Traditionally, Newtonian mechanics is recognized as an approximate theory which is said to arise from quantum mechanics (apparently a more detailed theory) in the large energy level limit (correspondence principle). We argue that both quantum mechanics and Newtonian mechanics are both approximate and both complement each other at any energy level. In the Newtonian case, quantum effects are so small that they are not visible, but in the quantum case, Newtonian effects are visible, we argue. We first note that exp(-iEt+ipx), the free particle wavefunction (complex probability) does not exist without the Newtonian expression x=vt. x=vt implies it is possible for dx->0, dt->0, while hbar/p and hbar/E show that there are limitations with respect to interactions. The only way to ascertain x=vt is through measurement and if these measurements have dt=hbar/E and dx=hbar/p uncertainties, one cannot have dx->0, dt->0 resolution. Nevertheless, the quantum free particle does move as x=vt (as long as one respects the uncertainties of dt,dx).The quantum exp(-iEt+ipx) does not exist without the Newtonian x=vt. In a single particle quantum bound state (time-independent), we suggest there are significant approximations. First, a conservation of energy equation exists in Newtonian physics, namely pp/2m + V(x) = E ((1)). This equation does not contain time and does not explicitly show x(t), but that does not mean that such motion does not physically exist. The quantum bound state solution follows from ((1)) which is an equation which does not display full information, even in the classical case, and so this loss of information carries down to the quantum solution. In particular, in the quantum case one solves a time-independent problem and so uses a(p)exp(ipx) + a(-p)exp(-ipx) with a(p)=a(-p) or a(p)= -a(-p). These conditions are equivalent to insisting that the local momentum average at each x (-i W* grad W + i W grad W) is 0. The fact that one is solving a quantum bound state using only energy considerations means that one averages out time which means averaging momentum. Thus, W*(x)W(x) describes a resolution spatial pattern for interaction, but W = W(x) exp(-iEt+ip x) where p=0 at each x. In other words, just because one does not see forward/backward motion in a particular quantum solution (based on energy alone) does not mean that it does not exist and that it cannot be measured, we suggest. For high energy levels, one should be able to measure such motion. The factor exp(-iEt+ipx) is a further approximation of the bound state which may be used if one wishes to describe a particle of mo in a potential V(x) created by a particle of mass M as an M+m+E rest mass system. Thus, one may see that various levels of approximation are used in the same quantum bound state solution. Furthermore, classical and Newtonian mechanics are still linked at any energy level because the quantum average KE (in the nonrelativistic case) must equal the classical kinetic energy. Thus, the notion of x= vrms(x) t is present and if hbar/(m vrms) << dx (where dx <<L with L being the system size), then one again has exp(-ipx+i KE t) and x=vrms(x) within each tiny dx, as an approximation.
Traditionally, Newtonian mechanics is recognized as an approximate theory which is said to arise from quantum mechanics (apparently a more detailed theory) in the large energy level limit (correspondence principle). We argue that both quantum mechanics and Newtonian mechanics are both approximate and both complement each other at any energy level. In the Newtonian case, quantum effects are so small that they are not visible, but in the quantum case, Newtonian effects are visible, we argue. We first note that exp(-iEt+ipx), the free particle wavefunction (complex probability) does not exist without the Newtonian expression x=vt. x=vt implies it is possible for dx->0, dt->0, while hbar/p and hbar/E show that there are limitations with respect to interactions. The only way to ascertain x=vt is through measurement and if these measurements have dt=hbar/E and dx=hbar/p uncertainties, one cannot have dx->0, dt->0 resolution. Nevertheless, the quantum free particle does move as x=vt (as long as one respects the uncertainties of dt,dx).The quantum exp(-iEt+ipx) does not exist without the Newtonian x=vt. In a single particle quantum bound state (time-independent), we suggest there are significant approximations. First, a conservation of energy equation exists in Newtonian physics, namely pp/2m + V(x) = E ((1)). This equation does not contain time and does not explicitly show x(t), but that does not mean that such motion does not physically exist. The quantum bound state solution follows from ((1)) which is an equation which does not display full information, even in the classical case, and so this loss of information carries down to the quantum solution. In particular, in the quantum case one solves a time-independent problem and so uses a(p)exp(ipx) + a(-p)exp(-ipx) with a(p)=a(-p) or a(p)= -a(-p). These conditions are equivalent to insisting that the local momentum average at each x (-i W* grad W + i W grad W) is 0. The fact that one is solving a quantum bound state using only energy considerations means that one averages out time which means averaging momentum. Thus, W*(x)W(x) describes a resolution spatial pattern for interaction, but W = W(x) exp(-iEt+ip x) where p=0 at each x. In other words, just because one does not see forward/backward motion in a particular quantum solution (based on energy alone) does not mean that it does not exist and that it cannot be measured, we suggest. For high energy levels, one should be able to measure such motion. The factor exp(-iEt+ipx) is a further approximation of the bound state which may be used if one wishes to describe a particle of mo in a potential V(x) created by a particle of mass M as an M+m+E rest mass system. Thus, one may see that various levels of approximation are used in the same quantum bound state solution. Furthermore, classical and Newtonian mechanics are still linked at any energy level because the quantum average KE (in the nonrelativistic case) must equal the classical kinetic energy. Thus, the notion of x= vrms(x) t is present and if hbar/(m vrms) << dx (where dx <<L with L being the system size), then one again has exp(-ipx+i KE t) and x=vrms(x) within each tiny dx, as an approximation.
