In the second quantization approach to quantum field theory, how I understand it, the field is decomposed in components of definite momentum which are treated as non-interacting harmonic oscillators, which are then quantized.

The spectrum of a quantum harmonic oscillator consists essentially of the non-negative integers, and it is natural to interpret the state corresponding to the eigenvalue $n$ to be the state that contains $n$ particles of the corresponding momentum. (Please let me know where you think that my understanding of second quantization is mistaken, if you think it is.)

In other formalisms, especially in the path integral formalism, of course we still have a photon field, fermion fields, etc, but the nicely countable discrete excitations are not explicitly present, for as far as I can tell.

Could we say that the concept of a particle in QFT as a discrete and countable quantity, in other words one that somewhat resembles a particle in our everyday experience, is something specific to second quantization (more generally: the formalism we're working in), rather than a concept inherent in QFT?

  • $\begingroup$ QFT is a mahtematical model, used in several frames.(as nuclear physics for example).In particle physics I think that QFT uses plane wave solutions of the corresponding quantum mechanical equation of the particles described: Dirac for fermions, Klein Gordon for bosons , quantized Maxwell for photons. The harmonic oscillator states are just used as an example of a field theory. As is well known plane waves cover the whole space time so the theory has to use a wavepacket to define a real particle localized in space and time, $\endgroup$
    – anna v
    Jul 23, 2019 at 3:44
  • $\begingroup$ afaik. If one is just interested in the scattering amplitude the formalism gives it correctly. $\endgroup$
    – anna v
    Jul 23, 2019 at 3:45
  • $\begingroup$ The validity of the particle concept in QFT is more a function of the specific physical problem being solved rather than the formalism used. You may find physics.stackexchange.com/questions/451492/… of interest. $\endgroup$
    – isometry
    Jul 23, 2019 at 15:27

2 Answers 2


Ultimately yes but there are subtleties preventing one from "counting" the photons: specifically, infrared divergences. These are relevant for massless fields (in particular for photons) but not for massive ones. The difficulty vaguely speaking is separating the situation where one has one photon at (close to) zero momentum from some number, possibly infinite, of photons also at zero momentum.

I can't do infrared divergences justice in an answer here so let's pretend we're talking about a massive scalar field instead (like the Higgs). You can ask most of your questions for such a field anyway.

Particles do exist in QFT regardless of the formalism one uses to calculate the S-matrix (which is one major goal of QFT). In particular, the LSZ argument for calculating the S-matrix from vacuum correlators (the latter can be calculated in any formalism one desires, e.g. using path integrals) fundamentally relies on the existence of asymptotic one-particle creation/annihilation operators; one uses the latter to construct a Fock space of multiparticle states, which decomposes into a sum of $n$-particle states for any $n$. In this sense particles exist independently of how one arrives at the vacuum correlators.

As you might have anticipated, the LSZ prescription doesn't quite work in the presence of infrared divergences. These can be taken care of, however, by modifying the definition of particle (to include a cloud of "soft particles" --- ones with small momenta as in the first paragraph --- alongside the "hard particle").


Could we say that the concept of a particle in QFT as a discrete and countable quantity

QFT is drastically different from (single particle or few particle) quantum mechanics: QFT states are NOT countable, while QM states are countable.

Let's take a look at the canonical case of QM: the harmonic oscillator. The states are ground state, first excitation, second excitation, etc. Therefore the states of the quantum harmonic oscillator are in one-to-one correspondence to non-negative integers (zero and positive natural numbers), which means countable.

Now let's turn to QFT. To simplify things a bit, let look at an infinite length chain of quantum spins, characterized by up and down states at locations $ i =1, 2, 3$, etc. The state at the location $i$ can be described as $n_i = 0$ or $n_i= 1$, corresponding to spin down or up.

Are the states of this quantum chain countable? let's construct a binary number (rather than a decimal number) $$ x = 0.n_1n_2n_3... = \frac{1}{2}n_1 + \frac{1}{4}n_2 + \frac{1}{8}n_3 + ... $$ The number $x$ is actually a real number ranging from 0 to 1! Unlike the natural number, real number is NOT countable according to Cantor's categorization of infinite sets.

In other words, QFT states are "more numerous" than QM states in Cantor's sense.

  • 1
    $\begingroup$ "QM states are countable" - not always. Counterexample: the set of position basis states $|x\rangle$. $\endgroup$ Jul 23, 2019 at 15:19
  • 1
    $\begingroup$ @probably_someone, there are different reasons that position ket $\lvert x \rangle$ does not define a state in the abstract sense. See for example here: physics.stackexchange.com/questions/351658/… $\endgroup$
    – MadMax
    Jul 23, 2019 at 15:30

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