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Quantum Mechanics and x=vt Versus Measurement of x and t Francesco R. Ruggeri Hanwell, N.B. Dec. 19, 2025 We try to consider free particle and single bound state quantum mechanics from the viewpoint of classical physics. We first assume that a free particle moves as x=vt. Secondly, we assume that dx and dt cannot be 0 as in Newtonian mechanics, but must be finite. This begs the question: How does one measure dx and dt? If one can measure a particle at a given x and t, then dx→0, dt→0. Thus, finite dx, dt means that a given p=constant and pp/2m=kinetic energy (nonrelativistic) interact with a force over a finite dx. In other words, one does not know where the interaction occurs within dx. As long as the free particle moves without interacting, then x=vt and there is no interest in any dx or interactions. For example, one may measure very large x and t ranges. Even if there are dx and dt effects at the start and end points, these are small compared to the large x and t values. As a result, one has a F(x,p) probability within dx, which repeats itself for the next dx (i.e. is periodic), where F is a probability for a p impulse hit at x within dx. At the same time, the particle moves with x=vt it seems within dx. In general, one might then think that F(x,p) and x=vt are decoupled. In previous notes, we have argued that one way to couple F(x,p) and x=vt is through special relativity, i.e. A = -Et+px. X and t values used would presumably satisfy x=vt, but for any x,t one has x+hbar/p and t+hbar/E as solutions yielding the same A value and so might consider dx= hbar/p and dt=hbar/E. The question we ask here is whether one may arrive at this conclusion without using special relativity. The answer seems to be that one can if one suggests that F(x,p) must contain information making it consistent with x=vt, i.e. with every x and t point having the same weight. Clearly, however, each x,t do not have the same weight within dx and dt, but are rather governed by F(x,p). This means, however, that F(x,-p) also represents a probability distribution and so if one has F(x,p)F(x,-p) there is no momentum at a given x point. In such a case, the distribution within dx must disappear and one must have a uniform distribution. This holds even if F(x,p) has the same dx value for any p. Thus, F(x,p)F(x,-p) = constant and must be periodic in x. This implies that F(x,p) = 1/F(x,-p) for constant=1. Given that F(x,p) =0 at the initial and final points of dx, then F(x,-p) = infinite at these points and that is not allowable. Thus, one is forced to use complex variables and try exp(i g(p) x). If g(p) is a function of p (i.e. not a constant), then conservation of momentum implies that exp(i g(p1) x) exp(i g(p2) x) = exp(i g(p1+p2) x), suggesting exp(ipx). What, however, if g(p) = constant? In such a case, one has issues creating an average kinetic energy: ( Sum over p a(p) exp(i constant x) pp/2m) / (Sum over p a(p) exp(i constant x) ((1)) This leads to a kinetic energy which does not depend on x and cannot be used with V(x) in a conservation of energy equation. Thus, we choose exp(ipx). Note: there is no reason to choose a(p,x) because the physics of finite dx and dt is restricted to exp(i constant x). Finally, we try to explain why the same dx=hbar/p and dt=hbar/E together with x=vt follow from special relativity.
Newtonian Mechanics Newtonian mechanics is based on the notion of a particle being measurable within dx→0 and dt→0. This idea is experimentally relevant to the conservation of energy equation: .5 mo v(x)v(x) + V(x) = E ((2)) as one must measure v(x) and each x. On the other hand, x=vt ((3)) does not require the notion of dx→0 and dt→0, for a free particle because one may choose x and t to be as large as one wants as long as the particle does not interact. We ask: What are the consequences of imposing a finite sized dx and dt into Newtonian mechanics? Finite dx and dt We focus on dx at first. If dx is finite, then one can only measure dx at x within this region. In other words, one should have a periodic probability distribution within dx to find a p at x. We use p because it is the variable associated with an impulse hit and a measurement within dx would act against p. We assign: F(x,p) as a probability distribution within dx which is periodic in subsequent dx units ((3)) If one is uncomfortable with p in F(x,p), one may consider F(x), but we show later that this leads to problems and so we keep F(x,p) as a more general value. The particle interacts in a strange way within dx, but in general its motion still follows x=vt ((3)) because this equation has nothing to do with interactions or measurements except at the beginning and end points. If x and t are very large, then any dx and dt effects are negligible. The crux of our argument seems to be that measurements are based on p impulse hits. In other words, we are considering a dynamic model, not a Newtonian picture of photons that may bounce off of a stationary particle without essentially affecting p. We think this is acceptable because dx is linked with measurements which would occur in a physical interaction of p with some force. Then, one may consider both F(x,p) and F(x,-p) ((4)) Thus: F(x,p) F(x,-p) = no motion case hence no interaction = constant ((5)) so F(x,p) = 1/F(x,-p) ((6))
