# In spontaneous symmetry breaking, can we define quanta of $\phi$ field?

A simple Lagrangian where spontaneous symmetry breaking is possible is that of a real scalar field $$\phi$$ with quartic interaction $$\mathcal{L}=\frac{1}{2}(\partial\phi)^2-\frac{\mu^2}{2}\phi^2-\frac{\lambda}{4}\phi^4$$ and $$\mu^2<0$$. We say that spontaneous symmetry breaking signals a nonzero VEV $$\langle \Omega|\hat{\phi}|\Omega\rangle=v\neq 0$$ of the quantum field $$\hat{\phi}$$.

• In this Lagrangian, can we declare $$\phi$$ to be a quantum field and replace it by $$\hat{\phi}$$?

• If yes, how will you define the quanta of $$\hat{\phi}$$-field? How will you define its creation and annihilation operators? $$|\Omega\rangle$$ does not seem to be the vacuum from which quanta of $$\phi$$ field can be created/defined.

Note 1 To be absolutely clear let me emphasize that I'm aware of the usual technique of redefining the field $$\hat{\phi}$$ by subtracting out the VEV $$\hat{h}\equiv\hat{\phi}-v$$ so that $$\langle \Omega|\hat{h}|\Omega\rangle=0.$$ Then quanta of $$h$$-field can be defined by $$\hat{h}^\dagger|\Omega\rangle$$. The question is not about this.

Note 2 When I said

"In this Lagrangian, can we declare $$\phi$$ to be a quantum field and replace it by $$\hat{\phi}$$ ?"

I didn't mean to ask if I can plug $$\hat{\phi}$$ in place of $$\phi$$ in $$\mathcal{L}$$. Sorry for the confusing language. I wanted to ask if we can quantize this theory without changing $$\phi$$ to $$h$$ i.e. do everything with $$\phi$$. I know that we can quantize $$\phi^4$$ theory. But with $$\mu^2<0$$, it's no longer a replica of the standard $$\phi^4$$ theory.

In this Lagrangian, can we declare $$\phi$$ to be a quantum field and replace it by $$\hat{\phi}$$?

Yes, absolutely. You can quantize any field you want, whether it's $$\phi$$, $$\phi - v$$, $$\phi - 2v$$, and so on.

If yes, how will you define the quanta of $$\hat{\phi}$$-field? How will you define its creation and annihilation operators? $$|\Omega\rangle$$ does not seem to be the vacuum of from which quanta of $$\phi$$ field can be created/defined.

Recall that this entire machinery works for free field theories. In order to get it to work in general, we use interaction picture, where the field operators are treated as free. In this context, the quantization of $$\phi$$ runs in exactly the same way as for a typical, free massive scalar field.

The only difference is in the state. Because the vev of $$\hat{\phi}$$ isn't zero, there are a very large number of $$\phi$$ excitations present (as defined through the previous paragraph), in the zero-momentum (spatially uniform) mode. So if you wanted to compute anything, such as the scattering cross section of two $$\phi$$ excitations with nonzero momentum, it would get very complicated because you would have to account for the possibility of these excitations interacting with the many $$\phi$$ particles already there.

That's the point of defining the field $$\hat{h} = \hat{\phi} - v$$. In the vacuum, there aren't any $$\hat{h}$$ particles present, so you don't get this problem.

Can you elaborate this a bit?

Note that the annihilation operator for the spatially uniform mode associated with the $$\hat{h}$$ field annihilates the vacuum. Therefore, the annihilation operator for the spatially uniform mode associated with the $$\hat{\phi}$$ field multiplies the vacuum by $$v$$. This is the definition of a coherent state, so there are a lot of $$\hat{\phi}$$-particles, but the number isn't determinate. (Of course, this is not very rigorous, because it's all in the naive interaction picture formalism.)

• "Because the vev of $\hat{\phi}$ isn't zero, there are a very large number of $\phi$ excitations present (as defined through the previous paragraph), in the zero-momentum (spatially uniform) mode." Can you elaborate this a bit? I think that $|\Omega\rangle$ is not a number eigenstate of $\hat{\phi}$ but that of $\hat{h}$ with zero $h$-type particles. I am not sure though. Feb 18 '20 at 20:16
• @mithusengupta123 Sure, added! Feb 18 '20 at 23:05
• "we use interaction picture, where the field operators are treated as free." Is this true? Aren't the fields are free only at $t\to \pm \infty$? Feb 21 '20 at 6:37
• @mithusengupta123 In the usual "adiabatic turn-off" picture, the Heisenberg picture fields at free only at $t \to \pm \infty$. But the interaction picture fields are always free no matter what, because that's just the definition of interaction picture. Feb 21 '20 at 6:38
• In the $\phi^4$ theory, the Heisenberg picture field $\phi_H$ cannot be expanded in creation and annihilation operators as $\phi_H\sim a+a^\dagger$. That I understand. Are you saying that the interaction picture field $\phi_I$ can be expanded as $\phi_I\sim a+a^\dagger$? Feb 21 '20 at 6:49