We always say that tree levels are classical but loop diagrams are quantum.

Let's talk about a concrete example: $$\mathcal{L}=\partial_a \phi\partial^a \phi-\frac{g}{4}\phi^4+\phi J$$ where $J$ is source.

The equation of motion is $$\Box \phi=-g \phi^3+J$$

Let's do perturbation, $\phi=\sum \phi_{n}$ and $\phi_n \sim \mathcal{O}(g^n) $. And define Green function $G(x)$ as $$\Box G(x) =\delta^4(x)$$


Zero order:

$\Box \phi_0 = J$

$\phi_0(x)=\int d^4y G(x-y) J(y) $

This solution corresponds to the following diagram:enter image description here

First order:

$\Box \phi_1 = -g \phi_0^3 $

$\phi_1(x)=-g \int d^4x_1 d^4x_2 d^4x_3 d^4x_4 G(x-x_1)G(x_1-x_2)G(x_1-x_3)G(x_1-x_4)J(x_2)J(x_3)J(x_4) $

This solution corresponds to the following diagram:enter image description here

Second order:

$\Box \phi_2 = -3g \phi_0^2\phi_1 $

$\phi_2(x)= 3g^2 \int d^4x_1 d^4x_2 d^4x_3 d^4x_4 d^4x_5 d^4yd^4z G(x-y)G(y-x_1)G(y-x_2)G(y-z)G(z-x_3)G(z-x_4)G(z-x_5) J(x_1)J(x_2)J(x_3)J(x_4)J(x_5) $ This solution corresponds to the following diagram:enter image description here

Therefore, we've proved in brute force that up to 2nd order, only tree level diagram make contribution.

However in principle the first order can have the loop diagram, such as enter image description here but it really does not occur in above classical calculation.

My question is:

  1. What's the crucial point in classical calculation, which forbids the loop diagram to occur? Because the classical calculation seems similiar with quantum calculation.

  2. How to prove the general claim rigorously that loop diagram will not occur in above classical perturbative calculation.

  • 6
    $\begingroup$ Classically we do not sum over infinite paths. Write the Feyman path integral keeping the constant $\hbar$ we you will see that a loop expansion is equivalent to an $\hbar$ expansion. $\endgroup$
    – Diracology
    Commented Jul 28, 2016 at 17:53
  • 9
    $\begingroup$ You just did the classical perturbative calculation and saw that they didn't arise. The reason is that one would have to force two of the $x_i$ to be the same. This doesn't happen classically but does in QFT through contact terms in the Schwinger-Dyson equations. $\endgroup$
    – knzhou
    Commented Jul 28, 2016 at 18:05
  • $\begingroup$ Your starting point was the equations of motion. They define the classical soltion. To get quantum corrections one must consider configurations which do not obey the classical equations of motion. In other words, one must not consider only minimal-action configurations. $\endgroup$
    – Rd Basha
    Commented May 23, 2022 at 8:53

3 Answers 3

  1. Perturbative expansion. OP's $\phi^4$ theory example is a special case. Let us consider a general action of the form $$ S[\phi] ~:=~\underbrace{S_2[\phi]}_{\text{quadratic part}} + \underbrace{S_{\neq 2}[\phi]}_{\text{the rest}}, \tag{1} $$ with non-degenerate quadratic part$^1$ $$ S_2[\phi] ~:=~\frac{1}{2} \phi^k (S_2)_{k\ell} \phi^{\ell} . \tag{2} $$ The rest$^2$ $S_{\neq 2}=S_0+S_1+S_{\geq 3}$ contains constant terms $S_0$, tadpole terms $S_1[\phi]=S_{1,k}\phi^k$, and interaction terms $S_{\geq 3}[\phi]$.

  2. The partition function $Z[J]$ can be formally written as $$\begin{align} Z[J] ~:=~~~& \int {\cal D}\frac{\phi}{\sqrt{\hbar}}~\exp\left\{ \frac{i}{\hbar}\left(S[\phi] +J_k \phi^k \right)\right\} \cr ~\stackrel{(1)}{=}~~~& \exp\left\{\frac{i}{\hbar} S_{\neq 2}\left[ \frac{\hbar}{i} \frac{\delta}{\delta J}\right] \right\}\cr &\int {\cal D}\frac{\phi}{\sqrt{\hbar}}~\exp\left\{ \frac{i}{\hbar}\left(S_2[\phi] +J_k \phi^k \right)\right\} \cr \stackrel{\text{Gauss. int.}}{\sim}&~ {\rm Det}\left(\frac{1}{i} (S_2)_{mn}\right)^{-1/2}\cr &\exp\left\{\frac{i}{\hbar} S_{\neq 2}\left[ \frac{\hbar}{i} \frac{\delta}{\delta J}\right] \right\} \cr &\exp\left\{- \frac{i}{2\hbar} J_k (S_2^{-1})^{k\ell} J_{\ell} \right\}, \end{align}\tag{3} $$ after a Gaussian integration. Here$^1$ $$ G^{k\ell}~=~-(S_2^{-1})^{k\ell} \tag{4}$$ is the free propagator. The RHS of eq. (3) represents the sum of all$^3$ Feynman diagrams built from vertices, free propagators and external sources $J_k$.

