Why $n-1$ point function vanishes in $D=0$ scalar theory?

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?

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, $$f(-\phi)=-f(\phi)$$.

• But maybe the function $G_{n-1}=0$ is true when $n$ is not necessarily a positive integer in non-Hermitian theories. Commented Mar 19 at 7:32
• 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. Commented Mar 19 at 7:45
• @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.) Commented Mar 19 at 8:10
• Sorry, I have ignored the sources of my question. I have updated. Commented Mar 19 at 9:18
• @XinranSu It is difficult to shoot at a moving target... Commented Mar 19 at 10:06
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 $$\phi^{\ast n}$$.