# Variations of S-matrix functional and Feynman diagrams in Weinberg QFT

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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

edited Aug 13, 2014

@dimension10 @Dilaton, the equation (2) looks truncated to me, is it just my browser's problem or do you see the same thing?

@JiaYiyang yes, I see it too.

I have now improved it. If this happens too often when importing question such that it becomes annoying, we can Polarkernel tell about it ...

@Dilaton, thanks. I vaguely remember seeing it before in another post with a long formula, I thought it could be my own problem and didn't bother to report it.

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1) is a misunderstood statement, Weinberg did not mean that,and connectedness of the diagrams is simply not a relevant issue.Hopefully it will be clarified by what I am inputting below.

The result is clear if you think in terms of Feynman diagrams, let's set $|\alpha\rangle$ and $|\beta\rangle$ to be vacuum states since they are of no essential relevance here. Now, think about what the difference is between the Feynman diagrams of, say,$\langle 0 | \hat{T}\left(e^{i\int \hat{V}(t)dt} \hat{o}_{a_{1}}(x_{1})\hat{o}_{a_{2}}(x_{2})\right)|0\rangle$and  $\langle p_1,a_1 | \hat{T}e^{i\int \hat{V}(t)dt} |p_2,a_2\rangle$? They are visually the same! Both are diagrams with two external lines from field types $a_1$ and $a_2$. The only difference is what we assign to these lines: for the former each external line contributes a propagator, which are something like $D_{a_1 a_m}(x_1-x_m)$ and $D_{a_2 a_n}(x_2-x_n)$, where $a_m,x_m$ and $a_n,x_n$ represent internal vertices that $a_1,x_1$ and $a_2,x_2$ connect to respectively; while for the latter each contributes a coefficient function, which are $u^*_{a_1}(p_1)$ and $u_{a_2}(p_2)$.

Now it's obvious about how to get the latter from the former: all we need to do is to strip $D_{a_1 a_m}(x_1-x_m)$ and $D_{a_2 a_n}(x_2-x_n)$ away, go to momentum space, and replace them with the coefficient functions $u^*_{a_1}(p_1)$ and $u_{a_2}(p_2)$! That's all what Weinberg meant. You can easily generalize it to $n$ external line cases, e.g. from $\langle 0 | \hat{T}\left(e^{i\int \hat{V}(t)dt} \hat{o}_{a_{1}}(x_{1})\hat{o}_{a_{2}}(x_{2}){o}_{a_{3}}(x_{3})\right)|0\rangle$ you can get $\langle p_1,a_1 | \hat{T}e^{i\int \hat{V}(t)dt} |p_2,a_2;p_3,a_3\rangle$ or $\langle p_1,a_1;p_3,a_3 | \hat{T}e^{i\int \hat{V}(t)dt} |p_2,a_2\rangle$ and etc., depending on how you do the replacement with the coefficient functions.

PS. The above discussion is only valid when $\hat{o}(x)$ represent a single-field operator, i.e. the argument fails if  $\hat{o}(x)$ is a product of more than 1 field operators. This is exactly what Weinberg assumed in the corresponding page.

answered Aug 13, 2014 by (2,640 points)
edited Aug 13, 2014

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