Tree-level QFT and classical fields/particles

Question

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 scattering 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?

Answer 1

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...

Answer 2

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.

Answer 3

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 $\hbar\to0$ 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 $\hbar\to0$ limit isn't so clear because of renormalization. However, if, as I vaguely recall, the various renormalization counterterms are of order $\hbar^n$ for $n > 0$, they don't matter in the formal $\hbar\to0$ limit. In that case, the formal $\hbar\to0$ limit yields the classical theory (as in the free field case).

Another interesting example is QED. With $\hbar=0$, the fermionic fields anticommute, which makes them zero in the context of a C*-algebra. So all of the fermionic fields vanish as $\hbar\to0$, 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 $\frac12(\partial\phi)^2-\frac12\nu^2\phi^2$. Note $\nu$ 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:

$$[a(k),a^\dagger(l)] = \hbar (2\pi)^2(2k^0) \delta^3(k - l)$$

$k$ and $l$ are not momenta. $\hbar k$ and $\hbar l$ are momenta. And the mass of a single particle is $\hbar\nu$.

The particle number operator $N$ is (with $\not \!dk = d^3k (2\pi)^{-3}(2k^0)^{-1}$):

$$N = \hbar^{-1}\int\not \!dk a^\dagger(k)a(k)$$

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

$$a(k)|f\rangle = f(k)|f\rangle$$

[I omit the expression for $|f\rangle$.] Note that:

$$\langle f| N |f\rangle =\hbar^{-1} \int\not \!dk |f(k)|^2$$

As $\hbar\to0$, $|f\rangle$ is composed of a huge number of very light particles.

$|f\rangle$ corresponds to the classical solution:

$$\Phi(x) = \int\not \!dk [f(k)\exp(ik⋅x) + \text{complex conjugate}]$$

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

$$\langle f|:φ(x)φ(y):|f\rangle = \Phi(x)\Phi(y)$$

Since the difference between $:φ(x)φ(y):$ and $φ(x)φ(y)$ vanishes as $\hbar\to0$, we have in that limit:

$$\langle f| φ(x)φ(y) |f\rangle\to\Phi(x)\Phi(y)$$

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

So, with coherent states, we can obtain all of the classical states in the $\hbar\to0$ 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 $\hbar$ in the diagram. Then, using standard results, we have:

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

So, if you set $\hbar\to0$, 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 $\hbar\to0$, but we do want to consider the theory up to, say, $O(\hbar^2)$. 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 $\hbar^n$ 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 $1/\hbar^{n-1}$ 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 $\hbar\to0$. I don't see any way to interpret them as classical.

Answer 4

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 theory, 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.

Answer 5

This is an excellent and very deep question.

Consider QED as an example: a classical electromagnetic plane wave has an energy density of $\frac{1}{2} |{\bf E}|^2$, while a gas of photons with frequency $\omega$ and number density $n$ has an energy density of $n \hbar \omega$. (Strictly speaking, the photon number density isn't well-defined because photon number isn't conserved, as photons are constantly splitting in virtual electron-positron pairs and recombining. But for large numbers of photons, these density fluctuations become tiny and the density becomes effectively constant.) Equating the two quantities, we find that a collection of a large number of collinear photons at the same momentum corresponds to an EM wave with amplitude $|{\bf E}| = \sqrt{2 n \hbar \omega}$.

So if we hold the number density constant and take $\hbar \to 0$, the corresponding classical wave vanishes entirely. The classical limit of any finite number of quantum particles vanishes; in order to get a well-defined classical limit, you need to take $n \to \infty$ and $\hbar \to 0$ in such a way that their product stays constant. This corresponds to including Feynman diagrams with more and more external legs. This agrees with the notes in Kyle's answer: the solutions to the Lagrangian's classical equation of motion are a sum of tree level Feynman diagrams with all possible numbers of external legs, because a completely classical wave packet corresponds to an infinite number of quantum particles.

In a QFT Feynman expansion, each vertex contributes a factor of the coupling constant $g$ and each loop contributes a factor of $\hbar$. The number of loops must clearly be less than the number of vertices, so a weak-coupling expansion where we only consider diagrams with a small number of vertices also turns out to be a semiclassical expansion where we only consider diagrams with a small number of loops (although the order of the two expansions doesn't always match up exactly). The converse is not true, because you can have diagrams with only one loop but many vertices and external legs. Such diagrams correspond to scattering processes which are "fairly classical" and are therefore important in a semiclassical expansion, but extremely weakly coupled and therefore unimportant in a weak-coupling expansion. But QFT is typically useful in contexts where we are concerned with scattering processes for small numbers of particles, so it's natural to keep the number of external legs fixed. In this case, although a Feynman QFT expansion is explicitly only a weak-coupling expansion, in practice it ends up being a simultaneous weak-coupling and semiclassical expansion.

