We have argued in previous notes that free particle quantum mechanics follows from -Et+px = constant, in that x,t and x+hbar/p, t+hbar/E are solutions with the same value of the constant. We further argued that this is equivalent to a probability exp(-iEt+ipx) which has the same phase (and hence value) as seen in different frames. In some notes, we argued that uncertainty in x follows from the uncertainty inherent in a rest particle with mo at x=0, t transforming to x’,t’ such that x’/t’ = v. Here we wish to consider this in more detail for the case of p and -p because we have argued before that quantum mechanics introduces a dynamical probability, i.e. exp(ipx) not= exp(-ipx) even though the motions are mirror images for the same mo. We try to show that special relativity leads directly to this dynamic feature of the probability.
We have argued in previous notes that free particle quantum mechanics follows from -Et+px = constant, in that x,t and x+hbar/p, t+hbar/E are solutions with the same value of the constant. We further argued that this is equivalent to a probability exp(-iEt+ipx) which has the same phase (and hence value) as seen in different frames. In some notes, we argued that uncertainty in x follows from the uncertainty inherent in a rest particle with mo at x=0, t transforming to x’,t’ such that x’/t’ = v. Here we wish to consider this in more detail for the case of p and -p because we have argued before that quantum mechanics introduces a dynamical probability, i.e. exp(ipx) not= exp(-ipx) even though the motions are mirror images for the same mo. We try to show that special relativity leads directly to this dynamic feature of the probability.
In Part 3, we started with the notion of a free particle probability linked to energy and momentum conservation in two body Newtonian elastic scattering. In particular, we argued that one needed a probability which shows an equal weight for any outcome (ei,ej) and (pi,pj) (momentum vector) set which conserves energy and momentum. This led to a first guess of exp(iC1 E) and exp(iC2 p), but in Part 3, we argued the momentum and energy are defined through a Lorentz transformation of mo at rest and such a transformation must also affect x,t → x’,t’. We concluded one cannot create any probability based on p and E without also considering x and t which transform according to the same transformation. This then allows one to create a Lorentz invariant probability. Here we take an opposite approach, i.e. one which does not involve p and E a priori. It is well known from classical physics that P(x)dx=dx/L, where L is an arbitrary length, describes a uniform distribution for particles at rest. One may, however, view these particles from two frames, one moving with constant v and the other with constant -v. This would seemingly introduce probabilities P(x1’,t1’,v) and P(x2’,t2’,-v) with no notion of energy or momentum present. It would seem that the product of the two for an x1’=x2’ and t1’=t2’ should equal the rest frame result. This suggests an exp(f(x,t,v)) solution, but there is no reason to have an ever increasing or decreasing exponential and so exp(i f(x,t,v)) seems like a solution. There is, however, a problem, because given P(x,t,v=0), and such a solution is then not P(x)= 1/L unless P(x,t,v)=1, which we reject as it assumes a moving particle is identical to one at rest in terms of probability.. In other words, there are two probabilities for the rest frame, P(x)=1/L and P(x,t,v=0) = exp(i f(x,t,v=0). Given that exp(i f(x,t,v) is based on Lorentz frames, one would also expect probability to be conserved and not be created or destroyed simply because a particle with rest mass is viewed from a rest frame or one moving at constant speed. This seems to force one to use relativistic variables linked to a Lorentz invariant, namely E,p as well as x,t. As a result, in Part 3 we argued that one cannot consider a probability of p,E without considering x,t as well and here we suggest that one cannot even consider a probability of x which is conserved in different constant moving frames without introducing t,p and E. Thus, P(x,t,v=0) which seemed like a “P(x)” type of probability in the rest frame is really a different probability, namely one in time exp(i f(x, v=0, t)), i.e. . for this to be Lorentz invariant, one has exp(-iEt+ipx) = exp(-iEt) for any x. In a sense, this is like P(x)=1/L in that each x carries the same weight, but there is still the time variable and the probability actually changes in time unlike P(x)=1/L. This seems to lead to a paradox because the given exp(-imocct) one may formally argue that exp(-px)=1 for p=0 which is the x part of the probability, yet above we argued that in order to obtain this value one should use P(x1’,t1’v)P(x2’,t2’,-v) at x1’=x2’ and t1’=t2’. The difference between P(x)=1/L is that the time measurement has infinite resolution, while exp(-imocc t) implies that it does not because otherwise information about x is the same. Thus, special relativistic considerations lead to the notion of uncertainty in time (hbar/E) and space hbar/p, but in the Newtonian scenario, these do not exist. We argue that the question then becomes: Why would one wish to change from a special relativistically related probability exp(-iEt+ipx) which shows physical uncertainty units of time and space, to one which shows infinite resolution for x and t, through exp(-iEt+ipx)exp(iEt+ipx)? We suggest that exp(-iEt+ipx) or exp(ipx) in time-independent problems is a dynamic probability needed to describe interactions. One may see that cos(px) and sin(px) may have positive and negative values, implying the removal of probability from one place and its addition somewhere else. One the interaction result has been obtained, one wishes to measure spatial results, i.e.count particles in bins in space. There is no need for the interaction information and it may be formally removed using exp(iEt-ipx) exp(-iEt+ipx).
