There seems to be much interest in the literature (1) in associating quantum mechanics (in particular the Schrodinger equation) with a classical joint phase space probability F(x,p,t) such as appears in classical Liouville theory etc. Here, we consider that P(x)dx = dx/L for a particle at rest and analyze the extra physics which appears when one considers a single particle bouncing back and forth in a box, which is what a person in a frame moving with -v would see (albeit with a Lorentz contracted L.) We suggest that the constraint L is important as it is linked with physical forces which appear when the particle moves. In other words, a particle at rest is linked with a static probability, while for motion, one must have a probability which is linked to more than x and t, but to force considerations in x and t, a dynamical probability. (Experiment bears this out as we discuss.) In particular, we argue that in the case of a moving particle, two physical concepts related to force appear, namely pressure which is an average over time and impulse hit which is only linked with space. Thus, two different particles may have the same p, but different v and pressure values. We thus suggest a physical grouping of (p,x) and (pressure, t). We then note that one needs to somehow preserve the notion of P(x)=1/L, i.e motion does not introduce a bias to x, nor to time. In a rest frame, there is no motion and so there is neither p nor pressure, but P(x)=1/L pertains. If one tries to define a probability with two groupings, one for pressure, t and the other for p,x, then if one multiples a p1 case by a p2 case, the p,x grouping should yield p1+p2, x. Furthermore, a p,x grouping should imply mixed terms, e.g. px etc, otherwise one does not really have an association between the two variables. This suggests a pf(x) form linked to the p-x grouping. Given the notion of special relativity, it seems that one may replace pressure with the notion of energy and create a Lorentz invariant probability. This would then lead to the free particle quantum wavefunction (probability) exp(-iEt+ipx), but we argue that the roots are in classical physics and a desire to physically describe the physical information present for a particle bouncing back and forth in a box.
There seems to be much interest in the literature (1) in associating quantum mechanics (in particular the Schrodinger equation) with a classical joint phase space probability F(x,p,t) such as appears in classical Liouville theory etc. Here, we consider that P(x)dx = dx/L for a particle at rest and analyze the extra physics which appears when one considers a single particle bouncing back and forth in a box, which is what a person in a frame moving with -v would see (albeit with a Lorentz contracted L.) We suggest that the constraint L is important as it is linked with physical forces which appear when the particle moves. In other words, a particle at rest is linked with a static probability, while for motion, one must have a probability which is linked to more than x and t, but to force considerations in x and t, a dynamical probability. (Experiment bears this out as we discuss.) In particular, we argue that in the case of a moving particle, two physical concepts related to force appear, namely pressure which is an average over time and impulse hit which is only linked with space. Thus, two different particles may have the same p, but different v and pressure values. We thus suggest a physical grouping of (p,x) and (pressure, t). We then note that one needs to somehow preserve the notion of P(x)=1/L, i.e motion does not introduce a bias to x, nor to time. In a rest frame, there is no motion and so there is neither p nor pressure, but P(x)=1/L pertains. If one tries to define a probability with two groupings, one for pressure, t and the other for p,x, then if one multiples a p1 case by a p2 case, the p,x grouping should yield p1+p2, x. Furthermore, a p,x grouping should imply mixed terms, e.g. px etc, otherwise one does not really have an association between the two variables. This suggests a pf(x) form linked to the p-x grouping. Given the notion of special relativity, it seems that one may replace pressure with the notion of energy and create a Lorentz invariant probability. This would then lead to the free particle quantum wavefunction (probability) exp(-iEt+ipx), but we argue that the roots are in classical physics and a desire to physically describe the physical information present for a particle bouncing back and forth in a box.
