Why Can't Detected Particles Be Virtual?

Bobhawke
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On the wikipedia page about virtual particles it says:

"If a single particle is detected, then the consequences of its existence are prolonged to such a degree that it cannot be virtual."

Could someone explain to me why this is true, ie why is it that if we detect a particle it cannot be virtual?
 
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Bobhawke said:
On the wikipedia page about virtual particles it says:

"If a single particle is detected, then the consequences of its existence are prolonged to such a degree that it cannot be virtual."

Could someone explain to me why this is true, ie why is it that if we detect a particle it cannot be virtual?

As I understand it at least, virtual particles are, by definition, particles which pop in and back out of existence between classical states. You shoot two classical electrons at each other. They exchange a photon and their momenta are affected. We collect the electrons a short while later when they hit a detector.

But the photon never hits a detector. The photon is never measured. It was not something we "put into" the system, not was a it a by product. It's a ghost. In fact, the electrons didn't even exchange a single photon. All possible interactions between the two electrons occurred, weighted by their appropriate amplitudes.
 
Read a little further in the Wikipedia article and it also says:

There is not a definite line differentiating virtual particles from real particles — the equations of physics just describe particles (which includes both equally). The amplitude that a virtual particle exists interferes with the amplitude for its non-existence; whereas for a real particle the cases of existence and non-existence cease to be coherent with each other and do not interfere any more. In the quantum field theory view, "real particles" are viewed as being detectable excitations of underlying quantum fields. As such, virtual particles are also excitations of the underlying fields, but are detectable only as forces but not particles. They are "temporary" in the sense that they appear in calculations, but are not detected as single particles.

So your quoted statement while accurate is subtle and easy to misinterpret. In classical physics the expalantion makes no sense.

An alternative way to think about it is that virtual particles appear in pairs which cancel...if you are able to detect one of the pairs, it's already real and it's partner annihilated...an example is right outside the horizon of a black hole...one particle is absorbed and disappears and presto chango it's partner becomes real and emitted as a bit of energy...
 
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Naty1 said:
An alternative way to think about it is that virtual particles appear in pairs which cancel...

I don't think that's a very good way to think about it. If an electron and a neutrino scatter through the exchange of a virtual W, there is no "pair".
 
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Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...

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