In previous notes, we argued that one could describe elastic two particle collisions, even in Newtonian mechanics, in a classical statistical manner. In particular, we argued that one could introduce a probability which depends on E,p, but at the same time has the same weight (in a sense) for any free particle. Product probabilities would be linked with a uniform distribution in the sense that exp(-iE1)exp(-iE2)= exp(-iE3)exp(-iE4) if E1+E2=E3+E4 and a similar expression for momentum. If one respects the special relativistic invariant -Et+px, then a possible solution is exp(-iEt+ipx), but even though this yields the free particle wavefunction (probability), this does not explain quantum mechanics. These are simply ideas of classical probability theory with the enforced special relativistic invariant -Et+px. Here, we argue that the real feature leading to quantum mechanics (the idea of hbar/p and hbar/E) arises from the notion of inertia (which we have argued in previous notes also gives rise to spin). We argue that a single force (inertia) in a rest frame becomes two independent interaction notions, E and p as well as creating x’ as a running variable. Thus, mo,t and E,t’ and p,x’ are intertwined. We suggest that there must be a physical mechanism which keeps these two concepts E,p independent and so one cannot simply rely on x=vt. Given the Lorentz invariant and the classical notion of probability exp(-iEt+ipx), one must go one step further and postulate that there are independent fluctuations in t’ and x’ in such a calculation such that hbar/E and hbar/p emerge. We argue that this is a key feature of quantum mechanics and that it does not immediately arise from exp(-iEt+ipx) if one only considers x=vt. One is forced to allow x,t to fluctuate independently which leads to a fluctuation in mocc in the rest frame as well. We argue that this feature is actually associated with special relativity as it allows a single force inertia to manifest itself as two independent interaction quantities E,p. As a result, we suggest that one may obtain exp(-iEt+ipx) without any notion of probability, but only by postulating the existence of a physical mechanism (fluctuation of x’,t’ or t in the rest frame) which keeps E,p independent because -Et+px does not ensure their independence if one only considers x=vt. One may then interpret exp(-iEt+ipx) as having the form of a classical probability which respects special relativity.
We have argued in Parts 1 and 2, that the free particle quantum form exp(-iEt+ipx), which we call a complex probability, arises from special relativity in order to allow for conservation of energy and momentum (in an elastic collision). One may see from single particle bound states how exp(-iEt) plays the part of a binding energy. We have noted in Parts 1 and 2, that exp(-iEt+ipx) follows from the Lorentz invariant -Et+px, but that one does not consider the deterministic x=vt equation which applies to the center-of-mass, but rather treats t and x as independent variables which force the independence of E and p (two different interaction variables) and also force separate E and p conservation. The question becomes: How is it possible to have x,t play two such different roles in special relativity? Here we try to answer this question. We argue that special relativity is based on determinism, i.e. x=vt. One may create a Lorentz transformation for a particle with rest mass mo at x=0 with t, a running variable. As seen from a frame moving with -v, one has x’=vt’. In other words, a separate x’,t’ is associated with each v. In Newtonian mechanics, it is possible to have mov1 and mov2 in the same frame (i.e. a frame with a ruler and clock), but in special relativity, each mo in the rest frame maps to a different x1’,t’ and x2’,t2’. This, however, seems to lead to a contradiction. If one considers two particles p1,E1 and p2,E2 (one dimensional motion) which elastically collide, then before the collision, they may be thought of as a composite particle (for more than a single x,t point) with energy E1+E2 and momentum p1+p2. Given the validity of special relativity, we argue that one may associate a Lorentz invariant -(E1+E2)t + (p1+p2) x with this composite particle. Furthermore, any outcome Ei+Ej = E1+E2 and pi+pj = p1+p2 would satisfy this same invariant. At this point