Photon is a "particle of light". Light is just a propagating EM field. Therefore photon is (at least intuitively) a localized EM field (i.e. wavepacket).

In quantum optics, the Hamiltonian of a single-mode light is given by $H= \hbar \omega ( a^{\dagger} a + \frac 1 2)$ where $a$ is the annihilation operator. Here, $a^{\dagger} a$ is the number operator, i.e. it gives the number of the photons with angular frequency $\omega$.

However, since wavepackets are not pure planewaves, they cannot have a well-defined angular frequency $\omega$ (although it could be defined upto some uncertainty).

This seems weird, since all the standard quantum optics book talk about "modes of (well-defined) angular frequency $\omega$" and their excitations i.e. photons.

Therefore, I am fundamentally misunderstanding some conceptual issue here. Could anyone point me out where am I wrong?

  • $\begingroup$ A photon is not localized as one easily can tell from Heisenberg's uncertainty principle. Particle physicist call it a particle but it's. not some. It's a quanta of the e.m. field. $\endgroup$
    – Jan Bos
    Jun 1 at 15:27

E&M field creation operators are labeled by a momentum $k$ and a helicity $\{ +, - \}$: $$ a^\dagger_{\pm, k}. $$ Let's denote a single particle photon state via $$ |k, \pm \rangle \equiv a^\dagger_{\pm, k}| 0 \rangle. $$ We can now "fuzz out" the momentum and make a positive helicity wave packet state as, say, $$ \int d^3 k' f(k - k') |k', + \rangle $$ where $f$ is a sharply peaked function around $0$. Note that the above state is a superposition of single particle states. In other words, this state contains only a single particle but its momentum is a bit uncertain.

People often talk in terms of photons as having a definite momentum as an idealization, which happens when you replace $f(k - k')$ with $\delta^3(k - k')$. However, if $f$ is a sharply peaked function, these two descriptions are qualitatively the same.

Edit: A plane wave solution to the classical E&M field takes the form

$$ A_\mu = \epsilon^\pm_\mu e^{- i k x} $$

where $k = (k_0, \vec{k})$. (This is in Lorenz gauge, where $\epsilon^\pm_\mu A^\mu = 0$ and we have a residual gauge symmetry $\epsilon^\pm_\mu + \lambda k_\mu \sim \epsilon^\pm_\mu$.) If you calculate the $\vec{E}$ and $\vec{B}$ fields arising from this vector potential (by calculating $F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ and taking the real part) you will see that this corresponds to a light wave.

Because the Hamiltonian for this field is a bunch of harmonic oscillators, two for every three dimensional momentum vector $\vec{k}$ (two because of the two helicity choices for $\epsilon^\pm_\mu$) you can give a simple basis of states by specifying the occupation number of each mode. We then say that these states correspond to "$n_1$ photons with momentum $k_1$ and helicity $+$, $n_2$ photons with momentum $k_2$ and helicity $-$, ... " etc.

QFT states are wavefunctionals of different possible input classical $A_\mu$'s on a time slice (with subtleties arising from gauge transformations in this case that I will not get into.) Different classical field configurations are more or less likely in a given state. A single particle state travelling with momentum $k$ means that, if you could somehow measure the $k^{\rm th}$ Fourier component of the field, $\widetilde{A}(k)$, it would on average have a higher value than all the other Fourier components. Indeed, the wavefunctional of these Fourier components just look exactly like the wave functions of the harmonic oscillator, but where the x-axis, instead of being position, is instead the value of the $k^{\rm th}$ Fourier component of the field $\widetilde{A}(k)$.

Now we come to the issue of if the whole field vibrates. Indeed, the classical solution I wrote above, $A_\mu = \epsilon^\pm_\mu e^{- i k x} $, is an infinite plane wave extending throughout all of space. This is quite unlike particles we experience in ordinary life, which don't seem to be extended infinitely throughout all of space.

However, this also happens in plain old QM, not even QFT. The wavefunction of a free particle with momentum p in quantum mechanics is $\psi(x) = e^{i p x}$. Note that this also describes an infinite plane wave. In Fourier space, $\widetilde{\psi}(k)$ is a delta function, $\widetilde{\psi}(k) = \delta(k - p)$. (With some 2pi's don't care about). However, if you smooth out the delta function in Fourier space from being infinitely sharp to being just some Gaussian, then the position space $\psi(x)$ will also be a Gaussian in position space. It'll be a "Gaussian wave packet." In general, the position uncertainty of this Gaussian wave packet will always be at least a few multiples of the wavelength, you can't do much better than that.

This truly is what a particle is. Particles aren't little points. They have Compton wavelengths. For instance, in an atom, the wave function of the electrons have some spatial spread, which is the Bohr radius of the atom.

Of course, one can point out that wave packets get larger over time. However, it seems as though, for instance, particles of air remain roughly "point like" over time. Why is that? Well, that has to do with the fact that the air molecules are all colliding with each other very frequently. This environmental entanglement "localizes" them via quantum decoherence, but that is a whole other topic of discussion.

  • $\begingroup$ I see... But how can the photon state $|k, + \rangle$ be interpreted physically? I don't think that it can be interpreted as "a single photon of momentum $k$ flying through space", since a photon is a particle so it must be somehow localized in space. Is $|k, +>$ just a mathematical convenience? $\endgroup$
    – Moca Aoba
    May 31 at 13:57
  • $\begingroup$ Please see my latest edit $\endgroup$ May 31 at 15:22
  • $\begingroup$ > This truly is what a particle is. Particles aren't little points. They have Compton wavelengths. -- You have jumped from photons and wave packets to ontology of massive particles and seem to suggest Compton wavelength of a particle is its size. That is misleading. Particles such as electron are much smaller than their Compton wavelengths. $\endgroup$ Jun 3 at 19:39
  • $\begingroup$ Particle wave functions can in principle be localized to an arbitrarily small region of space (which would be a delta function) but it would mean the particle would have to have an arbitrarily large energy. A "typical" size however would be roughly the Compton wavelength. You are right that the spread of the wave function (which gives the position uncertainty) and the "size" of the particle, in usual parlance, are different concepts. However, I would submit that the "size" of the particle is meaningless in QFT. There is no "particle radius" parameter that has been set to 0. $\endgroup$ Jun 4 at 3:38

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