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If we consider a $D=0$ theory with the Lagrangian: $$\mathcal{L}[\phi]=g\phi^n+J\phi$$ And its Green functions: $$G_n=\langle\phi^n\rangle_{J=0}=\frac{1}{Z[0]}\frac{\delta^nZ[J]}{\delta J^n}|_{J\rightarrow0}.$$ An infinite change of the integration variable $\phi$ leads to the equation of motion: $$\langle d\mathcal{L}/d\phi\rangle_J=gn\langle\phi^{n-1}\rangle_J=\langle J\rangle_J.$$ Then we get $G_{n-1}=0$ when $J\rightarrow0$.

My question is: Do we have any explicit insight about why the $n-1$ point correlation function vanishes?

My understanding of the correlation function is about an amplitude between two states. I know that the one-point function $\langle\phi\rangle$ vanishes if the system has some symmetries. So does the $n-1$ point function vanish also for symmetries?

Errorbar
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2 Answers2

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In the absence of space-time coordinates ($D=0$), the functional integral of the (Euclidean) "field" theory with the Lagrangian $g \phi^n$ reduces to an ordinary integral and the generating functional becomes the generating function $$Z(J) = \mathcal{N} \! \int\limits_{-\infty}^\infty \! d\phi \, e^{-g \phi^n+J\phi}, \quad g\gt 0,\tag{1} \label{1}$$ where the normalization constant $\mathcal N$ is determined by the condition $Z(0)=1$. It is obvious that the existence of the integral in \eqref{1} requires $n$ to be even. As a consequence, $$\langle \phi^{n-1} \rangle:= \frac{\partial^{n-1} Z(J)}{\partial J^{n-1}}{\Large|}_{J=0}= \mathcal{N} \! \int\limits_{-\infty}^\infty \! d\phi \, \phi^{n-1}\, e^{-g \phi^n} \tag{2} \label{2}$$ vanishes because of the trivial fact that the integrand $f(\phi) = \phi^{n-1} e^{-g \phi^n}$ is an odd function, $f(-\phi)=-f(\phi)$.

Hyperon
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  • But maybe the function $G_{n-1}=0$ is true when $n$ is not necessarily a positive integer in non-Hermitian theories. – Errorbar Mar 19 '24 at 07:32
  • For non-integer $n$, $\phi^n$ has a branch cut. In such a case you have to specify precisely how your integral is defined at all. – Hyperon Mar 19 '24 at 07:45
  • @XinranSu By the way, the second equation in your question implied implicitely that $n$ was supposed to be a non-negative integer. (Otherwise the $n$-th derivative would not make sense.) – Hyperon Mar 19 '24 at 08:10
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    Sorry, I have ignored the sources of my question. I have updated. – Errorbar Mar 19 '24 at 09:18
  • @XinranSu It is difficult to shoot at a moving target... – Hyperon Mar 19 '24 at 10:06
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  1. More generally, for a complex scalar $\phi^n$ theory$^1$ where $n\in\mathbb{N}$ with a global $\mathbb{Z}_n$-symmetry $\phi\to e^{2\pi i/n}\phi$, then the $m$-point function $\langle\phi^m\rangle_{J=0}=0$ vanishes whenever $m\notin n\mathbb{N}_0$ is not a multiple of $n$, cf. e.g. my Phys.SE answer here.

  2. For a real scalar $\phi^n$ theory where $n$ is even, there is a global $\mathbb{Z}_2$-symmetry $\phi\to -\phi$, so that all odd $m$-point function $\langle\phi^m\rangle_{J=0}=0$ vanish by a similar argument.

$^1$ To make the action real, add a hermitian conjugate interaction term $\phi^{\ast n}$.

Qmechanic
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