Netwonian mechanics and quantum mechanics seem to be very different theories, even though Newtonian concepts like momentum, energy and their conservatioon carry into quantum mechanics. The difference is not even due so much to the case that Newtonian mechanics is deterministic and quantum mechanics, statistical, we argue. We suggest that quantum mechanics utilizes a turly dynamic probability which treats forwards and backwards motion differently thorugh exp(ipx) and exp(-ipx), while Newtonian mechanics treats them in essentially the same manner, with the exception of the direction. We thus suggest that Newtonian mechanics is really based on the notion of stationary objects. In particular, x(t) means that a particle is at x in dx at t, as if it were stationary with the exception that it is translating with a fixed velocity. This stationary viewpoint might describe why the P(x)=1/L probability for stationary objects is sometimes also applied to objects moving with constant speed. The x(t) approach really means at t, the particle is frozen at x for dt before appearing frozen at x+dx at t+dt with a new velocity. There is of course the notion of acceleration, but for a particle to be at x at t, one must measure it and this means that it is as if it is stationary for an instant. The velocity is then considered to be constant at x (i.e. a translation) with a new translation value at x+dx. As a result, the notion of quantum mechanics does not really follow directly from this Newtonian stationary viewpoint. One has to think in terms of a very different idea, namely that exp(ipx and exp(-ipx) are not merely translations in different directions. A truly dynamic theory is needed. As we have argued in previous notes, P(x)=1/L represents a stationary object and so if one had an AND condition for p and -p (i.e not speed), then exp(ipx)/sqrt(L) exp(-ipx)/sqrt(L) = 1/L. Freezing out speed using quantum dynamic probability exp(ipx) (exp(-ipx)) leads to a stationary picture, but the actual physics of exp(ipx) is very different from anything in Newtonian mechanics. This leads to the question: What happens in the high energy (i.e. classical limit) if Newtonian mechanics and quantum mechanics are so fundamentally different? Quantum mechanics differs from classical physics in that there are fixed uncertainties of dt = hbar/E and dx=hbar/p for a particle with E,p, while Newtonian mechanics uses infinite resolution. One may note that exp(ipx) and exp(-ipx) are very different, but that is only if one can resolve x regions on the order of hbar/p. If one has worse resolution, one would not notice that exp(ipx) and exp(-ipx) differ. They are both possibly present in a dx which is the resolution limit. Thus, it is actually not the infinite resolution of Newtonian mechanics that is manifest, but rather the approximate finite resolution which means that on classical scales, a tiny dx is actually larger than hbar/prms where prms is the root means square momentum of a quantum bound solution. The quantum bound state solution is based on finding energy levels and a wavefunction W(x) = Sum over p a(p)exp(ipx). It includes exp(ipx) and exp(-ipx), but physically in the classical limit, the particle moves to the right for some time and then later to the left. Quantum mechanics cannot change this measured situation. As noted, exp(ipx)*exp(-ipx) =1 and this is linked with a stationary object even though exp(ipx) and exp(-ipx) both represent dynamic probabilities. The dynamic probabilities allow one to correctly solve problems at scales of hbar/p. W*(x)W(x) is an attempt to create a stationary distribution which is consistent with the kind of measurements made in the classical world. It is W(x), however, that is actually used in solving problems. As the energy level increases, the peaks of W*(x)W(x) become increasingly narrow and so even this stationary-quantum approach is replaced by the simpler, approximate exp(i prms x) * exp(-i prms x), leading to a stationary free particle distribution. This is why we suggest that Newtonian mechanics is really a stationary free particle solution with the particle then translating from one x (within dx) to the next in dt. In other words, x(t) means that the particle is at rest in t (for an instant as it must be measured). The v(x) = dx(t)/dt is called the constant speed in dx (as it must be measured) even though there is acceleration occurring. It seems that Newtonian mechanics is a static kind of viewpoint which is very different from the dynamic exp(ipx) and exp(-ipx).
