Let me try to outline a possible explanation of how certain numerical parameters originating in calculations of quantum-mechanical probabilities turn into what we call the mass, the energy, and the momentum of a particle.
An important calculational tool is the (single-particle) propagator, which is used to calculate the probability with which a particle detected at spacetime point A will be detected at spacetime point B. It can be obtained by summing over all possible paths leading from A to B. Each path contributes a phase factor.
Imagine a particle that travels along one of these paths. Further imagine that the particle "ticks" whenever the phase associated with the path segment already traveled increases by 2π. Except for the units, the rate at which the particle ticks is the particle's mass. In other words, m = N/T, where N is the number of ticks produced during a proper time interval T. Multiply this by Planck's constant h to convert it into energy units. When expressed in energy units, we call it the particle's rest-energy. Divide this by the square of the speed of light c to convert it into mass units. When expressed in mass units, we call it the particle's (rest)mass.
While there is no mass other than rest mass, there is a difference between the particle's rest-energy (a.k.a. mass) and its energy. The former essentially equals ticks per units of proper time, while the latter essentially equals ticks per units of coordinate time, provided that inertial coordinates are used.
Now consider an infinitesimal path segment at (t,x,y,z) with components (dt,dx,dy,dz). It is convenient to measure the corresponding phase difference dα in units of action. The corresponding infinitesimal action then is : dS = ħ dα.
If the particle is moving freely, then dS will have no explicit dependence on (t,x,y,z). The particle will behave the same no matter where and when. If dS does not (explicitly) depend on t, dS defines a constant, and this is what we call the particle's energy. If dS does not explicitly depend on x,y,z, then dS defines another constant, the particle's momentum.
Thus by the particle's energy we mean the particular quantity which is constant because dS does not (explicitly) depend on t, and by the particle's momentum we mean the particular quantity which is constant because dS does not (explicitly) depend on x,y,z. Note that these definitions imply that the energy and the momentum of a free particle are necessarily constant.
If there are freely moving particles (that is, if we ignore spacetime curvature), and if we are interested in the classical limit, in which quantum mechanics degenerates into classical mechanics, then the only possible way we have to mathematically describe effects on the motion of a particle is to add to the action differential dS of a freely moving particle a term that is linear in the coordinate differentials:
$$-V(t,x,y,z) dt + A(t,x,y,z) dx + B(t,x,y,z) dy+ C(t,x,y,z) dz.$$
In the classical limit, in which the particle follows the path that minimizes the action, V becomes the particle's potential energy, while A,B,C become the components of the particle's potential momentum. The path that minimizes the action satisfies the Lorentz force law. This contains the components of the classical electromagnetic field, which are defined in terms of the fields V,A,B,C. If we use the Newtonian gravitation potential V (with A,B,C = 0), we similarly obtain Newton's law of gravity.