# 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?

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.
• If I have a particle in initial state, do I get diagrams with line of initial particle terminating at $o(y)$ vertex? – Blazej Sep 18 '16 at 19:30
• If I understood correctly the procedure of further calculations, initial and final states are eventually reduced just to several additional external operators. For example, if the initial state is a one-particle excited state like $|\alpha\rangle=\int dx\:f(x)\Psi^+(x)|0>$, then we get the "external" $\Psi^+(x)$. If $o(y)$ is also a one-particle operator, then the pairing $\langle To(y)\Psi^+(x)\rangle$ (corresponding to the line you are talking about) will result in a disconnected diagrams, which are usually cancelled. – Alexey Sokolik Sep 18 '16 at 20:44
• You are right that pairing of $o(y)$ with initial or final state will result in a disconnected diagram, unless we are talking about one particle to vacuum (or vice versa) transition. However, Weinberg explicitly states in his book that he is also considering the case of multi-particle operators. I think these should give genuine contributions to the connected part of $S$-matrix. – Blazej Sep 18 '16 at 23:27
• Honestly, I don't understand why do we need to use arbitrary initial and final states $|\alpha\rangle$ and $|\beta\rangle$ instead of averaging over vacuum or Fermi sphere (as in solid state physics) if we already introduced the operators $o(y)$ which can play the same role of external field sources at initial, final or intermediate times. Maybe it is more convenient for further calculations. – Alexey Sokolik Sep 19 '16 at 22:38