The Feynman path integral approach to calculating a quantum propagator K(x,x1,t) such that W(x,t) = Integral dx1 K(x,x1,t) W(x1,0) involves a purely classical Lagranian. The form exp(i L(x,dx/dt)) and integration over all possible classical paths between x1,0 and x,t taken as tiny dt slices. For the free propagator, this calculation may be simplified by taking the Fourier transform of K(x,x1,t) which introduces exp(ikx) into the problem. We argue that taking the Fourier transform is not simply a math procedure as p represents physical (classical) momentum. Thus, exp(ipx) appears in this approach as associating physical momentum with a wavelength. Furthermore, it seems that backward forward motion is allowed in each dt slice because constant momentum is a classical solution and any positive or negative momentum may be used with the slice. Thus. “quantum mechanical” features appear already in what seems to be a classical calculation through exp(ipx). In an earlier note, we argued that exp(ipx), (the wavefunction of a free particle) may itself be treated as a conditional probability and obtained by maximizing Shannon’s entropy WlnW subject to the constraint xW i.e. allowing the particle to move back and forth even at time t=0. From exp(ipx), one may desire a Lorentz invariant wavefunction so px becomes px-Et. Then, one may construct the free particle propagator directly from: K(x,x1,t) = Integral dk exp(ik(x-x1)) exp(i kk/2m t) which matches the Feynman approach. In both cases, one uses classical physics and the idea of maximization of entropy. In the second approach, one considers a specific constant momentum p when maximizing W. In the Feynman approach, all momenta are included for a free particle because given that one has particle move form x1,0 to x,t any initial (constant) momentum is possible. Thus, we argue that the Feynman propagator for a free particle follows the same statistical approach as the maximization of a free particle conditional probability (i.e. wavefunction) subject to the constraint xW. This constraint, which seems to represent some forward backward motion directly, appears indirectly in the Feynman approach as exp(ipx) does not appear unless one uses a Fourier transform (which represents a physical wavelength).
The Feynman path integral approach to calculating a quantum propagator K(x,x1,t) such that W(x,t) = Integral dx1 K(x,x1,t) W(x1,0) involves a purely classical Lagranian. The form exp(i L(x,dx/dt)) and integration over all possible classical paths between x1,0 and x,t taken as tiny dt slices. For the free propagator, this calculation may be simplified by taking the Fourier transform of K(x,x1,t) which introduces exp(ikx) into the problem. We argue that taking the Fourier transform is not simply a math procedure as p represents physical (classical) momentum. Thus, exp(ipx) appears in this approach as associating physical momentum with a wavelength. Furthermore, it seems that backward forward motion is allowed in each dt slice because constant momentum is a classical solution and any positive or negative momentum may be used with the slice. Thus. “quantum mechanical” features appear already in what seems to be a classical calculation through exp(ipx). In an earlier note, we argued that exp(ipx), (the wavefunction of a free particle) may itself be treated as a conditional probability and obtained by maximizing Shannon’s entropy WlnW subject to the constraint xW i.e. allowing the particle to move back and forth even at time t=0. From exp(ipx), one may desire a Lorentz invariant wavefunction so px becomes px-Et. Then, one may construct the free particle propagator directly from: K(x,x1,t) = Integral dk exp(ik(x-x1)) exp(i kk/2m t) which matches the Feynman approach. In both cases, one uses classical physics and the idea of maximization of entropy. In the second approach, one considers a specific constant momentum p when maximizing W. In the Feynman approach, all momenta are included for a free particle because given that one has particle move form x1,0 to x,t any initial (constant) momentum is possible. Thus, we argue that the Feynman propagator for a free particle follows the same statistical approach as the maximization of a free particle conditional probability (i.e. wavefunction) subject to the constraint xW. This constraint, which seems to represent some forward backward motion directly, appears indirectly in the Feynman approach as exp(ipx) does not appear unless one uses a Fourier transform (which represents a physical wavelength).
In a series of notes (1), we argued that the time-independent Schrodinger equation may be considered as a statistical conservation of average energy equation. We further argued that V(x)=Sum over k Vk exp(ikx) delivers stochastic potential hits which create a momentum distribution W(x)= Sum over p a(p) exp(ipx) at each x. The form exp(ipx) describes a free particle and a statistical distribution is made of such free particles. The form exp(ipx) has a modulus of 1 indicating that P(x)=constant. d/dx exp(ipx) = ip suggesting the change in conditional probability with space is related to classical momentum. Given a momentum distribution, each p at x occurs at a different time. Thus, one does not follow a classical particle in time x(t), nevertheless an average energy conservation equation applies. Taking the time-independent Schrodinger equation E W = -1/2m d/dx d/dx W + V(x) W and writing EW = i d/dt exp(-iEt), one obtains a time-dependent equation. We argue that if the time-independent case represents a classical conservation of average energy, the time-dependent equation should be linked to Newton’s equation of motion. This immediately leads to a problem because we argue one does not want x(t) for a statistical system. As a result, x in the time-dependent Schrodinger equation, which represented spatial position before, now represents a spatial variable that is independent of time. We consider the oscillator case and note that an amplitude is such a candidate where x(t)=XA(t). Thus, the time-dependent Schrodinger equation ultimately becomes an equation in A(t) with the amplitude cancelling, at least in the free particle and oscillator case. From such a consideration, one may work “backwards” and find a function for which the time-dependent Schrodinger equation (equating powers of X) becomes Newton’s second law. From this, one may show that exp(i Action) where Action = Integral dt1 (0,t) L where L is the classical Lagrangian is this solution with d/dX applying to X which is an amplitude, not x(t).
