Steven Weinberg on p.167 in his second volume of QFT's book (in section about spontaneously broken symmetries, in subsection about Goldstone bosons) states (which we paraphrase as)

If we have linear transformation of fields$$ \varphi^{m} \to \varphi^{m} + i\varepsilon \sum_n t^{mn}\varphi_{n}, \tag{19.2.1}$$under which classical action and path integral measure is invariant, then quantum effective action$\Gamma [\varphi ]$ is also invariant under$(19.2.1)$.

Then he says that let's consider only case when theory is translational-invariant and fields are constant in space-time. The next calculations are based on this assumption.

But did he discuss only this case? Many results about Goldstone bosons which are obtained in case of constant fields be generalized in an arbitrary case? If no, why Weinberg assumes only this case?

Edit. It seems that the answer is following. We consider the vacuum state of the theory, i.e., stationary point of corresponding action. Since theory is translation invariant, stationary point doesn't depend on coordinates.

Reference: Weinberg QFT Vol. 2 p. 167 eq. (19.2.1).

1 Answer1

Start asking to get answers

Find the answer to your question by asking.

Explore related questions

See similar questions with these tags.