# Tree level QFT and classical fields/particles

It is well known that scattering cross-sections computed at tree level correspond to cross-sections in the classical theory. For example the tree level cross-section for electron-electron scaterring in QED corresponds to scattering of classical point charges. The naive explanation for this is that the power of hbar in a term of the perturbative expansion is the number of loops in the diagram.

However, it is not clear to me how to state this correspondence in general. In the above example the classical theory regards electrons as particles and photons as a field. This seems arbitrary. Moreover, if we consider for example phi^4 theory than the interaction of the phi-quanta is mediated by nothing except the phi-field itself. What is the corresponding classical theory? Does it contain both phi-particles and a phi-field?

Also, does this correspondence extend to anything besides scattering theory?

Summing up, my question is:

What is the precise statement of the correspondence between tree-level QFT and behavior of classical fields and particles?

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The classical limit of a quantum field theory is classical field theory. The limit you discuss (photons are encoded in the EM field, electrons are particles) seems to be the non-relativistic limit. Generally particles are only well-defined objects in such a limit -- contrary to common belief, there is really no such thing as a particle, but some excitations of massive free fields can look like particles in the appropriate limit. –  user566 Nov 26 '11 at 21:11
The statement "the tree level cross-section for electron-electron scaterring in QED corresponds to scattering of classical point charges" is somewhat wrong. In CED any scattering is accompanied with EMW radiation which is totally absent on the tree level of QED. This is a severe drawback of QED that should be "repaired" with summation of all soft diagrams. It means the initial approximation in QED is too far from a good one. –  Vladimir Kalitvianski Nov 26 '11 at 22:24
Diagrams contributing to the amplitude, including soft emission of any number of photons in initial/final states, are still tree level; when squaring the smplitude to get a cross section one has to involve some limit of one loop diagrams as well. This way one can reproduce the CED scattering cross section as one should (see chapter 6 of Peskin and Schroeder for details). This resummation of soft photons is essentially the reason the radiation has to be treated as a field always, and does not have a particle limit. –  user566 Nov 26 '11 at 23:57
@Squark, effective action in the tree-level approximation coincides with the classical action. That explains how to take the limit QFT -> classical field theory. –  Pavel Safronov Nov 27 '11 at 3:21
@Moshe: You are right, contribution of soft photons emitted from internal lines is much smaller. In my response I meant the first non vanishing approximation of a tree level, to be exact. –  Vladimir Kalitvianski Nov 27 '11 at 9:47

## 4 Answers

This was something that confused me for awhile as well until I found this great set of notes: homepages.physik.uni-muenchen.de/~helling/classical_fields.pdf

Let me just briefly summarize what's in there.

The free Klein-Gordon field satisfies the field equation

$(\partial_{\mu} \partial^{\mu} +m^2) \phi(x) = 0$

the most general solution to this equation is

$\phi(t, \vec{x}) = \int_{-\infty}^{\infty} \frac{d^3k}{(2\pi)^3} \; \frac{1}{2E_{\vec{k}}} \left( a(\vec{k}) e^{- i( E_{\vec{k}} t -\vec{k} \cdot \vec{x})} + a^{*}(\vec{k}) e^{ i (E_{\vec{k}} t- \vec{k} \cdot \vec{x})} \right)$

where

$\frac{a(\vec{k}) + a^{*}(-\vec{k})}{2E_{\vec{k}}} = \int_{-\infty}^{\infty} d^3x \; \phi(0,\vec{x}) e^{-i \vec{k} \cdot \vec{x}}$

and

$\frac{a(\vec{k}) - a^{*}(-\vec{k})}{2i} = \int_{-\infty}^{\infty} d^3x \; \dot{\phi}(0,\vec{x}) e^{-i \vec{k} \cdot \vec{x}}$

Introducing an interaction potential into the Lagrangian results in the field equation

