# Why can the $1$-point correlation function be made to vanish? [duplicate]

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 $$\langle \Omega | \phi(x)| \Omega \rangle=0~?$$

The 1-point function is constant in spacetime because of translation invariance, i.e. $\langle \phi(x)\rangle = \phi_0\in\mathbb{R}$ for all $x\in\mathbb{R}^4$. Obviously, the 1-point function of $\phi'(x) := \phi(x) - \phi_0$ is zero since the expectation value is linear. So $\phi\mapsto \phi' = \phi + \phi_0$ gets rid of the non-zero 1-point function. This works for all Poincaré-invariant Lagrangians.
• Why does translation invariance mean that the $1$-point function is constant in spacetime? Also, what is $\phi_{0}$? Lastly, how can you prove that the shift $\phi\rightarrow \phi'=\phi+\phi_{0}$ does not affect the Lagrangian? Nov 17, 2016 at 15:07
• @failexam 1. $\phi_0$ is just the value of $\langle \phi(x)\rangle$. 2. Translation invariance means it's constant because $\langle \phi(x)\rangle = \langle \phi(x+a)\rangle$ (the fields $\phi(x)$ and $\phi(x+a)$ have the same Lagrangian, so the same expectation values). 3. It does affect the Lagrangian, I never said it didn't. After the redefinition, there will be no linear terms in $\phi$ left in it because $\phi_0$ corresponds to the classical minimum of the potential to first order, so the redefinition is equivalent to expanding around the minimum, where linear terms are absent. Nov 17, 2016 at 15:26
• You first define $\phi'(x)$ in your answer as $\phi'(x)=\phi(x)-\phi_{0}$ but later in your answer as $\phi' = \phi+\phi_{0}$. Is this valid? Also, there may not be a linear term in $\phi$ in the Lagrangian after the transformation of $\phi$, but there's still a new constant term in the Lagrangian. Why won't the new constant term not affect the correlation functions? Nov 18, 2016 at 10:58