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Here notation used functional matrix notation. The spacetime variables $x_1,x_2,...$ will be denoted by $1,2,...$. For example $\int dx_1$ will be denoted by $\int_1 $ and the fermion propagator $S(x_1,x_2)$ will be denoted by $S_{12}$ and so on.

Let $Z$ be the partition function for qed, $\mathcal A$ the action and $S^{-1}$ the inverse fermion propagator.

In this paper Recursive Graphical Construction of Feynman Diagrams in Quantum Electrodynamics its showed that

$$\int \mathcal{D}\bar{\psi}\mathcal{D}\psi\mathcal{D}A\Bigg\{\delta_{12}+\int_3{\bar{\psi_2}S^{-1}_{13}\psi_3}-e\int_{34}{V_{134}\bar{\psi_2}\psi_3}A_4\Bigg\}\exp{(-\mathcal A})=0 \tag{4.2}$$

Substituting the field product $\bar{\psi_2}$ $\psi_3$ by functional derivatives with respect to the electron kernel $S^{-1}_{13}$ they arrived to

$$\delta_{12}Z +\int_3{S^{-1}_{3} \frac{\delta}{S^{-1}_{23}}Z} -e\int_{34}{V_{134}\frac{\delta}{S^{-1}_{23}}}\bigg[\langle\hat A_4\rangle Z\bigg]=0 \tag{4.3}$$ My question is why they substituted the field $A_4$ by the expectation value $\langle\hat A_4\rangle$?

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I think the notation in OP is a bit mixed up.

If you stay with the partition function $Z$ framework, you should add local and bi-local source terms such as $$ \bar\psi j + \bar j\psi + \bar\psi\eta \psi + k_\mu A^\mu, $$ then in the functional integration you may substitute $$ \bar{\psi_2}\psi_3 \rightarrow \frac{\delta}{\delta\eta_{23}} $$ or $$ \bar{\psi_2}\psi_3 \rightarrow \frac{\delta}{\delta j_{2}} \frac{\delta}{\delta \bar j_{3}} $$ and $$ A^\mu \rightarrow \frac{\delta}{\delta k_\mu}. $$

If you really want to leverage the language of functional dirivatives on propagator $S(x_1,x_2)$ such as $\frac{\delta}{\delta S(x_1,x_2)}$, you have to do a proper Legendre transformation and use the language of effective action $\Gamma[S]$ and such, rather than the original partition function $Z$.

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Let $Z=\exp (W)$, since $\langle\hat A\rangle=\frac{\delta W}{\delta j}$ we have that $$\frac{\delta Z}{\delta j}=\frac{\delta W}{\delta j}Z \tag 1$$

on the other hand $$\frac{\delta Z}{\delta j}=AZ \tag 2$$

Comparing (1) and (2) we arrive at $$\langle\hat A\rangle=\frac{\delta W}{\delta j}=A$$

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