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 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 23.14/p, one obtains i 23.14/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 physics (including classical physics), conservation of momentum is always considered, but one must be more careful with conservation of energy as it may include forms like sound etc. As a result, there is a tendency to focus on conservation of momentum all of the time, and conservation of particle energy only in elastic collisions. In addition to this, one may think of a particle striking a target, in which case two equal momentum (but different energies due to different v speed and mo rest mass) cause the same impact. In the case of an interaction with a potential, however, it is strictly energy considerations which are important classically, i.e one uses p dot p /2m + V(x) = E. Knowing momentum without rest mass is not sufficient. One may consider it unusual that one may consider two different scenarios, one which focuses only on momentum (impulse) or conservation of momentum in collisions while only energy is the focus in classical bound states or motion in V(x). Given the classical focus on momentum, imagine that one wishes to describe all free particles with different momenta as a uniform probability distribution such that a product of two probabilities for p1,p2 is the same as for p3,p4 if p1+p2=p3+p4. (We use 1-dimension for simplicity.). One cannot use P= 1/n because p is absent and n->infinite. In one sense, all particles must have the same probability, but at the same time must contain extra information related to p for the conservation of p part of the problem. The notion of a uniform distribution is independent of conservation of momentum, hence of p momentum. Conservation of momentum requires an explicit presence of p. As a result, there seems to be a paradox. One may, however, resolve this paradox by arguing that a probability is a mathematical way of presenting a subset of the information present. If one argues for a uniform probability, then it does not need to contain p, and one would have 1 as an unnormalized probability for any free particle p. A different probability (representing a different subset of information) which respects all free particles as having the same weight, but includes p can also be constructed, namely exp(i C p). C is a constant introduced for unit issues. In this case, the modulus gives all free particle probabilities with p the same weight, but the presence of p allows one to have conservation of momentum. Given that momentum is the key variable that is conserved in classical physics, one should be satisfied with exp(i C p). One might argue that there is no need to consider energy E. We note, however, that if one has p, E1 and -p, E2 in a frame, then in a frame moving with -v, one has gp+gvE1= p1’ and p2’ = -gp + gvE2. Adding yields gv(E1+E2) and so energy automatically appears and cannot be discarded. Furthermore for p, E1 and -p, E2, exp(i C 0) is exp(i C gv(E1+E2)) as seen in the moving frame and so one is forced to consider conservation of energy together with momentum and include both p and E in the more detailed probability. In other words, special relativity automatically links p and E in a 4-vector, leading to the idea that both are important. We argue that one cannot simply consider exp(i C p) due to special relativity. Both E and p must be considered together and this is a key idea, we argue, because if one wishes to have a Lorentz invariant probability, it must be in E and p, i.e. -Et+px. Instead of having a probability for an interaction with p given by exp(i C p) , one has exp(-iEt+ipx) and we argue that this must have the same value in all frames for a particular interaction. One may note that E and p are, on the one hand, separated allowing for impulse hit versus E related interactions, but that overall the probability for interaction includes both and it is the full -Et+ipx which must be invariant. This further suggests that if one does not artificially force an interaction to occur at the center-of-mass, then x,t and x+hbar/p, t+hbar/E are equally good points for the interaction. We argue that there is no need to force x=vt because exp(-iEt+ipx) * exp(iEt+ipx) already shows that one has uniformity, not only for any p,E, but also for any x, t, as there are now four variables. Thus, there is no reason to enforce x=vt in exp(-iEt+ipx), when considering an interaction, we argue. This means that one has two possible interactions at different x,t values, i.e. x,t and x+hbar/p, t+hbar/E. If one is unable to follow the interaction in time, but may follow it in space as in the case of 2-slits about hbar/p apart or for a potential V(x), then one may use the exp(ipx) part of the interaction probability piece by itself. Thus, we argue that special relativity does not allow one to treat p independently of E. For example, for p values that are the same in one frame, but have E1, E2, will be different in another frame. One must consider E and p together and so instead of creating a probability for p alone, one must create one for E,p and then impose Lorentz invariance which leads to different x,t pairs yielding the same probability value for an interaction, i.e. free particle quantum mechanics.
