In 'classical' quantum mechanics, a wave packet is a (more or less) localized particle. The wave packet can be expanded in a superposition of plane waves, each with a defined momentum and energy. This superposition is again still a wavefunction of one particle, with its physical interpretation being the probability amplitudes in space.

If we move up to quantum field theory, the quantized field is also a superposition of plane waves, which each represent a possible excitation (particle) of the field with well-defined momentum and energy. So let's say the electromagnetic field has many different quanta, created through several creation operators acting on the vacuum of the E-M field. They interfere with each other and again form a total 'wave packet' in configuration space. Should I interpret this as multiple photons, or can I think about it as if there is only one photon, more localized in space, but less localized in momentumspace?


The wave packet is a superposition of different photon number states.

A wave packet is just a superposition of a bunch of different single-frequency plane waves whose amplitudes interfere destructively everywhere except around the wave packet peak. Quantum mechanically, each of these plane wave components is best represented by a coherent state, which is itself not a state of definite photon number but a superposition of all the photon number states. So if you measure the photon number of one mode in a wave packet, there's actually a miniscule probability that you'll find arbitrarily many photons in the field! The presence of so many states is required to produce the interference that gives the wave packet.

  • $\begingroup$ Incidentally, you are right that the more localized the wave packet is in space the less well-defined its momentum will be. However, this is because sharp localization in position requires use of a wide frequency distribution. So you need more modes, each of which is an independent harmonic oscillator, to make your wave packet. Note that this is a little different from the uncertainty principle derived from the first quantization commutator, since in QFT position is a parameter not an operator. $\endgroup$ – Munthe Aug 23 '18 at 16:55
  • $\begingroup$ I should also say that it is possible to produce a one-photon wave packet, i.e. a superposition of different one-photon states where all but one of the modes are in the vacuum state. However, I don't think this will result in a spatially localized pulse like it seems you are picturing. Rather, each of the orthogonal eigenstates has a completely delocalized photon in some mode of the field. $\endgroup$ – Munthe Aug 23 '18 at 17:03

First, just to clear up some confusion perhaps. A superpositon of one-particle states $$ |\psi\rangle = \sum_n |\phi_n\rangle c_n , $$ is still a one-particle state. To get a state with multiple particles from one particle states one needs to form a tensor product$^*$ $$ |n\rangle = \frac{1}{n!}|1\rangle|1\rangle|1\rangle ...|1\rangle . $$

Now, what is a wave packet and how does it relate to particles? In general, a wave packet is the same thing as a wave function. However, the notion of a wave packet is related to the idea that it is almost like a particle in that it is somewhat localize.

If we now think of a wave function, then one can think of a complex field with the physical meaning that it can tell us what the probability is to observe a particle. In that sense, the wave functions (or wave packet) represents a single excited (a single particle).

When one allows creation operators to act multiple times, then one will end up with multiple particles. If they are all in the same state (assuming they are bosons) then one can use the same wave function to represent the multi-particle state. Otherwise, the wave function would need to be a function of multiple sets of of vaiables, one for each of the particles.

Different particles don't actually interfer with one another is the same way that photons woudl interfere in an optical interferometer. However, one can get quantum interference where the different terms in the expanson of a multi-particle state can cancel one-another, as in the Hong-Ou-Mandel effect.

Hope this helps.

$^*$ Often people don't like to think of it like this, but rather as multiple excitations in some Hilbert space, which is generated by multiple applications of creation operators.


An experimental particle physicists answer, who has been measuring particles for decades:

Quantum field theory is used to get numbers to be compared with experiments, (crossections, decays, mass distributions ), the calculations using Feynman diagrams. The particle fields are assumed to be plane wave solutions of the corresponding quantum mechanical equations (Dirac, Klein Gordon, quantized Maxwell). These wavefunctions enter the integrals for the calculations.

It is not necessary to model a single elementary particle running along in order to validate the theory (standard model in this case). Nevertheless, as is well know a plane wave cannot be localized, it goes from -infinity to +infinity in space time, so the need of a wavepacket solution arises. This solution is consistent with the Heisenberg uncertainty principle which has to hold when one measure a single particle of a given momentum, and the probabilistic nature of quantum mechanical entities.

The wavepacket solution describes a single particle when measured by itself in a consistent quantum mechanical probabilistic manner. It is one electron in the picture: turning in the magnetic field of the chamber


If one wants to describe its wavefunction in QFT it will be a wavepacket , interacting and ionizing consecutively the bubble chamber medium.

  • $\begingroup$ "The particle fields are assumed to be plane wave solutions of the corresponding quantum mechanical equations" - i hope you don't mind me asking here instead of another question, and I hope I'm not "going all philosophical" here, but... in classical wave mechanics we can "consider" a wave to be the superposition of many plane waves. We can also consider it to be an infinity of spherical waves. We don't consider either "to be the wave", do we? But that seems to be what is happening here, it seems we are ascribing a mathematical short cut to "be real"? $\endgroup$ – Maury Markowitz Aug 24 '18 at 14:33
  • $\begingroup$ @MauryMarkowitz In quantum field theory each particle has a field on which a creation operator creates a particle and an annihilation operator annihilates it, and theoretically it should propagate.BUT it is plane waves that are used which have no meaning for a path in space time as in the picture. Thus the wavepacket. In quantum mechanics the waves are probability waves . A plane wave's Ψ*Ψ (probability distribution) has the information that the particle is somewhere from -infitnity to +infinity in space time when the creation operator works on it. $\endgroup$ – anna v Aug 24 '18 at 15:35
  • $\begingroup$ This is not what is happening in the picture. Fortunately there is the Heisenberg uncertainty principle and the wavepacket formalism to solve the conuundrum. $\endgroup$ – anna v Aug 24 '18 at 15:35

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