$(\partial^{\mu} \partial_{\mu} + m^2) \phi = -V'(\phi)$

choosing a phi-4 theory $V(\phi) = \frac{g}{4} \phi^4$ this results in

$(\partial^{\mu} \partial_{\mu} + m^2) \phi = -g \phi^3$

Introduce a Green's function for the operator

$(\partial^{\mu} \partial_{\mu} + m^2) G(x) = -\delta(x)$

which is given by

$G(x) = \int \frac{d^4k}{(2\pi)^4} \; \frac{-e^{-i k \cdot x}}{-k^2 + m^2}$

now solve the full theory perturabtively by substituting

$\phi(x) = \sum_{n} g^n \phi_{n}(x)$

into the differential equation and identifying powers of $g$ to get the following equations

$(\partial^{\mu} \partial_{\mu} + m^2) \phi_0 (x) = 0$

$(\partial^{\mu} \partial_{\mu} + m^2) \phi_1(x) = -\phi_0(x)^3$

$(\partial^{\mu} \partial_{\mu} + m^2) \phi_2 (x) = -3 \phi_0(x)^2 \phi_1(x)$

the first equation is just the free field equation which has the general solution above. The rest are then solved recursively using $\phi_0(x)$. So the solution for $\phi_1$ is

$\phi_1(x) = \int d^4y\; \phi_0(y)^3 \, G(x-y)$

and so on. As is shown in the notes this perturbative expansion generates all no-loop Feynman diagrams and this is the origin of the claim that the tree level diagrams are the classical contributions...

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OK, so classical perturbation theory can be identified in some sense with quantum tree-level perturbation theory. This is nice (I'd upvote but I reached the 30 votes limit today). However, I still don't understand how to generalize the statement about scattering amplitudes. –  Squark Dec 2 '11 at 17:47

There is a very easy way to see this and it is through an $\hbar$ series. This claim can be traced back to Sydney Coleman and states that in the ultraviolet one is doing an expansion with $\hbar$ going to zero. A previous answer cited these lectures on classical fields but I would like to start from the generating functional of the scalar field theory and try to understand the classical limit:

$$Z[j]=\int[d\phi]e^{\frac{i}{\hbar}\int d^4x\left[\frac{1}{2}(\partial\phi)^2-\frac{1}{2\hbar^2}m^2\phi^2-\frac{\lambda}{4\hbar}\phi^4+j\phi\right]}.$$

Our aim is to recover perturbation theory for the classical fields at tree level as this will prove Coleman's claim. Indeed, the above generating functional can be rewritten in a different form as

$$Z[j]=e^{-i\hbar^2\frac{\lambda}{4}\int d^4x\frac{\delta^4}{\delta j(x)^4}}e^{\frac{i}{2\hbar}\int d^4xd^4yj(x)\Delta(x-y)j(y)}.$$

Now, let us focus on the two-point function, being the argument the same for the other correlation functions. We will get

$$\left.(-i\hbar)^2\frac{1}{Z}\frac{\delta^2Z}{\delta j(x)\delta j(y)}\right|_{j=0}=i\hbar\Delta(x-y).$$

From these equations it is not difficult to recover the first quantum correction at one loop that is given by

$$-i\hbar^4\frac{\lambda}{4}\int d^4\tilde x \frac{\delta^4}{\delta j^4(\tilde x)}\frac{\delta^2}{\delta j(x)\delta j(y)}\left(-\frac{1}{3!8\hbar^3}\int d^4x_1d^4y_1d^4x_2d^4y_2d^4x_3d^4y_3\right.$$ $$\left.j(x_1)\Delta(x_1-y_1)j(y_1)j(x_2)\Delta(x_2-y_2)j(y_2)j(x_3)\Delta(x_3-y_3)j(y_3)\right)$$

and this will be proportional to $\hbar$. This is the conclusion we aimed to that gives evidence for Coleman's claim. A similar analysis can be carried out using effective potential. This proof completes the previous answer but starting from quantum field theory.

