Classically, it is known that the probability density P(x)=1/L for a static particle confined randomly in a length L. If one has a moving particle, one sometimes sees that argument that the particle spends delta t =constant in each constant dx for a constant speed and so P(x)=1/L as well. In such a picture, there is no quantum W(x)= exp(ipx)/sqrt(L) or W(x)= exp(-ipx)/sqrt(L), accounting for motion in both directions. On the other hand, traditional quantum mechanics uses P(x) = W*(x)W(x) so one would have 1/sqrt(L) in this case for both p and -p. The interesting point is that exp(ipx) and exp(-ipx) are both dynamic (differ for p and -p), while W*(x)W(x) does not. We first argue that by considering dt in dx, one considers an event rather than describing a particle at an instant. dt implies there is an t(initial) and t(final) and one subtracts the first from the second. This is a description of an event and not a description of the moving particle we argue. Even in classical physics, motion is state with the particle being at x at t with v(x,t). Motion to the left or right represent different physical states (described by p and -p) and there is no reason that an instantaneous probability capturing the state of motion should not depend on this direction of motion as well. Furthermore, it should depend on p as this is a variable used in special relativity. The function W*(x)W(x) = 1/L is an AND statement of probability and removes the state of motion from the probability by combining p and -p (as an AND condition). This leaves a stationary state which obeys classical probability, i.e. the probability of a static object with a uniform probability to be anywhere within L. Thus, one has the classical equation: 1/L = F(-v) * F(v), with two solutions: F(v)=constant and F(v) = exp(-iEt+ipx). We describe in detail how we obtain the second solution using invariance of probability under a Lorentz transformation. One might argue that F(p,x)=constant means that the dynamic probability (the probability of the moving particle) is the same for p and -p when in fact the particles are in different states. In such a case, one has F((p,x) = 1/sqrt(L) and all dynamic features are lost. One would expect that such a probability would be useless in trying to describe a dynamic problem. (We discuss this in more detail). The alternative is exp(-iEt+ipx) which suggests that there is periodicity in x and t based on E and p. At first, this is equally strange from a classical point of view, one might argue. We suggest, however, that for x+dx and x-dx, exp(ipx) clearly defines the direction of motion which is essential in a dynamic problem. In other words, we argue that Newtonian mechanics uses dynamic variables x(t), v(x,t) etc to solve a problem. We suggest introducing a dynamic probability which considers instantaneous probaiblies (just as Newton considers instantaneous variables). The big question, from a classical point of view, is then: Why would one use probabilities at all when Newtonian mechanics describes deterministic physics? The answer, we argue, is that P(x)=1/L is an acceptable classical probability and one is allowed products in probability theory. We suggest that there is no reason why one cannot consider a dynamic probability F(p,x) such that F(p,x)*F(-p,x) = 1/L. Unless one can refute the existence of F(p,x) in classical theory, then mathematiaclly one has F(p,x) = exp(ipx) (because F(p,x)=constant makes no sense). The question then becomes whether such an F(p,x) may be used in solving any physical problem with the answer being yes. This formalism solves the 1-dimensional reflection-refraction problem of photons, assuming the photon to be a kind of particle moving at speed c. Newtonian mechanics does not solve this problem, it seems.
Classically, it is known that the probability density P(x)=1/L for a static particle confined randomly in a length L. If one has a moving particle, one sometimes sees that argument that the particle spends delta t =constant in each constant dx for a constant speed and so P(x)=1/L as well. In such a picture, there is no quantum W(x)= exp(ipx)/sqrt(L) or W(x)= exp(-ipx)/sqrt(L), accounting for motion in both directions. On the other hand, traditional quantum mechanics uses P(x) = W*(x)W(x) so one would have 1/sqrt(L) in this case for both p and -p. The interesting point is that exp(ipx) and exp(-ipx) are both dynamic (differ for p and -p), while W*(x)W(x) does not. We first argue that by considering dt in dx, one considers an event rather than describing a particle at an instant. dt implies there is an t(initial) and t(final) and one subtracts the first from the second. This is a description of an event and not a description of the moving particle we argue. Even in classical physics, motion is state with the particle being at x at t with v(x,t). Motion to the left or right represent different physical states (described by p and -p) and there is no reason that an instantaneous probability capturing the state of motion should not depend on this direction of motion as well. Furthermore, it should depend on p as this is a variable used in special relativity. The function W*(x)W(x) = 1/L is an AND statement of probability and removes the state of motion from the probability by combining p and -p (as an AND condition). This leaves a stationary state which obeys classical probability, i.e. the