Newtonian mechanics presents a simultaneous description of dynamics x(t) and interactions (F=dp/dt) because it assumes infinite resolution, i.e. dx->0, dt->0. We have argued in a previous note (1), that quantum mechanics follows from free particle quantum mechanics, which in turn follows from special relativity. Special relativity uses the notion of x=vt and provides formulas for E=mocc/sqrt(1-vv/cc) and p=mov/sqrt(1-vv/cc) even with no free particle quantum mechanical notions. We have also argued in previous notes, however, that these classical concepts are present in the Lorentz invariant -Et+px which we have argued (2) acts as a kind of free particle potential. This potential, however, has the same value for x, t and x+hbar/p, t+hbar/E and so we have suggested that free particle quantum mechanics is comprised of two parts. The first is a complex probability exp(-iEt+ipx) which describes possible interactions with a free particle and the second is x=vt. In other words, interactions and dynamics are split, but the interaction probability depends on classical notions such as x=vt, p and E. In the limit of hbar/p and hbar/E <<1, the interaction resolution becomes the Newtonian dx->0, dt.->0. Thus, as in (1), we argue that quantum and classical mechanics are intertwined. In (3), entitled Unifying Quantum and Classical Dynamics, a similar argument for the intertwining of quantum mechanics and classical mechanics is presented, but in a different manner. The basic argument of (3) seems to be that quantum mechanics, written in the Heisenberg approach, yields operator equations which are identical in form to classical equations, in particular, d/dt X(t) = P(t) and d/dt d/dt X(t) = -d/dx V( X(t)). Here X(t) and P(t) are operators given by X(t) = exp(iH t) X(0) exp(-iHt) (and a similar expression for P(t). We try to show that given that free particle quantum mechanics is directly linked to special relativity and hence to the acceptance of x=vt, that in the free particle case, any operator description of d/dt X(t) = P(t) cannot yield any result other than a classical one because no interaction is occurring. In other words, quantum mechanics is directly linked to classical mechanics at the free particle level. Later, one takes a superposition (OR probability addition) of exp(ipx)s, as in the case of bound states. In such a case, one deals with average interactions and so one must be careful when interpreting d/dt d/dt X(t) = -d/dx V( X(t)) as the system description is now an average one.
We have suggested in previous notes that there is a tendency to consider quantum mechanics as a kind of hybrid probabilistic-deterministic theory because it is supposed to ultimately “merge’ with classical mechanics for high energy levels (tiny wavelengths). Newtonian mechanics is deterministic and so somehow it seems quantum mechanics should also be seen as deterministic, hence the sense of surprise at tunneling when KEave(x)=0, even though vrms(x)=0 does not mean that actual values creating the probability are 0. Here, we wish to argue that exp(-iEt+ipx) (or the exp(ipx) part for spatial interactions) is really a momentum probability scheme which may possibly bypass many aspects of the actual physics to describe a problem in a simpler form in terms of a momentum conserving probability. A key idea is that exp(ipx) is actually the form to use when one has randomness as we will argue for reflection-refraction from an n1-n2 index of refraction junction. We argue that this idea is already known for a particle in an infinite well for which one has sin(pave x), i.e. 1/2i (exp(i pave x) - exp(- i pave x)). In other words, one has the form exp(i pave x), even though it is linked with an average p. Given that one is not considering deterministic physics, there is nothing wrong with exp(i pave x) being composed of Sum over p a(p) exp(ipx), where exp(ipx) holds in all space, not just within the infinite well potential. exp(ipx) is a momentum conserving probability, but on average momentum must also be conserved by a free statistical entity within the well. Similarly, W(x) for a single particle bound state consists of many exp(ipx)s, but represents a fixed E ave and on average this Eave must be conserved in time, hence the time dependent exp(-i Eave t). As a result, one may see that the quantum mechanical exp(-iEt+ipx) is an attempt to describe a system in a simple manner in terms of momentum conserving probability. To see this in more detail, we consider (1) which describes reflection and refraction of a photon at an n1-n2 junction in depth. In a previous note, we have already pointed out that in n2, c/n2 simply means that the photon is interacting on average with atoms/molecules in n2, but has c in between such interactions. It is only on average that one has c/n2 and even E and p2 = p n2. Nevertheless, one may use exp(ip2 x) as a probability which describes average properties of the system, just as it does in the examples in the above paragraph. The question then becomes: Do the microscopic details of the photon physics (the actual interactions) dictate the probabilities to reflect/refract or does the simplified exp(ipx) set linked to momentum conservation do so? Based on (1), it seems that reflection is completely independent physically from “refraction”, the latter requiring re-radiation and affecting all glass molecules. As a result, it appears that it is difficult to suggest that it is the microscopic interactions which dictate probability because reflection and refraction are microscopically so different. On the other hand, reflection-refraction is not a .5-.5 probability random process as experimental results show. The question then becomes: If some kind of systematic probability driver is not supplied by the microscopic interaction, what provides it? We argued in previous notes that one may introduce the probability exp(ipx) into Newtonian scattering to account for an initial p1,p2 (momentum vector set) having equal probability to produce any pi, pj (outcome set) as long as momentum is conserved. We suggest here that it is exp(ipx) that one must use for random interaction if one insists on conserving momentum. Thus, exp(ipx) bypasses microscopic details of interaction and provides an “ any outcome pair of momenta” is equally likely if momentum is conserved. This is a major simplification of the problem, but the question is: What constraints does it impose on physical results? For a free particle, it suggests that it may interact probabilistically with both slits of 2-slit apparatus if the slits are about hbar/p apart. For randomness, one would expect equal probability for each slit and this is the case as long as one also considers exp(ipx) for each slit to a the same point on a screen. In the case of reflection-refraction at an n1-n2 index of refraction junction, we try to see how the simplification exp(ipx) can be consistent with the microscopic details of the interactions and at the same time dictate the probabilities to reflect and refract. We try to consider why an exp(ipx) scheme predicts how nature behaves while a random one with no momentum conservation considerations does not. In particular, we argue that exp(ipx) represent randomness in a two body scattering problem in which momentum is conserved and so this same exp(ipx) must be carried over into problems which involve interactions instead of simply suggesting that one has .5 probability to reflect and .5 to refract because the two processes are independent and hence random. They are random, but with conserved momentum and hence an exp(ipx) scheme is required, we argue.