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.
In Part 1, we argued that Newtonian mechanics essentially contains the notion of probability found in the free particle wavefunction exp(-iEt+ipx). The difference is that the wavefunction/probability quantifies dx=hbar/p and dt=hbar/E. The standard Newtonian approach is to have dx->0 dt->0 in order not to deal with the lengths of these intervals. This limiting feature may work in the classical world (but not for photons reflecting and refracting through windows), as the actual sizes of dx and dt may in general be very small in the classical world, but they are not small in the case of atomic or nuclear lengths. The question is: In Newtonian mechanics, without any reference to exp(-iEt+ipx), how would one obtain finite values for dx and dt? Furthermore, why should one wish to obtain dx,dt values associated with x,t in x=vt. We suggest that because one deals with a moving object, and not a stationary one, any x,t on a x=vt trajectory should be associated with a dx,dt to verify its existence. Unlike the stationary case, for which one may strike the particle and make it move, dx,dt for the moving particle should be dynamical and associated with Lorentz invariance. We suggest that it is meaningless to speak of x,t for a moving particle without a corresponding dx,dt. If one can obtain dx,dt values, then one can argue, as we did in Part 1, that Newtonian mechanics contains the notion of an uncertainty dx about x and dt about t for a free particle. This is essentially the uncertainty which allows the 2-slit interference to occur and solves the probabilistic reflection-refraction problem at an n1-n2 index of refraction interface. Of course, if probability must exist within dx and dt, one requires a math expression for such a probability (i.e. exp(-iEt+ipx). One cannot simply use Newtonian mechanics because it does not make use of such a probability, but at least it predicts that such a probability should exist, we argue. Here we show that one may directly use special relativity to show that dx=hbar/p and dt=hbar/E in Newtonian mechanics, such that one has a dx and dt associated with the same value of the classical action as x,t for which x=vt. The classical action describes, for various x,t values describes the x(t)) trajectory in spacetime and so having built-in dx,dt values in this same action ( which is the Lorentz invariant -Et+px both nonrelativistically and relativistically) shows that Newtonian mechanics must be associated with finite dx,dt. The reason that we consider a dx and dt in the first place is that v=dx/dt. The Newtonian approach is to take the limit dx->0 and dt->0. This, we argue, is associated with a coarse-grained equation x=vt which holds both relativistically and nonrelativistically. This is associated with a coarse-grained classical action.Each x,t on the trajectory x=vt yields a different action value for the action (x/v*.5mvv or -x/v mo sqrt(1-vv/cc) or -Et+px where x/t=v. One may consider x=vt and x+dx1 and t+dt1 as the next points in the trajectory. According to Newtonian mechanics, one may make these values arbitrarily small. Furthermore, x=vt holds both nonrelativsitically and nonrelativisitically. How then can one propose a finite valued dx and dt? We argue that the manner in which this may be achieved is to have a dx,dt associated with x,t which does not change the action value for x,t, i.e. does not change -Et+px. This set dx,dt is finite valued even though the next x,t considered may have a dx1, dt1 value such that dx1<<dx and dt1<<dt. If, however, dx and dt are to be taken seriously, the dx1 and dt1 less than dx,dt are simply math values, not physical ones. In other words, one needs a dx,dt to justify the existence of x,t and if x=vt allows any x+dx1 (dx1<<hbar/p), then this is a coarse-grained equation. The coarse-grained action, however, allows for the defining dx=hbar/p and dt=hbar/E even if the action itself is coarse-grained.
