I'm looking for expressions for the electromagnetic fields (preferably $E$ and $B$) of a typical photon which is localised in space to some extent (i.e. I'm not interested in the infinite plane wave solution of Maxwell's equations).

  • $\begingroup$ like the radiation emitted by an oscillating dipole? $\endgroup$ – Phoenix87 Feb 9 '15 at 13:58
  • $\begingroup$ like the fields associated to a photon in an electromagnetic cavity? $\endgroup$ – glS Feb 9 '15 at 14:21

The short answer is $\vec E=\vec B=0$. Strange as it may sound, this is only one of the numerous counter intuitive facts of the quantum realm. But first let me try to clarify, without going into mathematical details, the context within which these equalities should be interpreted.

Perhaps one of the easiest ways to approach the quantum nature of the electromagnetic field in the vacuum is to study the field in a perfectly conducting cavity.

Classically, such cavity supports distinct, countably infinite modes, each with its own specific frequency and spacial profile. The electromagnetic field attributed to each mode is completely defined by the shape of the cavity, up to a multiplying constant, which represents the energy contained in each mode.

In the quantum field approach, the values of the energy each mode contains are quantized and equally spaced, given by the expression $$E = \hbar\omega (n+1),$$ with $n$ being equal to "the degree of excitation", or number of photons of each mode.

The number of photons is an observable physical quantity related to the operator $\hat{N}$, but in quantum optics, as well as in single particle quantum mechanics, not all operators of all observable quantities commute. In particular $\hat{N}$ does not commute with $ \hat{\mathbf{E}}$ and $\hat{\mathbf{B}}$, the operators of the electric and magnetic field respectively. In other words, one cannot know the number of photons and the value of the electric field simultaneously.

When the cavity is in an eigenstate of $\hat{N}$, which is usually called a number state, or Fock state, the expectation value of the electric field operator and the magnetic field operator is equal to zero. This is what the expression at the beginning implies. On the other hand, the fluctuations are proportional to the classical values of the field and increase with increasing numbers of photons.

States where the expectation value of the electric field takes its classical value, are called coherent states and are considered to be the ones closest resembling classical light more closely. In such states, the exact number of photons is not known.

To rephrase my initial statement, it is impossible to attribute some field value to a photon, except in the sense of the expectation value in a 1-photon-number state, where it is always equal to zero.

  • $\begingroup$ Ok, I understand that, it is a bit similar to an electron inside potential well. But can you expalain to me how it looks like, when we consider photons in free space. Are they localized like an electron wave packet? $\endgroup$ – Jakub Skórka Apr 20 '17 at 13:03
  • $\begingroup$ Not exactly. An electron wave packet may exhibit some uncertainty in energy and momentum, but it is still interpreted as one electron. Photons, on the other hand, being massless, have their energies sharply defined. They can only be in an energy eigenstate and they "look like" the corresponding classical eigenmode, of which they are quantum excitations. In free space, such modes take the form of plane waves. A wave packet is formed by a superposition of them, so that uncertainty is no longer a one-particle quality, but an uncertainty of the number and kind of photons that are actually present. $\endgroup$ – zap Apr 20 '17 at 13:47
  • $\begingroup$ Ok, so when we imagine a wave packet it is never a single photon but superposition of many (infinitely many) photons? Also, do I understand correctly, that if I have a single photon mode inside a cavity, I can not calculate the spatial distribution of electric and magnetic field of that mode? I know we can easily do this in case of classical fields insied conducting cavity and it seams strange that I can not do this in QED. $\endgroup$ – Jakub Skórka Apr 20 '17 at 15:01
  • $\begingroup$ Needing infinitely many photons would imply needing infinitely large amounts of energy, which is kind of unphysical. What you need is to abandon the requirement that you are in a number state, same as with the electron wavepacket, where you don't have exact knowledge of the energy. Similarly, if you are in a number state, you no longer have a fixed value for the fields themselves(not all measurements are going to return the same value), but you know the statistics. The mean value is always going to be zero, but the fluctuations are going to be proportional to the classical field. $\endgroup$ – zap Apr 20 '17 at 15:36

Light, the classical electromagnetic field , is built up/emerges from a large number of photons, not in a simple manner.

Photons are elementary particles and therefore can only be described within a quantum mechanical framework. They have a wave-function that obeys the potential form of Maxwell's equations turned into operators which operate on the photon wave function. In this link a path is shown of how the classical wave emerges from the quantum state. Hand-waving my understanding of this: a photon, in addition to its spin, has information connected with A, the electromagnetic potential which information builds up the corresponding potential of the classical wave, and thus the macroscopically observed electric and magnetic fields of light.