there is no contradiction. The contradiction arises if one considers a single t and x for a composite particle, while internally there exist t1’,x1’ and t2’, x2’. The composite particle, in terms of special relativity, is not compatible with each -E’t’+p’x. The energies and momenta, however, add for each particle (in conservation equations) and so we argue that one should have a relativistically compatible approach which allows one to link each particle to a composite particle and see conservation of E and p. We note that it is precisely x=vt for each particle which leads to x1’,t1’ and x2’,t2’ and so argue that if one gives up this notion of determinism for x and t and introduces independent x, t values (fluctuations), then one can indeed retain the -Et+px form for each particle and use it to create composite particles. This seems to give rise to an AND situation which does not exist for both p,x and E,t in the deterministic relativistic approach. In that approach, one can only think of p’s and E’s adding separately, without any linking of x’s and t’s. In the fluctuation approach, it is possible to have the same x and t fluctuation points for each particle. We argue that it is this kind of thinking which leads to the probabilistic form exp(-iEt+ipx) which still respects the Lorentz invariant form and at the same time allows for conservation of E,p with the same t,x fluctuation points allowed. x=vt does not allow one to combine x1’,t1’ and x2’,t2’ for a composite particle because these may coincide for only one x,t point and one wishes to describe a composite particle with an x,t which exists over more than one x,t. This is why we suggest the physical fluctuation scenario which exists together with x=vt. x=vt describes the deterministic motion of the center-of-mass, while exp(-iEt+ipx) describes probabilistic hits in x and energy in t. It is special relativity which forces one to associate time fluctuations with E and spatial ones with p. As a result, we argue that special relativity gives rise to both probabilistic and deterministic versions, both of which respect the form -Et+px. The deterministic version is linked with x=vt center-of-mass motion, while exp(-iEt+ipx) represents probabilistic appearances of E in t and p in x, i.e. there is a spatial probability for impulse hit delivery.
May 1, 2026 Correction: In the example of = dW/dx / W for W= Sum over p>=0 a(p)exp(ipx), Real is finite when W(x)=0, but Im( ) tends to infinite if one uses l'Hopital's rule. In the note, only the real part of is considered, but the Im part leads to 0/0. In previous notes, we argued that the Lorentz invariant -Et+px, allows for x,t and x+hbar/p, t+hbar/p solutions. We suggested this might give rise to the quantum mechanical probability exp(-iEt+ipx). In other notes, we sought a probability which conserves energy and momentum and respects Lorentz invariance, again leading to exp(-iEt+ipx). In a couple of notes, we argued that the uncertainty in x arises from a rest mass at x=0 jumping to x’/t’ = v as seen in the moving frame. This uncertainty in x forces one to introduce probabilities. Here we combine the latter idea with the notion that special relativity yields formulas for the interaction variables E and p. We argue that one may replace the notion of a force reaction as well as the idea of frames moving at constant speed relative to each other by demanding that an interaction be represented by an AND (product) of probabilities and this must necessarily conserve energy and momentum. This probability is in the location and time of interactions. This leads to a skewed dynamic probability exp(-iEt+ipx) because exp(ipx) and ex(-ipx) are not the same. Furthermore, as we have noted before, one uses exp(ipx) as an interaction probability which leads to complex average kinetic energy (nonrelativistic case) and momentum if one only uses p>=0. The Schrodinger equation (which is based on conservation of energy) bypasses this problem by combining p>=0 and p<0. Nevertheless, we note that there can exist W(x) = Sum over p>0 a(p)exp(ipx) such that if W(x) = 0 dW/dx and d/dx dW/dx are also zero at these points even though the average kinetic energy and momentum are complex. We finally show that a complex average can be used in a physical calculation, namely one dimensional reflection-refraction.