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 have argued that one may obtain exp(ipx) (the spatial free particle probability) from the maximization of P1(x) ln(P1(x)) + i px P1(x), and a similar expression for P2(t). This yields exp(-iEt+ipx). If one integrates i Integral px exp(ipx) dx from px=0 to 2*3.14/p, one obtains i 2*3.14/p as a spatial entropy and i2*3.14/E as a temporal one associated with the complex probability exp(-iEt+ipx). If one considers classical motion of a free particle in space, then there is no uncertainty in time or space and so one might create a classical level entropy of: -E dt + p dx, where dt=hbar/E and dx=hbar/p.This yields -Edt+pdx=0 and one might equate this with a classical entropy dS as done in (1). This overall classical entropy is different from the spatial and temporal wave entropies dx and dt. If one does not wish to work at the classical level, one might try to use the hbar/p, hbar/E entropies in a problem requiring exp(-iEt+ipx), such as 1-D reflection-refraction at an n1-n2 index of refraction junction (x=0). In such a case, it seems that one does not use entropies, but rather their reciprocals which indicate the amount of resolution information present, i.e. L/dx = Lp/hbar. As a result, conservation of E and p may be reformulated in terms of a balance of resolution information which is the reciprocal of dx, i..e p (in the spatial case). For example, a particle at rest may break into two pieces with p and -p. The resolution is proportional to p initial =0 and after one has p and -p (i.e. conservation) of momentum which may be argued to be equivalent to conservation of resolution information. Using the notion of a balance of the reciprocal of entropy (i.e. p which is linked to conservation of momentum), one might interpret: Ap - Bp = C p2 where p2=n2 p in terms of resolution and not momentum conservation, i.e. A-B = C p2/p where p2/p = n2 is the greater relative resolution of the p2 photon. The idea of both a wave entropy (both temporal hbar/E and hbar/p) and its reciprocal resolution information seem to arise as well an overall classical entropy -Edt+pdx = 0 if dt=hbar/E and hbar/hp = 0. dt=hbar/E and dx= hbar/p We have argued in previous notes that one may obtain: dt=hbar/E and dx=hbar/p ((1)) as independent dx and dt fluctuation from the Lorentz invariant. One may show that -Et+px is equivalent to the classical relativistic or nonrelativistic action Lt if Lt=.5movv t (nonrelativistic) or Lt = -mo sqrt(1-vv/cc) t (relatvistic) and x/t=v ((2)) The reason that one recasts Lt as -Et+px is to use relativistic variables. The notion of an action remaining unchanged under a fluctuation delta x(t) , i.e. at t one considers x1 instead to x=vt, leads to: 0 = -E delta t + p delta x = 0 which has a solution of ((1)). In (1), however, the notion of an overall entropy is introduced such that: dS = 0 = -E delta t + p delta x ((3)) Given that one really has a probabilistic scenario (free particle quantum mechanics) associated with ((1)), i.e. exp(-iEt+ipx) = probability which ensures ((1)) and also spatial and temporal invariance by the unit modulus (less information than exp(-iEt+ipx)), then one might consider the notion of an overall entropy.
