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