Quantum Bound State Statistics From Classical Free Particle Action?

   In a previous note (1) we argued that both the relativistic A=  -mo t sqrt(1-vv) and nonrelativistic A=t .5mvv classical free particle actions may be written with v=x/t and x and t varied independently to yield:  dA/dx partial = p and dA/dt partial = -E. Thus A is equivalent to -Et + px for a free particle with x and t being treated as independent. We argue in this note that such an assumption is equivalent to a statistical picture for a free particle even at the classical level because x and t are treated as independent. Thus a constant p may occur at any x because no time is linked to x. Given that all space points are the same for constant p  one expects a constant probability at all x points. Knowing p exactly means not knowing anything about which x one has.

   Secondly, one may create an ensemble of free particle p classical actions for a situation containing a potential V(x). Given that x and t are independent, there cannot be any notion of  single p accelerating, although a vrms from an ensemble may. At a given x (x known exactly) one does not know which p is present. Thus the ideas of uncertainty of p and x follow immediately from this picture. 

    If one considers a single particle and an ensemble of free particles in the x-t independent scenario, then each free particle action p is linked to A=-Et+px where E=pp/2m. This ensemble would arguably be stirred up by a potential V(x). If one wishes to consider an equilibrium scenario for which there is an overall En for the bound system one would expect an overall Ent for the system and for -pp/2m t pieces to disappear. Thus we postulate that pp/2m +{p part} G(p’s k’s) = En for all p where {p part}G(p’s, k’s) is an unknown function and V(x)=Sum over k Vk exp(ikx). 

   Next one may consider the spatial situation for which V(x)  is the classical potential for a single particle. The kinetic energy follows from an ensemble of free particles with the overall -Et removed. From classical considerations one would expect to have a KE(x). Thus one expects: KE(x) = {Sum over p a(p) f(px) } / {Sum over a(p)f(px) } with f(px), an orthonormal basis set, because the action contains px. (It is orthonormal because one wishes the identity of p to exist if not at a point, then over all of x.)

  We try to argue in this note that f(px)=exp(ipx), the momentum eigenstate which follows from A=-Et+px. We find G(p’s, k’s) which implies entanglement or mixing of a given p state with other p and k states from the potential  V(x)=Sum over k Vk exp(ikx). Thus one expects an entropy linked to this momentum entanglement. Given the form exp(ipx), finding a set of ap’s which satisfy pp/2m + G(p’s, k’s)=En  leads to a spatial distribution W(x)=Sum over p a(p)exp(ipx) so there should be an entropy contribution from this spatial part as well. In classical statistical mechanics one has both momentum and spatial contributions to entropy. 

   We argue for S=Sp+Sx, but note that even though px has x, p symmetry as does the Heisenberg uncertainty principle var x var p >= hbar/2, Sp and Sx may not be symmetric in terms of important parameters such as n, energy level. In other words, Sp the entanglement entropy with V(x) may be the driver entropy sensitive to energy level changes while Sx the resulting spatial density may not be as sensitive. Furthermore the overall entropy Sp+Sx gives information about uncertainties in x and p as is already known in the literature where bound state Sp+Sx bounds are given.