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.
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 classical physics, a particle traveling to the right is the mirror image of one traveling to the left. In terms of a spatial density (the time spent in each constant dx), the value is the same, as is the energy. A similar notion of such a spatial invariance is in the classical conservation of energy equation for a bound particle: p(x)p(x)/2m + V(x) = E. Quantum mechanics seems to be different because exp(ipx) and exp(-ipx) are distinctly different spatial probabilities, i.e. they are dynamic. It is not just that case that a particle moves to the right for exp(ipx) and to the left for exp(-ipx). The impulse hit probabilities (represented by exp(ipx) and exp(-ipx)) are distinctly different. If one wishes to create an OR situation and write: W(x) = Sum over p a(p)exp(ipx) and use this in a conservation of energy equation, then simply including only positive or negative p values in W(x) leads to a complex average kinetic energy -1/2m d/dx dW/dx / W which cannot be used with V(x) and E. Furthermore, W_ (only negative p) and W+ (only positive p) values do not yield the same average kinetic energy. Fortunately, given that this is a statistical average for a localized object (somewhat analogous to a gas), one may combine W+ + W- and obtain a real kinetic energy. We have argued in previous notes that there exists a quantum-classical duality. For example, for a free particle one has two independent notions. One is x=vt which describes the center-of-mass of the particle. We take this to represent a state description which treats v and -v as mirror images. exp(ipx), on the other hand, represents probabilistic impulse hits and does not define the state by itself, we argue, only its interactions. Interactions are physical and define what happens in an interacting system. One may, however, measure a system, i.e interact with exp(ipx), and this should define a measurement state ( a second state ensemble average). For example, exp(ipx)exp(-ipx) = 1 which is equivalent to the constant dt in each dx associated with x=vt. Thus, the two states coincide, but exp(ipx)exp(-ipx) means that one may place the 2-slit apparatus at any x point and is not about the center-of-mass, as x=vt is. In some cases, the two states are quite different. W*(x)W(x) represents an interaction state, but vrms(x) from .5mvrms(x)vrms(x) = -1/2m d/dx dW/dx / W represents a center-of-mass state description. (Note: in these cases a state means an average state, i.e. an average over many ensemble entries.) If one takes W*(x)W(x) and integrates over x, then interference terms vanish and one obtains a classical counting of distinct p value states with their weight a*(p)a(p).This, we argue, leads to the notion of orthogonality of probabilities. In classical probability, there is no such notion, but there is the notion of AND (multiply) probabilities. Given exp(ipx) and exp(-ipx), if one multiplies one removes p, i.e. the interactions and only sees the state left. W*(x)W(x) yields a set of peaks which represent a pattern of probabilistic hits in space which define a state which one would measure in space. The interference pattern, however, must be one such that if one integrates over x, one only obtains a classical sum of probabilities a*(p)a(p)’s as mentioned above. This leads to the notion of orthogonal exp(ipx)’s (which follows mathematically anyway). If one considers different bound En states, then these two are orthogonal because W*n1(x)Wn2(x) mean n1 and n2 are associated with different energies and integrating over x cannot show a result which looks like a state result, where W*n(x)Wn(x) shows a state result. In order to achieve this orthogonality (linked directly to the notion of a state) one must have that Wn(2) usually has 0 points and is positive and negative with the possible exception of a ground state. In other words, if Wn1*(x)Wn2(x) integrated over x is not 0, one violates conservation of energy. As a result, we argue an average probability for a bound state must statistically include +p and -p contributions and must create a function which has positive and negative peaks (except possibly for the ground state), i.e. W(x) must have 0 points. This W(x) is linked to a measurement state W*(x)W(x), but this is strictly considering exp(ipx)s. We argue that there are two pieces of information for a free particle, x=vt and exp(ipx). There must also be an average state linked to x=vt and this given by KEave(x) = .5m vrms(x)vrms(x) + V(x) = E which shows that the particle may exist and have average kinetic energy even at W(x)=0 points. When measuring a system, one interacts with the exp(ipx)s and so the W*(x)W(x) state is the one which is traditionally presented, but whether one measures the center-of-mass motion or not, one also has the notion of the x=vt, i.e. vrms(x) state (average) as well, we argue.