In previous notes, we suggested that quantum bound states may be analysed in terms of the conditional probability P(p/x)= a(p)exp(ipx)/W(x) (where W(x) is the wavefunction) which allows for the calculation of average kinetic energy at each point x i.e. [Sum over p pp/2m P(p/x)]/ W(x). This added to V(x) yields E (average energy) at each x and yields the time-independent Schrodinger equation). We argued that probability is handled at the level of W(x)=Sum over p a(p) exp(ipx). Measurements (even internal) may knock a particle from one p1 state to another so another set of probabilities (P(x) and P(p)) arise, P(x) being based on: a(p1)exp(i p1x) exp(i p2x). Averages of f(p,x) follow from P(x) f(p,x) P(p/x). Thus, there are two sets of probabilities in quantum mechanics. Entropic dynamics (1) appears to begin with the probability P(x) and considers a transition probability P(x1 |x) with x1-x= an infinitesimal quantity. A product of transition probabilities is then taken to allow for finite x1-x differences. It is stated in (1) that there exists information about x but not p. The maximization of entropy density may be used to determine the form of P(x1 | x). A “drift velocity” is also included (which is later shown to be related to the phase of the wavefunction). Ultimately, the Schrodinger equation is also obtained. We argue one may use P(p/x) to also calculate P(x1 | x) and compare this result with that obtained through the maximization of entropy density. A type of drift velocity appears naturally in this approach. We argue that the form a(p)exp(ipx) already incorporates statistical ideas (i.e. maximum entropy) in that the modulus of exp(ipx) is identical for all x, endowing all x points with a “kind” of equivalent status. exp(ipx) is a function of p and x because one must calculate average kinetic energy at each x and this must change with x i.e. KEave(x) + V(x) = E. Thus, we think there are strong similarities between entropic dynamics and the approach of using P(p/x). There are also some differences in that the coefficient A of delta(x)delta(x) in P(x1 | x) = 1/Z exp{ A delta(x)delta(x) + B d Phi/dx delta(x)} is treated as independent of x in entropic dynamics, but is a specific function of x (related to d ln(W)/dx d ln(W)/dx + 1/W d/dx d/dx W) in the P(p/x) approach.
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 (1), the notion of a stochastic classical action is introduced and a probability is linked to exp( constant * action). The objective of (1) is not to derive quantum mechanics, but to derive stochastic dynamics based on maximum entropy in action space. We have argued in a number of previous notes (2) that one may use the classical action to create a quantum free particle probability. This classical action, however, is not chosen a priori, but rather arises due to other considerations. We suggest that stochasticity is present in Newtonian mechanics at some time scale because dt is finite in dp/dt= Force (Newton’s second law) and so postulate that force delivery at some x,t is probabilistic within some dt and dx. A priori, dt and dx are not known, but if there is stochasticity, there should be a free particle probability, i.e. a probability(x,t) where x,t represent an x, t interaction point. Furthermore, the notion of dx and dt should arise naturally. We note that the form exp(i function(x,t,p,E) may be introduced because free particles have no distributed variable (like energy in the Maxwell-Boltzmann case). exp(i function(x,t)) immediately introduces periodicity, but we have argued that stochasticity can only exist within tiny, fixed dx, dt linked to dp/dt=F. In particular, one classical Newtonian scales this stochasticity is not seen. The periodicity should be directly in x and t space, suggesting exp( i A1 t + i A2 x). This expression, however, should be Lorentz invariant, and A1 and A2 should be constants, at least for a fixed mo and v. The reason that the one must have periodicity is that the dx and dt ranges of stochasticity in dp/dt=Fr must appear naturally. At this point, one is led to exp(-iEt+ipx) with -Et+px being a Lorentz invariant which satisfies the above criteria. This Lorentz invariant happens to be the classical action in both the relativistic and nonrelativistic cases, hence giving rise to action as being a stochastic variable (of (1)) because it is a function of the two scholastic variables x and t. Alternatively, one may choose the exp(iA1x+iA2t) due to its conservation properties. One may then argue that A1 and A2 could also be conserved for x,t held constant, i.e. A1 and A2 could be linked to Newtonian energy and momentum in elastic collisions. This allows one to solve for the special relativistic g(v) =1/sqrt(1-vv/cc). Given the idea of a conserved quantity form for a probability, one may argue that exp(i Action) satisfies a maximization of Shannon’s entropy subject to a conserved constraint (which is again a link with (1).) The question then becomes: Why should the Lorentz invariant A= -Et+px be the classical action.One may note that x,t usually refer to trajectory values and so for a free particle, one has x=vt. This means that one may factor out t, i.e. A= t( -E+pv), but this is automatically the form of the Lagrangian with E being the Hamiltonian. This holds relativistically and non-relativistically with the form: dL/dv partial = p. The remaining question is: Why should the classical action be Lorentz invariant? We consider E(v)-p(v) v = -L such that dE/dv - v dp/dv -p = -dL/dv partial. This means that dE/dv - v dpd/v = 0 ((1)). A Lorentz transformation, however, changes mo in a rest frame into p=mov/sqrt(1-vv/cc) and E=mocc/sqrt(1-vv/cc). One may see that this solution may be obtained from ((1)) showing a direct link between the classical action approach and special relativity. The connection may be carried further as Lt = -Et+px is a Lorentz invariant, but also the classical action (in both the relativistic and nonrelativistic cases). We argue that the use of the action in (1) as a stochastic variable is directly related to special relativity. In summary, from finite dt in dp/dt=F, t and x are stochastic variables within some naturally appearing dx and dt. One may write a linear combination of x and t and to create a stochastic Lorent invariant variable which should appear in exp(i Lorentz invariant) and create the dx and dt units. It may then be shown that this Lorentz invariant in E,p,x,t should be equivalent to the classical action.