Is a single photon a wave plane or a wave packet?

Question

According to the definition a photon is monochromatic, so it has a unique frequency $\omega$ and thus it can be expressed as

$\psi(x,t)=\exp i(kx-\omega t)$.

This suggests that a photon is a plane wave which occupies the whole space at the same time.

But why we can say a photon transports one place to another? In ordinary thinking a photon is more like a wave packet, and its probability density has a non-uninform distribution in the space.

So what the photon indeed is?

Answer 1

Mode expansion of the EM field usually uses modes that are delocalized in space. Photons are basically* quanta of amplitude of corresponding modes. This means that they indeed correspond to delocalized excitations of the EM field.

Just as with electrons in quantum mechanics, we can localize a single photon to a wave packet by superposing several single-photon states with different excited modes. This will no longer be the "pure" photon with a definite frequency that we discussed above. But it's the cost of localization. We see the same result when forming a wave packet from electron's definite-momentum states.

But, just as electrons, in interactions the photons, however delocalized their wavefunction is, act as point-like objects. Whenever something absorbs or scatters a photon, this happens (from a classical observer's point of view) at a particular spot, and only a single spot for a single absorption event. Examples of such interaction are registration of a photon by a pixel on a photographic sensor, or polymerization of a patch of photoresist in a lithographic process.

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^{*I discuss this in a bit more detail in the last section of this answer, that, although talking about phonons, is also relevant to photons.}

Answer 2

In the standard model of particle physics, which is the lowest frame of mainframe physics from which all other frames should emerge, a photon is an elementary point particle , no extension in space time, as are all other particles in the table.

The photon has zero mass, spin +1 or -1, and energy $E=hν$ where $ν$ is the frequency a large number of same energy photons will build up to a classical electromagnetic wave.This answer of mine may help understand that the wave nature of the photon is in the probability of its location.

In general for all particles the plane wave solutions of the basic quantum mechanical equations used in calculating the Feynman diagrams for interactions are not useful in describing the path of a single particle in space time, as you observe. One has to use the wave-packet solutions but that is not really necessary because there is no need to describe the incoming tracks or outgoing tracks in experiments, the wavepacket is included in the errors of the incoming particles four vectors as far as comparing data and theory.

Here is a bubble chamber event and a charged pion decaying into a muon and an electron:

The main interaction happens at the vertex on the top. That has the specific wave function which the experiment is studying, i.e. measuring multiplicity, and finding energy and momentum by using the imposed magnetic field and accumulating data to compare with theory.

The pion and the other tracks can be described in a wavepacket but the measurement errors are much larger and the classical approximation of $Bqv=mv^2/r$ is adequate to give the momentum of this pion, using a wavepacket would be overkill.

Answer 3

Photon may mean different things, depending on the context.

• EM field quantization When quantizing electromagnetic field, we expand this field in its eigenmodes and define photons as the quanta of excitation of an eigenmode. If the quantization is done in free space, these eigenmodes are plane waves, and photon can be though to be associated with a plane wave (although saying that photon is a plane wave would be technically incorrect).
• Photon emission Emission of a photon, e.g., by an atom, happens over finite time and in a finite volume, determined by the environment, the density-of-state of the field, how the field is measured, the age of the universe, etc. Thus, no real emitted photon is an excitation of a plane wave, but rather a packet of waves (i.e., wave packet in an OP parlance, although the plane waves remain the modes, not the quanta).
• Phase-number uncertainty a salient point in the previous two bullets is that photon is not the field. In fact, photon number does not commute with the phase of the electromagnetic field, i.e., a state with a single photon is characterized by random EM field, which fits neither definition of plane wave, nor that of a wave packet.

The list above is by no means exhaustive - e.g., @annav yet mentions a different meaning of photons in particle theory.

Answer 4

One can derive the longitudinal and lateral intrinsic fields of a photon by equating the expectation values of the field operator in terms of the electric and magnetic field expectation values with the expectation value obtained with the wavefunction of two consecutive number states.
One can have a monochromatic temporal pulse by using the mixed time-frequency representation which involves time-varying spectra, also known as the Wigner spectrum. This localizes the energy of the photon without the difficulties of the Fourier limit relation of time spread multiplied by the frequency spread. More mathematical and physical details can be found by searching online for "Instantaneous Quantum Description of Photonic Wavefronts and Applications"

Answer 5

All you need is some mode with a definite freqency so that in that mode the EM field dynamics is that of a harmonic oscillator having that frequency. In a conducting cavity there are many such modes and a photon can occupy any of them. Similarly there are many confined modes in an optical fibre or waveguide and any of them can be occupied by any number of photons.

Answer 6

A photon is a point-sized particle whose position is described by an extended, wave packet-shaped probability distribution. This means that there is a degree of limited information in the position, which stands at a trade-off relationship with the information in the momentum, so that if you have a situation in which you can attribute full information to the momentum, there must be zero information in the position, i.e. what you are talking about with a wave plane. If, however, you admit partial information about both, you can have a localized wave packet, but interestingly, inevitably it can never be perfectly localized, i.e. there must always be some "variance" in the position distribution.

This is the part that is often missed in these "is it a particle or a wave/field/etc." back-and-forths: position is a relationship of an object versus other objects or else to a fictitious coordinate grid. It is not the same as shape of that object. In fact, we can apply quantum mechanics to hypothetical hard spheres just as we can apply Newtonian mechanics to them, though there's no evidence real-life particles are, in fact, hard spheres of a finite size. And what happens is the information describing this relationship becomes restricted in quantum mechanics.

In terms of the EM field, you can think of it as like a single point where the E/B field is not definitely zero, but the spatial location of this point is underdetermined in the same way that the position of a particle in ordinary QM is underdetermined. That is, the EM field for a 1-photon state is a superposition of configurations with a single excited point, weighted by the probability requirements corresponding to the photon position-space wave function.

The impossibility of perfect localization can be understood by comparing to the analogous situation with a "phonon" (note the "n") in a finite crystal lattice. A phonon is to sound what photons are to light, hence the name (as in "telephone", or "phonograph", and the like). There, one can derive that, specifically when the lattice is finite, while one can construct pointlike excitations of arbitrarily good positional localization, they only "count", in the sense that the number operator "recognizes" them as single-phonon states, as single phonons so long as you don't make them "too localized", which basically means you don't add up too many Fourier modes (how many you can add up depends on the lattice size, with more naturally becoming possible the larger the lattice gets and thus "approximates infinity") to make the $\psi_x$ wave packet that you're talking about. And in this paper:

https://arxiv.org/abs/math-ph/0607044

it is described how that for relativistic quantum fields, which the EM field is an example of, the situation is directly analogous to the finite crystal lattice case. In effect, try to localize a phonon or photon "too well", and it transforms into multiple phonons or photons or, even more accurately, a fuzzy (in the same sense of lacking information as in the position) amount of such.

Indeed, this doesn't just apply to photons, but to all relativistic particles including even electrons, where the scale of maximal localization is the Compton wavelength, $\frac{h}{mc}$, about 2.4 pm for an electron (compare against the diameter of a $\mathrm{H}$ atom, about 106 pm). Note again, this is not the "size" of the electron, only the scale of maximal compression of the wave function of position, or if you like, the absolute maximum amount of "precision" that can be put into the position "variable". The shape or avatar of the electron is unchanged.

Answer 7

A photon is not a wave, but a point particle. The behaviour of this particle is described in a probabilistic manner by the electromagnetic wave.

See also this answer: https://physics.stackexchange.com/a/161056/186017