In previous notes, we argued that one may introduce a complex probability into Newtonian mechanics to describe elastic scattering. In particular, even in exercises in Newtonian mechanics, given an elastic two-body collision, energy and momentum are conserved and so one accepts that any E, p sets that satisfy this conservation have equal probability. In other words, deterministic mechanics is not deterministic enough in this case to predict the exact outcome of a collision of two (e1,e1), (p1,p2) vectors. This justifies the introduction of energy and momentum probabilities. In previous notes, we suggested exp(ip) as the momentum probability. A complex number with unit modulus is used because each free particle has the same weight. Here we point out a problem with this definition which was not mentioned in previous notes. Given the periodicity of exp(ip), exp(ip) = exp(i (p+2*n*3.14) which means that p has the same probability as p+ 2*n*31.4. One, however, wishes to have a unique probability. In previous notes we went on to argue that one ultimately requires exp(-iEt+ip dot r) as this is Lorentz invariant and exp(ipx) yields the same value for the x-axis pointing in one direction or the other. If one considers exp(ipx), however, the same problem mentioned above occurs, because any px1+ 2*n*3.14 yields the same value for exp(ipx) and so any pnew= p + 2*n*3.14/x1 has the same probability as p when what one desires is a unique probability. We note that for changing x1 values, the p(new) values change. In fact, one may obtain a uniqueness for exp(ipx) if one considers a range of x values, i.e. the wavelength=hbar/|p|. This leads to the notion of orthogonality, i.e. Integral dx exp(-ip1x) exp(ip2x) = 0 if p1 not= p2. By sampling x-space, i.e by taking the notion of wavelength seriously, one may have exp(ipx) represent a unqiue p value over a range of x because in such a case one has a unique number, namely the wavelength hbar/|p|. Thus, it is the wavelength and the positive and negative values of sin(px) and cos(px) which in fact define the probability associated with p. P represents a momentum impulse hit, but in terms of probability, it is actually the wavelength and the adding and subtracting of probability in space which defines p, we argue. This probability in space with addition and removal of probability then becomes important in a calculation of a probabilistic interaction with a 2-slit apparatus in which the slits are separated by about a wavelength.
In previous notes, we argued that one may introduce a complex probability into Newtonian mechanics to describe elastic scattering. In particular, even in exercises in Newtonian mechanics, given an elastic two-body collision, energy and momentum are conserved and so one accepts that any E, p sets that satisfy this conservation have equal probability. In other words, deterministic mechanics is not deterministic enough in this case to predict the exact outcome of a collision of two (e1,e1), (p1,p2) vectors. This justifies the introduction of energy and momentum probabilities. In previous notes, we suggested exp(ip) as the momentum probability. A complex number with unit modulus is used because each free particle has the same weight. Here we point out a problem with this definition which was not mentioned in previous notes. Given the periodicity of exp(ip), exp(ip) = exp(i (p+2n3.14) which means that p has the same probability as p+ 2n31.4. One, however, wishes to have a unique probability. In previous notes we went on to argue that one ultimately requires exp(-iEt+ip dot r) as this is Lorentz invariant and exp(ipx) yields the same value for the x-axis pointing in one direction or the other. If one considers exp(ipx), however, the same problem mentioned above occurs, because any px1+ 2n3.14 yields the same value for exp(ipx) and so any pnew= p + 2n3.14/x1 has the same probability as p when what one desires is a unique probability. We note that for changing x1 values, the p(new) values change. In fact, one may obtain a uniqueness for exp(ipx) if one considers a range of x values, i.e. the wavelength=hbar/|p|. This leads to the notion of orthogonality, i.e. Integral dx exp(-ip1x) exp(ip2x) = 0 if p1 not= p2. By sampling x-space, i.e by taking the notion of wavelength seriously, one may have exp(ipx) represent a unqiue p value over a range of x because in such a case one has a unique number, namely the wavelength hbar/|p|. Thus, it is the wavelength and the positive and negative values of sin(px) and cos(px) which in fact define the probability associated with p. P represents a momentum impulse hit, but in terms of probability, it is actually the wavelength and the adding and subtracting of probability in space which defines p, we argue. This probability in space with addition and removal of probability then becomes important in a calculation of a probabilistic interaction with a 2-slit apparatus in which the slits are separated by about a wavelength.
