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

Qmechanic
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nightmarish
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1 Answers1

4

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.

ACuriousMind
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