  3. Euler-Lagrange (EL) equations$^4$ $$ - J_k ~\approx~\frac{\delta S[\phi]}{\delta \phi^k}~\stackrel{(1)+(2)}{=}~ (S_2)_{k\ell}\phi^{\ell} +\frac{\delta S_{\neq 2}[\phi]}{\delta \phi^k} \tag{5}$$ can be turned into a fixed-point equation$^5$ $$\phi^{\ell}~\approx~-(S_2^{-1})^{\ell k}\left( J_k + \frac{\delta S_{\neq 2}[\phi]}{\delta \phi^k} \right),\tag{6}$$ whose repeated iterations generate (directed rooted) trees (with a $\phi^{\ell}$ as root, and $J$s & tadpoles as leaves), as opposed to loop diagrams, cf. OP's calculation. This answers OP's questions.

Finally, let us mention below some hopefully helpful facts beyond tree-level.

I) The linked cluster theorem. The generating functional for connected diagrams is $$ W_c[J]~=~\frac{\hbar}{i}\ln Z[J]. \tag{7}$$

For a proof see e.g. this Phys.SE post. So it is enough to study connected diagrams.

II) The $\hbar$/loop-expansion. Assume that the $S[\phi]$ action (1) does not$^6$ depend explicitly on $\hbar$. Then the order of $\hbar$ in a connected diagram with $E$ external legs$^7$ is the number $L$ of independent loops, i.e. the number of independent $4$-wavevector$^8$ integrations.

Proof. We follow here Ref. 1. Let $I$ be the number of internal propagators and $V$ the number of vertices.

On one hand, for each vertex there is a 4-wavevector Dirac delta function. Except for 1 vertex, because the external legs already satisfy total wavevector conservation. (Recall that spacetime translation invariance implies that each connected Feynman diagram in wavevector space is proportional to a Dirac delta function imposing total 4-wavevector conservation.) The $V$ vertices therefore yield only $V-1$ constraints among the $I$ wavevector integrations. In other words, the number of independent loops is$^9$ $$L~=~I-(V-1). \tag{8}$$ On the other hand, it follows$^{10}$ from the RHS of eq. (3) that we have one $\hbar$ for each internal propagator, none for each external leg, and one $\hbar^{-1}$ for each vertex. There is also a single extra factor of $\hbar$ from the rhs. of eq. (7). Altogether, the power of $\hbar$s of the connected diagram is $$ \hbar^{I-V+1}~\stackrel{(8)}{=}~\hbar^{L},\tag{9}$$ i.e. equal to the number $L$ of loops. $\Box$

III) In particular, the generating functional of connected diagrams $$W_c[J]~=~W_c^{\rm tree}[J]+W_c^{\rm loops}[J]~\in~ \mathbb{C}[[\hbar]]\tag{10} $$ is a power series in $\hbar$, i.e. it contains no negative powers of $\hbar$. In contrast, the partition function $$Z[J]~=~\underbrace{\exp(\frac{i}{\hbar}W_c^{\rm tree}[J])}_{\in \mathbb{C}[[\hbar^{-1}]]}~\underbrace{\exp(\frac{i}{\hbar}W_c^{\rm loops}[J])}_{\in \mathbb{C}[[\hbar]]}\tag{11}$$ is a Laurent series in $\hbar$.