We have argued in previous notes that exp(-iEt+ipx) is a modulus 1 (representing a uniform distribution under reduced information) complex probability. It represents the spatial probability for hits in space by the inherent force. Given that this complex probability may interfere, it has an effect on average kinetic energy: -1/2m d/dx d/dx W/W where W= Sum over p a(p)exp(ipx). One has various free particle interaction probabilities as the particle is interacting with a potential. Both the numerator and denominator of the average kinetic energy are related to the interaction spatial density, but by dividing the two one obtains the classical kinetic energy. Thus, the classical picture is present alongside the quantum one. The interference of exp(ipx)s in a time independent problem can give rise to interaction points with zero or very low probability. This, however, does not mean that the velocity is 0 at these points. For example, a particle in an infinite well has KEave(x)= -1/2m d/dx d/dx sin(pave x) / sin(pave x) = pave pave/2m. This result matches x=vrms t which is the motion of the particle because it is not interacting with a potential. Nevertheless, the interaction spatial density is W*W = C sin(pave x) sin(pave x).We have argued that exp(-iEt+ipx) is a probability which conserves energy and momentum in a two body elastic collision. Energy conservation should hold for a 1-dimensional single particle bound state as one creates an energy resonance when creating a bound state. This is important because we argue that transitions from one energy level to another should be linked with energy considerations, in particular < W1 V W2>. We first note that the Schrodinger equation, which is a conservation of nonrelativistic energy, establishes an energy resonance. V(x) is the classical energy which may be considered to be a number at x, one has E on the RHS of the equation and average kinetic energy: {-1/2m Sum over p a(p)pp/2m exp(ipx)} / {Sum over p a(p)exp(ipx)) on the LHS. Thus, it requires two pieces of information {-1/2m Sum over p a(p)pp/2m exp(ipx)} and V(x)W(x) to determine if one has a resonance energy state. In principle, if one only had V(x) W(x), this is a math function which may be written as Sum over i an Wn(x), where n is the energy level. We argue that given V(x)W(x), one should first calculate -1/2m d/dx dW/dx / W to see if one has an energy resonance, because this is a steady solution (if no V perturbation exists). Otherwise, V(x)W(x) may be seen as a sum of states an Wn(x) and should not depend on the energy of W(x). Given that V(x)W(x) represents an interaction between W(x) and V(x), Sum over n an Wn(x) represents the weighted bound state amplitudes and < W1 V(x) W2 > * <W2 V(x) W1> should be proportional to the transition probability and not depend on any energy of W1 or W2. One may create units of probability/per sec by dividing by 1/hbar and multiplying by density(E) which has units of 1/E. Thus, we argue that it is the single particle bound state resonance based on conservation of energy which gives rise to Fermi’s Golden rule. We note that (1) derives Fermi’s Golden Rule in a somewhat different manner. We suggest that -Et+px, a Lorentz invariant, is akin to a potential for a particle with rest mass and so exp(-iEt+ipx) is linked with a potential. This also holds for an electrostatic potential, while a photon has cos(-Et+px) apply directly to the inherent force and not a potential. A single particle Schrodinger equation, however, deals with a particle with rest mass mo and so it is the potential energy which is the key idea for quantum interactions. We suggest that quantum mechanics is inherently a potential based theory (with the exception of the photon) and this is why Fermi’s Golden Rule is based on <W1 V W2>. In other words, in quantum mechanics, force follows from a potential and not the other way around as in Newtonian mechanics. In quantum mechanics, the rest mass of a particle represents an inherent force which is linked with a probabilistic hit pattern in space given by exp(-iEt+ipx). Thus, the inherent force gives rise to a probabilistic potential which in turn gives rise to a term proportional to force -grad (-iEt+ipx) = p. This seems to explain why quantum mechanics equations deal with potential instead of force. The probabilistic wavefunction is linked with a potential related to the inherent force for particles with rest mass.
In Part I we noted that in quantum mechanics, and even in the case of special relativity, there is an arbitrariness associated with x and t. For example, momentum in one dimension is exactly 0. One cannot arbitrarily shift it. If x=0, however, this is arbitrary, it may be shifted. We argued that this arbitrariness and the independence of x and t in a rest frame mo, x=0, t=to should carry over into the results seen in a moving frame, but in that case x’/t’=v because a Lorentz transformation links (x=0,t=0) to (x’=0, t’=0). Thus it appears that the arbitrariness of x,t and their independence have been broken, but we argue that a special Lorentz invariant -Et+px restores it by creating a time period hbar/E and wavelength hbar/p. In other words, a mathematical point is an abstraction representing a wavelength region defined by p through a complex probability exp(ipx) i.e. the wavefunction (and a similar exp(-iEt for time). In this note, we focus on these points in further detail. A point x is associated with p or -p in one dimension, but a distribution, say cos(px), combines a fixed p with probabilities and gives the impression of positive and negative momentum values when the slope is positive or negative. Thus cos(px) by itself shows no overall translational motion, but contains regions of positive or negative momentum i.e. zero point motion. One must superimpose on this zero point motion (which exists for all energies) actual “translation” to the right or left. Given that positive and negative p’s already exist in one dimension which have nothing to do with translation, one is forced to move into a second dimension to create the translational motion which is associated with a probability such that each x point has the same value. This then leads to exp(ipx). Next we consider a bound state situation. In the case of exp(ipx), cos(px) and sin(px) are linked to the positive/negative zero point type energy which has nothing to do with translation, while p has everything to do with translational motion. Thus if one is only interested in the translational motion (as Newton was) one may map exp(ipx) to p at x. An interesting feature, however, that appears in quantum mechanics is that is the zero point type motion which creates a very specific kind of averaging combines with pp/2m values to create the notion of translation(to the right and left at the same time so to speak because time is removed from the bound problem. This is very different from the quantum free particle which has p and pp/2m at each x point and is due to the two dimensional nature of exp(ipx) which leads to pexp(ipx) - pexp(-px) not equal 0. Thus the nature of “acceleration” in a bound quantum system is closely linked to zero point energy (i.e. the back and forth action-reaction momentum) which is already present in a free particle. In fact, for a bound system, this zero type motion is linked to the average -idW/dx / W (W=wavefunction) which is not the same as prms(x) obtained from -1/2m d/dx dW/dx /W.