In Part 1, we argued that P(x)dx=dx/L (where L is an arbitrary length) applies to a particle at rest. If one views such a particle from a frame moving at constant -v, it is moving. We argued that in such a case, the relevant probability is not simply linked to the position x, but also to the momentum p at x because this delivers an impulse hit. In particular, probability is presumably to be used in a physical problem with interactions and probability and a dynamical particle requires a dynamical probability describing both x,t and E,p. This led to exp(-iEt+px) which actually predicts intervals in space hbar/p and time hbar/E. In other words, a probabilistic view point leads to these results. Here, we ask the question: What if one simply wishes to use deterministic physics to describe a moving particle? After all, Newtonian mechanics is deterministic and one does not really need to focus on P(x)=1/L or its modification for a moving particle. In such a case, a particle moving with constant v (starting at x=0, t=0) is described by x/t=v both in Newtonian mechanics and special relativity. There is no uncertainty in x and t here which begs the question: Why does it appear in the probabilistic formalism of Part 1? In particular, exp(-iEt+ipx) derived in Part 1 implies physical regions: dx = hbar/p and dt=hbar/E and these seem to be absent in x/t=v and this is a problem. We suggest that a probabilistic approach is buried in the Lagrangian formalism which leads to the result d/dt v = 0, i.e describes the particle’s motion in space. This motion is actually x/t=v, but this equation only presents a subset of the description of a particle that is moving with constant speed, even in a deterministic Lagrangian/Action approach. We have already pointed this out in (1). We briefly explain it again here, but the real question is why does one need to consider the Lagrangian in order to describe x/t=v? In other words, the probabilistic approach associated with v=x/t only emerges in the Lagrangian formalism using a relativistic action Lt = -Et+px, with x/t=v. Ignoring the action and Lagrangian leads to an absence of hbar/p and hbar/E and one is only concerned with x/t=v. We note that the Lagrangian is primarily linked to the result: p = dL/dv partial and so argue that one is not simply interested in motion (x/t=v), but also in interaction because p is associated with an impulse hit. Thus, the Lagrangian approach is more than simply a mechanism which describes x,t motion, it is associated with p and E (through Hamiltonian = E = pv-L.) Using the Lagrangian approach suggests that it does not suffice to consider x and t by themselves when p and E are present. One must consider all four variables x,t,E,p which is the same point made in Part 1. The physics of the problem dictates that it is not enough to simply think in terms of x,t,v. The Lagrangian approach, which is constructed to ultimately yield x/t=v, when written in terms of E, p, i.e. Lt = -Et+px demonstrates uncertainty in x, hbar/p and t, hbar/E as shown in (1). This uncertainty must then be written in terms of a probability which is Lorentz invariant and leads to exp(-iEt+ipx). The point is that one would not even consider a probability if one did not consider E,p and x,t together as a complete treatment of a moving particle. In other words, x/t=v is an incomplete description of the state of a particle moving with constant speed, even in a deterministic framework.. At first this might seem surprising, but it is well known that energy and p are linked to interactions. In special relativity, one does not consider x,t,v alone and then compute p and E from these, but considers four vectors and Lorentz invariant equations. It is possible to create a Lorentz invariant equation involving all four state variables x,t,E,p and it turns out that this is the relativistic Lagrangian which describes motion, so this approach is more general than simply stating x/t=v. Thus, delta x = hbar/p and delta t= hbar/E appear in both the probabilistic approach of Part 1 and the deterministic one of the Lagrangian used here.
In 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 Part 1, we considered Newtonian elastic scattering and proposed a probability exp(-iEt+ip dot r) to allow for equal product probability weights for any (ei, ej) (energy) and (pi, pj) (momentum vector) outcome pair. We insisted on the probability being a Lorentz scalar and argued that all free particles have the same real value weight, hence the modulus of 1. This notion is based on probability as it is applied to possible outcomes. This means that one does not have a fully deterministic situation because otherwise one would know the exact ei,ej, pi,pj outcome and would not require probabilities in the first place. In Part 2, we argued that a deterministic approach based on x=vt for a free particle contains no notion of probability and so this seems to contradict exp(-iEt+ipx) with its physical intervals: delta t = hbar/E and delta x = hbar/p. We then noted that these intervals and hence probability are found in the Action= Lagrangian * t = -Et+px with v=x/t. We concluded that x=vt, a solution of d/dt dL/dv partial - dL/dv = 0 loses some of the information present and that examining the action reveals more information because one is forced to deal with x,t, E and p at the same time. Here, we return to Newtonian elastic two-body scattering and assume that one has the following deterministic case: e1,e2, p1, p2 → e3,e4, p3,p4. One would then assume that no probability of any kind is present. We suggest that momentum is conserved due to action-reaction, something that Newton already proposed without any notion of probability required. Energy is conserved because it cannot be ultimately created or destroyed and we consider special elastic collisions. We then consider the Newtonian idea of a particle being acted upon by a force -dV(x)/dx, where V(x) is a potential. The interaction occurs for a dt, dx, but Newtonian physics assumes that dx→0, dt→0. Thus, Newtonian mechanics allows for a minimum dx->0 for conservation of momentum. By going into the center-of-mass frame of two particles which collide elastically, one has p and -p in this frame. When the two particles come together at x,t, for a tiny time one has a joint particle with no net momentum which is interacting and breaking apart. Then, in the 1D case, one particle moves out with p1 and the other with -p1 in the cm-frame. If one considers A= -Et+px for one particle, then Newton’s dt->0, dx->0 as well as x=0, t=0 yield A=0. It is known from special relativity, however, that there is a jump in x for a particle viewed from a moving frame. In particular, if the particle is at x=0 at t in the rest frame (cm frame), then in the moving frame: x’ = g(v) v t where g(v)=1/sqrt(1-vv/cc).This is linked to the person in the moving frame (-v) measuring the particle which seems to move to him/her. This measurement takes some minimum amount of time. Thus, it seems that one must reject dx→0 on the grounds of special relativity. The question then becomes: What is the smallest dx and dt one may have based on special relativity? We suggest that given the A=0 case, one may have delta x =hbar/p and delta t = hbar/E. Even in a deterministic elastic interaction, the notion that dt and dx cannot be 0 leads to minimal values for these which are probabilistic, i.e. they are based on special relativity and not on the actual interaction. If one has initial uncertainty in x and t, even if the particle then moves with x=vt, this uncertainty should also be propagated along as well, because the dx, dt uncertainty cannot simply vanish. We suggest that the notion of exp(-iEt+ipx) emerges even if one removes the initial probabilistic assumption of elastic scattering with unknown outcomes. Known outcomes still lead to uncertain dx=hbar/p and dt =hbar/E in keeping with the uncertainty contained in Lt= -Et+px, which is used for deterministic calculations. Thus, it is special relativity which places a constraint on how small dx and dt may be in a physical reaction and one cannot simply use the Newtonian dx→0, dt→0.
