J(x_1)} \left[i\int D\phi e^{i(S+\int d^dJ(x)\phi(x)}\int d^d\delta(x-x_2)\phi(x)\int d^dx \left(\frac{\delta S}{\delta\phi(x)}+J(x) \right)\delta \phi(x)\\+i\int D\phi e^{i(S+\int d^dJ(x)\phi(x)}\int d^dx \delta^d(x-x_2)\delta \phi(x) \right]_{J=0}$$

$$=\left[i\int D\phi e^{i(S+\int d^dJ(x)\phi(x)}\left( \left(\frac{\delta S}{\delta \phi(x)}+J(x)\right)\delta\phi(x)\phi(x_1)\phi(x_2)+\delta\phi(x_1)\phi(x_2)+\phi(x_1)\delta\phi(x_2) \right)\right]_{J=0}$$

$$=i\int d^dx\left\langle \frac{\delta S}{\delta\phi(x)}\delta\phi(x)\phi(x_1)\phi(x_2)\right\rangle +\left\langle \phi(x_1)\delta\phi(x_2)\right\rangle+\left\langle \delta\phi(x_1)\phi(x_2)\right\rangle$$ Comparing this result with the book I see that there are some unwanted $$\delta\phi(x_i)$$'s as well as an integral that is also not in the equation in the book. I am almost certain that this result is true, but there is a missing final step that I am not realizing. I have tried taking the functional derivative $$\frac{\delta}{\delta(\delta\phi(y))}$$, which would at the very least give a dirac-delta function in all the terms as well as getting rid of all the shifted fields $$\delta\phi(x_i)$$. $$0=\frac{\delta}{\delta(\delta\phi(y))}\left[i\int d^dx\left\langle \frac{\delta S}{\delta\phi(x)}\delta\phi(x)\phi(x_1)\phi(x_2)\right\rangle +\left\langle \phi(x_1)\delta\phi(x_2)\right\rangle+\left\langle \delta\phi(x_1)\phi(x_2)\right\rangle\right]$$ $$=i\int d^dx\left\langle \frac{\delta S}{\delta\phi(x)}\delta^d(y-x)\phi(x_1)\phi(x_2)\right\rangle +\left\langle \phi(x_1)\delta(y-x_2)\right\rangle+\left\langle \delta(y-x_1)\phi(x_2)\right\rangle$$

Howvever its still doesn't completely work as I am now missing a field $$\phi(x_i)$$ in the second and third term. Where am I going wrong here?

The first equation in the question holds for arbitrary infinitesimal $$\delta\phi(x)$$, so it implies $$0 = \int D\phi\ e^{i\left(S+\int J\phi\right)} \left(\frac{\delta S}{\delta\phi(x)}+J(x)\right). \tag{1}$$ If you calculate functional derivatives of this with respect to $$J$$, then everything should work out.
By the way, equation (1) is just the fundamental theorem of calculus: $$0 = \int D\phi\ \frac{\delta}{\delta\phi(x)} e^{i\left(S+\int J\phi\right)}. \tag{2}$$