I) I know that virtual-photons are known to be the force-carriers for the Electromagnetic force, and that they are called "virtual" because the Energy-Time-inequality version of the Heisenberg Uncertainty Principle allows for particles that are high enough energy that they are very difficult to observe (because higher energy means a smaller possible time-scale for observation).

But I also know that photons are the quanta of EM radiation; i.e. they emitted from atoms at some point in space, and absorbed at other points in space as a means of transmitting radiation energy.

My question is this: are the photons that act as the force carrier of the Electromagnetic force the "same" photons (i.e. the exact same particle) as the photons that act as the quanta of EM radiation?

Is it just that the photons emitted as virtual particles have high enough energy that they act as a force carrier? If so, what causes charged particles to emit photons of such high energy?

II) As an add-on question: I'm being introduced loosely to Electro-weak Unification and the idea that at high enough energy, the EM- and Weak forces become indistinguishable from one another (and, I believe, that the difference between the EM-force and the Weak force, at low energy, is that the W and Z bosons that mediate the Weak force are massive, and therefor act at low range, whereas photons are massless and therefor act at long ranges). And subsequently, that the Higgs Boson helps to explain what gives W and Z bosons mass.

But what is the difference between the W and Z bosons and the photon that makes them interact with the Higgs mechanism, and the photon remain unaffected?

I hope these questions make sense.


2 Answers 2


There is only one kind of photon.

Indeed, when we describe elementary interactions between two electrons for example, we call the photon "virtual" as opposed to a physical photon that might exist outside of this process.

Still, these are the same particles, i.e. excitations of the same fundamental field, as the photons that make up light for example.

Again, virtual photons can only appear in the context of a direct interaction between charged particles, while real photons are the electromagnetic waves send out e.g. by excited atoms. Macroscopic (constant) electric and magnetic fields are coherent states of virtual photons.

Regarding the electroweak unification you seem to have a misconception. In the unified theory there is no electromagnetism any more, but only the electroweak force, which has four force carriers: The $W^\pm, W^0$ and $B$.

The Higgs field couples to all of those, giving mass to the $W^\pm$ and to a linear combination of $W^0$ and $B$, which we call $Z = \cos\left(\theta_W\right) W^0 + \sin\left(\theta_W\right) B$, while the orthogonal linear combination $\gamma = -\sin\left(\theta_W\right)W^0 + \cos(\theta_W)B$ remains massless.

So the photon is defined as the boson that remains massless after electroweak symmetry breaking.

  • $\begingroup$ Wow, excellent response, thank you. If you don't mind I have a few follow-up questions: (1) Is it incorrect to still think of virtual photons being "sent out" from charged particles, as real photons are sent out from excited atoms? If it is not, is there a property of charged particles that causes them to send out such high-energy/virtual photons instead of real photons? (2) What is the significance of the photon being orthogonal to the Z bozon, and how does that relate to whether or not the bosons have mass? (I have a relative familiarity with orthogonal functions, so I'm very interested). $\endgroup$
    – D. W.
    Commented Jun 5, 2014 at 15:38
  • $\begingroup$ Also, (3), when you say macroscopic Electric and Magnetic fields are Coherent States of real photons, is there any connection to standing waves and resonance? Is it incorrect to think of constant electric fields as standing-photon-waves with the electric and magnetic fields that describe their propagation "lining up" in such a way as to appear constant as propagating waves in a string "line up" to create a standing wave with nodes? $\endgroup$
    – D. W.
    Commented Jun 5, 2014 at 15:42
  • $\begingroup$ (1) Yes, you can think of a virtual photon as a photon that is immediately catched by another particle, leading to a sudden change in that other particle's momentum. (2) The significance is that while the Higgs boson couples to all four electroweak gauge bosons, there is one linear combination of the (as it turns out) electrically neutral ones to which it couples 'with full strength' and one to which it does not copule at all. That they are orthogonal leads to the $Z$ boson being electrically neutral, i.e. no $Z - \gamma$-coupling. And (3): yes :) $\endgroup$
    – Neuneck
    Commented Jun 5, 2014 at 17:32
  • $\begingroup$ Re (1) - Do you really mean "...*immediately* caught by another particle"? Wouldn't it need to still move at the speed of light? $\endgroup$
    – D. W.
    Commented Jun 5, 2014 at 18:02
  • 1
    $\begingroup$ No. Since they are not required to have energy and momentum on the mass shell (see rob's answer), other limits do not apply either. One naive way of picturing this is that the interaction between two electrons via a photon takes place as a small lightning bolt, connecting the two electrons for an instant, allowing momentum to be exchanged. After the lightning bolt disappears the electrons change their direction and are oblivious of each other again. $\endgroup$
    – Neuneck
    Commented Jun 6, 2014 at 6:25

A major difference between real and virtual photons is that virtual particles are not required to have energy and momentum on the "mass shell". That is, virtual photons may have $E^2-p^2 \neq m^2$, while real photons must obey $E^2-p^2=m^2=0$.

My memory disagrees with Neuneck (v1): I think that a coherent superposition of real photons is a laser, while static electric and magnetic fields are composed of virtual photons. For instance, consider the field between two point charges with opposite sign. Real photons moving along the path between the two charges can produce only transverse electric fields, but the static field is entirely longitudinal. A virtual photon with $m\neq0$, however, isn't restricted to the two transverse polarization states.

Neuneck's description of the weak mixing angle is spot-on. I'd add that you do recover a "pure" electroweak force in the limit where the energy $E$ involved in the interaction is much larger than the $W$ and $Z$ masses, $E\gg m_W$. In that case you can use the approximation that all four electroweak bosons $W^\pm,Z^0,\gamma$ have the "same" mass $m\approx0$.

  • $\begingroup$ Is there any usefulness in thinking about virtual photons as "moving" from one charged particle to another? I know in your link it says that particles that are off shell are not required to follow the normal equations of motion in classical dynamics, so I guess not... But I'm having trouble thinking about how a photon moving in any way from one charged particle to another would create a longitudinal electric field. Wouldn't the photon have to be moving perp. to the line connecting the two? (I guess since it doesn't have to travel normally, thinking about it this way is useless...?) $\endgroup$
    – D. W.
    Commented Jun 5, 2014 at 17:31
  • 2
    $\begingroup$ Nice post, also +1 for pointing out my mistake! $\endgroup$
    – Neuneck
    Commented Jun 5, 2014 at 17:34
  • $\begingroup$ Re photons "moving": You have think of all the virtual photons exchanging momentum in both directions, which is a little hairy. $\endgroup$
    – rob
    Commented Jun 5, 2014 at 17:35
  • $\begingroup$ Re polarization: The photon has unit spin, and so in principle it has three circular polarization states; however, since the photon moves at $c$ only the completely aligned and anti-aligned helicity states (or superpositions, like plane polarization) can be observed. Off-shell photons have an additional orthogonal polarization state. $\endgroup$
    – rob
    Commented Jun 5, 2014 at 17:39
  • $\begingroup$ @rob: "A major difference between real and virtual photons is that virtual particles are not required to have energy and momentum ...". They are not required or not allowed to have energy and momentum? $\endgroup$ Commented Jun 5, 2014 at 17:56

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