Quantum Wavelength and Frequency Linked to the Classical Action?

  In previous notes (1) we discussed how (p=0, E=mocc,)  (x=0, to) transforms to p’, E’, x’,t’ such that x’/t’=v. We noted that (x=0, to) is a special (x,t) vector linked to a stationary frame. We noted that a general vector (x,t) = (0,t) + (x,0) and that (x,0) Lorentz transforms such that the Lorentz invariantt -Et+px =0. We also noted that E scales t and p scales x and argued that this was the basis for the frequency and wavelength for a quantum free particle. 

   In this note we investigate this idea further. In particular we argue that given -Et+px, it is not only the case that (x,0) transforms into (x1,t1) such that -E’x1+p’t1=0, but that 1/E=t and 1/p=x also yields -E 1/E + p/p=0. In other words there is freedom in both t and x proportional to 1/E and 1/p even though the special vector (0,to) defines the velocity i.e. x’/t’=v.  This freedom may be associated with uncertainty in position and time even though there is a well defined x’/t’=v. This uncertainty has important consequences if interactions occur i.e. interference effects.These are usually associated with a wave which has the dispersion relationship E=pv, but we argue that they also hold for a particle with rest mass i.e p=E(v) v because it is the uncertainty in t and x which keeps -Et+px=0 which is the source. This also happens to hold for a photon which is a wave in the sense that it has no rest mass.The uncertainty itself yields the velocity i.e. x/t = 1/p / (1/E) = E/p = c.

  Given the periodic nature of -E 1/E + p 1/p=0 i.e.  -E n/E + p/p = 0 where n=1,2,3… we argue that a periodic function be sought such that it describes a kind of probability due to uncertainty in x,t, but this uncertainty disappears when t=n/E and x=n/p.