In Part 1, we argued that P(x)dx=dx/L (where L is an arbitrary length) applies to a particle at rest. If one views such a particle from a frame moving at constant -v, it is moving. We argued that in such a case, the relevant probability is not simply linked to the position x, but also to the momentum p at x because this delivers an impulse hit. In particular, probability is presumably to be used in a physical problem with interactions and probability and a dynamical particle requires a dynamical probability describing both x,t and E,p. This led to exp(-iEt+px) which actually predicts intervals in space hbar/p and time hbar/E. In other words, a probabilistic view point leads to these results. Here, we ask the question: What if one simply wishes to use deterministic physics to describe a moving particle? After all, Newtonian mechanics is deterministic and one does not really need to focus on P(x)=1/L or its modification for a moving particle. In such a case, a particle moving with constant v (starting at x=0, t=0) is described by x/t=v both in Newtonian mechanics and special relativity. There is no uncertainty in x and t here which begs the question: Why does it appear in the probabilistic formalism of Part 1? In particular, exp(-iEt+ipx) derived in Part 1 implies physical regions: dx = hbar/p and dt=hbar/E and these seem to be absent in x/t=v and this is a problem. We suggest that a probabilistic approach is buried in the Lagrangian formalism which leads to the result d/dt v = 0, i.e describes the particle’s motion in space. This motion is actually x/t=v, but this equation only presents a subset of the description of a particle that is moving with constant speed, even in a deterministic Lagrangian/Action approach. We have already pointed this out in (1). We briefly explain it again here, but the real question is why does one need to consider the Lagrangian in order to describe x/t=v? In other words, the probabilistic approach associated with v=x/t only emerges in the Lagrangian formalism using a relativistic action Lt = -Et+px, with x/t=v. Ignoring the action and Lagrangian leads to an absence of hbar/p and hbar/E and one is only concerned with x/t=v. We note that the Lagrangian is primarily linked to the result: p = dL/dv partial and so argue that one is not simply interested in motion (x/t=v), but also in interaction because p is associated with an impulse hit. Thus, the Lagrangian approach is more than simply a mechanism which describes x,t motion, it is associated with p and E (through Hamiltonian = E = pv-L.) Using the Lagrangian approach suggests that it does not suffice to consider x and t by themselves when p and E are present. One must consider all four variables x,t,E,p which is the same point made in Part 1. The physics of the problem dictates that it is not enough to simply think in terms of x,t,v. The Lagrangian approach, which is constructed to ultimately yield x/t=v, when written in terms of E, p, i.e. Lt = -Et+px demonstrates uncertainty in x, hbar/p and t, hbar/E as shown in (1). This uncertainty must then be written in terms of a probability which is Lorentz invariant and leads to exp(-iEt+ipx). The point is that one would not even consider a probability if one did not consider E,p and x,t together as a complete treatment of a moving particle. In other words, x/t=v is an incomplete description of the state of a particle moving with constant speed, even in a deterministic framework.. At first this might seem surprising, but it is well known that energy and p are linked to interactions. In special relativity, one does not consider x,t,v alone and then compute p and E from these, but considers four vectors and Lorentz invariant equations. It is possible to create a Lorentz invariant equation involving all four state variables x,t,E,p and it turns out that this is the relativistic Lagrangian which describes motion, so this approach is more general than simply stating x/t=v. Thus, delta x = hbar/p and delta t= hbar/E appear in both the probabilistic approach of Part 1 and the deterministic one of the Lagrangian used here.
Newtonian mechanics presents a simultaneous description of dynamics x(t) and interactions (F=dp/dt) because it assumes infinite resolution, i.e. dx->0, dt->0. We have argued in a previous note (1), that quantum mechanics follows from free particle quantum mechanics, which in turn follows from special relativity. Special relativity uses the notion of x=vt and provides formulas for E=mocc/sqrt(1-vv/cc) and p=mov/sqrt(1-vv/cc) even with no free particle quantum mechanical notions. We have also argued in previous notes, however, that these classical concepts are present in the Lorentz invariant -Et+px which we have argued (2) acts as a kind of free particle potential. This potential, however, has the same value for x, t and x+hbar/p, t+hbar/E and so we have suggested that free particle quantum mechanics is comprised of two parts. The first is a complex probability exp(-iEt+ipx) which describes possible interactions with a free particle and the second is x=vt. In other words, interactions and dynamics are split, but the interaction probability depends on classical notions such as x=vt, p and E. In the limit of hbar/p and hbar/E <<1, the interaction resolution becomes the Newtonian dx->0, dt.->0. Thus, as in (1), we argue that quantum and classical mechanics are intertwined. In (3), entitled Unifying Quantum and Classical Dynamics, a similar argument for the intertwining of quantum mechanics and classical mechanics is presented, but in a different manner. The basic argument of (3) seems to be that quantum mechanics, written in the Heisenberg approach, yields operator equations which are identical in form to classical equations, in particular, d/dt X(t) = P(t) and d/dt d/dt X(t) = -d/dx V( X(t)). Here X(t) and P(t) are operators given by X(t) = exp(iH t) X(0) exp(-iHt) (and a similar expression for P(t). We try to show that given that free particle quantum mechanics is directly linked to special relativity and hence to the acceptance of x=vt, that in the free particle case, any operator description of d/dt X(t) = P(t) cannot yield any result other than a classical one because no interaction is occurring. In other words, quantum mechanics is directly linked to classical mechanics at the free particle level. Later, one takes a superposition (OR probability addition) of exp(ipx)s, as in the case of bound states. In such a case, one deals with average interactions and so one must be careful when interpreting d/dt d/dt X(t) = -d/dx V( X(t)) as the system description is now an average one.