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 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.
Newtonian mechanics describes forward and backwards motion by x=vt and x= -vt. In order to understand this motion, one needs to create dx and dt units such that a particle with v moves from x,t to x+dx, t+dt. X and t are necessarily coupled and dx and dt are not zero (even in the dx->0, dt->0 limit). They are just very small. For a particle with v=0 (at rest), x and t are decoupled. Given that the particle may be at rest at any x in L, it is natural to define a probability P(x)dx = dx/L, even in Newtonian mechanics. For x=vt and x=-vt, however, x and t must be decoupled, but given different initial points, one can have different sets of x,t measurements. The key point we stress is that motion is linked with dx and dt and these do not have a fixed numerical value in Newtonian mechanics, but nevertheless the concept exists. In principle, Newtonian mechanics is probabilistic because one does not really know what occurs within dx and dt. It is this essential probability, we argue, that is manifested in quantum mechanics by defining actual dx and dt values, although in essence it is already present in Newtonian mechanics. Given that probability P(x) is well-defined and applies to v=0, one may postulate a dynamic probaiblity P1(v) as we have done in previous notes. If one wishes probability to be Lorentz invariant, one may write P1(-Et+px). At first, based on Newton’s x=vt and x=-vt, one would expect that x and t must be coupled through x=vt. The problem with this is that one may form an AND (product) scenario in probability theory. Furthermore, P1(-Et+px) does not contain initial point information.If one writes P1(v)*P1(-v), one should obtain P(x)=1/L because there is no motion. P(x) decouples x,t and so P1 must do the same. In other words, P1(-Et+px) contains decoupled x,t. This differs from Newton’s coupled x(t). If P1 does not increase or decrease indefinitely in x,t, then a solution is P1(-Et+px) = exp(-iEt+ipx). The question we ask is: How does exp(-iEt+ipx) contain more information than x=vt, x=-vt from the Newtonian scenario? In the Newtonian picture, in order to have a v and -v particle at the same x,t, they must be at different x,t values before and after. This suggests a disconnect in dx (and dt) for x and t. This disconnect is present in exp(ipx) and exp(-ipx), at least in the sin(px) term. The only new information contained in exp(-ipx-iEt) is that dx=hbar/p and dt=hbar/E. The uncertainty in x and t (due to dx and dt) is already present in Newtonian mechanics, but one ignores it by considering dx->0 and dt->0. The notion that dx and dt may be finite does not arise experimentally in classical calculations, but becomes apparent on atomic and nuclear scales. As a result, we argue that the probabilistic nature of a moving particle exp(-iEt+ipx) is already present in Newtonian mechanics, but one does not know how to fix dx and dt. As a result, it is not really the case that Newtonian mechanics is so greatly different from quantum free particle mechanics. The dx and dt lengths may be associated with uncertainty even in Newtonian mechanics and if this is considered, then Newtonian mechanics would not be dissociated with 2-slit interference and 1-D reflection-refraction at an n1-n2 index of refraction interface.