probability of a static object with a uniform probability to be anywhere within L. Thus, one has the classical equation: 1/L = F(-v) * F(v), with two solutions: F(v)=constant and F(v) = exp(-iEt+ipx). We describe in detail how we obtain the second solution using invariance of probability under a Lorentz transformation. One might argue that F(p,x)=constant means that the dynamic probability (the probability of the moving particle) is the same for p and -p when in fact the particles are in different states. In such a case, one has F((p,x) = 1/sqrt(L) and all dynamic features are lost. One would expect that such a probability would be useless in trying to describe a dynamic problem. (We discuss this in more detail). The alternative is exp(-iEt+ipx) which suggests that there is periodicity in x and t based on E and p. At first, this is equally strange from a classical point of view, one might argue. We suggest, however, that for x+dx and x-dx, exp(ipx) clearly defines the direction of motion which is essential in a dynamic problem. In other words, we argue that Newtonian mechanics uses dynamic variables x(t), v(x,t) etc to solve a problem. We suggest introducing a dynamic probability which considers instantaneous probaiblies (just as Newton considers instantaneous variables). The big question, from a classical point of view, is then: Why would one use probabilities at all when Newtonian mechanics describes deterministic physics? The answer, we argue, is that P(x)=1/L is an acceptable classical probability and one is allowed products in probability theory. We suggest that there is no reason why one cannot consider a dynamic probability F(p,x) such that F(p,x)*F(-p,x) = 1/L. Unless one can refute the existence of F(p,x) in classical theory, then mathematiaclly one has F(p,x) = exp(ipx) (because F(p,x)=constant makes no sense). The question then becomes whether such an F(p,x) may be used in solving any physical problem with the answer being yes. This formalism solves the 1-dimensional reflection-refraction problem of photons, assuming the photon to be a kind of particle moving at speed c. Newtonian mechanics does not solve this problem, it seems.
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 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.
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.
We try to argue that the idea of probability associated with motion may have classical beginnings in that the center-of-mass (cm) of a piece of mass moving with constant velocity v is present within a dx at some time, but moves out of dx. Flux is constant, but at one time the particle cm leaves dx and arrives at the adjacent dx. Constant flux suggests one does not need to consider time. In such a case, one might associate a periodic function showing probability being removed from a region dx (at x) (i.e. a negative distribution) followed by it being added (a positive distribution) to a region x+dx. This suggests a negative-positive cycle, but this labels certain x regions as negative and others as positive which is inconsistent with all x regions having the same weight based on a constant v. Given that time is removed, it seems this cycling may have nothing to do with velocity v or perhaps anything else for that matter. As a result, one may postulate an Aexp(ibx), where b is a constant, for motion. We will argue for a phase linear in x. AAexp(-ibx)exp(ipx) yields a constant AA, which is flux, and treats all x points the same. At this point, one cannot go farther unless one considers other issues.We consider two such issues. First we ask what happens if the particle is viewed from a frame moving at constant v? The velocity of the particle will not be the same, but we have argued that exp(ibx’) may be the form, yielding no information about b when (x,t) changes to (x’,t’). If on the other hand, the phase cannot change if the particle is viewed from the moving frame, one requires a Lorentz invariant. This suggests that the phase may actually be -Et+px. (We also show that it is not necessary to use special relativity to obtain the form exp(ipx). Rather one may consider ideas of pressure balance associated with continuity of exp(ibx).) As a second consideration, we examine a one dimensional reflection-refraction problem at a point xo for which a probability equation 1= Probability(reflect)+Probability(refract) exists. If exp(ipx) for constant E describes motion, this motion changes with changing p, yet one wishes probability to be a continuous function at xo. This suggests different weight coefficients for different exp(ipx) which seem to be associated with P(reflect), and P(refract). If x=0, one has an equation in the weights, with A+B (x<xo) matching C (x>xo) at xo. If one stops at this point, exp(ipx) does not seem to have any impact. One may note, however, that in general xo is not zero and that one wishes at the very least to have the derivatives match. This immediately links the A,B,C values to the p (momentum) discontinuity. Thus using special relativity to create an exp(ipx) has practical implications because it allows one to deal with p discontinuities. We also show though that special relativity, i.e the use of -Et+px, may be bypassed and the form Aexp(ipx) obtained as a result of its first derivative yielding p which is needed in a pressure balance equation. This then suggests A is the square root of flux.
