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.

  • 4
    $\begingroup$ 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 Jul 28 '16 at 17:53
  • 4
    $\begingroup$ 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 Jul 28 '16 at 18:05
  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 $$ Z[J] ~:=~ \int {\cal D}\frac{\phi}{\sqrt{\hbar}}~\exp\left\{ \frac{i}{\hbar}\left(S[\phi] +J_k \phi^k \right)\right\} $$ $$\stackrel{\text{Gauss. int.}}{\sim}~ {\rm Det}\left(\frac{1}{i} (S_2)_{mn}\right)^{-1/2} \exp\left\{\frac{i}{\hbar} S_{\neq 2}\left[ \frac{\hbar}{i} \frac{\delta}{\delta J}\right] \right\} \exp\left\{- \frac{i}{2\hbar} J_k (S_2^{-1})^{k\ell} J_{\ell} \right\}, \tag{3} $$ after a Gaussian integration. Here $$ (S_2^{-1})^{k\ell} \tag{4}$$ is the free propagator.

  3. Euler-Lagrange (EL) equations$^3$ $$ - J_k ~\approx~\frac{\delta S[\phi]}{\delta \phi^k}~\stackrel{(1)+(2)}{=}~ (S_2)_{k\ell}\phi^{\ell} +\frac{\delta S_{\neq 2}[\phi]}{\delta \phi^k} \tag{5}$$ can be turned into a fixed-point equation$^4$ $$\phi^{\ell}~\approx~-(S_2^{-1})^{\ell k}\left( J_k + \frac{\delta S_{\neq 2}[\phi]}{\delta \phi^k} \right),\tag{6}$$ 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.

  1. 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.

  2. The $\hbar$/loop-expansion. Assume that the $S[\phi]$ action (1) does not depend explicitly on $\hbar$. Then the order of $\hbar$ in a connected diagram with $E$ external legs is the number $L$ of independent loops, i.e. the number of independent $4$-momentum 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-momentum Dirac delta function. Except for 1 vertex, because the external legs already satisfy total momentum conservation. (Recall that spacetime translation invariance implies that each connected Feynman diagram in momentum space is proportional to a Dirac delta function imposing total 4-momentum conservation.) The $V$ vertices therefore yield only $V-1$ constraints among the $I$ momentum integrations. In other words, $$L~=~I-(V-1). \tag{8}$$ On the other hand, it follows from 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. $\Box$

  3. 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.

$^2$ To be as general as possible, one could formally allow quadratic terms in the $S_{\neq 2}$ part as well. 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 $\approx$ symbol means equality modulo eqs. of motion.

$^4$ 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).


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.