Special relativity seems to be associated with linking physical relations as seen in one frame with those as seen in a frame moving at a speed relative to such a frame. A constant speed is often used for the moving frame. This suggests that classically there is a special frame for a particle with rest mass called a rest frame. In such a case, one simply introduces a parameter for rest mass and does not worry about the complicated motion and interactions which actually create the rest mass. It seems that even quantum mechanics, which considers a particle to be moving as one does not know its location, one uses the idea of a potential V(x) which in principle is associated with a very large rest mass, presumably at rest as no variables of such a rest mass appear in the Klein-Gordon or Dirac equations. It seems that one uses a preferred frame when solving relativistic quantum mechanical equations. Furthermore, given V(x) and a bound state with exp(-i E bound t), E applies to the overall bound system, not to a possible exp(ipx) constituent. The special relativistic invariant equation seems to exist at the system level, like with the rest mass and one may apparently calculate E in the special rest frame of the large mass and hence find a rest mass of the bound system, i..e large rest mass*cc + mo cc + E, mo is the rest mass of the single bound particle. As a result of the use of the special rest mass frame for the V(x) producing large rest mass, it seems that one does not have to worry about special relativistic constraints on the interaction with V(x), as only one frame is involved. In other words, a bound calculation is about finding a new rest mass for an overall new system. Then this system may be used in Lorentz boosts. One simplifies by working in the rest frame of the mass creating V(x). Perhaps surprisingly then, we note that exp(-iEt+ipx) for a free particle is Lorentz invariant and in fact must be Lorentz invariant for the following reason. Imagine that one wishes to create a probability for free particles which is linked with conservation of momentum. Any p must have the same real value weight because for free particles there is no bias for one p or the other. There is, however, a bias for a p1 and p2. It should have the same probability as a p3 and p4, if p1+p2 = p3+p4. (We use one dimension for simplicity.) As a result, exp(ip) seems to be a probability which respects these conditions. The problem is that one may have p=m1v1=m2v2 hence (E1,p) and (E2,p). With respect to momentum, it is irrelevant if one particle has E1 and the other E2, but not if one does a Lorentz boost. Then, in one case: p→ g(v) p + g(v) v E1 and in the second, p→ g(v) p + g(v)vE2. Here g(v) =1/sqrt(1-vv/cc). In other words, p is the same for each particle in one frame and not in the other which violates conservation of momentum because in frame 1 both particles can complete the same conservation of momentum equation, but in the second frame this is not the case, because each has a different momentum. Thus, exp(ip) cannot hold. One requires a linear form in p which is Lorentz invariant, i.e. exp(-iEt+ipx), we argue (and have done so in previous notes). We see now that momentum probability is linked with space in exp(ipx). The idea seems to be that for free colliding particles there is no preferred frame and the probabilities used must reflect this Lorentz invariance. As a result, we argue that a free particle probability in E,p arises because of Lorentz invariance, but then apply it to a V(x) problem and no longer worry about Lorentz invariance at the interaction level. V(x) with no associated rest mass variables assumes a rest frame for the V(x) producing mass. A Lorentz boost which then invalidates exp(ip), would only apply to the total rest mass of the bound particle system which has no overall momentum. It seems that one does apply the Lorentz transformation to the interaction of exp(ipx)’s with V(x). One is only interested in the overall bound system as a single entity for special relativity. As a result, one may only impose momentum considerations as V(x) is a spatial function and one must link it with exp(ipx). In fact, one may write V(x) = Sum over k Vk exp(ikx) to associate it with momentum hits linked with Vk energies, which already moves away from the idea of special relativity as Vk and k are not linked as a 4-vector. The idea of exp(ip1x)exp(i k1x) = exp(ip2x)exp(ik2x), however, seems to still hold (as it does for free particles), linking these two values (i.e. Vk1 and Vk2) with (p1+k)(p1+k) exp(i (p1+k)x). Lorentz invariance for a particle with rest mass seems to be linked with the idea that one ultimately has a rest frame for the particle. One then establishes a relation between the rest mass in one frame and its new E and p as seen by another frame. One does not, however, need to impose Lorentz invariance conditions on the interaction which creates the rest mass it seems. For example, exp(-iEt+ipx) is the Lorentz form used to describe a free particle, but exp(ipx) is the only portion used to describe its interaction with V(x) in the rest frame of the large mass creating V(x). Furthermore, exp(-iEt+ipx) ensures conservation of energy and p for a free particle, but in an interaction case, that does not happen as the potential may absorb energy.
