I have just been introduced to the concept of on/off-shell particles, and to my understanding, on-shell particles are those that verify: $$E^2=(pc)^2+(mc^2)^2$$ Free particles verify this equation but, if I am not mistaken, the energy of a bounded state/particle is such that $E^2<(pc)^2+(mc^2)^2$. If that is the case, bounded particles are not on-shell, and since they cannot be detected by instruments due to their bound status, how are they different from virtual particles?

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    $\begingroup$ Yes, it means the bound particle becomes virtual when freely propagates between moments of interaction with the external field. $\endgroup$ – Alexey Sokolik Jun 6 '17 at 17:42
  • $\begingroup$ So, bounded particles are off-shell and there is no effective difference between them and virtual particles. $\endgroup$ – Mario Jun 6 '17 at 17:51
  • $\begingroup$ I think bound and virtual particles are not the same. Virtual particle is a particle in some intermediate state (for example, as an internal line of a Feynman diagram). Bound particle exists in a "real" state (it has definite energy), but it can be considered as a superposition of freely propagating states. The latter are virtual, because the particle spends only a fraction of time in each of them, then scattering on the external potential and going to a state with different momentum. $\endgroup$ – Alexey Sokolik Jun 6 '17 at 18:04

It is a very good question. I will try to give an explanation based on section 7.1 Field-Strength Renormalization of An Introduction to Quantum Field Theory (Peskin & Schroeder). Due to the limit of my knowledge, my answer is far from comprehensive.

The major comments:

  1. The physical states correspond to the singular points of the Feynman propagator in momentum space $\mathcal{D}_F(p) = \int d^4x e^{ipx}\langle \Omega | T \phi(x)\phi(0) | \Omega \rangle$. The virtual states correspond to regular points of the propagator.
  2. One-particle states $p^2=m^2$ correspond to an isolated pole of the propagator. We often call them on-shell, because they are the major contribution to the LSZ reduction formula, which is used to calculate the cross section and decay rate of physical process. So we usually assume that ingoing and outgoing particles in a scattering experiment are all on-shell.
  3. States of two or more free particles give a branch cut (non-isolated singularities) for the propagator. They do not contribute to LSZ reduction formula.
  4. Bound states give additional poles. Although we usually do not call them on-shell, they are physical states, not virtual states. Study of their physical effect is a rich and complex subject, but one that lies beyond the scope of a first course of QFT. In this stage, they can be neglected in most cases.

For a free scalar quantum field theory, the Lagrangian is $$\mathcal{L} = \frac{1}{2}\partial_{\mu}\phi \partial^{\mu}\phi-\frac{1}{2}m_0^2\phi^2$$ The Feynman propagator of the free field theory is $$D_F(x-y) = \langle 0 | T \phi(x)\phi(y) | 0 \rangle = \int \frac{d^4p}{(2\pi)^4} \frac{i}{p^2-m_0^2+i\epsilon} e^{-ip(x-y)}$$ In momentum space, we have $$D_F(p) = \frac{i}{p^2-m_0^2+i\epsilon}$$ When we say a particle is on-shell, we mean the four-momentum of the particle is the isolated singularity of the Feynman propagator $D_F(p)$.

However, for a quantum field theory with interaction, the case is much more complicated. A detailed analysis can show that $$\langle \Omega | T \phi(x) \phi(y) | \Omega \rangle_C = \int_0^{\infty} \frac{dM^2}{2\pi} \rho(M^2) D_{\rm F}(x-y;M^2)$$ with $$\rho(M^2) \equiv \sum_{\lambda} (2\pi) \delta(M^2-m_{\lambda}^2)|\langle \Omega | \phi(0) | \lambda_0 \rangle|^2.$$ Here,$m_{\lambda}$ is the mass of one particular state. It is defined as the energy of the state in an inertial reference frame where the total momentum of the state is $0$. The formular is called the Kallen-Lehmann spectral representation. In momentum space, we have $$\mathcal{D}_F(p) = \int_0^{\infty} \frac{dM^2}{2\pi} \rho(M^2) D_{\rm F}(p;M^2)$$ We know that $p^2=M^2$ is the singularity of $D_{\rm F}(p;M^2)$. The singularity of the $\mathcal{D}_F(p)$ is totally determined by $\rho(M^2)$. Picture from Peskin's book As we can see, the one-particle state is an isolated singularity of the propagator. So, $p^2=m^2$ is on-shell. States of two or more free particles give a branch cut and must be off-shell. Bound states give additional poles. Picture from Peskin's book

In LSZ reduction formula, which is used to calculate the cross section or decay rates, only isolated singularities (on-shell states) can contribute. The effect of branch cut can be neglected. As for the effect of bound states, it is a rich and complex subject, but one that lies beyond the scope of a first course of QFT. The section 5.3 of An Introduction to Quantum Field Theory (Peskin & Schroeder) discuss this topic briefly.

  • $\begingroup$ This seems a very sound answer, and I'm upvoting it, but I'm afraid I'm too early in my journeythrough QFT to fully understand it. I hope you understand if I don't close the question just yet. $\endgroup$ – Mario Jun 7 '17 at 19:32
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    $\begingroup$ This answer is far from comprehensive and my knowledge to QFT is also limited. I am looking forward to a better answer to the question as well. Thanks for your support! $\endgroup$ – Eric Yang Jun 8 '17 at 5:12

My experimentalist's answer is that in quantum mechanically bound states as the total invariant mass is smaller than the sum of the constituent masses, the particles are off mass shell, as you state. This is deduced, one cannot measure individual particles in bound states as one cannot measure virtual particles, as you say.

The difference is given in the other answer, virtual particles are a mathematical construct pertinent to internal lines in Feynman diagrams. Different tools are needed to study bound states with different mathematics, thus one just calls them off mass shell, and not virtual particles.

  • $\begingroup$ thanks for the answer above - I've asked a similar question in other places and been told that the off shell notion should be totally ignored (or doesn't apply) to bound states. $\endgroup$ – asimo Sep 22 '17 at 20:35
  • $\begingroup$ However, as you mention above, the particles can be considered to be off shell - is this due to binding energy? What are the implications for QFT - aren't all 'real' particles supposed to be on-shell? $\endgroup$ – asimo Sep 22 '17 at 20:42
  • $\begingroup$ QFT is a particular tool, based on the postulates of quantum mechanics and, in particle physics , (there exist QFTs in other disciplines) based on perturbative free particle solutions of the basic quantum mechanical equations (Dirak, Kline Gordon, Maxwell). For closed solutions with potentials there are unique wave functions, but they are few, as for the hydrogen atom. QFT is necessary to calculate the effect of potentials using the Feynman diagram formalism where the potential is in the exchanged dp/dt between particles which is carried by the internal lines which have the quantumnumbers $\endgroup$ – anna v Sep 23 '17 at 4:55
  • $\begingroup$ of the named particle but are off mass shell and called virtual. Virtual particles are a mathematical construct and identified by being off mass shell. In that sense, mathematically, since the sum of the masses of electron and proton is less due to the binding energy, the electron and proton are off mass shell: they have the quantum numbers of the electron and proton but the mass is off for the exact solution. An infinite number of Feynman diagrams would have to be added in order to get the exact solutions of the hydrogen atom . QFT is the tool when no exact solutions exists $\endgroup$ – anna v Sep 23 '17 at 5:00
  • $\begingroup$ But the electron and proton in the hydrogen atom are not virtual (although they may exchange virtual particles). The quantum state defining the hydrogen atom has 8 degrees of freedom, yes? $\endgroup$ – asimo Sep 23 '17 at 8:46

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