# Goldstone bosons when SSB potential has two fields

A theory consists of two complex scalar fields $$φ_0$$ and $$φ _1$$ with a symmetry-breaking potential $$V(|φ_0|^2 + |φ_1|^2).$$ How many Goldstone particles will there be in the theory?

• – Cosmas Zachos May 4 at 11:57
• Makes sense? ..... – Cosmas Zachos May 7 at 21:14

There are four degrees of freedom, $$\Re\varphi_0=\phi_1, \qquad \Im\varphi_0=\phi_2, \qquad \Re\varphi_1=\phi_3, \qquad \Im\varphi_1=\phi_4.$$ Your potential then amounts to
$$V(\vec{\phi}\cdot \vec{\phi}),$$ so, then, it has SO(4) symmetry, with 6 generators. Doesn't it look like the peerless paradigm of the $$\sigma$$-model you were taught in your QFT course?

The presumption, if there is to be SSB, is that this potential is minimized off of the field origin, e.g., $$V\propto ( \vec{\phi}\cdot \vec{\phi} -v^2)^2,$$ so at the vacuum, at this minimum, $$\langle \vec{\phi}\cdot \vec{\phi}\rangle =v^2$$.

Make the immaterial choice $$\langle \phi_1 \rangle=v$$, so you've now declared this the $$\sigma$$. The rest of the components have $$\langle \phi_{2,3,4} \rangle=0$$. So there is an unbroken SO(3) connecting these three components: doesn't it look like unbroken isospin acting on a triplet of pions?

By contrast, the remaining three generators must be SSBroken: they are the ones connecting these three components with the $$\sigma$$, as in low energy meson physics (the "broken chiral directions").

Redefining $$\phi_1'\equiv \phi_1-v$$, your potential now reads $$V\propto (\phi_1^{'~2}+ 2v\phi_1' +\phi_2^{2}+\phi_3^{2}+\phi_4^{2})^2.$$ It should be evident the $$\phi'_1$$ is massive, and the other three components are massless and Goldstone bosons: the three "pions" of $$SU(2)\times SU(2)/SU(2) \sim SO(4)/SO(3)$$ in low energy QCD, upon flavor chiral symmetry breaking, the prototype of what you are studying.

• In the impossibly pompous general treatment of formal QFT texts, you note $$\langle \delta \phi_1\rangle=0$$, but $$\langle \delta \phi_{2,3,4}\rangle\propto v$$, so the latter three components, $$\phi_{2,3,4}$$, transform to the nonvanishing null-eigenvectors of the Goldstone matrix $$\bbox[yellow]{\langle \delta ^2V/\delta \phi_i \delta \phi_j \rangle }$$ under the broken symmetries, thereby specifying the three goldstons found above: the v.e.v. of the transforms of the goldstons under the broken symmetry must be nonvanishing.

Section 1 of this might be helpful.

https://arxiv.org/pdf/1804.05664.pdf