# Spontaneous symmetry breaking, massless bosons and the equations of motion

I am currently studying spontaneous symmetry breaking, and I don't entirely understand the implications of what we are doing at certain places. Consider the standard complex scalar field with the $$\phi^4$$ term, so that the Lagrangian density $$\mathcal{L}$$ looks like this: $$\mathcal{L} = \partial_{\mu} \phi\partial^{\mu}\phi^{*} - \mu^2\phi \phi^* - \lambda(\phi\phi^*)^2$$ It is clear that this lagrangian has a $$U(1)$$ symmetry. Further, if $$\mu^2 < 0$$ it can be shown that the potential term $$V(\phi) := \mu^2\phi \phi^* + \lambda(\phi\phi^*)^2$$ has minima in a ring around the origin in the $$\phi-\phi^*$$ plane at a distance $$\sqrt{-\mu^2/2\phi}$$. Most authors conclude here that there is some sort of spontaneous symmetry breaking, because the Lagrangian when expanded around one of these minima will not be $$U(1)$$ symmetric any further. Further, there is the appearance of a massless boson in such an expansion.

However, at the same time, the field obeys the Euler-Lagrange equation $$\frac{\partial \mathcal{L}}{\partial \phi} = \partial_{\mu} \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)}$$ and so our $$\phi$$ is actually evolving.

So my question is what does it mean for us to have a massless boson appear in our lagrangian? Let us say our field starts off in the initial state of one of the minima (the ground state); then the Euler-Lagrange equation causes the field to evolve so that it is no longer at this minima, and is, say at some other point where the symmetry is not broken. What happens to this masless boson term?

• – Cosmas Zachos Jan 24 at 15:22
• The equations of motion will make the field oscillate up and down the walls of the trough and hence the vacuum, but these are the massive "Higgs/σ excitations", and will not restore symmetry. However, the symmetry "generators" will shift your original vacuum to its degenerate neighbor having pumped a Goldstone boson into it, traveling at the speed of light. These shifts commute with the goldston equations of motion and were not generated by them. – Cosmas Zachos Jan 24 at 15:31
• Also linked. – Cosmas Zachos Jan 24 at 15:36
• @CosmasZachos Thank you so much for your reply and links :-) – Rahul Arvind Jan 24 at 17:27
• Done! sorry i was busy with some report I had to finish up. – Rahul Arvind Jan 27 at 7:27

It is easiest to see this in the polar parameterization of the fields, $$\phi (x) = (\sigma (x)+v)e^{i\theta (x)/v}, \qquad v\equiv\sqrt{-\mu^2/2\lambda}~~,$$ for real fields θ and σ canonically normalized. The lagrangian is now easier to understand, $${\cal L}= \partial \sigma \cdot \partial \sigma -4\lambda v^2 \sigma^2 +\partial\theta \cdot \partial \theta \\ +\partial\theta \cdot \partial \theta \left ( 2\frac{\sigma}{v} +\frac{\sigma^2}{v^2} \right )- \lambda (\sigma^4 -v^4 + v\sigma^3),$$ where you separated the kinetic/mass terms from the interaction terms on the second line. It is evident the σ is massive and the goldston θ is massless, and, crucially couples to "everybody" via derivatives ("Adler zeros"), so the couplings vanish at zero momenta and energies.
Under the U(1) symmetry transformation, the σ is invariant, and, by E-L, it oscillates up and down the walls of the potential, around the vacuum; whereas the goldston shifts by a constant along the trough, $$\delta \theta(x)= \epsilon,$$ the hallmark of the nonlinear realization involved in SSB: $$\langle \delta \theta(x) \rangle=\epsilon\neq 0$$, even though $$\langle \sigma (x)\rangle = \langle \theta (x)\rangle = 0\rangle$$. The conserved Noether current is, naturally, proportional to the gradient of the goldston, $$J_\mu= \partial_\mu \theta + \partial_\mu \theta \left ( 2\frac{\sigma}{v} +\frac{\sigma^2}{v^2} \right ),$$ so the leading term will pump goldstons into and out of the vacuum. (Some picture this symmetry-varying vacuum state as some sort of a coherent state.)