Consider an electromagnetic wave of frequency $\nu$ interacting with a stationary charge placed at point $x$. My question concerns the consistency of two equally valid quantum-mechanical descriptions of the electromagnetic wave. First I will describe the classical picture, then the two quantum mechanical descriptions, then I will ask experts for a conceptual unification of the two quantum mechanical descriptions. For parsimony I'll assume an extremely low frequency wave, but this is not strictly necessary.

Classical description 1: At point $x$ the electromagnetic wave contributes a slowly varying electric field, and a slowly varying magnetic field. The charge at $x$ experiences a force due to the electric field, and begins to move. As the charge begins to move it experiences a force due to the magnetic field. Using the right-hand-rule it is easy to see that the net force on the charge is in the direction of motion of the electromagnetic wave.

Quantum mechanical description 1: In the quantum mechanical description of the same electromagnetic wave, real photons are traveling with momentum $\frac{h\nu}{c}$ (in the direction of motion of the above electromagnetic wave), and are absorbed by the charge at $x$, causing it to recoil in the above electromagnetic wave's direction of motion (due to conservation of momentum).

Quantum mechanical description 2: (I will be assuming that in the quantum mechanical description of the electromagnetic field, the force due to the electric/magnetic fields between two moving charges is due to the exchange of virtual photons). At point $x$ the electromagnetic wave contributes a slowly varying electric field, and a slowly varying magnetic field. The charge at $x$ experiences a force due to the electric field, due to the exchange of virtual photons with the charge that produced the electric field. Similarly for the magnetic field. In other words, there are no real photons -- just the virtual photons mediating between the charge at $x$ and the charge whose movement created the electromagnetic wave in the first place.

Finally, my question: how are descriptions 1 and 2 reconciled? In description 1, the origin of the electric and magnetic fields (a charge), and those fields' descriptions in terms of virtual photon exchange, is completely ignored. On the other hand in description 2 there are no real photons, and the virtual photons are longer-ranged (do they self-interact?). Are the two descriptions equivalent? If so, it must be that a real photon can be written in terms of "virtual-photon" basis states. What is such a decomposition called, and can someone point me to a discussion of it?


3 Answers 3


first of all, the quantum mechanical description 1 in its current form is impossible. A charged particle can't simply absorb a real photon. This is most easily seen in the final particle's rest frame. There is no photon left so the total energy is just the rest mass of the particle, times $c^2$, but the initial state has a higher energy of the particle in this frame, because the particle was moving, and also an additional positive energy of the photon. So the energy couldn't have been conserved in this process.

A charged particle may only absorb a real photon if it emits another one in another direction. So microscopically, it's always a process composed out of the Compton scattering subprocesses. If the number of real photons in the same state is large, they may be described as a classical wave. You may partially quantize the system, keeping the electromagnetic field classical and quantizing just the particle, or otherwise. The reason why this description agrees with the full quantum description in the classical limit is manifest.

Also, I don't quite follow the difference between your classical description and the quantum mechanical description 2. Concerning the usage of virtual photons, well, if you want to study the whole process using the tools of quantum field theory - including the pre-history when no electromagnetic wave existed - then of course the electromagnetic wave had to be created at some point, and some of the photons were absorbed by the charged particle. The charged particle became virtual for a little moment as well, before it emitted another photon which is needed for the energy conservation, as I have explained.

So the photon that was absorbed by the charged particle was virtual - it only existed for a finite amount of time. However, the electromagnetic wave was probably propagating for such a long time that even this photon may be called "real". There is a simple relationship between virtual and real particles - real particles are the virtual ones that happen to sit exactly on the mass shell, so they satisfy $E^2-p^2=m^2$. This identity may be exactly checked only when $E,p$ are measured totally accurately - which means that the particles must exist indefinitely. If they don't exist indefinitely, then they're always "virtual" to some extent, but chances are that if they exist for a long time, you may also imagine that they're "real".

The "virtuality" of a particle may be defined as the difference $E^2-p^2-m^2$ - the distance from the physical mass shell. If the virtuality is low, the virtual particle may exist for a long time and look "real".

Finally, there are no "virtual photon states" in the Hilbert space. The Hilbert space only contains real particles. Virtual particles are an object that appears in the calculation of probability amplitudes for various processes - in the Feynman diagrams. Virtual particles are internal lines of Feynman diagrams, given by propagators that determine the 2-point function (correlator) of a quantum field. But they don't correspond to any physical states. There are no off-shell physical states in the Hilbert space.

So if you have a history in which some particles exist for a finite amount of time, so that they're strictly speaking virtual from the Feynman-diagrammatic viewpoint, it is still true that at every moment, there must exist some real particles that are actually present. However, it is tough, misleading, ambiguous, and unnecessary to calculate the "exact intermediate states" in quantum field theory. Such objects - wave functionals - would also depend on the field redefinitions (of the quantum fields), renormalization schemes, and other things. It's actually very useful to avoid these things when they're not necessary and only talk about the things that can be measured - the cross sections that may be calculated from the scattering amplitudes.

A problem with the "wave functionals" of the intermediate states is that they're only well-defined with respect to a reference frame - but virtually all regularizations we know to calculate the loop diagrams rely on the Lorentz symmetry. Because the Lorentz symmetry is obscure by the foliations of the spacetime, it becomes harder to "regulate" the exact wave functional at the loop level. Of course, at the classical or semiclassical level, one may describe very accurately what's going on.

In your particular situation, there was no real problem because all the photons in the problem were really on-shell, and you may present them as real photons if you wish.

