Tree level QFT and classical fields/particles

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...

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.

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.