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Timeline for The proof of Goldstone's theorem

Current License: CC BY-SA 3.0

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Aug 16, 2017 at 0:55 comment added GotchaP When this symmetry is spontaneously broken, $\phi_0^i$ is fixed at one point of the surface(Actually they choose coordinates so that $\phi_0^i$ points in the $N$th direction.). So at first I thougt that when the symmetry is not spontaneously broken, $\Delta^a (\phi_0)\ne 0$ and when the symmetry is spontaneously broken, $\Delta^a (\phi_0)= 0$.
Aug 16, 2017 at 0:55 comment added GotchaP @AccidentalFourierTransform : In the same section of this book, they discuss the linear sigma model which involves a set of $N$ real scalar field $\phi^i(x)$: $$ \cal{L}=\frac{1}{2}(\partial_{\mu}\phi^i)^2 + \frac{1}{2}{\mu}^2(\phi^i)^2 - \frac{\lambda}{4}[(\phi^i)^2]^2 , \tag{11.5}$$ This Lagrangian has an $O(N)$ symmetry. When this symmetry is not spontaneously broken, $\phi_0^i$(a constant field which has lowest-energy) moves along the surface of the sphere $(\phi_0^i)^2=\frac{\mu^2}{\lambda}$ where the potential is minimized.
Aug 14, 2017 at 5:38 answer added flippiefanus timeline score: 2
Aug 14, 2017 at 0:31 answer added Cosmas Zachos timeline score: 5
Aug 13, 2017 at 8:57 comment added AccidentalFourierTransform Is OP perhaps asking what an invariance of $\phi_0$ has to do with an invariance of $|\Omega\rangle$, the ground state vector? SSB is [perhaps sometimes confusingly] stated in terms of invariance of the minimum of the potential, and invariance of the vacuum vector. Might this be the source of confusion here?
Aug 13, 2017 at 8:50 comment added ACuriousMind I don't know what sort of answer you're looking for: The definition 11.12 of $\Delta^a(\phi_0)$ means that $\Delta^a(\phi_0) = 0$ if and only if $\phi_0$ is invariant under the symmetry. The definition of SSB says that SSB occurs when $\phi_0$ is not invariant under the symmetry. There are no further steps, it's literally just comparing the two definitions.
Aug 13, 2017 at 7:48 history asked GotchaP CC BY-SA 3.0