We try to consider free particle and single bound state quantum mechanics from the viewpoint of classical physics. We first assume that a free particle moves as x=vt. Secondly, we assume that dx and dt cannot be 0 as in Newtonian mechanics, but must be finite. This begs the question: How does one measure dx and dt? If one can measure a particle at a given x and t, then dx→0, dt→0. Thus, finite dx, dt means that a given p=constant and pp/2m=kinetic energy (nonrelativistic) interact with a force over a finite dx. In other words, one does not know where the interaction occurs within dx. As long as the free particle moves without interacting, then x=vt and there is no interest in any dx or interactions. For example, one may measure very large x and t ranges. Even if there are dx and dt effects at the start and end points, these are small compared to the large x and t values. As a result, one has a F(x,p) probability within dx, which repeats itself for the next dx (i.e. is periodic), where F is a probability for a p impulse hit at x within dx. At the same time, the particle moves with x=vt it seems within dx. In general, one might then think that F(x,p) and x=vt are decoupled. In previous notes, we have argued that one way to couple F(x,p) and x=vt is through special relativity, i.e. A = -Et+px. X and t values used would presumably satisfy x=vt, but for any x,t one has x+hbar/p and t+hbar/E as solutions yielding the same A value and so might consider dx= hbar/p and dt=hbar/E. The question we ask here is whether one may arrive at this conclusion without using special relativity. The answer seems to be that one can if one suggests that F(x,p) must contain information making it consistent with x=vt, i.e. with every x and t point having the same weight. Clearly, however, each x,t do not have the same weight within dx and dt, but are rather governed by F(x,p). This means, however, that F(x,-p) also represents a probability distribution and so if one has F(x,p)F(x,-p) there is no momentum at a given x point. In such a case, the distribution within dx must disappear and one must have a uniform distribution. This holds even if F(x,p) has the same dx value for any p. Thus, F(x,p)F(x,-p) = constant and must be periodic in x. This implies that F(x,p) = 1/F(x,-p) for constant=1. Given that F(x,p) =0 at the initial and final points of dx, then F(x,-p) = infinite at these points and that is not allowable. Thus, one is forced to use complex variables and try exp(i g(p) x). If g(p) is a function of p (i.e. not a constant), then conservation of momentum implies that exp(i g(p1) x) exp(i g(p2) x) = exp(i g(p1+p2) x), suggesting exp(ipx). What, however, if g(p) = constant? In such a case, one has issues creating an average kinetic energy: ( Sum over p a(p) exp(i constant x) pp/2m) / (Sum over p a(p) exp(i constant x) ((1)) This leads to a kinetic energy which does not depend on x and cannot be used with V(x) in a conservation of energy equation. Thus, we choose exp(ipx). Note: there is no reason to choose a(p,x) because the physics of finite dx and dt is restricted to exp(i constant x). Finally, we try to explain why the same dx=hbar/p and dt=hbar/E together with x=vt follow from special relativity.
We have argued in previous notes that both Newtonian mechanics and free particle quantum mechanics arise from special relativity. Thus, special relativity gives an expression for free particle energy and momentum and these are related to concepts of work. The Lorentz invariant -Et+px shows that x,t and x+hbar/p, t+hbar/E give the same solution to this potential-like term showing that there is stochasticity in the position and time delivery of interaction hits. In other words, even though the center-of-mass moves as x=vt, a momentum hit (proportional to rest mass mo) does not necessarily act at the center-of-mass. This gives the impression of shifts in probability, kinetic energy etc in the case of interference, i.e. interactions. For a single free particle, there is no interaction and -1/2m d/dx d/dx exp(ipx) = pp/2m exp(ipx). We argue that exp(ipx) is a probability and that in general one may add independent probabilities in an OR situation as in a 2-slit interference case. A bound state, however, is different because even though one adds weighted exp(ipx)s, they are part of a resonance or equilibrium and this is no longer an independent OR situation. We suggest that one uses a conditional probability when computing a Newtonian nonrelativistic conservation of energy equation to try to remove as much of the quantum features from this equation. As a result, one has: KEclassical(x) + V(x) = E = Sum over p pp/2m P(p/x) + V(x). One cannot bypass the fact that a quantum particle interacts differently from a classical one and that this involves a probabilistic granularity (wavelength), but classical notions of energy and energy conservation still exist. As a result, the bound state wavefunction Wn(x) = Sum over p an(p) exp(ipx) is not a usual OR sum, but is rather 1 over a normalization factor for an equilibrium picture. As a result, one cannot use Wn1(x)Wn2(x) = W(n1+n2) (x) as one would for independent probabilities. Furthermore, given that one has Wn(x) and Wn1(x)+Wn2(x) which are both of the form Sum over p a(p)exp(ipx), and both represent average energies, one needs a way to distinguish a resonance/equilibrium from a sum of resonance/equilibrium normalization factors. We suggest that the orthogonality of resonance Wn(x)’s provides this feature. The standard textbook explanation is based on Hermitian operators, but in the nonrelativistic case, we argue that one may obtain the result based on integration by parts based on the form of the p operator = -grad. Thus, it is ultimately the form exp(ipx) which leads to orthogonal bound state Wn(x)’s, together with the boundary condition that Wn(x)-->0 of x-> +infinite and -infinite.
