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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?

Qmechanic
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1 Answers1

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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.
Cosmas Zachos
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