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I know loop contributions are proportional to hbar. The question is not about proving it. It's about the physical interpretation of this arithmetical fact. –  Squark Dec 2 '11 at 17:44
@Squark: Writing down something in places like this implies some kind of effort and the hope is that OP should be able to understand properly the content of what one is writing. As this is not the case and having seen this repeated again and again. This is my last experience with stackexchange et similia. Good luck and goodbye! –  Jon Dec 2 '11 at 21:06

The classical analogue of quantum $\Phi^4$ theory is classical $\Phi^4$ theory, with the same action. There are no particles, but there is still scattering of waves! The correspondence between tree-level QFT and classical fields is on the level of fields only. (Particles make their appearance in classical field theory only in the limit where geometric optics is valid. Even in quantum field theoy, the particle picture is not really appropriate except in the geometric optics regime.)

Feynman diagrams arise in any perturbative treatment of correlations of fields, even classically. Indeed, Feynman diagrams are just a graphical notation for writing products of tensors with many indices summed via the Einstein summation convention. The indices of the results are the external lines, while the indices summed over are the internal lines. As such sums of products occur in any multipoint expansion of expectations, irrespective of the classical or quantum nature of the system. No connection with particles is implied, unless one imposes it.

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What is the precise statement of the correspondence between tree-level QFT and behavior of classical fields and particles?

What follows are four discussions about the connection between quantum and classical fields, viewed from various angles. This will interest people to varying degrees (I hope). If you care only about the loop expansion, skip down to C.

[An initial point: Many people, myself included, would like to see a (relativistic) interacting theory of quantum fields approximated by a (most likely nonrelativistic) theory of quantum particles. The question above may have been posed with this approximation in mind. But I've never seen this approximation.]

A. The one framework that I know of that includes both classical and quantum physics is to view the theory as a mapping from observables into what is known as a C*-algebra. A state maps elements (of the algebra) to expectation values. Given a state, a representation of the algebra elements as operators on a Hilbert space can be obtained. (I'm speaking of the GNS reconstruction.)

Now let's consider a free scalar field theory.

In the quantum case, there will be a vacuum state, and the GNS reconstruction from this state will yield the the usual field theory. (There will also be states with nonzero temperature and nonzero particle density. I mention this simply as one advertisement for the algebraic approach.)

In the classical case, there will also be a vacuum state. But the reconstruction from this state will yield a trivial, one-dimensional Hilbert space. And the scalar field will be uniformly zero. [I'm suppressing irrelevant technical details.]

Fortunately, in the classical case, there will also be states for every classical solution. For these, the GNS representations will be one-dimensional, with every operator having the same value as the classical solution.

So, in the formal ħ → 0 limit, the algebra becomes commutative, it has states that correspond to classical solutions, and its observables take on their classical values in these states.

In the case of an interacting theory, the formal ħ → 0 limit isn't so clear because of renormalization. However, if, as I vaguely recall, the various renormalization counterterms are of order ħⁿ for n > 0, they don't matter in the formal ħ → 0 limit. In that case, the formal ħ → 0 limit yields the classical theory (as in the free field case).

Another interesting example is QED. With ħ = 0, the fermionic fields anticommute, which makes them zero in the context of a C*-algebra. So all of the fermionic fields vanish as ħ → 0, and we're left with free classical electrodynamics.

You may or may not derive any satisfaction from these formal limits of C*-algebras. In either case, we're done with them. Below, we talk about ordinary QFT.

B. Let's now consider a free Klein-Gordon QFT. We'll choose a "coherent" state and obtain an ħ → 0 limit. Actually, this will be a sketch without proofs.

The Lagrangian is ½(∂ᵤφ∂ᵘφ - ν²φ). Note ν instead of m. m has the wrong units, so you see a frequency instead. (c = 1.)

We have the usual free field expansion in terms of creation and annihilation operators. These satisfy:

[a(k),a†(l)] = ħ (2π)³(2k⁰) δ³(k - l)


k and l are not momenta. ħk and ħl are momenta. And the mass of a single particle is ħν.