Quantum mechanics is often compared with classical mechanics because both may deal with a single particle subject to a potential. Classical mechanics is said to be deterministic describing x(t) and its derivatives i.e showing how a particle moves and accelerates while bound state quantum mechanics produces a real wavefunction W(x) such that W(x)W(x) yields a spatial density profile. The comparison is carried further with the correspondence principle which states that at high energy levels quantum mechanics should mimic classical mechanics. Classical mechanics does not simply describe x(t) and its derivatives i.e. deterministic motion. It also includes two body scattering, force and torque balances for a stationary object etc. In the case of the classical statistical mechanics of an equilibrium ideal gas, the Maxwell-Boltzmann (MB) distribution in the absence of a potential is C1 exp(- pp/(2mT) ). If a potential V(x) is added, one may find P(x) two different ways using Newtonian ideas. The first is to consider the gas as a collective entity like a tiny macroscopic classical object at x. Then one may argue that equilibrium means the system is stationary and apply Newtonian statics i.e. a balance of pressures and -dV/dx to obtain exp(-V(x)/T) as is often done in textbooks. Alternatively, one may consider a single particle and apply Newtonian mechanics differently, namely use a conservation of energy equation: p1p1/2m + V(x1) = p2p2/2m + V(x2). Again one obtains exp(-V(x)/T). The spatial part of probability may be based on Newtonian mechanics indicating that it is not stochastic. The stochastic function is C1 exp(-pp/2mT). Interestingly, however, the two Newtonian approaches are different with respect to a single particle. A collective nonmoving set of particles acts as a stationary object, yet within it each particle seems to be “bouncing” off of an imaginary wall creating pressure. Thus one has an impulse free particle picture yet obtains the correct result exp(-V(x)/T). Alternatively one may think of a single particle accelerating which it does if it is not colliding with other particles. This is a very different picture of the single particle than of one bouncing back and forth hitting an imaginary wall. In fact, there is usually a wall holding the gas together and pressure and bouncing particles are the real picture there, but within the gas one would normally think of a single particle accelerating. Nevertheless there are two scenarios which yield the same result. The idea of pressure balance is more in keeping with an overall statistical picture of the gas. In other words, one may actually ignore the idea of following a single particle as it accelerates. (That is not to say that single particle acceleration does not exist - it does.) Quantum mechanics we argue is a statistical theory. In the case of a single bound particle one has a momentum distribution W(x)=wavefunction = Sum over p a(p)exp(ipx). One may apply conservation of average energy at each x to obtain the time-independent Schrodinger equation. In such a picture, like with the statistical mechanical gas, we argue that one may imagine impulse hits with the potential instead of considering acceleration. The quantity vrms(x) from KEclassical= .5m vrms(x)vrms(x) = -1/2m d/dx dW/dx / W does display acceleration, but this is a mathematical average. As energy levels become higher the correspondence principle is said to apply, but a classical particle exhibits acceleration. The quantum bound state does not, but like in the case of the statistical mechanical gas it may be that two scenarios lead to the same result. In other words, the quantum bound W(x) still has crests and troughs, but with low resolution one may only see an envelope roughly touching W(x)W(x) which matches a classical C2/v(x) where v(x) is classical velocity. Given that quantum mechanics is a statistical theory it may not be necessary to have it explicitly display acceleration.