There exists also this preprint.

Properties of six-component electromagnetic field solutions of a matrix form of the Maxwell equations, analogous to the four-component solutions of the Dirac equation, are described. It is shown that the six-component equation, including sources, is invariant under Lorentz transformations. Complete sets of eigenfunctions of the Hamiltonian for the electromagnetic fields, which may be interpreted as photon wave functions, are given both for plane waves and for angular-momentum eigenstates

In this preprint there is explicit an E and B field in the expression of the photon wavefunction.( One should always keep in mind that the wave function squared gives a probability distribution).

  • $\begingroup$ One should also keep in mind that position eigenstate of photon is an ill-defined concept, so probability distribution is not too good of an interpretation of photon's wavefunction, at least not in a simple way. $\endgroup$ – Ruslan Feb 9 '15 at 15:20
  • $\begingroup$ Is it true that light is never wave? Light is just photons,, single particles do not produce any wave pattern - the pattern that looks like interference is only statistical. physics.stackexchange.com/questions/163082/can-light-emit-light/… $\endgroup$ – Paul Feb 9 '15 at 16:43
  • $\begingroup$ @Paul When one looks at the quantum mechanical level it is statistics=probability distributions. When the number and dimensions become large one can model continuous functions of spacetime on the wave, actually true for all waves if one goes to what builds them up in the quantum mechanical level. We do not see photons, but large ensembles of photons. $\endgroup$ – anna v Feb 9 '15 at 18:31
  • $\begingroup$ Thanks for the preprint, which states "there is not yet a consensus on the form a photon wave function should take or the properties it should have." So I suppose the answer to my question is: nobody knows? $\endgroup$ – kotozna Feb 10 '15 at 12:17
  • $\begingroup$ the answer is that there are alternate ways of modeling a wave function for a photon, which are consistent with the emergence of classical electromagnetic waves macroscopically. On the individual photon, one photon at a time experiments the electric field does not have the function of the classical field, since it is connected with a probability function, ( needs statistical accumulation) whereas the classical electric field is a vector measurable in three dimensional space. In the collective of photons the classical fields are built up. $\endgroup$ – anna v Feb 10 '15 at 12:52

The answer is: you can't ask. That is, the question is counter-factual, in the quantum-mechanical sense.

In quantum electrodynamics, your goal is to include electromagnetism into the same type of quantum mechanics that you use things like the motion of an electron around the atom or a particle undergoing diffraction in a double-slit experiment. In quantum mechanics, you take the dynamical quantity of interest (say, the position of an electron around an atom) and you build wavefunctions on top of that dynamical quantity: $$ \begin{array}{c} \text{classical mechanics}\\ x(t) \end{array} \ \mapsto\ \begin{array}{c} \text{quantum mechanics}\\ \psi(x,t) \end{array} $$ This means that you can no longer speak about "the" value of the particle; instead, if you decide to measure the position of the particle at a time $t$, then you can talk about the distribution of possible values and their probabilities (given by $|\psi(x,t)|^2$), as well as things as the average value, the width of the distribution, and so on.

In quantum electrodynamics, your variables of interest are essentially the electric field amplitudes* $E(t)$, so you do the same thing: you knock out the "having-a-value-ness" of that dynamical variable, and you change it for a wavefunction: $$ \begin{array}{c} \text{classical electrodynamics}\\ E(t) \end{array} \ \mapsto\ \begin{array}{c} \text{quantum electrodynamics}\\ \psi(E,t) \end{array} $$ This means, therefore, that you can no longer speak about "the" value of the electric field; it has become an operator and no longer has a value. Instead, you have some distribution.

Given some reasonable assumptions, the probability distribution for the amplitude of a given mode (i.e. some travelling or standing wave) when the field is in a one-photon state looks more or less like this:

Mathematica graphics

In this state, you can have a reasonably good idea of the value of the square of the amplitude, but most importantly you have no information about the phase of the oscillations of the node. Among other things, this means that the average value of the field from that mode at any given point will be zero.

There's plenty of additional subtleties to be had, but this is a reasonable starting point.

This obviously hides plenty of nuance, but the essentials stand. In reality, your dynamical variables are the amplitudes of given modes, i.e. $E(t)$ is the amplitude of some standing or travelling wave. There's also trouble in dealing with gauges and so on, but once you've done all that it still looks much the same.

Mathematica code for the image through Import["http://halirutan.github.io/Mathematica-SE-Tools/decode.m"]["http://i.stack.imgur.com/334QY.png"].


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