Weinberg on p. 287 of his QFT vol. 1 introduces the extended interaction operator:

\(\tag 1 \hat{V}(t) \to \hat{V}(t) + \sum_{a}\int d^{3}\mathbf x \hat{o}_{a}(\mathbf x ,t)\varepsilon_{a}(x).\)

Here

\(\hat{S} = \hat{T}e^{i\int \hat{V}(t)dt}, \quad \hat{o}_{a}(\mathbf x, t) = e^{i\hat{H}_{0}t}\hat{o}_{a}(\mathbf x , 0)e^{-i\hat{H}_{0}t}.\)

Then he says that S-matrix arbitrary element $S_{\beta \alpha} = \langle \beta | \hat{S}| \alpha \rangle$ after extension $(1)$ becomes the $\varepsilon $-functional, and after that he introduces generalized Feynman rules by adding new vertexes corresponding to $\hat{o}_{a}$ with $n_{a}$ lines (the number $n_{a}$ coincides with the number of fields in $\hat{o}_{a}$) and c-factor $\varepsilon_{a}$.

After that he introduces variational derivative

\( \tag 2 \left( \frac{\delta^{r}S_{\beta \alpha}}{\delta \varepsilon_{a_{1}}(x_{1})...\delta \varepsilon_{a_{r}}(x_{r})}\right)_{\varepsilon = 0} = (i)^{r}\langle \beta | \hat{T}\left(e^{i\int \hat{V}(t)dt} \hat{o}_{a_{1}}(x_{1})...\hat{o}_{a_{r}}(x_{r})\right)|\alpha\rangle\)

and notices that all $n_{a_{1}}, ..., n_{a_{r}}$ lines correspond to $\hat{o}_{a_{1}},...,\hat{o}_{a_{r}}$ respectively are internal, i.e. in case when $n_{a_{1}} = ... = n_{a_{r}} = 1$ they are compared to the propagators.

Finally, he says, that if we want to get Feynman diagram with $r$ external lines with types $a_{1},...,a_{r}$ in momentum representation we need to do following with $(2)$:

1) to throw out of propagators $D_{a_{1}a_{r}}(x_{1} - x_{r})$,

2) to apply the Fourier transformation,

3) to add corresponding coefficient functions $u_{a_{1}},...$.

Here is the question: сould you make the sense of introduction of mechanism 1)-3) clearer for me? Why do we need additional $r$ external lines which with corresponding vertexes which aren't connected to other vertexes (so the diagram is non-connected), as I think?

This post imported from StackExchange Physics at 2014-08-13 08:20 (UCT), posted by SE-user Andrew McAddams