Newton’s law F= dp/dt is considered to be associated with deterministic kinematics. The problem, we argue, is that dt is not really zero (only very small on Newtonian scales). If, for the sake of argument, one considers a finite dt, one must then consider the workings of force in a very tiny time and there is no reason to believe this is a deterministic situation. It is more likely that there is some stochasticity at the level of dt and dx and so t and x within dt and dx are stochastic and associated with probability. On Newtonian scales, this would not be noticed, but on other scales (quantum ones), it could be. A key point seems to be that even if one has x=vt ( a deterministic equation), one does not really know the x and t values exactly, there is a stochastic uncertainty of dt, dx (because the particle was put in motion by an interaction). This means that even a free particle should have this inherent stochasticity in x and t, even if p and E are considered exact. Now expressions for p and E formally follow from special relativity and so if there is some dx,dt stochasticity and it has a certain range (i.e. a maximum value for dx and dt in a sense), this range should appear in special relativity as one must know dx,dt and dx’, dt’. We have suggested in previous notes that possible dx,dt ranges in fact appear in the Lorentz invariant A = -Et+px because for x,t (center of mass description) yielding a certain. Values, off center-of-mass values, x1=x+hbar/p and t1=t+hbar/E also yield the same A. We suggest that these are the “maximum” dx and dt values with uncertainty because the momentum or energy interaction can occur with probability within dx and dt and need not occur at the center-of-mass. It is dx and dt which differ in different frames, e.g. hbar/p’ and hbar/E’, but there is uncertainty in these ranges, i.e. a probability distribution. This uncertainty is associated with a free particle and should manifest itself in an interaction (e.g. 2-slit interference), but it also means that two body elastic scattering cannot ever be considered fully deterministic. There is still the stochastic, uncertain dx, dt. If one wishes to consider conservation of energy and momentum, however, one uses the notion of a free particle x=vt. Even though actual interaction hits may occur at different places, if one has two particles P(p1,x1) P(p2,x2) in an AND situation, one must have x1=x2 in order to show that momentum is conserved at the same x. This leads to the notion of exp(ipx) as a probability, or more fully exp(-iEt+ipx) which is consistent with the special relativistic condition on the range of dx and dt. If one were to consider special relativity, then two particles together have E1+E2 and p1+p2 and one uses the same x and t and so dx= hbar/(p1+p2) and dt= hbar/(E1+E2). This matches the P(p1,x)P(p2,x) scheme for ranges of dx and dt. Alternatively, one may consider that one has a uniform distribution for any outcomes of E1,E2, p1,p2 elastic scattering which conserve energy and momentum. This leads to exp(iC1E) and exp(iC2p), but these must display the dt, dx ranges and so one has exp(-iEt+ipx) in order to have a Lorentz invariant probability. The above arguments have been based on Newton’s second law dp/dt=F and special relativity which means that the two must exist together, which in fact they do. In particular, special relativity is not simply about viewing particles from different frames, but must also be linked with interactions which create the equivalent motion. As shown in (1), one may start with number and vector concepts E,p without knowing their explicit forms and then use special relativity to calculate them. In (1), however, this involved the notion of dp = Fdt and dke= Fdx even for the form of p unknown. Thus, dp/dt =F and special relativity are directly intertwined and the notion of a finite stochastic dt and hence dx in dp/dt=F should have effects on special relativity. One, however, has to directly impose this stochasticity in dt (and hence dx) on dp/dt=F by arguing that dt has a finite value with probabilistic t value within and so one must also look for the stochastic dx and dt range in special relativity, i.e. by considering A= -Et+px as described above. The stochasticity is an assumption, but the finite dt and t+hbar/E, x+hbar/p are not.
Quantum mechanics is concerned with measurement and we suggest that in a bound problem impulses due to momentum are the form of measurement which occurs in x space. The Lorentz invariant A= -Et+px, which is also the free particle relativistic or nonrelativistic action with v=x/t, associates energy = pp/2m with time and p with x. Creating eigenvalue equations yields id/dt exp(-iEt) = E exp(-iEt) and -id/dx exp(ipx) = p exp(ipx). P (momentum) is associated with physical impulses and with its own two dimensional probability distribution exp(ipx). E is associated with t and has its own distribution in time exp(-iEt) which implies that the same E holds for each x point without any notion of x probability associated with it. In classical physics, a momentum p exists in a tiny dx (dx→0). For exp(ipx), a p impulse may occur within hbar/p with different probabilities, so one cannot restrict the measurement of p to a tiny dx unless dx is of the order of hbar/p. Nevertheless energy exists at all x points. Given that p measurements occur through specific impulses at different x points with exp(ipx) probabilities, one must calculate an average energy using {Sum over p pp/2m a(p) exp(ipx)} / {Sum over p a(p)exp(ipx)} using the probability associated with the measurement of p in space i.e. exp(ipx). This adds to V(x) to yield a constant E at each x point. We suggest that weighted exp(ipx)s are used to calculate average energy at each x, because it is through impulse measurements that one determines p at different x values and hence the average energy. One may note that one does not calculate {Sum over p a(p)exp(ipx) p } / {Sum over p a(p)exp(ipx)}. An impulse hit is associated with a certain probability and this measurement in x yields information about the corresponding energy pp/2m. Each impulse hit also yields information about a time probability exp(-i ppt/2m), but only if the particle is knocked out of the bound system i.e. only if pp/2m is the only energy present, which is not the case for a single particle bound system.
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.
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.
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.