In (1) it is argued that a quantum particle’s probability P(x,t) satisfies a Nelson stochastic equation: d/dt partial P(x,t) + d/dx partial (b(x,t)P(x,t)) = D d/dx d/dx partial P. ((1)) where b(x,t) = hbar/m d/dx partial S(x,t) + 2D d/dx partial ln(R(x,t)) ((2)) with W(x,t)=wavefunction= R(x,t) exp(i S(x,t). Thus both a quantum wavefunction (satisfying the Schrodinger equation) and a separate P(x,t) are present. If P(x,0)= W*(x,0)W(x,0) ((3)), i.e. Born’s rule, then this rule holds for later times. It is, however, suggested in (1) that if P(x,0) not= W*(x,0)W(x,0) then the system may relax over time into a situation for which ((3)) holds. In this note, we argue that for a free particle moving with constant p in one direction, P(x,t)=constant, but there is an underlying two dimensional wavefunction exp(ipx). We ascribe this to two x-probability distributions associated with motion. In particular: v= (xfinal - xintial) / delta t. If xfinal and xinitial are both distributions such that on average v = constant, one needs to give a probability for both i.e. have two dimensions or an ordered pair. The two probability distributions, however, are not independent as they must describe a constant velocity i.e all points of space being the same. This leads to exp(ipx) such that exp(-ipx) exp(ipx) =1 . In other words, one may constrain two probability distributions to yield a deterministic type result for a single particle which appears as a probabilistic result for a steady stream.Born’s Rule automatically applies to a free particle and is not a “postulate” as argued in (1). The form exp(ipx) follows from the Lorentz invariant -Et+px we argue. We further argue that if one creates a linear superposition of exp(ipx)s i.e. W(x)=Sum over p a(p)exp(ipx) then Integral dx W*(x)W(x) is always linked to a steady stream probability picture ie. Sum over p a*(p)a(p) = 1. This is why given a p, one does not know its x point- it is associated with a steady stream. As a result, we argue that there is always a meaning for Born’s rule W*(x,t)W(x,t). As a result P(x,t=0) not= W*(x,t=0)W(x,t=0) must describe a probability which differs from the steady stream scenario linked with exp(ikpx).
In (1) it is argued that a quantum particle’s probability P(x,t) satisfies a Nelson stochastic equation: d/dt partial P(x,t) + d/dx partial (b(x,t)P(x,t)) = D d/dx d/dx partial P. ((1)) where b(x,t) = hbar/m d/dx partial S(x,t) + 2D d/dx partial ln(R(x,t)) ((2)) with W(x,t)=wavefunction= R(x,t) exp(i S(x,t). Thus both a quantum wavefunction (satisfying the Schrodinger equation) and a separate P(x,t) are present. If P(x,0)= W*(x,0)W(x,0) ((3)), i.e. Born’s rule, then this rule holds for later times. It is, however, suggested in (1) that if P(x,0) not= W*(x,0)W(x,0) then the system may relax over time into a situation for which ((3)) holds. In this note, we argue that for a free particle moving with constant p in one direction, P(x,t)=constant, but there is an underlying two dimensional wavefunction exp(ipx). We ascribe this to two x-probability distributions associated with motion. In particular: v= (xfinal - xintial) / delta t. If xfinal and xinitial are both distributions such that on average v = constant, one needs to give a probability for both i.e. have two dimensions or an ordered pair. The two probability distributions, however, are not independent as they must describe a constant velocity i.e all points of space being the same. This leads to exp(ipx) such that exp(-ipx) exp(ipx) =1 . In other words, one may constrain two probability distributions to yield a deterministic type result for a single particle which appears as a probabilistic result for a steady stream.Born’s Rule automatically applies to a free particle and is not a “postulate” as argued in (1). The form exp(ipx) follows from the Lorentz invariant -Et+px we argue. We further argue that if one creates a linear superposition of exp(ipx)s i.e. W(x)=Sum over p a(p)exp(ipx) then Integral dx W*(x)W(x) is always linked to a steady stream probability picture ie. Sum over p a*(p)a(p) = 1. This is why given a p, one does not know its x point- it is associated with a steady stream. As a result, we argue that there is always a meaning for Born’s rule W*(x,t)W(x,t). As a result P(x,t=0) not= W*(x,t=0)W(x,t=0) must describe a probability which differs from the steady stream scenario linked with exp(ikpx).