The particle number operator N is (with đk = d³k (2π)⁻³(2k⁰)⁻¹):

N = (1/ħ) ∫đk a†(k)a(k)


And for some nice function f(k), we define the coherent state |f> by:

a(k)|f> = f(k)|f>


[I omit the expression for |f>.] Note that:

<f| N |f> = (1/ħ) ∫đk |f(k)|²


As ħ → 0, |f> is composed of a huge number of very light particles.

|f> corresponds to the classical solution:

Φ(x) = ∫đk [f(k)exp(ik⋅x) + complex conjugate]


Indeed, for normal-ordered products of fields, we have results like the following:

<f|:φ(x)φ(y):|f> = Φ(x)Φ(y)


Since the difference between :φ(x)φ(y): and φ(x)φ(y) vanishes as ħ → 0, we have in that limit:

<f| φ(x)φ(y) |f> → Φ(x)Φ(y)


If we reconstruct the theory from these expectation values, we obtain a one-dimensional Hilbert space on which φ(x) = Φ(x).

So, with coherent states, we can obtain all of the classical states in the ħ → 0 limit.

C. Consider an x-space Feynman diagram in some conventional QFT perturbation theory. Let: n = the number of fields being multiplied. P = the number of arcs (ie, propagators). V = the number of vertices. L = the number of independent loops. C = the number of connected components. Finally, let H be the number of factors of ħ in the diagram. Then, using standard results, we have:

H = P - V = n + L - C > 0


So, if you set ħ = 0, all Feynman diagrams vanish. All fields are identically zero.

This is reasonable. The Feynman diagrams contribute to vacuum expectation values. And the classical vacuum corresponds to fields vanishing everywhere.

D. Suppose that we don't want to take ħ → 0, but we do want to consider the theory up to, say, O(ħ²). But what is "the theory"? Let the answer be: the Green functions. But all of the connected Feynman diagrams with n > 3 have H > 2. In order to retain these diagrams and their associated Green functions, we need to ignore the factor ħⁿ that is part of every n-point function.

And that is what people do. When people define, say the generating functional for connected Green functions, they insert a factor of 1/ħⁿ⁻¹ multiplying the n-point functions. With these insertions, the above equation sort-of-becomes:

"H" = L


In particular, all of the (connected) tree diagrams appear at O(1) in the generating functional.

But recall that all of these diagrams vanish as ħ → 0. I don't see any way to interpret them as classical.

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Thank you for your answer Greg but note that tree-level scattering does correspond to classical scattering –  Squark Apr 28 '12 at 12:28
According to whom and by what argument? If you really mean "classical", where do the particles come from, given that a classical field theory has no particles? If you mean the particles of nonrelativistic QM cerca 1926, do you now how to obtain that theory from QFT? (I don't, and I've looking for it for a long time now.) –  Greg Weeks Apr 29 '12 at 15:46
@GregWeeks: Feynman diagrams arise in any perturbative treatment of correlations of fields, even classically. Indeed, Feynman diagrams are just a graphical notation for writing products of tensors with many indices summed via the Einstein summation convention. The indices of the results are the external lines, while the indices summed over are the internal lines. As such sums of products occur in any multipoint expansion of expectations, irrespective of the classical or quantum nature of the system. No connection with particles is implied, unless one imposes it. –  Arnold Neumaier Apr 29 '12 at 19:50
@ArnoldNeumaier: I don't follow your tensors and indices description. But I agree that Euclidean Feynman diagrams give you the expectation values of products of suitably randomized fields (in 4-d). But even those become trivial in the ħ → 0 limit, as do their Minkowski space counterparts. The calculation was given above. –  Greg Weeks May 2 '12 at 2:00
@GregWeeks: Think of momentum space as being discrete, and the momenta as indices. Then the prescription for evaluating Feynman diagrams in momentum space is just a big sum that condenses to a product of tensors in Einstein notation. You'd get a perturbation expansion in terms of such tensors also from any finite-dimensional state space. The lines have nothing to do with particles; associating them with ''virtual particles'' is traditional but without any but visual support. –  Arnold Neumaier May 2 '12 at 11:39