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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

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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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  • 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? – nightmarish Nov 17 '16 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. – ACuriousMind Nov 17 '16 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? – nightmarish Nov 18 '16 at 10:58
  • @failexam No, that's a typo. Also, of course the redefinition affects the correlation functions - that's the whole points of the redefinition, we did it to make the 1-point function vanish. I don't get what exactly your problem is. – ACuriousMind Nov 18 '16 at 14:37