Relativistic Rest Mass Energy Arising From Nonrelativistic Quantum Mechanics?

Correction April 4, 2024:  Equation ((4)) should be: -P(p,x)ln{P(p,x)} - P(-p,x)ln(P(-p,x) + b1 ( xP(p,x)) + b2 xP(-p,x). P(p,x) is a dyanmical probability.

  It is sometimes noted in the literature (1) that a constant velocity classical particle has a probability P(x)dx=dx/L ((1)). This statement seems to be based on the idea that the amount of time spent by the particle in each dx region is a constant dt. A deterministic particle, however, does not appear to be associated with any uncertainty and hence probability. In order to introduce the notion of probability, one must not follow the particle classically, i.e, in x,t. Furthermore, probability is supposed to be associated with maximum entropy (subject to constraints) and all possible bias should be removed. A particle which moves in one direction has bias in that direction and classical probability should remove all bias. Thus, we argue that the statement ((1)) only makes sense if one has velocity (or momentum) to both the right and left. This is not an OR statement, it is an AND statement in probability. Thus, 1/L = P(p,x)P(-p,x) with p and -p being time reversed (movie playing backward) scenarios (2).

    Furthermore, a lack of bias is associated with the maximization of entropy subject to constraints. In previous notes, we suggested that one obtain exp(ipx) by maximizing  P(p,x) ln(P(p,x)) + ipx P(p,x). Here we try to use maximize Shannon’s entropy using probability= P(p,x)P(-p,x) subject to two constraints:  P(-p,x)P(p,x)=1/L and <x for p> = - <x for -p>. This, we argue, leads to exp(ipx). 

   At present, all arguments are performed for in the nonrelativistic situation although the above could just as well apply to relativistic p. Playing a moving backwards to obtain -p from p  is the same as considering x→-x and this suggests that one consider t→-t. We argue that the conditions which require P(-p,x)P(p,x)=1/L should hold for P(E, t)P(E,-t) =1/T. If this is the case, then one has:  exp(-iEt+ipx), where E=pp/2mo and p=mov. One may, however, view the system from a moving frame in which p=0. If one uses E=pp/2m, then E=0 and probability is 0 in this frame, but nonzero in the other. This cannot be as probability is invariant and so one cannot use E=pp/2m, but must find an energy which is nonzero even when the particle is not moving. This, we argue, leads to the notion of rest mass with -Et+px  being an invariant in all frames.