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.