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 argued that one may introduce a probability into Newtonian elastic scattering if one assumes that for an incident (e1,e2) energies and (p1,p2) momenta vectors, any (ei,ej), (pi,pj) which satisfy conservation of momentum and energy have equal probability. We then found that for a single particle, the relevant probability is exp(-iEt+ip dot r). This is a unit modulus complex number which is Lorentz invariant, and exp(i p dot r) is time reversal invariant, i.e. p→-p and r→-r yield the same value. The question then becomes: How does one use exp(-iEt+ipx) or for time-independent problems, exp(ipx)? This is not a usual probability linked to quantity, e.g. N P(reflect) = the number of particles which reflect. exp(ipx) is a probability to have a certain momentum, but there is no reason it cannot carry a weight, i.e. a math factor. The question is: What does such a factor mean physically? Secondly, exp(ipx) depends on x (space), but exp(ipx) is not the probability to be at x, but it is somewhat related in that for any x and a given p, the modulus has the same value, suggesting equal probability to be at any x. This mapping, however, really only holds for a single exp(ipx). It seems that one can say two things about exp(ipx). First, if it is a probability for a momentum p, then one should be able to write p exp(ipx) or function(p)exp(ipx). Secondly, one should be able to write AND (multiply) and OR (add) statements. The idea of an OR or addition of exp(ipx)s is important, because one may be able to write an equation for constant kinetic energy: -1/2m grad dot grad W(x) = numberW. The general solution should be a W(x) which contains a sum of all exp(ipx) type solutions which have this kinetic energy. (In the relativistic case, one may consider d/dt d/dt W - cc grad dot grad W = -momocccc W.) The question we consider in this note is: Given that one may write W and grad W, what can only really do with these? They do not directly represent physical observables. Furthermore, if one wishes to impose continuity on W and gradW, which is reasonable mathematically, one should have a physical argument to justify it, and we suggest that this is a little complicated. Ultimately, we try to argue that one may consider continuity of W as well grad W in certain cases based on the association with a classical quantity probability equation. We also suggest one may relate W* grad W which may be generalized to W* grad W + (grad W*) W, as discussed in Part 3 and provides a link to classical probabilities.
In a number of previous notes, we argued that free particle quantum mechanics, i.e. exp(-iEt+ipx), follows from the Lorentz invariant A = -Et+px. For t=0,x=0, A=0, but we argued that there is another solution, dx = hbar/p and dt=hbar/E (we use intervals because x=0, t=0 is a trajectory point). We then noted that this solution is linked with the probability exp(-iEt+ipx) which is Lorentz invariant and creates the above intervals in t and x. Finally, we remarked on the fact that exp(-iEt) conserves energy and exp(ipx), momentum, when one uses these for 2-body collisions, i.e. AND situations. Here we point out that there are numerous dt= f(p,E) and dx=g(p,E) (f, g are arbitrary functions) which yield A=0. For example, dt=p and dx=E is an example. Only dx=hbar/p and dt =hbar/E, however, lead to a probability form which conserves momentum and energy and this is something which we did not stress clearly in previous notes. Thus, there seem to be a number of issues involved in obtaining a free particle probability exp(-iEt+ipx) and these are independent requirements, which then lead to a particular solution. We argue that these requirements are: ((1a)) One wishes to preserve Lorentz invariance ((1b)) One desires discrete dx, dt intervals instead of Newton’s intervals which tend to 0. ((1c)) These intervals need to be linked with a particle related probability which defines them. This probability is critical physically because if the theory is to define photons and particles equally well ( as special relativity ((1a)) applies to both), one must account for the probabilistic effects of 2-slit interference and 1-D reflection-refraction at an n1-n2 index of refraction junction. ((1d)) One must have conservation of momentum and energy. We argue that it is the combination of these requirements that ultimately leads to the free particle probability exp(-iEt+ipx).