As is easily checked, fields linear in creation and annihilation operators (and hence amenable to a particle interpretation) have zero vacuum expectation value. Thus the $\phi$ field with its nonvanishing vacuum expectation value cannot be given a particle interpretation. But the field $\psi=\phi-v$ has such an interpretation as its vacuum expectation value is zero. This works only if $v$ is the vacuum expectation of $\phi$.
Note that the field $\phi$ is and remains massless; it is the field $\psi$ that had acquired a mass term.
The 1-loop approximation to a quantum field theory is given by the saddle-point approximation of the functional integral. For that you have to expand around a stationary point, and for stability reasons this stationary point has to be a local minimizer. If the local minimum is not global, the vacuum state is metastable only; so one usually expands around the global minimizer.
A mass term breaks the scaling symmetry of a previously scale-invariant theory. It may or may not break other symmetries. In the above case, the symmetry $\phi\to-\phi~~~$ of the action is broken in the stable vacuum.
First of all, classically (neglecting loop corrections), we obviously want to expand around the true minima that are found in the vacuum, which means around one of the states $|0_\pm\rangle$ – these two are really equivalent to one another due to the gauge symmetry. The state around $\phi=0$ is a maximum of energy, not a minimum, so Nature doesn't spend much time over there before it rolls down to the true minimum.
If we expanded around a shifted scalar field, there would be, assuming that the Lagrangian is Taylor-expanded, also first-order terms in the scalar field. They would produce Feynman "vertices" with one external line connected to a "cross" (in which a Feynman diagram ends like an external line but isn't associated with an actual external particle). These simple decorations of Feynman diagrams could be resummed and their effect would be simple – effective shift the scalar field back to the minimum.
When loop corrections are taken into account, the true minimum isn't exactly given by the classical approximation and the vev isn't exactly as the location of the minimum, too. All these things get quantum corrections – suppressed formally by positive powers of $\hbar$ and more quantitatively by the small dimensionless values of the coupling constants.
The calculations in a quantum field theory shift the scalar field to the new, true minimum (incorporating quantum corrections) automatically. Do you remember I took about one-external-leg vertices of Feynman diagrams that we could eliminate by a clever shift? Well, when you consider loop diagrams, there is a similar subtlety – tadpole diagrams such as this one:
It's a one-loop diagram and the loop at the end plays the same role as the small "cross" indicating the direct Feynman diagram vertex from the beginning of the discussion. But even if we eliminate the linear terms from the action we start with, quantum loops effectively generate their own tadpoles that may be attached through other vertices to any Feynman diagram and that effectively shift the scalar field to the right minimum corrected by the corrections suppressed by powers of $\hbar$.
The LSZ formula doesn't break at all. If you correctly add all the Feynman diagrams, there will also be the Feynman diagrams with the tadpoles attached. They fixed the "internal" part of the Feynman diagram so that it knows about the true minimum. The external part of the LSZ formula is also fixed automatically because the particles in LSZ are connected with the creation action by a near-mass-shall mode of the scalar fields. And it doesn't matter whether you pick a Fourier component of $\phi$ or $\phi-c$ for a constant $c$ – only the actual field $c$ will get magnified near $k^2=m^2$ while the constant term is annihilated by $(k^2-m^2)$, anyway.
So:
- Yes, there is a linear term generated by quantum effects.
- All such quantum corrections are accounted for by summing over all the Feynman diagrams automatically.
- There is always some freedom about how you parameterize the fields etc. Aside from the simple linear shifts above, you may consider nonlinear redefinitions of the scalar fields or redefinitions that depend on derivatives of $\phi$ (and, therefore, indirectly on the energy-momentum vector of the quanta). These different approaches are all possible and they give you different "renormalization schemes", if I use the most general buzzword for such choices. All the measurable physical predictions will ultimately be independent of the renormalization scheme if you adjust the parameters correctly in a given scheme (the correct way depends on the scheme).
So you shouldn't worry about any of these things. The calculation has intermediate steps that are not unique but if you carefully follow your conventions, you don't have to do anything special and the summation over all the Feynman diagrams produces the right physical answers without any extra "cures".
Best Answer
We have the functional of the external source $J$, which gives us v.e.v.s of field operators, by functional differentiation: $$e^{-iE[J]} = \int {\cal{D}}\phi\, e^{iS[\phi]+iJ\phi} $$ $$\phi_{cl}=\langle\phi\rangle_J = -\frac{\delta E}{\delta J}$$ Where $\langle\phi\rangle_J$ is the v.e.v of $\phi$ in presence of external source $J$. That could be considered as a visible "response" of the system on the source and it usually denoted as a new variable, called the "classical field". We would like to find it when there are no external sources: $J=0$.
For that, one then does the Legendre transform trick, arriving at the effective action: $$\Gamma[\phi_{cl}] = - E - J\phi_{cl}\quad\quad\frac{\delta\,\Gamma}{\delta \phi_{cl}} = - J$$ Remembering our goal to find $\phi_{cl}$ at $J=0$, we arrive at the equation. $$\frac{\delta\,\Gamma}{\delta \phi_{cl}} = 0$$ Adding an extra assumption that $\phi_{cl}$ is space and time independent: $\phi_{cl}(x) = v$, the effective action functional $\Gamma[\phi_{cl}]$ is then reduced to effective potential $V_{eff}(v)$ and the equation becomes. $$\frac{dV_{eff}}{dv} = 0$$ Now, as David Vercauteren correctly pointed out, $V_{eff}(v)$ is not the same function as $V(\phi)$. But usually it is a good first approximation, because we usually consider systems where the "real" quantum field fluctuates weakly around its vacuum: $\phi(x)=v+\eta(x)$ with $\eta$ being small.