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 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.
In previous notes, we presented a simple argument for obtaining the quantum free particle wavefunction exp(-iEt+ipx). We suggested that there are situations for which one cannot predict the outcome of an elastic collision and that given initial e1,e2 (energies) and p1,p2 (momentum vectors in 1-direction for simplicity), any ei,ej and pi,pj which conserve energy and momentum have the same product probability. Even in Newtonian elastic collisions of free particles, one sees calculations which allow any outcome which conserves momentum and energy. If the probability has the same real value for any free particle as no variable is being distributed and one wishes to impose Lorentz invariance, then exp(-iEt) and exp(ipx) represent the probabilities. This automatically gives rise to units in time and space, hbar/E and hbar/p (the hbar comes from considerations linked to a photon). One may ask, however: What is the real physics behind dx and dt? We argued in (1) that this physics is associated with force, in particular an intrinsic force. We considered a particle with rest mass mo and argued that this rest mass represents intrinsic force as it is proportional to Newton’s inertia and the force needed to set it in motion. If viewed from a moving frame, (0, mocc) maps to (E,p) with E containing the intrinsic inertial force as well as enhancements due to motion and p containing the intrinsic mo as well as the velocity vector. For a particle with rest mass, we suggest that a key idea is this notion of intrinsic force. We argued that one may consider the Lorentz invariant -Et+px = -mocc t1 and introduce dx=hbar/E, dt=hbar/p, dt1=hbar/(mocc) and dx1-> infinite as p->0. This leads to -E dt + p dx =0 or -E/dx + p /dt = 0. We then argued in (1) that p/dt is like an average force and from this we introduced exp(-iEt+ipx). Thus, dt and dx are linked with the notion of an average force. We suggest that this notion of intrinsic force is the reason behind linearizing -EE + p dot p = -momo cccc ((1)). -Edt+pdx=0 is too simple to show the geometry linked with rotational invariance etc in ((1)), yet at the same time ((1)) is not linear in the intrinsic force mo. This is the physical reason why ((1)) should be linearized via Dirac’s process, giving rise to spin ½. As a result, we postulate that a particle has an intrinsic force, in this case rest mass, which is directly linked with the probability exp(-iEt+ipx) and that one should be able to write an equation containing operators and the intrinsic force (in linear form) which also shows geometry. It is this equation which gives rise to spin. At this point, this is a postulate, i.e. speculation. It would be good if one could find an example in which these ideas arise naturally, i.e. a case in which one did not have to linearize an equation like ((1)) or make special arguments in order to obtain exp(-iEt+ipx). In that case, an equation linear in the intrinsic force would exist a priori and a direct link to exp(-iEt+ipx) would also naturally exist. We suggest that such an example does exist, namely for a photon. One may use the special relativistic equations developed for a particle with rest mass and set mo=0 to obtain E=[p] c for a photon. This is a fine equation, but unlike the rest mass case mo (proportional to the intrinsic force), there is no rest mass and no intrinsic force present for the photon. Certainly p (momentum) is linked with a force as both a particle with rest mass and a photon impart an impulse on a target, but momentum is not an intrinsic force, it is a consequential one. In the case of a rest mass, it arises because the rest mass (the intrinsic force) is seen to be moving. For a photon, one would then expect the existence of an intrinsic force which gives rise to momentum. We suggest that the electric field (together with the magnetic) represents this intrinsic force and that Maxwell’s equations allow one to directly write a linear equation in El (electric field) (no linearization required) which shows the geometry in the system. One may also explicitly show that El behaves as exp(-iEt+ipx) if one extends cos(-Et+ipx) to a more general complex number. In other words, the postulates made for the particle with rest mass mo seem to naturally arise in the case of the photon. This, we argue, is why the procedures for finding spin for a spin ½ particle with rest mass mo and a photon are so different even though both a photon and rest mass particle satisfy special relativity and are associated with exp(-iEt+ipx). It is the difference in the intrinsic force which is the key to finding spin, we speculate.