  1. C. Itzykson & J.B. Zuber, QFT, 1985, Section 6-2-1, p.287-288.


$^1$ We use DeWitt condensed notation to not clutter the notation. If we spell out eq. (2) in its full glory it reads $$S_2[\phi]~=~\frac{1}{2}\int\!d^dx\int\!d^dy~\phi^{\alpha}(x)~(S_2)_{\alpha\beta}(x,y)~\phi^{\beta}(y)\tag{12}$$ after possible integration by parts. Here the integration kernel is typically of the form $$(S_2)_{\alpha\beta}(x,y)~=~\delta_{\alpha\beta}~(\Box-m^2)\delta^d(x-y)\tag{13}$$ with the $(-,+,...,+)$ Minkowski sign convention. If we impose appropriate boundary conditions in eq. (4), the inverse integration kernel $$(S_2^{-1})^{\alpha\beta}(x,y)~=~\delta^{\alpha\beta}~(\Box-m^2)^{-1}\delta^d(x-y)~=~-G^{\alpha\beta}(x,y)\tag{14}$$ is minus the Greens function $$(-\Box+m^2)G^{\alpha\beta}(x,y)~=~\delta^{\alpha\beta} ~\delta^d(x-y)\tag{15}$$ with Fourier transform $$\widetilde{G}^{\alpha\beta}(k)~=~\frac{\delta^{\alpha\beta}}{k^2+m^2-i\epsilon}.\tag{16}$$

$^2$ If we split the action $$S[\phi] ~=~\underbrace{S_1[\phi]+S_2[\phi]}_{\text{free part}}+\underbrace{S_{\neq 12}[\phi]}_{\text{the rest}}\tag{17}$$ (by including tadpoles in the free part), then the propagator factor on the RHS of eq. (3) becomes $$\exp\left\{- \frac{i}{2\hbar} (S_{1,k}+J_k) (S_2^{-1})^{k\ell} (S_{1,\ell}+J_{\ell}) \right\}.\tag{18}$$ Conversely, we could formally allow quadratic terms in the $S_{\neq 2}$ part as well, e.g. if we want to treat a mass term as a 2-vertex-interaction. This would of course ruin the logic behind the subscript label of the notation $S_{\neq 2}$, but that's an acceptable prize to pay:)

$^3$ The Gaussian determinant factor ${\rm Det}\left(\frac{1}{i} (S_2)_{mn}\right)^{-1/2}$ (which we normally ignore) is interpreted as Feynman diagrams built from free propagators only with no vertices, although the precise interpretation is quite subtle. E.g. note that if we reclassify the mass-term in the free propagator as a $2$-vertex-interaction, the mass-contribution shifts from the determinant factor to the interaction part in eq. (3).

$^4$ The $\approx$ symbol means equality modulo eqs. of motion.

$^5$ In fact, eq. (6) can be viewed as an operad. A bit oversimplified, while an operator has one input and one output, an operad may have several inputs, but still only one output. Operads may be composed together and thereby form (directed rooted) tree (with the lone output being the root).

$^6$ In order to keep the action $S$ without explicit $\hbar$-dependence, we might have to appropriately redefine mass parameters $m^{\prime}=\frac{mc}{\hbar}$, coupling constants $e^{\prime}=\frac{e}{\hbar}$, etc. If the interaction terms in the action $S$ do depend on $\hbar$, then a diagram will contain the usual loop-power $L$ of $\hbar$s plus a number of powers of $\hbar$ from the corresponding vertices.

$^7$ We assume that the sources $J_k$ are either stripped from the Feynman diagram or are delta functions in wavevector space so that the external legs carry fixed 4-wavevectors.

$^8$ In order not to introduce extra factors of $\hbar$ when we Fourier transform, let us work with 4-wavevector $k$ rather than 4-momentum $p=\hbar k$.

$^9$ If the Feynman diagram is planar, then it is a polygon mesh of a disk, i.e. its Euler characteristic is $\chi=1$. Comparing with eq. (8), we see that the number $L$ of independent loops is then the number of faces.

$^{10}$ The RHS of eq. (3) yields that a propagator attached to $n$ sources contributes with a factor $\hbar^{1-n}$, where $n\in\{0,1,2\}$.