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 the literature, stochastic approaches employed to derive the Schrodinger equation seem to focus on the idea of a classical spatial density which depends on time and space ie. density(x,t). As a result, d/dt (partial) density appears in various equations such as the continuity and Focker Planck equations. At one point in the derivation the quantum wavefunction is introduced “mathematically” as W(x,t)=sqrt(density(x,t)) exp(i S(x,t)) where S is related to a velocity field through u(x,t)=velocity= d/dx S. From quantum mechanics, -Et becomes a term of S the velocity field. The stochastic/hydrodynamical approaches employ the term du/dt partial = d/dt (partial) d/dx S and then “pull out”” d/dx leaving d/dt (partial) S which yields -E average energy from the -Et term. In this note we argue that for a bound quantum state density(x) so d/dt (partial) density = 0, yet the statistical ideas of quantum mechanics should be present in such a case. Furthermore for the free particle wavefunction exp(ipx-iEt), density=1. In fact, the Schrodinger equation should follow for a free particle even the case of density=1. Stochastic processes seem to be based on equations which consider a change in classical spatial density at a point x in time due to flow in or out or stochastic changes. Quantum processes seem to deal with periodic probability changes from one dimension to another exp(ipx) (2 dimensions) which are somehow related to a physical process of momentum flow. Ultimately one links this process to an energy conservation which does not seem to be the same as a classical stochastic/hydrodynamical approach. In this note we argue that a free particle with a constant momentum p has periodic stochastic processes associated with it through its complex probability change in the region of its wavelength, but that these are linked to creating the momentum p. Furthermore momentum is linked to energy so we argue that the Schrodinger equation is a stochastic equation linking energy and momentum and that time and space are separated i.e. there is no situation of following a change in spatial density in time for a free particle with momentum p. For a bound state, one has a distribution of free particles, but again time is removed from the problem in the sense that it only appears in exp(-iEt) in the wavefunction to again establish a relationship between energy and average momentum and potential. There is again no nonzero d/dt (partial) density term which seems to be key in stochastic equations. Only if one takes a linear combination of energy eigenstates Sum over En an(En) exp(-iEnt) Wn(x) does one obtain the form density(x,t), but we argue that quantum features should already appear in exp(ipx) and the bound state scenario with a zero d/dt (partial) density(x,t). We argue, however, that the Schrodinger equation and fundamental stochastic ideas of quantum mechanics should already be present at the level of exp(ipx).
In a previous note (1) we argued that the free particle classical action A (relativistic or nonrelativistic) may be written in terms of v=x/t and then varied in x and t independently. One finds that dA/dx partial = p and dA/dt partial = -E implying A= -Et+px with x and t independent. This is the opposite of the classical mechanical idea of x(t). Creating an eigenvalue equation yields -id/dx partial exp(iA) = pexp(iA) and id/dt partial exp(iA) = Eexp(iA). exp(iA) plays the role of an unusual type of probability we argue which describes two dimensional periodicity (cos(px) and sin(px) ). We have argued that the quantum particle is the medium for a two dimensional wavelike motion of an energy-momentum wave. A wave type structure is needed to describe dynamics (like a classical wave), but unlike a classical wave two dimensions are required because there is no preference of any x point for a free particle. Both cos(px) and sin(px) both treat x in a preferential manner as their values vary with x. For a classical wave this is fine because the creation point of the wave imposes boundary conditions, but there is no such boundary condition imposed on a free particle. exp(ipx-iEt0 must somehow then link to P(x)=constant. One may note that exp(-iA)exp(iA) = 1 for the free particle. This may be argued to be linked to spatial invariance i.e. P(x)=constant where P(x) is not exp(ipx). Here we argue that exp(-iA)exp(iA) represents a product of two unusual probabilities , one for motion in a certain direction and the other for the time reversed scenario. If the quantum particle is the medium of an energy-momentum wave exp(-iEt+ipx), then the product of the regular and time reversed unusual probabilities cancel or undo each other. In other words one does not see any change in space as long as one has a noninteracting exp(-iEt+ipx) . Given that -id/dx is a linear operator this product yields a momentum of 0. We consider this idea and its implications on a momentum distribution Sum over p a(p)exp(ipx) which may represent a bound state or other states. In particular, it seems that P(x)=W*(x,t)W(x,t) which yields a classical type of probability (P(x) real, positive and normalized to 1) is linked to a product (merging) of forward and time reversed situations so that any spatial density which remain due to the quantum particle acting as a medium for an energy-momentum wave (as well as the time reversed part) would represent an actual change in spatial density. exp(ipx) already contains two periodic spatial functions (cos(px) and sin(px)), but does not manifest it unless there is an interaction.
