Mathematical proof for virtual particle can be an off-shell particle We have the Feynman propagator$(x^0>y^0)$
as
$$\Delta_F(x-y)=\int {d^4p\over2\pi^4}\frac{\imath}{p_\mu p^\mu-m^2}e^{-\imath p\cdot(x-y)},$$ integrating through the $p^0$ with the help of contour integration will give
$$\Delta_F(x-y)=\int {d^3p\over2\pi^3}\frac{1}{2E_p}e^{-\imath p\cdot(x-y)}$$
Where $E_p=\sqrt{{\bf p^2}-m^2}$.
If we take the Feynman propagator$(x^0>y^0)$ as
$$\Delta_F(x-y)=\int {d^4p\over2\pi^4}\delta({p_\mu p^\mu-m^2})\Theta(k^0)e^{-\imath p\cdot(x-y)}=\int {d^3p\over2\pi^3}\frac{1}{2E_p}e^{-\imath p\cdot(x-y)},$$
The Delta function($\delta({p_\mu p^\mu-m^2})$) implies the particles associated with the Feynman propagator $(x^0>y^0)$  fall on On-shell.
Then why the virtual particles associated with the propagator is termed as Off-shell?
To become an off-shell particle, I think the mass need to vary, but there is no integral over mass and we can fix the mass and can vary $p^0$ and $\bf p$ over the mass shell. so anyone please give a mathematical proof that virtual particle can be off-shell?
 A: The propagator I mention is for free theory. Free particles are always on the shell. We can neither create free particles nor detect them because both processes require interaction. In other words, the propagator in my question is an asymptotic particle that is on-shell. So integral only has an on-shell contribution. Virtual particles arise in interacting theories, where momenta of virtual particles away from mass-shell contributes to the propagator. More mathematically, the momenta that we are integrating over appears in more than one propagator, so if by contour integration we choose an on-shell pole from one propagator, the other won’t be.
A: 
The propagator I mention is for free theory. Free particles are always on the shell. We can neither create free particles nor detect them because both processes require interaction.

It is a common misunderstanding; a free particle is by definition a real particle in a weak external field - so weak that the kinetic term dominates the other ones in the particle equation.
If you imply really non interacting particles, you may not write any equation for them nor assign specific properties like masses, spins, shapes, whatever.
A virtual particle is a retarded (or not) solution that gets into equation written for another particle. A Coulomb field is a virtual patricle of this sort.
A propagator is an intermediate part of calculations; it has no other meaning.