We have argued in Parts 1 and 2, that the free particle quantum form exp(-iEt+ipx), which we call a complex probability, arises from special relativity in order to allow for conservation of energy and momentum (in an elastic collision). One may see from single particle bound states how exp(-iEt) plays the part of a binding energy. We have noted in Parts 1 and 2, that exp(-iEt+ipx) follows from the Lorentz invariant -Et+px, but that one does not consider the deterministic x=vt equation which applies to the center-of-mass, but rather treats t and x as independent variables which force the independence of E and p (two different interaction variables) and also force separate E and p conservation. The question becomes: How is it possible to have x,t play two such different roles in special relativity? Here we try to answer this question. We argue that special relativity is based on determinism, i.e. x=vt. One may create a Lorentz transformation for a particle with rest mass mo at x=0 with t, a running variable. As seen from a frame moving with -v, one has x’=vt’. In other words, a separate x’,t’ is associated with each v. In Newtonian mechanics, it is possible to have mov1 and mov2 in the same frame (i.e. a frame with a ruler and clock), but in special relativity, each mo in the rest frame maps to a different x1’,t’ and x2’,t2’. This, however, seems to lead to a contradiction. If one considers two particles p1,E1 and p2,E2 (one dimensional motion) which elastically collide, then before the collision, they may be thought of as a composite particle (for more than a single x,t point) with energy E1+E2 and momentum p1+p2. Given the validity of special relativity, we argue that one may associate a Lorentz invariant -(E1+E2)t + (p1+p2) x with this composite particle. Furthermore, any outcome Ei+Ej = E1+E2 and pi+pj = p1+p2 would satisfy this same invariant. At this point there is no contradiction. The contradiction arises if one considers a single t and x for a composite particle, while internally there exist t1’,x1’ and t2’, x2’. The composite particle, in terms of special relativity, is not compatible with each -E’t’+p’x. The energies and momenta, however, add for each particle (in conservation equations) and so we argue that one should have a relativistically compatible approach which allows one to link each particle to a composite particle and see conservation of E and p. We note that it is precisely x=vt for each particle which leads to x1’,t1’ and x2’,t2’ and so argue that if one gives up this notion of determinism for x and t and introduces independent x, t values (fluctuations), then one can indeed retain the -Et+px form for each particle and use it to create composite particles. This seems to give rise to an AND situation which does not exist for both p,x and E,t in the deterministic relativistic approach. In that approach, one can only think of p’s and E’s adding separately, without any linking of x’s and t’s. In the fluctuation approach, it is possible to have the same x and t fluctuation points for each particle. We argue that it is this kind of thinking which leads to the probabilistic form exp(-iEt+ipx) which still respects the Lorentz invariant form and at the same time allows for conservation of E,p with the same t,x fluctuation points allowed. x=vt does not allow one to combine x1’,t1’ and x2’,t2’ for a composite particle because these may coincide for only one x,t point and one wishes to describe a composite particle with an x,t which exists over more than one x,t. This is why we suggest the physical fluctuation scenario which exists together with x=vt. x=vt describes the deterministic motion of the center-of-mass, while exp(-iEt+ipx) describes probabilistic hits in x and energy in t. It is special relativity which forces one to associate time fluctuations with E and spatial ones with p. As a result, we argue that special relativity gives rise to both probabilistic and deterministic versions, both of which respect the form -Et+px. The deterministic version is linked with x=vt center-of-mass motion, while exp(-iEt+ipx) represents probabilistic appearances of E in t and p in x, i.e. there is a spatial probability for impulse hit delivery.
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.