Vacuum expectation value of scalar field in QFT I have some confusion about the vacuum expectation value of scalar field in QFT.
I know that the one-point function
$$
\langle \Omega | \phi(x) | \Omega \rangle = \langle \Omega | \phi(0) | \Omega \rangle =: v
$$
is a constant, which follows immediately by translation properties of $\phi$, that is
$$
\phi(x) = e^{i P x} \phi(0) e^{-iPx}.
$$
Q1: If one would like to determine $v$, should one just calculate the one-point function (that means the tadpole diagrams) in perturbation theory? Or is there an easier (or exact) way to get it?
Q2: Also I have heard that one can set the one-point function to zero by making a field redefinition $\phi \rightarrow \phi - v$ (see e.g. Why can the $1$-point correlation function be made to vanish?) and therefore one can safely consider it zero while happily continuing with the same theory. I see that it sets $v \rightarrow 0$, but it also massively changes the theory, since is may generate additional interaction terms in the Lagrangian? I need some clarification.
 A: *

*Yes, in order to determine $v$ in the original field's formulation you would need to calculate the $1$-point correlation function, and in the interacting theory this would mean summation over all diagrams.

*However, you are free to perform the field redefinition $\phi \rightarrow \phi - v$. Fields by themselves are not observable quantities, the choice of whether to use field $\phi$ or some other $\chi$ is purely a matter of convenience and convention. The physically observable is $S$-matrix, which encodes in itself amplitudes of all scattering processes $i \rightarrow f$, where $i$ is the initial state, and $f$ is the final.

Field redefinition can slightly change the action $S^{'}[\phi] = S[\phi + v]$, vertices with a certain power of the field may emerge of disappear, but this would be the only change since replacement $\phi \rightarrow \phi + v$ has a unit Jacobian and measure in the functional integration is preserved.
For some practical purposes, this redefinition can be inconvenient, however for theoretical purposes one can assume that $\langle \phi \rangle = 0$ without loss of generality.