Quantum mechanics utilizes a wavefunction W(x)=Sum over p a(p)exp(ipx) (in time independent cases) with exp(ipx) being what we call a free particle complex probability. In such a case, W(x) is an OR scenario of these various probabilities. We have argued in previous notes that exp(ipx) is the probability for a p impulse hit at x, even if the center-of-mass of the particle is not at x. This means that impulse hits may occur probabilistically in space as in the 2-slit example which requires adding two probabilities as an OR case for each slit. The point then seems to be that W(x) (or even its momentum frozen form W*(x)W(x) gives a profile in space of the number of interactions (various p impulse hits) at each x. Classically, the kinetic energy (nonrelativistic) is pp/2m. In a classical bound problem this kinetic energy is considered to exist (in calculations) at the center-of-mass, but forces are also said to be rendered at this point so pp/2m, the center-of-mass and the force hit point all coincide. In quantum mechanics, a p impulse hit and the center-of-mass do not coincide. If p is said to hit at a probabilistic x, then it seems that pp/2m should also be seen to exist at this same hit point. A physical interaction point verifies in a sense the location of pp/2m in a probabilistic manner, i.e. a(p) exp(ipx)/W(x) is the probability used to calculate KEaverage and this is based on impulse hit locations. By normalizing using W(x), one removes the relative number of average hits at each x and obtains a pure energy value at x. (This may then be used in a KEave(x)+V(x)=E equation.) On average W(x) has zero points, which means that if one considers W*(x)W(x) to represent the motion-frozen-out spatial distribution of hits, then no hits are rendered at certain x points (on average). One might argue then that the particle cannot exist at x, but the center-of-mass can exist anywhere. One might next argue that there is no kinetic energy at x, but that cannot be true because 1/2m d/dx d/dx W has the same 0 points as W and the two divide. Thus, there is average kinetic energy at all bound points and this represents an average based on impulse hits (exp(ipx)). Classically, if one has many particles in a potential, one may write density(x)V(x) to find the potential energy of the particles. This implies that the particles interact with V(x), e.g. neutrinos usually do not interact. Density(x) in a classical sense means that the center-of-mass and force hit point are both at x. In quantum mechanics, exp(ipx) represents the probability for a force hit point, W(x) represents an average and W*(x)W(x) represents a motion-frozen average hit point. W*(x)W(x)V(x) seems to represent the average potential energy of the particle at x because W*(x)W(x) is linked with interaction. Even if a particle is not present, however, there is energy V(x) in space. It seems that if W*(x)W(x)=0 at some x, then there is no average particle interaction at this point, but there is still average kinetic energy Sum over p a(p)pp/2m exp(ipx) / W(x) based on the probability for average probability hits. The key idea is that one normalizes probability by dividing by W(x) in the average kinetic energy calculation. W*(x)W(x) is normalized not at x, but integrating over dx and setting this value to 1. One has relative hit at x values represented by W*(x)W(x). For exp(-ipx)exp(ipx)=1, each dx has the same average hit value (with one not being able to distinguish details in hbar/p, i.e. this is a motion frozen result. The question then becomes: Why would one write: W*(x)W(x) KEave(x)? One might argue that this term occurs if one takes the quantum mechanical equation: KEave(x) +V(x) = En and multiplies by W*(x)W(x). One may ask: Does V(x)W*(x)W(x) have physical meaning? V(x) is the potential energy at x. The average interaction probability with V(x) at x should be W(x)V(x) and not W*(x)W(x)V(x). The latter is a classical mechanical idealization. If W*(x)W(x) is 3 at x1 and 2 at x2, then it is as if the particle interacts 1.5 times more at x1 and so if V(x1)=V(x2), then this approach assumes 1.5 the interaction energy at x1. KEave(x), however, is simply free particle energy (average) at x and one does not need to take this average value and multiply it by the relative number of interactions at x. W(x) has zero points meaning that there is no interaction at such a point on average. KEave(x) on the other hand is the average energy at x based on the particle kinetic energy presence pp/2m being at the point of an impulse hit. This does not mean that an interaction is occurring with V(x) at this point in this formalism. Even a free particle has an exp(ipx) probability when it is not interacting. The sum Sum over p a(p)exp(ipx) means that all of the interactions linked with x have occurred and one is making an OR calculation. KEave(x) is free particle kinetic energy at x and V(x) is potential energy (not interaction potential energy with the particle). Multiplying by W*(x)W(x) is a classical minded approach to creating a classical equation: density(x) (KEave(x)+V(x)) = E density(x) ((1)) Here KEave(x) would represent the average kinetic energy of a single particle and density(x) would be the average number of particles at x. density(x)KE(x) has a clear meaning. W*(x)W(x) KEave(x) on the other hand is different. KEave(x) is actually the classical kinetic energy value at x and so it is not clear why one would multiply it by W*(x)W(x). W*(x)W(x) represents the probability for an impulse hit at x in a p-frozen out picture. The notion of exp(ipx) (i.e. probability for a p hit at x) is already buried within the quantum calculation of KEave(x) (which yields the classical value). In order to find the physical meaning of W*(x)W(x) multiplied by KEave(x) or V(x) it seems that one has broken dx into equally sized segments which correspond to different detection hits of the particle. This is then somehow equated with the presence of the particle per unit time in dx which is then somehow rendered as the average energy at x multiplied by the detected presence at x through a steady rate of measurement/time. Thus, multiplying by W*(x)W(x) is not multiplying by a physical density.