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(I'm just wondering if some of the concepts in QFT can be exported back to QM, especially the concept of virtual particle)

So I've just read Arnold Neumaier's description about virtual particle in Are W & Z bosons virtual or not?

How wrong is it to claim that we can only see virtual particles appearing in our equations when we have at least 2 quantum objects in our system? In normal 'semiclassical' QM e.g. quantum electron in classical potential, the electron is on-shell all the time.

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1. A single, stable particle is always on-shell. Thus your claim is trivially correct if there is nothing else around. But this is a physically dull situation.

2. More interesting is already an electron in an external electromagnetic field. In quantum field theory, perturbation theory produces in lowest order a sum of infinitely many tree diagrams with comb-like trees having a stem representing the electron and wiggly teeth representing soft photons. The free ends are real particles, everything else is virtual.

An interaction vertex with an onshell electron and an onshell photon must result in an offshell electron because of energy conservation, which is strictly enforced in Feynman diagrams. (There is no room in the formalism for a lack of energy conservation that would enable virtual particles to exist with "borrowed energy", as wikipedia would like to have it!.)

2a. If diagrams of quantum field theory are taken wholly serious (and interpreted in the meaningless but popular space-time fashion), the electron is on-shell at time $-\infty$ and time $-\infty$, but sligthly of-shell at all finite times, as it interacts with an infinitude of on-shell photons representing the electromagnetic field, and an interaction vertex with an onshell electron and an onshell photon must result in an offshell electron because of energy conservation (which is strictly enforced in Feynman diagrams).

2b. But if particles are called real if they have a state that changes with time then an electron in an external field is always real and on-shell (as off-shell states do not even exist formally). This is consistent with the usual quantum mechanical description of the electron, and this description can be justified from quantum field theory in the appropriate semiclassical approximation (the e/m field being generated by many far away particles).

3. With two electrons, the only dynamically valid description is again in terms of real electrons only (as in the nonrelativistic case). In contrast, the standard Feynman diagrams with their internal lines aka virtual particles only describe the scattering situation (real particles at time $\pm\infty$), with complete disregard of the details that happen at finite times.

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Thanks. What if the two electrons are used as a glue (as in H2 molecule) -> covalent bond? This is kinda related to the tunneling question, where the the electron is probably virtual in the forbidden region? –  pcr Mar 16 '12 at 22:39
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Two bound electrons have a state, hence are onshell, hence are real. - A tunneling electron is also real, though it may be bilocal (in both wells with a significant probability). Because of the exponential decay of the wave function in the barrier, it will not be there (except perhaps very close to the boundary). –  Arnold Neumaier Mar 16 '12 at 22:42
    
Okay, got it, my mistake. So I got mixed up in thinking that a particle in a virtual energy level (e.g. physics.stackexchange.com/questions/7166/unstable-energy-levels -> although it is already a QED problem) is momentarily a virtual particle. But then the definition of virtual particle is stricter because it exists only during the interaction process (in scattering), and won't exist in a stable, bound state system. –  pcr Mar 16 '12 at 23:24
    
So, if only 1 kind of field is quantized (only electrons), every state is real and there is no need for virtual particles. There is no way I can construct a virtual state by using only quantized electrons. Hmm nvm, I'll think more about it. (even in the bound QED system, there are also virtual photons) –  pcr Mar 16 '12 at 23:33
    
Note: by "virtual state" I don't mean that it exists as a real state. I think my main question has been answered. The rest is just my own messy thought. –  pcr Mar 16 '12 at 23:46

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