In classical field theory we don't need to worry about loops, which makes things easier. But on the other hand, in the classical context it isn't natural to hold the number of external legs fixed, for the reason described above (any wave scattering process gets contributions from Feynman diagrams with all numbers of external legs), which makes things harder. Of course, in practice, in a perturbative expansion you eventually stop after adding up all Feynman diagrams with some maximum number of vertices, which necessarily also have some maximum number of external legs. In a semiclassical context where $\hbar$ is small but positive, this corresponds to waves with small amplitude. So unlike in the QFT context, where a weak-coupling expansion automatically ends up being a semiclassical expansion as well, in the classical context a weak-coupling expansion automatically ends up being a small-wave-amplitude expansion as well. Scattering between large waves would receive contributions from Feynman diagrams with many external legs and therefore a huge number of vertices, which would be impractical to calculate in a weak-coupling expansion.

Here's another way to think about that last point. In a linear theory, the amplitude of the outgoing waves is proportional to the amplitude of the incoming waves. So if you send small waves in, small waves come out. But in an interacting theory the classical equations of motion are nonlinear, and you can get feedback loops. It's therefore possible that you can send small waves in, but they combine nonlinearly and large waves come out. A weakly coupled theory should be "fairly linear," so this should be unlikely. So the Feynman diagrams with small numbers of both ingoing and outgoing external legs should be the most important. But in the strongly nonlinear regime, the fact that small incoming waves can produce large outgoing waves means that Feynman diagrams with few incoming but many outgoing external legs can be important - limiting the usefulness of the expansion.

TLDR: the Feynman expansion of a classical field theory is only useful when the field coupling is weak and the scattering waves have small amplitudes.

Answer 6

Hopefully this point of view complements what the other answers here have already stated. I don't think it I have seen it explicitly written in textbooks. But it is definitely implied in, e.g. the discussion of the effective action of Weinberg. Long story short, the sum of all tree level diagrams is the classical action.

Before going through the proof, it is interesting to check it in a toy model. For example, consider a scalar field at a single spacetime point (so that the field is just a number) with action $S(\phi)=\frac{1}{2}\phi m^2 \phi-\frac{\lambda}{3!}\phi^3-J\phi$. The diagrams corresponding to this action have trivalent vertices associated to $\lambda$ and vertices of single valency associated to $J$. One can attempt to solve the classical equation of motion $m^2\phi_c=J+\frac{\lambda}{2}\phi_c^2$ perturbatively $\phi_c=\sum_{n=0}^\infty\lambda^n\phi_n$, following Kyle's answer. If one plugs the solution back into the action, one obtains to order $\mathcal{O}(\lambda^6)$ an action $$S(\phi_c)=\frac{J^2}{2m^2}+\frac{J^3\lambda}{6m^6}+\frac{J^4\lambda^2}{8m^{10}}+\frac{J^5\lambda^3}{8m^{14}}+\frac{7J^6\lambda^4}{48m^{18}}+\mathcal{O}( \lambda^5),$$ as one can check using this Mathematica code:

S[[Phi]_] := 1/2 m^2 [Phi]^2 - [Lambda]/3! [Phi]^3 - J [Phi]; Collect[ReplaceAll[-S[[Phi]], AsymptoticSolve[ D[S[[Phi]], [Phi]] == 0, {[Phi]}, {J, 0, 3}][1]], [Lambda]]

adapted from this post https://mathematica.stackexchange.com/questions/26938/solution-of-equation-with-power-series-perturbation.

On the other hand, these correspond to the sum of the following diagrams:

2 symmetries

$6=3!$ symmetries

$8=2^3$ symmetries

$8=2^3$ symmetries

$8=2^3$ symmetries

$48=2^3\times 3!$ symmetries

Note how the last two diagrams both contribute $\frac{J^6\lambda^4}{m^{18}}$, and their symmetry factors combine to $\frac{1}{8}+\frac{1}{48}=\frac{7}{48}$. It is also interesting to note that had $J=0$, there would not be any tree level diagrams. But then again, in that case $\phi_c$ is a solution to the equations of motion and $S(\phi_c)=0$. On the other hand, if $\lambda =0$, only the first diagram would survive, which corresponds to the minimum of the action $S(\phi_c)=\frac{J^2}{2m^2}$ that one would obtain just like one obtains the minimum of any other parabola.

