Off the mass shell and external fields

[ismaili]

In Weinberg's book, he said "Feynman graphs with lines off the mass shell are just a special case of a wider generalization of the Feynman rules that takes into account the effects of various possible external fields."

Suppose the external fields are treated as a c-number, $$\epsilon_a(x)$$, and they coupled to the current operators $$o_a(t)$$ in this way that the potential looks like,
$$V_\epsilon(t) = V(t) + \sum_a\int d^3x\epsilon_a(\mathbf{x},t)o_a(\mathbf{x},t)$$
where $$V(t)$$ is the interaction used in the Dyson series.
Now he argued something that I don't understand:
"It follows then that the rth variational derivative of the scattering matrix $$S_{\beta\alpha}[\epsilon]$$ with respect to $$\epsilon_a(x), \epsilon_b(y)\cdots$$ at $$\epsilon=0$$ is given by position space diagrams with $$r$$ additional vertices, to which are attached some internal lines, and NO external lines."
I don't understand why there are no external lines attached to the vertices formed by $$o_a(x)$$? I think there is no reason why the external lines are ruled out?

Any help would be appreciated.
These materials are in the page 287 of Weinberg's book.
Thanks.

 

[olgranpappy]

In Weinberg's book, he said "Feynman graphs with lines off the mass shell are just a special case of a wider generalization of the Feynman rules that takes into account the effects of various possible external fields."

Suppose the external fields are treated as a c-number, $$\epsilon_a(x)$$, and they coupled to the current operators $$o_a(t)$$ in this way that the potential looks like,
$$V_\epsilon(t) = V(t) + \sum_a\int d^3x\epsilon_a(\mathbf{x},t)o_a(\mathbf{x},t)$$
where $$V(t)$$ is the interaction used in the Dyson series.
Now he argued something that I don't understand:
"It follows then that the rth variational derivative of the scattering matrix $$S_{\beta\alpha}[\epsilon]$$ with respect to $$\epsilon_a(x), \epsilon_b(y)\cdots$$ at $$\epsilon=0$$ is given by position space diagrams with $$r$$ additional vertices, to which are attached some internal lines, and NO external lines."
I don't understand why there are no external lines attached to the vertices formed by $$o_a(x)$$? I think there is no reason why the external lines are ruled out?

Any help would be appreciated.
These materials are in the page 287 of Weinberg's book.
Thanks.

hmm... it seems like you are right. Maybe Weinberg is assuming that the fields in the
$$o_a(x)$$
operators are never in the incoming or outgoing states.

 

[denian]

I would like to clarify and explain the concept of "off the mass shell" and external fields in the context of Feynman graphs and Feynman rules, as discussed by Weinberg in his book.

Firstly, "off the mass shell" refers to the virtual particles that are involved in the scattering process. These particles are not on their mass shell, meaning they do not have their usual energy and momentum values, but rather they are "off" their usual values due to their interaction with other particles. This is a common occurrence in particle interactions and is taken into account in Feynman diagrams.

External fields, on the other hand, refer to the external forces or interactions that are present in the system. These can be treated as c-numbers (complex numbers) and are represented by the term \epsilon_a(x) in the potential V_\epsilon(t). The external fields can be thought of as perturbations to the system, affecting the behavior of the particles involved in the scattering process.

Weinberg's statement about Feynman graphs with lines off the mass shell being a special case of a wider generalization of the Feynman rules is referring to the fact that the Feynman rules can be extended to include the effects of external fields. In other words, the Feynman rules can be modified to take into account the presence of external fields, in addition to the usual interactions between particles.

Now, let's consider Weinberg's argument about the rth variational derivative of the scattering matrix S_{\beta\alpha}[\epsilon] with respect to \epsilon_a(x), \epsilon_b(y)\cdots at \epsilon=0. This means taking the derivative of the scattering matrix with respect to the external fields at the point where the external fields are set to zero. In this case, the potential V_\epsilon(t) reduces to just the interaction term V(t).

Weinberg then states that the resulting Feynman diagrams will have r additional vertices, to which are attached some internal lines, but no external lines. This is because at \epsilon=0, the external fields are zero, meaning there are no external forces acting on the system. Therefore, the external lines, which represent these forces, are not present in the Feynman diagrams.

In summary, the reason why there are no external lines attached to the vertices formed by o_a(x) in the rth variational