#### [SOLVED] Virtual photons, what makes them virtual?

By Lucas

The electromagnetic force can be described by the exchange of virtual photons.

The virtual photon thing baffles me a little. I get that virtual particles are supposed to be short lived, but as photons live for zero units of proper time I can't see how their lifetime can be used to distinguish between virtual and non-virtual.

Another idea I had was that virtual photons are only those associated with the electromagnetic field, non-virtual ones are not. But in this case, I could not see what was wrong with this: If I have a photon detecting instrument it is just detecting the force carrying particles of the electromagnetic interactions between it and the thing I am using it to observe? (even if that thing is a long way away)

Are virtual photons just photons that you don't observe? Or, is there some kind of photon that is not connected with the electromagnetic field? Or something else? Or perhaps there is no concrete distinction to be made?

#### @anna v 2013-06-23 17:09:34

Virtual particles appear when one wants to calculate cross sections and branching ratios for elementary particle interactions. This is done with the prescription of Feynman diagrams:

Feynman Diagram of Electron-Positron Annihilation

In the above diagram the external "legs" are real particles with the quantum numbers given in the standard model table, including the mass.

The red line between the dots (interaction vertices) can be considered either a virtual electron going to the right and at the interaction point becomes a positron, since a backwards in time electron is a positron. Or a positron going to the left and becoming an electron (since a backwards in time positron is an electron). You may not believe it, but this translates into a mathematical formula which gives the value for the cross section for e+ e- annihilation into two gamma.

The virtual exchanged particle has the quantum numbers of the real particle, that is why the name carries, but it can be off mass shell: its four momentum does not dot into the mass squared.

Here are different Feynman diagrams which enter into calculating scattering cross sections.

The exchanged particles between the vertices are virtual. They have all the quantum numbers of their name, except the mass which is off mass shell. The photon is characterized by spin=1 and charge=0. The m=0 is not an attribute of a virtual photon.

#### @Selene Routley 2014-06-15 11:47:34

Here's an answer from a non-particle physicist to complement what (former) professional particle physicist Anna V has written.

"Real particles" enter and leave Feynman diagrams. Therefore, in principle, they can be detected in an experiment - they are the "terminals" of a Feynman diagram: ports through which we can "see" the system within.

In contrast, the path of a virtual particle begins and ends within a Feynman diagram. It has no "free ends" dangling over the "boundaries" of the diagram and is therefore not directly measurable. We can't detect them in experiment.

None of this is likely new to you. You're still left wondering what reality we can ascribe to virtual particles, if we can't directly detect them. You can think of virtual particles more literally as Feynman liked to do, or you might try this approach: I personally like to think of them a little more abstractly as simply as mathematical terms in a perturbation series.

A good starting point to visualise this gist is the kinds of ideas explored in the following papers:

as well as the works of the late Hilary Booth of the Australian National University. This is not standard QED and it is very specialised and contrived: think of it as an illustrative "Baby QED" for someone (like me) who hasn't mastered quantum field theory. We consider here the system of one electron, a proton (the latter thought of as a classical particle, simply setting up an inverse square electrostatic field in a Hydrogen atom and the "virtual photons" that are swapped between them. The electron in the classical potential is of course simply described by the first quantised Dirac equation. Now we add the electromagnetic field by adding Maxwell's equations and coupling the system as follows:

$$\gamma^\mu\left(i \partial_\mu - q A_\mu\right) \psi + V \psi - \psi = 0$$

$$\partial_\nu F^{\nu\,\mu} = q\,\bar{\psi} \gamma^\mu \psi$$

$$F_{\mu\,\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$$

with the Lorenz Gauge

$$\partial_\mu A^\mu = 0$$.

The first equation is the Dirac equation, the second Maxwell's equations with a charge / current (4-current) distribution determined by the probability density of the Dirac electron. The third relates the Maxwell tensor (containing the $\vec{E}$ and $\vec{B}$ fields) to the four-potential, which couples back into the Dirac equation through the "gauge covariant derivative". So we have a rather elegant, but thorny to solve, coupled nonliear system.

In the papers, the equations lead to a fixed point problem $X=F(X)$ of a certain integro-differential operator $F$, which is contractive, so the solution is the limit of the sequence:

$$X_0,\, F(X_0),\, F^2(X_0),\,\cdots$$

and can thus be solved nonperturbatively, by the contraction mapping principle and it gives an infinite series of terms corresponding to virtual pairs too. It yields an exact solution which is an infinite series, what a mathematician would call the Peano-Baker series (see Baake and Schlaegel, "The Peano Baker Series" and it is what a theoretical particle physicist would call (I believe) the Dyson series.

Now the terms in this infinite series are $X_0$: Dirac's solution of the Hydrogen atom and the higher order terms are iterated integral operators: these iterations can be thought of as the perterbations wrought by one "virtual photon", the next term involves virtual photons and virtual pair production followed by virtual pair annihilation and so forth.

The "Virtual particles" in this viewpoint can be thought of simply as an evocative "mnemonic" to the structure of the mathematical terms in the infinite series.