I never learnt QFT and I apologize for my (probably) elementary question. Somebody told me that in QFT a particle is viewed as an irregularity in the field.

On the other hand, in an article in Wikipedia I see the sentence "A QFT treats particles as excited states of an underlying physical field, so these are called field quanta."

Which one of the true is a better description? The 1st description hints that the particle is a localized phenomenon inside a field that maybe occupies a big region in space. The 2nd description speaks of an "underlying" field. So, is there a field and in addition there is a particle? If it is, then what is the occupation number of that "underlying" field?

None of these approaches is clear to me, I know the approach in QM, and none of them resembles the QM.

The motivation behind my question is a certain similarity that I find between the above descriptions and the Bohm interpretation of QM, (i.e. the background field - in Bohm's interpretation there is a background quantum potential - and a particle floating in it.)

In all, is a particle treated in QFT as a localized phenomenon inside a field occupying a wider volume? I would appreciate a simple and direct answer.

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    $\begingroup$ In rough terms, in QFT, elementary particles are associated with fields. Then a particle in a given state is a mode of "vibration" of this field. The characteristic picture to have in mind is a vibrating string, where the state can be described by the frequency content of the oscillations. $\endgroup$ – Phoenix87 Feb 6 '15 at 13:38
  • $\begingroup$ @Phoenix87 thanks a lot, but I will see your answer later. I am now breaking my head with a "cub" that asked for help, but didn't explain the definitions of the quantities with which he works. $\endgroup$ – Sofia Feb 6 '15 at 13:50
  • $\begingroup$ @Phoenix87 : why don't you post an answer and get points for your explanation? Now, tell me please, isn't there in the QFT a concept parallel to the wave-packet that we have in QM? A wave-packet is something localized in space - it occupies a finite volume. $\endgroup$ – Sofia Feb 6 '15 at 17:31
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    $\begingroup$ my comment doesn't really convey anything concrete. I think ACuriousMind has given plenty of details and it looks to me like a good answer to your question as it is, but I understand that perhaps it is a bit too technical for your needs. $\endgroup$ – Phoenix87 Feb 7 '15 at 14:05
  • $\begingroup$ @Phoenix87 : yes, you understand correctly. I have to stop at a couple of points and ask questions. But, I will ask, both you and him. And please, don't be shy. If you explain things, you deserve points. I like to express my appreciations for being given help. $\endgroup$ – Sofia Feb 7 '15 at 15:12

Somewhat surprisingly, the "generic" particle of QFT is in fact totally delocalized.

More precisely, particles are thought to come from the mode expansion of free fields. Since every free relativistic field $\phi$ fulfills the Klein-Gordon equation $(\partial^\mu\partial_\mu - m^2)\phi = 0$, a Fourier transform shows that it can be expanded as

$$ \phi(x) = \int \frac{\mathrm{d}^3p}{(2\pi)^3}\frac{1}{\sqrt{2p^0}}(a(\vec p)\mathrm{e}^{\mathrm{i}px} + a^\dagger(\vec p)\mathrm{e}^{-\mathrm{i}px})$$

where Lorentz invariance is not manifest, but can nevertheless be shown. A quantum field is operator-valued, and the operator valued objects $a(\vec p),a^\dagger(\vec p)$ fulfill exactly the correct commutation relations to be interpreted as creation and annihilation operators. The $n$-particle state of particles that are associated with the field $\phi$ is now defined as

$$ \lvert n;p_1,\dots,p_n \rangle := a^\dagger(p_1)\dots a^\dagger(p_n)\lvert \Omega \rangle$$

where $\lvert \Omega \rangle$ is the (mostly) unique vacuum state. In this way, you first create all particle states that are sharply localized in momentum space (and hence completely delocalized in position space) and you can build localized particle states by the usual building of "wavepackets" with fuzzy momentum out of the sharp momentum states:

A QM wavepacket of width $\sigma_x$ localized at $x_0$ is constructed out of the pure momentum states $\lvert \vec p \rangle$ as something like $$\lvert x_0,\sigma_x\rangle = \int \frac{\mathrm{d}^3 p}{(2\pi)^3}\mathrm{e}^{\mathrm{2i\sigma_x^2(x - x_0)^2}}\lvert p \rangle$$It works exactly the same for localized QFT particles, except that one should multiply the measure with $\frac{1}{\sqrt{2p^0}}$ to have a Lorentz invariant integration, and, of course, $\lvert p \rangle = a^\dagger(p)\lvert \Omega \rangle$.

The idea that "particles are local excitations of the fields" comes from the observation that this mode expansion is almost completely analogous to a classical field fulfilling a wave equation like the Klein-Gordon equation, where the $a(\vec p),a^\dagger(\vec p)$ would directly represent an excitation of the field of wavenumber $\vec p$. It cannot be made precise in the context of QFT because the quantum field is operator-valued and has no definite values, so it is wholly unclear what rigorous sense could be given to it being "excited". It is a nice picture, but nothing you should take too literally.

Also, take note that this is for the free field. The true interacting field of a QFT cannot be mode expanded in this way, and particle states are (through the LSZ formalism) only obtained in the asymptotic past and future (when they were far enough apart for interactions to be effectively non-existent) of the theory - the Hilbert space (and hence any states you could or could not identify as particles) of interacting QFTs is essentially unkown.

Furthermore, more mathematical methods of constructing QFTs often first construct the $a,a^\dagger$ and the Fock space of particle states, and then define the field out of it - then, the roles of particle and field as "fundamental" and "derived" are somewhat reversed.