Best wishes Lubos

  • $\begingroup$ Thanks. I think my question distills to this: A stationary charge produces an electric field, whose effect on another charge can be described by the exchange of virtual photons. If the charge is wiggled, the same electric field exists: virtual photons are still exchanged. However, the movement of the wiggling charge adds or subtracts energy from the emitted and received virtual photons, which can therefore "push" them on-shell: a wiggling charge effectively emits on-shell photons in addition to virtual, whereas a stationary charge only "emits" virtual photons. Does the above make sense? $\endgroup$
    – user1247
    Commented Jan 15, 2011 at 11:18
  • $\begingroup$ Hi user, yes, it makes sense, but I don't think it's true. The photons emitted by a wiggling charge are always on-shell, at least at the accuracy you can experimentally detect. And the energy (and momentum) of each photon is given by its frequency, so it is unaffected by the wiggling of the source. This is an important point and I would like to encourage you to think about it: the virtual photons responsible for the electrostatic attraction between two static objects have zero energy - it's just the difference between the first electron's final and initial energy and its frequency is 0, too. $\endgroup$ Commented Jan 15, 2011 at 12:19
  • $\begingroup$ You may be confusing the intensity of an electric field - or electromagnetic wave - with the energy of one photon. They are not the same thing at all. For the static field, the "virtual photons", if it makes sense to talk about them at all, have a vanishing energy (and frequency). For an electromagnetic wave, the energy of one photon is $E=hf$ and the total energy of the wave is much, $N$ times larger because it contains many more photons, $N$. $\endgroup$ Commented Jan 15, 2011 at 12:21
  • $\begingroup$ OK, I do not think you understood my last attempt at clarification. Let me try again. If you have a stationary charge, it is "emitting and receiving" virtual photons (they mediate the coulomb force). As you say, these virtual photons have zero energy. But if you wiggle the charge, we see that in addition to the continued virtual photon exchange, there also appear on-shell photons that carry energy away. $\endgroup$
    – user1247
    Commented Jan 15, 2011 at 14:26
  • $\begingroup$ In a consistent picture, I would expect that the same process (virtual photon exchange) involved in the mediation of the coulomb force is involved in the creation of on-shell photons, since in the classical picture (as per my "description 1") the view is similarly unified: an electromagnetic wave at time t is merely some set of electric and magnetic fields, and the forces on a charge can be calculated using the laws of interaction with a static field. So I am asking: is the same true in the quantum picture? The charge is wiggled, the virtual photons doppler shifted --> some on-shell photons? $\endgroup$
    – user1247
    Commented Jan 15, 2011 at 14:33

Lubos's answer is 100% correct, but it is missing the subtle error in OP's thinking.

The OP is imagining that if you have two charges that repel according to Coulomb's law, and you slowly shake one, the response of the other is as if it were repelled from the retarded position of the charge. If this were true, then the real photons would just be related to virtual photons, because the actual propagating signal would just be the location of the place to feel a force from.

But this is not at all what happens when you wriggle a charge. The outgoing wave-part is a 1/r field, which is completely separate from the Coulomb repulsion. In Dirac gauge, you can consider the Coulomb repulsion as to the instantaneous current position of the other particle, plus a propagating field which is 1/r. The propagating field fixes up the causality--- this doesn't actually transmit forces faster than light, but the propagating field is not related in a simple way to the Coulomb field.

The two fields, Coulomb and wave field, really are separate even classically, and it is miraculous that Feynman could combine them in quantum mechanics using virtual states. Lubos's answer covers the rest, in particular his discussion of field wave-functionals.


Feynman analyzes this situation in Chapter 20 of his "The Theory of Fundamental Processes".

  1. Working in the Lorenz gauge, he writes the electromagnetic scattering amplitude between two charged particles, $a$ and $b$, as a sum over four photon polarizations: timelike (axis 4), longitudinal (3), and the two transverse polarizations (1&2), with photon four-momentum q= {$\omega$,$Q$} and polarization $\epsilon$. $$ M=\frac{j_4^a j_4^b} {\omega^2 - Q^2} -\frac{j_3^a j_3^b} {\omega^2 - Q^2}-\frac{j_2^a j_2^b} {\omega^2 - Q^2}-\frac{j_1^a j_1^b} {\omega^2 - Q^2} $$
    "The last two terms are the expected contributions of the two transversely polarized photons. What is the meaning of the first two terms?"

  2. He then uses charge-current conservation to relate those "first two" timelike and longitudinal terms: $$ \omega j_4 = Q j_3 $$

  3. Substituting: $$ M=-\frac{j_4^a j_4^b} {Q^2} -\sum_{trans}\frac{(j^a \cdot \epsilon)(j^b \cdot \epsilon)} {\omega^2 - Q^2} $$ "If the photon transferred is real, $\omega \cong Q$. Then the contribution of longitudinal plus timelike photons to M (first term) vanishes, compared to that of transverse photons. However, in general, the virtual longitudinal and timelike photons cannot be neglected and, in fact, play a very important role."

  4. He then exhibits that role: Integrating that first term over frequency and momentum gives the instantaneous coulomb interaction between the two particles.

  5. "The total interaction, which includes the interchange of transverse photons, then gives rise to the retarded interaction."

So the coulomb interactions are due to the non-transverse polarization components of the virtual photons.

  • $\begingroup$ This is true, but it is not OP's confusion. $\endgroup$
    – Ron Maimon
    Commented Apr 13, 2012 at 7:48

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