Feynman diagram representation of variational derivative of S-matrix For quite some time I am struggling to understand section 6.4 in Weinberg volume 1. He observes there that if interaction hamiltonian density is extended by coupling to c-number fields $\epsilon$,
$$ \mathcal H_{\mathrm{int}}(x) \mapsto \mathcal H_{\mathrm{int}}(x) + \epsilon(x) o(x), $$
where $o$ are some operators in the interaction picture then S matrix becomes a functional of $\epsilon$. This functional can be calculated using Feynman rules. That is quite clear. However he then claims that if we calculate variational derivative with respect to $\epsilon$ at $\epsilon=0$ we get sum of terms represented by Feynmann diagrams with only internal lines meeting at $o(x)$ vertices. I don't understand what is the reason for discarding diagrams with external lines flowing into the $o(x)$ vertices. Explicitly, I obtained the formula (which is also written down one page later in Weinberg)
$$ \left. \frac{\delta^r S_{\beta \alpha}[\epsilon]}{\delta \epsilon(y_1)...\delta\epsilon(y_r)} \right|_{\epsilon=0} = \sum_{n=0}^{\infty} (-i)^{n+r} \langle \beta | T \left\{ \prod_{i=1}^n \left[ \int \mathrm d x_i \mathcal H_{\mathrm{int}}(x_i) \right] o(y_1)...o(y_r) \right\} |\alpha \rangle . $$
It seems to me that field operators in $o(y)$ can be contracted with initial and final states, just as these in $\mathcal H_{\mathrm{int}}$. What is the difference here?
 A: In Feynman diagrams in coordinate representation, external lines are those with one end fixed (i.e. having fixed coordinate which does not take part in integrations) and the other end being internal vertex.
In your formula, if the operators $o(y_i)$ are of one-particle nature (i.e. contain $\Psi$ or $\Psi^+$ but not their products), then you have $r$ external lines starting from $y_1,\ldots,y_r$. See Fig. 1: it is an example for $r=4$, external lines are blue.
When the operators $o(y_i)$ are two-particle (for example, current operators like $\Psi^+\hat{J}\Psi$), we have rather external vertices with coordinates $y_1,\ldots,y_r$, each of them being a source for two external lines (see Fig. 2, external lines are blue).
As for the initial and final states $|\alpha\rangle$ and $\langle\beta|$: if they depend on its own coordinates, this can introduce additional external vertices to the diagram. For example, if $|\alpha\rangle=\Psi(z_\alpha)|0\rangle$, $|\beta\rangle=\Psi(z_\beta)|0\rangle$, you will get additional external vertices with the coordinates $z_\alpha$ and $z_\beta$. If $|\alpha\rangle=\Psi^+(z_\alpha)\Psi(z_\alpha)$, it will correspond to two-particle vertex, and so on.

