2
$\begingroup$

Suppose I study a quantum field theory in which among other fields a shift symmetric scalar field appear: $$\phi\rightarrow\phi+c$$ with $c$ a real constant.

Can this always be interpreted as a Nambu-Goldstone boson for some spontaneous symmetry breaking?

I would say that shift symmetry is a necessary condition to have the non linear realization of the spontaneously broken group: it corresponds to the case in which the parameter of the transformation is taken to be a constant and therefore the gauge fields do not vary, but I am not a lot sure this is exactly how it works, if it works.

Is there any weaker or stronger statement that can be generally made about shift symmetric scalars in order to connect their symmetry to larger symmetries of the theory?

$\endgroup$
1
  • $\begingroup$ if you allow another scalar $a$ with the same shift symmetry, you can add a potential, e.g. $V=(\phi+a)^2$. Then the goldstone boson is not $\phi$ but rather a combination $\phi-a$. But I suppose $\phi$ in one way or another is always related to the would be goldstone. $\endgroup$
    – Kosm
    May 18, 2019 at 16:14

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.