quantum field theory - Symmetry Breaking and Vacuum Expectation Values

Consider a real scalar field $\phi$ in a theory with a Lagrangian $$\mathcal{L}:=-\frac{1}{2}\partial _\mu \phi \partial ^\mu \phi -V(\phi ),$$ where $$V(\phi ):= -\mu ^2\phi ^2+\frac{\lambda}{4!}\phi ^4,$$ where both $\mu$ and $\lambda$ are positive real numbers.

We see that the potential has a couple of non-zero minima: $$V'(\phi )=-2\mu ^2\phi +\frac{\lambda}{3!}\phi ^3=0\Rightarrow \phi =0,\pm 2\mu \sqrt{\frac{3}{\lambda}}=:\pm V_0$$ (It turns out that $\phi =0$ is a local max, and $\phi =\pm V_0$ are local mins; check the second derivative.)

As the usual story goes, we must define a new field $\psi :=\phi -V_0$ and re-write the theory in terms of this $\psi$ to get the appropriate Feynman rules of the quantum theory. If I did my algebra correctly (the details aren't exactly relevant here anyways), this substitution gives us $$\mathcal{L}=-\frac{1}{2}\partial _\mu \psi \partial ^\mu \psi -2\mu ^2\psi ^2+\mu \sqrt{\frac{\lambda}{3}}\psi ^3+\frac{\lambda}{4!}\psi ^4-6\frac{\mu ^4}{\lambda}.$$ (Our Lagrangian no longer admits the symmetry $\psi \mapsto -\psi$, hence the term "symmetry breaking".)

Th question arises: Why is this substitution special? This form of the Lagrangian has some nice properties (namely that the potential has a local min at $0$), but surely there are some other substitutions that could given us some other nice properties as well. What about those?

My understanding of this was the following: The LSZ Reduction Formula, among other things, requires a priori that the fields one is working with have vanishing vacuum expectation value. Thus, when applying the LSZ formula, we must be working with $\psi$, not $\phi$, and so the appropriate Feynman rules can be read off only when the Lagrangian is written in terms of $\psi$. I have just recently discovered a problem with this explanation, however.

Before, I was under the impression that $\langle 0_\pm |\phi |0_\pm \rangle =\pm V_0$ (this theory evidently has two physical vacuums, whatever that precisely means), so that the definition of $\psi$ forces $\psi$ to have vanishing expectation value, so that the LSZ formula can be applied. However, I recently learned that $\pm V_0$ is only an approximation to $\langle 0_\pm |\phi |0_\pm \rangle$, which implies that $\psi$ only approximately has vanishing vacuum expectation value, which means that LSZ doesn't technically apply.

It seems that the proper substitution is in fact $\psi :=\phi -\langle 0_+|\phi|0_+\rangle$. There are several problems I see with this:

1. 
   

   The Lagrangian re-written in terms of $\psi$ should have a small, but non-zero, linear term in $\psi$.

   
   


2. 
   

   The Feynman rules I've been using all along that arise from the substitution $\psi :=\phi -V_0$ are only approximation.

   
   


3. 
   

   The coefficients that arise from the 'proper' substitution $\psi :=\phi -\langle 0_+|\phi |0_+\rangle$ are going to be written in terms or something that can (to the best of my knowledge) only be calculated perturbatively (namely $\langle 0_+|\phi |0_+\rangle$), but we need to know these coefficients to obtain the Feynman rules to begin with (resulting in a 'circularity' problem).

   
   

How does one go about resolving all these issues?

(Disclaimer: I asked a very similar question here not quite a year ago, but my understanding of the situation has improved since then, and as is usual, my improvement of understanding has only brought forth many more questions regarding this, so I felt it was appropriate to address the issue once again.)