F(x,p) is periodic in x and is 0 at the beginning and endpoints of dx. This means that F(x,-p) is infinite at such points which is unacceptable. If one considers F(x,p) = F(x), then one still has problems. We suggest a complex probability: exp(i g(p) x) ((7)) If g(p) is not a constant, then: exp(i g(p1) x ) exp(i g(p2) x) = exp(i g(p1+p2) x) ((8)) so we suggest exp(ipx). What if g(p)=constant? We argue a finite dx forces one to reevaluate conservation of energy ((2)) because a particle does not interact at dx→0. Nevertheless, average energy may still be conserved at each x point. This suggests calculating an average kinetic energy. We first consider: exp(i constant x) ((9)) KEave(x) = { Sum over p a(p) pp/2m exp(i C x) } / {Sum over p a(p)exp(iCx) } ((10)) Unfortunately, ((10)) is not a function of x which is problematic. Furthermore, one cannot introduce a(p,x), because the dx features of x should appear in F(p,x). Thus, we adopt exp(ipx) = F(p,x). Special Relativity In the above arguments, we obtained finite dx=hbar/p and dt= hbar/E and exp(ipx) simply by considering Newtonian mechanics (conservation of energy on average) with finite dx and dt. This begs the question: Why do not finite dx and dt arise from a generalization of Newtonian mechanics? We suggest that they do as we have pointed out in previous notes. In particular, dx=hbar/p and dt=hbar/E follow from the Lorentz invariant of special relativity: A = -Et+px ((11)) For a given x,t on the trajectory x=vt, A is left unchanged by: x+hbar/p and t+hbar/E ((12)) The question is why should one obtain finite size dx and dt values based on p and E when x and t can take on any value (consistent with dx and dt)? We argue that x,t in ((11)) is linked to x=vt (free particle) in which no interaction occurs. There is no notion of interaction and there is no reason to consider finite dx and dt which are only interaction specific. The point is that E and p in ((11)) are interaction specific. Thus, one may any x,t in x=vt together with the finite dx and dt required for interactions and both should be part of the general Lorentz invariant as they are.
The arguments made in the above sections seem to follow directly from special relativity because given ((12)), one may construct the probability exp(-iEt+ipx) to account for the periodic intervals. This, however, also requires the assumption of probability F(p,x)F(-p,x) = constant because one cannot use sin(px) or cos(px) as F(p,x) even given ((12)) from special relativity. Conclusion In conclusion, we consider Newtonian mechanics with the notion of a finite dx and dt. We argue that such a finite dx,dt means that it must be detected by interaction with the particle. We base such an interaction on p as this represents an impulse hit. In other words, we do not consider photons bouncing off of a heavy particle and leaving p untouched. We argue that Newtonian conservation of energy must hold, but this involves both interactions and dx->0 (time-independent case). We argue that dx cannot be 0, and so conservation can only occur at each x on average. Thus, one requires F(x,p). Now, we argue that dx only occurs because of the presence of p which is needed for an interaction, i.e. we do not consider photons bouncing off a mass, doing nothing to p). Then F(x,p)F(x,-p) = constant as there is no longer a dx as there is an interaction. Then F(x,p) = 1/F(x,-p) for constant = 1. Given that F(x,p) is periodic in x (i.e. 0 at initial and endpoints), this implies F(x,-p) can be infinite which is unphysical. We choose a complex probability exp(i g(p) x). If g(p) not= constant, then exp(i g(p1)x) exp(i g(p2)x) = exp(i g(p1+p2)x) so exp(ipx). On the other hand, if g(p)=constant, then an average kinetic energy does depend on x due to the uncertainty of the interaction in dx, i.e. see ((10)). Thus, we choose exp(ipx) as a probability for interaction even though x=vt holds. Finally, we argue that the theory of special relativity combines the idea of x=vt (which is consistent with dx=0, dt=0 because there is no interaction), but also involves E,p which only arise due to interactions. We ask: Could the theory of special relativity contain finite dx, dt while still upholding x=vt? We argue that there is no reason it cannot. Given A = -Et+px, one sees that one may choose any x,t satisfying x=vt (suggesting dx=0, dt=0), but when one considers the presence of E and p, one sees that A remains unchanged for x+hbar/p and t+hbar/E, suggesting that even though x=vt describes center-of-mass motion, interactions may occur in a dx interval about this trajectory which causes one to revisit interactions of the particle. We still conclude that Newtonian conservation of energy occurs on average at each x.
