# Classical version of tree-level QFT correlation functions

Having read this previous question (and its answers) about the relation between tree-level quantum field theory and classical field theory I can see two facts that support the (perhaps too vague) statement that they're the same:

• When solving perturbatively the equations of motion of a classical theory, an expansion in tree-level diagrams arises. These have the same Feynman rules as the ones we would obtain if it was a quantum theory.

• The tree level is the leading order of an expansion in powers of $\hbar$. Because $\hbar\to 0$ is known to be the classical limit, we can call the tree level "classical".

However, I feel that the answer to the following question is missing to have a complete understanding of the subject. Consider a QFT that has a classical analog. In the QFT we have the $n$-point functions $\left<\phi(x_1)\cdots \phi(x_n)\right>$. If we can do a loop expansion, we can compute the tree level approximation of these functions: $$\left<\phi(x_1)\cdots\phi(x_n)\right>_{\text{tree}}.$$ Is there any object in the classical field theory that corresponds to $\left<\phi(x_1) \cdots \phi(x_n)\right>_{\text{tree}}$ or is somehow related to it?

• Interesting question. I always thought that the correspondence is $\langle\phi(x_1)\cdots\phi(x_n)\rangle_\text{tree}\sim \phi_c(x_1)\cdots\phi_c(x_n)$, where $\phi_c$ is the classical field. I don't have a proof of this (and perhaps it is not even true), but it seems reasonable. – AccidentalFourierTransform Dec 4 '17 at 22:09
• @AccidentalFourierTransform I'll probably add something tomorrow but right now think of the simplest example. For vacuum you usually have $\phi_c(x)\equiv\langle\phi(x)\rangle=0$ and your formula will then give zero for any higher-point correlator. However actually the 2-point correlator gives you certain propagator i.e. some Green function of the free theory operator. – OON Dec 5 '17 at 0:40
• @OON fair enough; perhaps you are right, I have to think about it. But do note that $\phi_c(x)\equiv 0$ only when you turn off the external source, $J(x)\equiv 0$. Otherwise, $\phi_c(x)\neq0$. – AccidentalFourierTransform Dec 5 '17 at 11:54
• @AccidentalFourierTransform If $\langle\phi(x)\rangle\neq 0$ there actually will be the contribution like the one you wrote. The point is, even at the tree level that's not the whole story – OON Dec 5 '17 at 12:45

You can make the connection between classical and quantum field theories through the functional methods. In QFT you have a generating functionals, $$Z[J]=\int \mathcal{D}\Phi e^{iS[\phi]+iJ\cdot\phi},\quad E[J]=i\ln{Z[J]},\quad J\cdot\phi=\int d^4x\,J(x)\phi(x)$$ with $E[J]$ generating connected correlators, $$(i)^{n+1}\langle\phi(x_1)\ldots\phi(x_n)\rangle_{con,J}=\frac{\delta^n}{\delta J(x_1)\ldots \delta J(x_n)}E[J]$$ Let's introduce $\phi_{cl}(x)\equiv\langle\phi(x)\rangle_J=-\frac{\delta E[J]}{\delta J(x)}$. Then we can perform the Legendre transform from function depending on $J$ to function depending on $\phi_{cl}$ introduce the quantum effective action, $$\Gamma[\phi_{cl}]\equiv -E[J]-J\cdot\phi$$ To go back you can always use $\frac{\delta \Gamma[\phi_{cl}]}{\delta\phi_{cl}(x)}=-J(x)$. The important thing for us is that at tree level, $$\Gamma[\phi_{cl}]\simeq S[\phi_{cl}]$$ From here we will make the connection we are looking for.
Start with the classical action $S[\phi]$ and supplement it with some external sources, $$S_{full}[\phi,J]=S[\phi]+J\cdot\phi$$ The classical analog of $E[J]$ is given simply by the value of this full action on the solution of the classical equations of motion $\phi_J$, $$\frac{\delta S[\phi]}{\delta\phi(x)}\Bigg\vert_{\phi=\phi_J}=-J(x),\quad E_{cl}[J]\equiv-S_{full}[\phi_J,J]$$ The variation of this functional will be a classical analogue of the connected correlators. You may ask yourself what it's good for. You may try to find the approximate solution of $\phi_J$ as a series in powers of $J$, $$\phi_J(x_1)=\phi_0(x_1)+\sum_{n=1}^{+\infty}\frac{1}{n!}\int d^4x_2\ldots d^4x_n\,G_n(x_1,x_2\ldots x_n)J(x_2)\ldots J(x_n)$$ As usual in the Taylor series, $$G_n(x_1,x_2\ldots x_n)=\frac{\delta^{n-1}\phi(x_1)}{\delta J(x_2)\ldots\delta J(x_n)}\Bigg\vert_{J=0}$$ On the other hand if you take the variation of $E_{cl}$ you will have, \begin{aligned} &\frac{\delta^n E_{cl}[J]}{\delta J(x_1)\ldots \delta J(x_n)}\Bigg\vert_{J=0}=\\ &-\frac{\delta^{n-1}}{\delta J(x_2)\ldots \delta J(x_n)}\left\{\int d^4y\frac{\delta S[\phi]}{\delta \phi(y)}\frac{\delta\phi_J(y)}{\delta J(x_1)}+\int d^4y\, J(y)\frac{\delta\phi_J(y)}{\delta J(x_1)}+\phi_J(x_1)\right\}\Bigg\vert_{J=0} \end{aligned} using the definition of $\phi_J$ two first terms cancel each other and we get, $$\frac{\delta^n E_{cl}[J]}{\delta J(x_1)\ldots \delta J(x_n)}\Bigg\vert_{J=0}=-\frac{\delta^{n-1}\phi_J(x_1)}{\delta J(x_2)\ldots\delta J(x_n)}\Bigg\vert_{J=0}=-G_n(x_1,x_2\ldots x_n)$$ I.e. those functions contracted with $J$ will appear in the decomposition of $\phi_J$. Even more usefully if you consider $\phi^n$ theory, $$-\partial_\mu\partial^\mu\phi(x)-m^2\phi(x)=\frac{g}{(n-1)!}\phi^{n-1}(x)-J(x)$$ then in the perturbation theory in $g$ the power of $g$ would also uniquely determine the power of $J$. Thus you may see that the $n$-th power in $g$ will be described by a specific tree level $N$-point connected correlation function which is mentioned in the answers on the question you've linked.