In any case, for the proof, one simply computes the limit as $\hbar\rightarrow 0$ of the generating function of connected diagrams $$\lim_{\hbar\rightarrow 0}W=-\lim_{\hbar\rightarrow 0}\hbar\log\left(\frac{1}{Z_0}\int\mathcal{D}\phi\, e^{-\frac{1}{\hbar}S(\phi)}\right).$$ The left hand side is of course just the sum of all tree level diagrams, as has been explained in the other answers. To understand the right hand part, one first notes that one can forget about $Z_0$, which is the Gaussian integral associated to the quadratic part of $S(\phi)$. After, say zeta function regularization, this $Z_0\propto\hbar^\#$ (as long as the space over which the path integral is being taken is truly a vector space), so that it contributes a term proportional to $\hbar\log\hbar\rightarrow 0$. The remaining integral on the right hand side can be evaluated by expanding the action around a solution to the classical field equations $$\int\mathcal{D}\phi\, e^{-\frac{1}{\hbar}S(\phi)}=\det\left(\mu S''(\phi_c)\right)^{-1/2}e^{-\frac{1}{\hbar}S(\phi_c)}(\cdots).$$ The $\mu$ on the $\det\left(\mu S''(\phi_c)\right)$ is a scale introduced in the zeta function regularization. At the end of the day, this factor won't contribute for the same reasons that $Z_0$ doesn't. The term $\left(\cdots\right)$ is a sum of all Feynman diagrams with vertices obtained by expanding $S$ around $\phi_c$. These are all proportional to $\hbar^{L-1}$ with $L$ the number of loops. Furthermore, they all have $L\geq 1$, since around $\phi_c$ there are no 1-valent vertices ($S'(\phi_c)=0$). Thus, their contribution goes like $\hbar\ln(\#+\hbar\#+\cdots)\rightarrow 0$. We are only left with the contribution of the classical action $$\lim_{\hbar\rightarrow 0}W=S(\phi_c),$$ thus proving our claim.

I think the connection between this answer and the one given by Kyle and the notes https://homepages.physik.uni-muenchen.de/%7Ehelling/classical_fields.pdf is the following. Consider the action obtained by adding a source to another action of interest $\tilde{S}(\phi)=S(\phi)-J\phi$. The solution $\phi_J$ of the classical equations of motion of $\tilde{S}$ then satisfies $S'(\phi_J)=J$. Furthermore, the claim in this post guarantees that $\tilde{S}(\phi_J)=S(\phi_J)-J\phi_J$ (which is the Legendre transform of $S$) is the sum of all tree diagrams coming from $S$ and the source term. Taking the derivative of this with respect to $J$ we obtain $$\frac{\operatorname{d}S(\phi_J)}{\operatorname{d}J}-\phi_J-J\frac{\operatorname{d}\phi_J}{\operatorname{d}J}=S'(\phi_J)\frac{\operatorname{d}\phi_J}{\operatorname{d}J}-\phi_J-J\frac{\operatorname{d}\phi_J}{\operatorname{d}J}=\phi_J$$ is equal to the derivative with respect to $J$ of each of these diagrams. This amounts to removing a factor of $J$ in all possible diagrams, leaving the attached edge as an external leg. In particular, if one is interested in $\phi_{J=0}$, which is the solution to the equations of motion of the original action of interest $S$, then one needs to set $J=0$ after taking this derivative. In an example like the one we had above, where the action of interest already has the current, one need not set $J=0$ at the end of course. The results from the notes are then obtained by noting that $\phi^{(0)}$ is the Green's function acting on $J$ (i.e. $J/m^2$ in our example).

It is interesting to note that this gives a really simple proof that the quantum effective action is the generating functional of 1PI diagrams (which is the one given in Weinberg). Indeed, above we mentioned that the Legendre transform of an action is the sum of tree level diagrams, with the conjugate variable at the 1-valent vertices. So, if the effective action $\Gamma$ is the Legendre transform of $W$, then $W$ must be the sum of the tree level diagrams coming from $\Gamma$ and a source. But the only way this can be true, given that $W$ is the generating function all connected diagrams, is if the vertices of $\Gamma$ are the sum of the 1PI diagrams.