Why can the ##1##-point correlation function be made to vanish?

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spaghetti3451
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The ##1##-point correlation function in any theory, free or interacting, can be made to vanish by a suitable rescaling of the field ##\phi##.

I would like to understand this statement.

With the above goal in mind, consider the following theory:

$$\mathcal{L} = \frac{1}{2}\left((\partial\phi)^{2}-m^{2}\phi^{2}\right)+\frac{g}{2}\phi\partial^{\mu}\phi\partial_{\mu}\phi.$$

What criteria (on the Lagrangian ##\mathcal{L}##) is used to determine the value of the field ##\phi_{0}## such that the transformation ##\phi \rightarrow \phi + \phi_{0}## leads to a vanishing ##1##-point correlation function ##\displaystyle{\langle \Omega | T\{\phi(x_{1}\phi(x_{2})\}| \Omega \rangle}##?
 
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radium said:
You should look at the saddlepoint solution for the field from the path integral. When they mean shift the field it corresponds to expanding about the different vaccua.

Right!

When you expand about the vacua, shouldn't you get a new constant term in the Lagrangian?

Doesn't this constant term affect the correlation functions?
 
Well I think you are referring to explicitly choosing a ground state writing an effective theory in term of fluctuations about the new ground state. This would correspond to picking a phase for a complex scalar with U(1) symmetry breaking for example (an effective theory for a superfluid). Once you have chosen this vacuum you must stick with your choice since the ground state is no longer invariant under whatever symmetry (remember the action still is, the spontaneous symmetry breaking is because \langle \phi \rangle \neq 0 for certain values of the parameters.
 
A 1-point correlation function is simply the vacuum expectation value of the field. Subtract this expectation value from the field operator, and you get a shifted (not a rescaled!) field operator whose vacuum expectation value vanishes.