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Edit: my question is not directly about Feynman diagram, but instead about the nature of off shell objects, were they as common as the on shell ones. I am trying to suggest that if an electron goes off shell in a suitable way, then it will be impossible to tell it apart from muon. Again, i have no intention to directly refer to internal lines in Feynman diagram.

A particle is said to be moving off shell if $E^2-p^2\neq m^2$, where $m$ is the known rest mass of the particle.

But how do we measure the rest mass of a particle for the first time?

We measure it by using the formula $E^2-p^2= m^2$. (We accelerate the particle in a known electric field, measure the acquired velocity and then calculate the mass using the equations of Special relativity.)

Now, thankfully, in everyday life, when we repeat this experiment, we get the same value for the mass, making us believe that the rest mass is a constant.

But in QFT, we come across the idea of off shell motion, in which, $E^2-p^2= (m+a)^2$, where $m$ is the 'known' rest mass and $a$ is an arbitrary number.

So, basically, a particle moving off shell will appear as a normal particle with a slightly different mass. So, for a person who is unaware of its known rest mass, a particle moving off shell would appear as a completely normal relativistic particle.

In other words, according to my argument, muon can be considered as an electron moving off shell.

Is this interpretation correct?

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    $\begingroup$ Virtual particles are internal lines in a Feyman diagram. That's it. There is no physical state in the theory where there are any virtual particles. You should not trust people who try to say that virtual particles "pop in and out of existence" or such, because they don't know what they are talking about. physics.stackexchange.com/questions/221842/… $\endgroup$ – Robin Ekman May 14 '17 at 12:43
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So, basically, a particle moving off shell will appear as a normal particle with a slightly different mass. So, for a person who is unaware of its known rest mass, a particle moving off shell would appear as a completely normal relativistic particle.

Is this interpretation correct?

Virtual particles are a mathematical shorthand in a specific method of calculating the fundamental interactions of particles. This Feynman diagram will make it clearer:

feynman diagram

Only lines entering or leaving the diagram represent observable particles. Here two electrons enter, exchange a photon, and then exit. The time and space axes are usually not indicated. The vertical direction indicates the progress of time upward, but the horizontal spacing does not give the distance between the particles

It is only the incoming and outgoing particles that are on shell and real and can be measured. The internal lines are under the integral which is implied by the Feynman diagram, and thus the four vector assigned to the internal line varies within the limits of integration. It is called a photon in this case, because it carries the quantum numbers of the photon, but not the mass, which is off shell. Particles keep the name when they become internal lines after a vertex, because of conserving quantum numbers,lepton, baryon, charge, etc.

There is no access to a virtual particle so that it can be measured. In the formulas for calculating crossections, the virtual particle is described by a propagator which has the mass of the named particle in the denominator, leading to resonances in the crossection.

Here is the crossection of e+e- scattering ( see link for details) showing the resonances seen.

e+e-

The effect of the propagator accompanying a virtual particle is seen clearly. At the mass of the Z there would be a delta function except for quantum mechanical indeterminacy, but there is a contribution of a Z virtual exchange ( we only see the external lines of the Z decay) even far off the mass of the Z. It does not turn into something else. Far away from the mass value of the peaks, the contributions of all these resonances add up, except they become fast very small. There is no confusion of masses.( I have used the Z as an example because it is an elementary particle, but resonances also like j/psi and rho etc can be mathematically described with a virtual line and a propagator).

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  • $\begingroup$ Thanks for the answer! Yes, I agree that all measurable entities in QED are real particles, and virtual ones are only used during the calculation. But given the definition of a virtual particle moving off shell, my argument is that there is no difference between a virtual particle and a real particle displaying different mass. For example, a muon could be an electron moving off shell, because other than mass difference, both of them are theoretically identical, as far as my argument in the question is concerned. But the question is, is there any error in the argument? $\endgroup$ – Prem kumar May 14 '17 at 12:07
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    $\begingroup$ @Raja if on-shell relation doesn't hold, why should $a$ be constant? $\endgroup$ – Kosm May 14 '17 at 12:33
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    $\begingroup$ The internal line corresponds to a propagator that for spin-1/2 fermions is $\frac{\gamma_\mu p^\mu + m_f}{p^2 - m_f^2 + i\varepsilon}$. You cannot arbitrarily change $m_f$ in this propagator without affecting observable cross sections. That the internal line corresponds to precisely this propagator follows from (i) the Dyson series for the $S$-matrix (ii) expanding the series using Wick's theorem (iii) using the free-particle wave equations. Step (ii), in QED, means you will conserve lepton number at each vertex, i.e., no electrons transforming into muons, and step (iii) means the mass in the $\endgroup$ – Robin Ekman May 14 '17 at 12:49
  • $\begingroup$ propagator will the mass appearing in the free-particle wave equation. $\endgroup$ – Robin Ekman May 14 '17 at 12:49
  • $\begingroup$ @Kosm All that i wanted to say is that if a particle goes off shell, then for some value of $a$, the particle would behave as if it is on shell, and it's rest mass has been changed to $m+a$. My question is not directly related to QED, but only to the investigation of the behaviour an off shell particle would assume were we to observe one. $\endgroup$ – Prem kumar May 14 '17 at 13:47

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