Are bounded particles off-shell? 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?
 A: 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:


*

*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.

*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.

*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.

*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 $\rho(M^2)$.

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.

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. 
A: I'm not sure about the answers given to be right.
Basically, I think since bound states are not limited to QFT and can be found also in non-relativistic QM, one has to either extend the meaning of off-shellness to non-relativistic QM or very superficially say that off-shellness has nothing "in particular" to do with a particle being bound or unbound as we know all particles are slightly off-shell due to infrared divergences.
I think that "being off-shell" is nothing more than tunneling in non-relativistic QM, namely: $E=T+V$ should be violated, in other words, particles should keep propagating even in the region where $E<V$.
But in the case of QFT and Relativistic QM, writing down the real equation for the energy of a particle due to a high degree of complexity is harder, which means writing $E$ in terms of $V$ and $T$ is not as trivial as NR QM, so one has to treat $V$ as a perturbation and treat the particles as freely propagating particles that are linked by vertices, which means each of these particles(lines inside a Feynman diagram) are supposed to obey the free relativistic particle equation $E^2=m^2 + p^2$.
But to take into account the "quantum tunneling" in this perturbative approach, one has to consider the possibility of violation of the aforementioned equation for each particle separately.
In my very perception, non-perturbative tunneling in RQM should look like, just the same as the NR QM, which is the dominance of the particle's Energy over the potential energy (and consequently exponential suppression in the coordinate space) but since we can't write down such an equation that differentiates $T$ and $V$ from each other easily(if there's any at all!), we take into account quantum effects(tunneling) in a perturbative manner(which considers all fields to be free ones, perturbatively).
Far off-shell particles do not propagate far from vertices and are exponentially suppressed(in time or in space) and only slightly off-shell particles can propagate and form the asymptotic states.
As one can see, the non-relativistic hydrogen atom, according to this definition can be off-shell or on-shell depending on $V(r)>E$ or $V(r)<E$.
The atom goes off-shell for $V(r)>E$ and it is quite on-shell for $V(r)<E$. As one can check it out, the energy eigenstates decay exponentially in space above a certain radius.
That radius is by the way the well-known Bohr radius.
Generally speaking, the atom is limited to the $E<max[V]=0$ region which is the definition of bound states (at rest) that finally results in the quantization of energy.
For me, the same applies to the case of the Hydrogen atom in QFT with the difference that there are contributions to the energy levels of the atom because of a new degree of freedom in which electron and proton can become off-shell. In other words, some fluctuations around the steady solutions of RQM that for example results in the de-excitation of an excited atom.
For sure the on-shell particles are quantum mechanical or classical depending on the theory, but being off-shell is a purely quantum mechanical feature.
A: 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. 