In previous notes, we argued that one may find the source of exp(-iEt+ip dot r) in Newtonian mechanics by considering two body elastic scattering. Even though Newtonian mechanics is deterministic, probability is routinely introduced in such elastic scattering events because one cannot deterministically predict the outcome in most cases. Rather, one relies on conservation of energy and momentum and any possible conforming solution has equal probability. We argued that a free particle does not have a real valued probability weight because each is independent of the other (unlike in an ideal gas). As a result, we suggested the probabilities exp(iE C1) and exp(ip C2), with the constants C1 and C2 accounting for units. We then went on to make arguments linked to time reversal symmetry for exp(ipC2) and overall Lorentz invariance to finally arrive at exp(-iEt+ip dot r). Here, we wish to examine these considerations in more detail as we think there is a more general principle present, namely that of a state probability. We first note that even though exp(iEC1) and exp(ipC2) are consistent with conservation of energy and momentum, i.e. exp(iC1 E1) exp(iC1 E2) = exp(iC1 E3) exp(iC1 E4) and a similar expression for p, one only requires such conservation if all particles are classically at the same x and t, i.e. the collision point. As a result, simply given exp(iC1 E1) and exp(iC1 E2), the particles with E1 and E2 may be far apart in space. As a result, it is not clear why one would create a product probability with them unless they were actually interacting at the same x,t classically, but mathematically there is nothing preventing this and claiming conservation even though there is no reaction. It is this idea which suggests that exp(iC1 E) and exp(iC2 p) should be linked with x and t.Then, conservation only occurs in products if x and t are the same for different particles, otherwise it does not follow. This might suggest: exp(i E * (f(x)+g(t)) ) and exp(i p * (f(x)+g(t)). Here p is taken as a one-dimensional number along the direction of momentum. It may be positive or negative.f and g are unknown functions. We next note that x is measured from an origin. Thus, x1 and x2 are distances measured from an origin as is x1+x2. If one writes exp(i E * (f(x)+g(t)), this seems to imply that x and t are both measured from zero points. If one multiplies by exp(i E * (f(x2)+g(x2)), it seems one would wish this result to be equivalent to exp(i E (f(x+x2)+g(t+t2)). This leads to f and g being linear functions in x and t. Now one may have noticed earlier that including E,p,x and t in the “conservation” probability for a free particle means that one has completely defined the state of the particle. This probability is a state probability and so even though it is complex, one may argue that this is a property which should not change through a Lorentz transformation, just like particle number. This then leads to the notion of exp(-iEt+ip dot r). Thus, one initially attempts to describe the probability to have a particular p and E such that conservation of momentum and energy holds. In such a case, there is no need to introduce x and t and a Lorentz transformation changes the probabilities, but not their conservation properties as is shown. Thus, it is the desire to enforce conservation of energy and momentum only when a classical reaction occurs with all particles at the same x and t which introduces x and t into the probability along with E and p making it a state probability which should be Lorentz invariant. We suggest that there is a further important consequence of introducing a state probability. Given that it does not change under a Lorentz boost, it is like a quantity and may be used in OR situations a priori, even though it does not represent an obvious physically conserved quantity. This allows one to immediately write: Aexp(ipx)+Bexp(-ipx)=Cexp(ip2x) at x=0 for 1-D reflection-refraction from an n1-n2 index of refraction junction. (One may write a second equation involving p as exp(ipx) represents the probability to have p at x.) This then leads to a quantum mechanical manner of dealing with an interacting free particle.
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.
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 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 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.
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.