Why must there always exist a real particle with the same mass of the virtual particle of a certain force field? I've tried to ask this question before, but I've never quite got a satisfying answer so I'm going to simplify my question.

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*As I understand it, virtual particles are just 'internal legs of a Feynman diagram' and are thus unobservable and we can in fact consider them purely as a 'convenient way of organising a perturbative expansion' and not as actual particles themselves.


*The propagator used for virtual particles is given by $\frac{1}{p^2-m^2}$ for the momentum $p$ (which is conserved on the Feynman vertices) and mass $m$ of the virtual particle.
I understand that virtual particles are 'off-mass-shell' such that $p^2 \neq m^2$, so by 'mass of virtual particle' I'm just referring to the quantity m being used in the propagator


*The mass of a virtual particle is related to its maximum range of its Yukawa potential. i.e. $m \propto \mu$ for $\mu$ in $U\propto \frac{e^{-\mu r}}{r}$
My question is, if virtual particles are in a sense a 'convenient fudge' to aid in perturbation calculations. Why does the variable 'm' being used in the propagator always seem to have the same value as a mass of a particle which we can detect in other situations as real and not virtual.
It seems like a massive coincidence to me that a mass of a virtual particle, which we we just defined as related to $\mu$ for convenience in studying interactions, would also always be able to be detected as an 'external leg' particle itself.
So ultimately my question is, why do we never have forces with $\mu$ that don't happen to be related to the mass of an actual real particle. It there some deep theorem to all this?
I'm guessing there might be, since I have heard explnations online like 'the Higgs boson has nothing to do with giving particles mass, the Higgs field does, and the fact that the field exists means the excitation (namely the Higgs boson) must exist'.
 A: 
Why must there always exist a real particle with the same mass of the virtual particle of a certain force field

Because physics is not perturbation theory or mathematical objects in general. It is about observing nature, measuring accurately defined for this reason variables, and then finding mathematical models that fit the measurements and also are predictive of (ideally) all new measurements.
So in discussing models of physics one should keep in mind that the mathematical format  is a tool defined in order to describe nature. In your case perturbation theory is a model to describe particle scattering and decays.
One very strong observation from the data, in addition with conservation of energy, momentum and angular momentum is the conservation of quantum numbers found in particle scattering experiments over the last almost 100 years. If you look at the table of elementary particles, you will see that each carries a number of quantum numbers, which have to be considered when calculating crossection and decays, their specific behavior under the different forces to be taken into account.
In the  Feynman diagram representation of the series expansion for calculating interactions these quantum numbers are carried by the lines clearly counting the conservation laws that apply at each vertex, so that the final outgoing particles have the correct quantum numbers.
This means, for example, that the quantum numbers of an electron accompany the line which has  the propagator with the mass of the electron as a pole . It is the brilliant representation of complicated calculations that Feynman discovered.
So the virtual particle is an effect, not a cause. Because the line has all the attributes of the particle except the mass, it is called  virtual electron,photon, up_quark, etc.
For every real particle , a virtual particle can be defined in the perturbation series expansion for calculating cross sections and decays, to keep track of quantum numbers in the terms of the expansion.