  • $\begingroup$ Notes for later: Generalization to transition amplitude/overlap: $\quad Z_{fi}[J]=\langle \phi_f,t_f|\phi_i,t_i\rangle_J=\exp\left\{\frac{i}{\hbar} S_{\neq 2}\left[\frac{\hbar}{i} \frac{\delta}{\delta J}\right] \right\}Z_{2,fi}[J].$ Stationary solution $\phi^k_2[J]=\phi^k_H+\phi^k_I$ to the quadratic action $S_2[\phi] +J_k\phi^k$, where $\phi^k_I:= -(S^{-1}_2)^{k\ell}J_{\ell}$ and $\phi^k_H$ solves the homogeneous problem $(S_2)_{k\ell}\phi^{\ell}_H=0$ with BCs $\phi^k_H(t_i)=\phi^k_i-\phi^k_I(t_i)$ and $\phi^k_H(t_f)=\phi^k_f-\phi^k_I(t_f)$, and hence must be linear in the BCs. $\endgroup$
    – Qmechanic
    Commented Oct 16, 2018 at 12:45
  • $\begingroup$ Notes for later. $\quad (S_2)_{k\ell}\phi^{\ell}_2[J]+J_k=0.$ Compare with Feynman & Hibbs, Section 7.4 p. 182. $\quad Z_{2,fi}[J] \stackrel{\text{Gauss. int.}}{\sim}$ ${\rm Det}\left(\frac{1}{i}(S_2)_{mn}\right)^{-1/2} \exp\left\{\frac{i}{\hbar}S_2[\phi_2[J]]+\frac{i}{\hbar} J_k\phi^k_2[J]\right\}$ $\sim\exp\left\{\text{bdry-bdry+bdry-bulk+bulk-bulk}\right\}$ $\sim\exp\left\{-\frac{i}{2\hbar}J_k(S_2^{-1})^{k\ell}J_{\ell} \right\}$ bulk-bulk. There are bulk & bdry sources. Momentum conservation is maintained. bdry-bdry & bdry-bulk can only be external legs. $\endgroup$
    – Qmechanic
    Commented Mar 26, 2020 at 16:58
  • $\begingroup$ Comment 19.04.21: 1. A vacuum bubble has an overall Dirac delta function $\delta^d(0)$ associated with total momentum conservation. 2. In particular, a vertex with only self-loops attached yields a momentum conservation $\delta^d(0)$. Tadpoles $\langle \widetilde{\phi}(k)\rangle_{J=0} \propto \delta^d(k)$ has zero wavevector. physics.stackexchange.com/q/629808/2451 arxiv.org/abs/2108.02276 $\endgroup$
    – Qmechanic
    Commented Apr 20, 2021 at 10:34
  • $\begingroup$ Large $N$ expansion: The 't Hooft coupling $\lambda=g_{YM}^2N$ is kept fixed. (i) Replace $1/\hbar\to 1/\hbar g_{YM}^2$ in front of $F^2$/Dirac/scalar/source action terms. $D=d-iA$ without $g_{YM}$. Refs: 1. H. Liu: MIT lecture 7. 2. D. Tong: Gauge theory lectures. 3. E. Witten: NPB160 (1979) 57. 4. S. Coleman, Aspects of symmetry; chap. 8. $\endgroup$
    – Qmechanic
    Commented Aug 21, 2021 at 14:43
  • 1
    $\begingroup$ Notes for later: 1. For a disconnected diagram with $\pi_0$ connected components, eq. (8) generalizes to $L=I-V+\pi_0$. 2. In string theory, we can replace $\hbar\to\hbar/T_0$, which is related to $\alpha^{\prime}$ and $g_s$. $\endgroup$
    – Qmechanic
    Commented Aug 18, 2023 at 14:29

My words won't help answering the question, but I just want to clarify sth on the analogy between classical perturbative field theory and QFT tree level amplitudes.

  • Tree-level quantum field theory, which describes scattering experiments of a very small number of quantum excitations around the vacuum, are highly quantum mechanical phenomena.

  • Classical field theory, on the other hand, describes the scattering between classical waves. The behavior of a large cluster of QFT excitations could be approximated by classical waves.

They are two completely different regimes of physics, there's no way that a classical field theory could give rise to quantum scattering experiments, although it gives rise to analogues of tree level diagrams.

This may only be my own little resolution of my own little confusion, but I've heard people carelessly interchanging the term classical and tree-level.

As an example of this disanalogy: While unitarity violation is a big concern for tree-level QFT, I don't think it is of any relevance for classical wave scattering, as long as the wave is classical (the wave energy is high enough to contain a lot of field quanta).(Sorry, this example is probably wrong)


The explanation by Qmechanic is crisp and precise. However, let me give a simpler but limited explanation :-

Starting with the path integral. we get the classical limit by taking h tending to zero limit. In this limit the leading order term that contributes to the generating functional is the classical action. The first order variation is zero, and we are ignoring the second order variation. Now, as the complete contribution is from the classical action, E.O.M is satisfied and external states obeys the usual E-p dispersion relation

However, in a loop integral, we integrate over all 4 components of the momenta and treat them as being independent i.e the momenta is off-shell, which as explained above can't happen if you have taken the classical limit.

Of course this argument is only limited to understand why we can't have loops on external legs in the classical limit. This argument doesn't restrict loops in the internal lines.

Can somebody point out if there is any major flaw in this argument and can it be modified to also make a statement about not having loops on internal lines.

The best way to understand this is Schwinger Dyson equations. Read from Matthew Schwarz.


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