  • $\begingroup$ My dear, I am trying to understand your explanation, part by part, as I am alien to QFT. The fields in QFT are relativistic fields? In simple terms, down to my understanding, the particles studied by QFT should be relativistic? I thought that QFT is a parallel to quantum optics, i.e. deals with the 2nd quantization for particles. But do they have also to be relativistic? $\endgroup$ – Sofia Feb 6 '15 at 17:00
  • $\begingroup$ @Sofia: It is not necessary to consider relativistic QFT - relativistic QFT is the quantization of classical relativistic field theory, and one can as well quantize non-relativistic classical field theory. My answer assumes a relativistic theory, though. It is also more natural to discuss particles in the relativistic setting, since most particle experiments are high energy and since, that way, you can get mass from the momentum squared (otherwise, it is quite unclear what a "mass operator" should be). $\endgroup$ – ACuriousMind Feb 6 '15 at 17:21
  • $\begingroup$ @Sofia, in chapter 1 of Brian Hatfield's Quantum Field Theory of Point particles and Strings, he '2nd quantizes' the non-relativistic Schrodinger equation, promoting the wavefunction to an operator valued field and showing how to, in the field operator formalism, construct a one-particle state described by a '1st quantized' wavefunction. If you can get your hands on a copy of this book (or a copy of chapter 1), I think you'll find it quite helpful. $\endgroup$ – Alfred Centauri Feb 6 '15 at 23:08
  • $\begingroup$ @AlfredCentauri : Hmmm! I will ask our librarian. Many thanks! $\endgroup$ – Sofia Feb 7 '15 at 0:08
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    $\begingroup$ In the wavepacket of width $\sigma_x$, should there be another integral over $d^3x$ and the factor $e^{-ipx}e^{-(x-x_0)^2/2\sigma_x^2}$ instead of $e^{2i\sigma_x^2(x-x_0)^2}$? $\endgroup$ – adipy Jun 26 '15 at 15:28

A good way to understand a particle in quantum field theory is through an analogy with solid state (condensed matter) physics.

Imagine a solid, there is a lattice of atoms, with a regular spacing. But the atoms can move, if one is displaced towards its neighbor, that neighbor moves a bit away thus influencing its neighbor and so on. So there is an underlying structure, the stationary lattice. And there are disturbances to that underlying structure, these various modes of vibration. You could imagine a small vibration, or a larger vibration. These vibrational modes are called phonons, and that's not a typo, they are like particles of sound. They aren't fundamental particles, they are modes of vibration of the lattice. But they obey certain rules, particularly if we look at the quantum nature of the lattice interactions.

Now instead of a lattice of atoms, imagine a single electron-positron field, filling all of space. It might have a vacuum state, and maybe there are disturbances or modes of vibration of that vacuum state, some of these modes are called electrons, some are called positrons. Some have momentum in one direction, some in others. Some have a spin of up, some of spin down. Each is a mode in the single unified electron-positron field that fills spacetime. Just as the lattice vibrations had quantized modes and we called them phonons, so these modes of the electron-positron field are quanta as well and they are electrons (or positrons).


If you look at a bunch of masses connected by springs, the normal modes are not localized at all. You can actually compute the eigen-modes for a 1d system of a few identical masses with identical springs, and it's a good exercise if you've never done it before. Similarly, the modes of a quantum field are not localized in the slightest and the field itself is everywhere.

It's similar to a wave having a Fourier transform, or a periodic wave having a Fourier series. When it has a Fourier version you can describe it as "made up" of a bit of this mode, and a bit of that mode. If later you have a very different Fourier transform then you can talk about dome of those bits having been destroyed or created. None of the parts, none of the modes, none of them are localized.

In QFT a particle is sometimes viewed as an irregularity in the field

If you think of the lattice as regular, then those vibrational modes can be thought of as an irregularity, but they are not localized in the slightest. One mode of vibration could be to mentally group the atoms into pairs of partner neighbors that then keep their center of mass constant as they move symmetrically towards their partner then away from their partner in a regular periodic fashion. That mode is not localized in the slightest. All the modes are qualitatively like that, they can be thought of as deviations from the regular lattice but they are deviations with their own regularity and are spread out everywhere, just as the original lattice was spread out everywhere.

So for QFT, instead of a lattice you have a vacuum state of a quantum field (such as the electron-positron field). The vacuum state is not zero, it is not trivial, and it has regularities. It is spread out everywhere. The higher energy modes of excitation are like deviations that themselves are regular and spread out over all space, just like the vacuum state was regular and spread out over all space. And while you are used to thinking of there being many electrons, there is only one field for every electron in the universe (and all the positrons share that field too). It is just allowed to have many excitations at once, like how a vibrating string can have many harmonics at once.

QFT treats particles as excited states of an underlying physical field, so these are called field quanta

The exact same thing, there is an underlying field, it can be excited in many ways, the excitations are what we call particles. They are like modes of vibration or deviations of the vacuum state. And all of them are spread out over all space.

  • $\begingroup$ thank you, but my question was about localization. Whether the particle in QFT is a spacely-localized concept. $\endgroup$ – Sofia Feb 8 '15 at 12:48
  • $\begingroup$ @Sofia Maybe you can rephrase your question then, you mentioned two descriptions that aren't actually different and neither of which implies localization in the slightest. So I gave you a valid conceptual picture so you know what is going on in QFT. I'll add more to my answer, but please edit your question if I'm not taking it in a useful direction. $\endgroup$ – Timaeus Feb 8 '15 at 16:28
  • $\begingroup$ A very pictorial answer. $\endgroup$ – buzhidao Jul 28 '15 at 2:55

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