In a classical bound state, conservation of energy: p(x)p(x)/2m + V(x) = E holds for all x. Examined in a statistical sense, one may consider each constant dx to have a probability density proportional to dt(x) where v(x)=velocity =dx/ dt(x). Thus the particle is followed in time. In quantum mechanics, even for a free particle exp(ipx), there exists a wavelength hbar/p. Within this region one cannot follow the particle in time. In fact, there are two probability distributions cos(px) and the shifted sin(px). The two dimensional nature of exp(ipx) is required to describe the direction of motion. There exists a mathematical abstraction or average which holds for any x within the wavelength and yields the classical x= p/m t. Thus there is a duality, but within a wavelength one cannot follow the particle exactly using x=p/m t. Rather there is no time and one thinks of probabilities to be at one x or another. The operator -id/dx yields the abstracted or average momentum at any x and (-id/dx)(-id/dx) average pp if one divides by probability. In other words: (-id/dx) exp(ipx)/ exp(ipx) = p and (-id/dx) (-id/dx) exp(ipx) / exp(ipx) = pp. These then apply to any x point. For a bound state, exp(ipx) is replaced with W(x)= Sum over p a(p)exp(ipx). Thus p and -p are included together. From a statistical point of view this is fine because a quantum bound state is described in terms of energy conservation (time-independent Schrodinger equation). Unlike the classical case, the probabilities of p and -p differ, i.e. exp(ipx) and exp(-ipx). The sum exp(ipx)+exp(-ipx) yields cos(px) which is the resolution probability distribution. Taking -d/dx d/dx of W(x) and then dividing by W(x) yields the average kinetic energy at any x within the wavelength. In other words this is the average kinetic energy which matches the classical result for any energy level. As a result, prms(x)prms(x)/2m + V(x) = En holds for any quantum level, but prms(x) is an rms average value which bypasses wave structure. That is why the exp(ipx) for p,-p and -id/dx must be used to calculate average quantities within a wavelength. Many different p,-p combinations are needed so that at each x within a wavelength the correct average kinetic energy value appears. Now consider a high energy level and choose a fixed dx length to represent a tiny region in the system, a cutoff for which there is no further resolution. p(x) is constant within this region. If hbar/ prms(x) <= dx, the wave structure of quantum mechanics does not play a role. (Note that for an oscillator, W(x) is a set of crests and troughs with different heights and widths.) As a result, there is no need for d/dx acting on W(x) to describe average kinetic energy within dx. The average kinetic energy is prms(x)prms(x)/2m and dx is treated like a point. Only if exp(ipx) is used does -id/dx yield the proper classical result. (An alternative scenario is that d/dx acting on a function divided by the function yields the same result as acting on exp(ipx) divided by exp(ipx) e.g. -d/dx d/dx cos(px)/cos(px) = -d/dx d/dx exp(ipx) / exp(ipx). If one has the quantum kinetic energy expression -1/2m d/dx dW/dx / W ((1)) and considers one momentum, namely prms(x), and uses W(x)= cos( prms(x) x), then with prms(x) constant at x, ((1)) yields prms(x) prms(x)/ 2m. Thus d/dx has not broken down even though it does not describe motion within dx which is now a point. A problem arises, however, if one considers Wn(x)= Wn1(x)Wn2(x) as this yields an equation containing dW1/dx dW2/dx/ (W1W2). Then the above procedure yields: dW1/dx/W1 (prms(x) sin(prms(x) x) / cos(prms(x) x). This does not match the classical value. Furthermore, dW2/dx/ W2 contains +prms(x) and -prms(x) with different weights while a classical scenario is -pP(-p,x) + pP(p,x) = 0 because P(-p,x)=P(p,x) while quantum mechanically it doesn’t. We analyze this situation and conclude that d/dx breaks down/ does not apply in the classical limit because it is used to calculate average values within a wavelength only. It only “works” if the effect of the derivative(s) on a function divided by the function is the same as the effect on exp(ipx) divided by exp(ipx). If a wavelength fits within dx, there is no need to calculate values at any lower x level and d/dx is not needed. In fact, there is a single prms(x) value in dx and the average time links different dx’s so that one follows the particle (on average in the sense that one ignores structure within each x) by x(t). As a practical consequence, using -id/dx together with the wavefunction W(x) can lead to problems in the classical limit because d/dx does apply. In short, using -id/dx operating on Wn(x) divided by Wn(x) leads to issues unless it produces the same effect as acting on exp(ipx) divided by exp(ipx).
A classical particle is considered to be a point described by x(t) i.e. x and t are both a set of points. In a one dimensional problem, however, dx= v(x) dt so if all dx tend to zero, then dt’s are small, but don’t tend to zero in the same way. In other words dx may be considered small, but of a finite size. To have a particle within dx, it should have a probability distribution that goes to zero at the endpoints of dx i.e. one could have a repeated set of these for a constant velocity. This, however, contradicts P(x)= C1 dt = C1 dx/v(x). Thus it seems first that there are two kinds of probability distributions and that the set of distributions fitting into each dx is incomplete because if this set is described by a function F(x), this function should have the same magnitude at F(x+dx) for a free particle. This suggests a phase function linear in x i.e. exp(i C1 x). P(x) = C1dx/v(x) is based on dynamics and no dynamics appears in exp(i C1 x). It seems that dynamics should occur leading to the possibility of exp(i p x). (This , however, is ad hoc.) We call this a dynamic probability as opposed to P(x). In quantum mechanics it is called a wavefunction. If one considers the classical free action (relativistic or nonrelativistic) with x and t independent, then A= -Et + px (1) and one may create eigenvalue equations -id/dx partial exp(iA) = p exp(iA) and id/dt partial exp(iA) = E exp(iA) so exp(ipx) also appears more formally. In other words a fixed resolution periodic distribution requires the same shifted distribution (i.e. cos(px) and sin(px)) kept as separate dimensions i.e. “i” in order to create a dynamic probability which is a phase function which may be linked to P(x). This phase function exp(ipx), however, consists of two classical-like waves showing that in each dimension probability is “carved” out of troughs and added to crests. The two dimensions allow for the dynamic probability to be consistent with P(x)=constant for a free particle.
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.