In a recent paper, (1) introduces a path related probability scheme in order to derive quantum mechanics probability. This probability is a real positive number between 0,1 and so one is classical, but the fact that it is linked to paths and not simply a position, time x,t suggests that this approach differs from strictly classical probability. In particular, the assumptions used by (1) are path related. (1)’s path based probability is P(nm) ( f(x,t)), where nm is a path and f(x,t), a Lorentz invariant. Bayesian composition is allowed and P(nm) = P(n)(path1) P(m)(path2). As a result, one may break a probability into a product of probabilities based on paths. For example, path1 might be from a starting point to a point P and path2 from P to a detector. For a linear path along x, this suggests the form exp(-iEt+ipx)=W(x,t). To obtain a real probability, one needs (1)’s assumption of time reversal invariant, i.e. P(-Et+px)=P(Et-px), so P = W*(x,t)W(x,t). We argue that (1)’s quantum form follows from the last two (of three assumptions). Formally, one suggests that quantum mechanics arises from (A) “Kolmogorov additivity restricted to one or two path contributions”, (B) time reversal symmetry and (C ) Bayesian composition. We go on to argue that the path related approach of (1) is equivalent to a momentum-energy conserving probability approach of (2).
In a recent paper, (1) introduces a path related probability scheme in order to derive quantum mechanics probability. This probability is a real positive number between 0,1 and so one is classical, but the fact that it is linked to paths and not simply a position, time x,t suggests that this approach differs from strictly classical probability. In particular, the assumptions used by (1) are path related. (1)’s path based probability is P(nm) ( f(x,t)), where nm is a path and f(x,t), a Lorentz invariant. Bayesian composition is allowed and P(nm) = P(n)(path1) P(m)(path2). As a result, one may break a probability into a product of probabilities based on paths. For example, path1 might be from a starting point to a point P and path2 from P to a detector. For a linear path along x, this suggests the form exp(-iEt+ipx)=W(x,t). To obtain a real probability, one needs (1)’s assumption of time reversal invariant, i.e. P(-Et+px)=P(Et-px), so P = W*(x,t)W(x,t). We argue that (1)’s quantum form follows from the last two (of three assumptions). Formally, one suggests that quantum mechanics arises from (A) “Kolmogorov additivity restricted to one or two path contributions”, (B) time reversal symmetry and (C ) Bayesian composition. We go on to argue that the path related approach of (1) is equivalent to a momentum-energy conserving probability approach of (2).
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 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).
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 2, we suggested the existence of a free particle probability P(x,t,E,p) and argued that it must be Lorentz invariant as equations and formulas should appear the same as written in terms of the variables of various rest frames moving at constant speeds relative to each other. This led to P(-Et+px) and we saw that energy and momentum conservation followed. Here, we consider the following. Imagine that one wishes to introduce a momentum and energy conserving probability and does not wish to make it Lorentz invariant. In previous notes, we argued that one would be led to exp(iC1 E) and exp(iC2 p) where C1 and C2 are constants related to units and p is along an x-axis. We argued that for time reversal invariance exp(iC2p) → exp(-iC2p) and suggested that this is unphysical. Here, we show that using exp(iC1 E) and exp(iC2 p) leads to problems if one compares results from mixed frames (v and -v) to a rest frame. Given that probability exp(iC1E) and exp(iC2 p) only contain E and p, one may formally consider such probabilities as being created using different frames. For example, one may take a rest particle and create p using a frame moving with -v (relative to the rest frame) and -p using a frame moving with v. exp(iC2 p) provides no information about a frame, it is only linked to a momentum consideration. This leads, we argue, to issues, and forces one to include x,t as well as E,p, i.e. use Lorentz invariance when dealing with such probabilities. As a result, we argue that there is no way to avoid introducing x,t into such a probability and as a result, quantum free particle features such as wavelength and frequency emerge. In other words, when one thinks only in terms of E and p, as in Newtonian mechanics, one forces a fixed x,t frame. If, however, one wishes to consider p and E as resulting from different frames and use probabilities, these must combine p,E and x,t to define the frame. Then one may actually mix probabilities linked with different frames.
In (1), a type of probability conservation of an anti-Bernouilli variable linked to diagonal motion along box diagonals in an x-t space with 1-mdt being the probability that x direction motion changes leads to a probability of exp(imt). This approach seems to involve a kind of time reversal symmetry as (1) proposes a t-x space consisting of grids of spacing e and possible motion along diagonals with both +/- e motion in t and x. In particular, (1) considers motion in the positive t direction from the origin to a some T (with +/-e motion possible) and then retraces the direction to return to the origin. In other words, (1)’s m is not an arbitrary number, but is linked to a conserved quantity, energy, i.e. probability steps are in terms of energy if one compares with the quantum exp(-Et), which (1) does. (1) obtains a kind of Dirac equation, but although we provide analysis of this, we only consider the form exp(imt) as relevant. Furthermore, exp(iEt) is consistent with conservation of energy and conservation of momentum follows from the x form of the solution exp(ipx) and so the probability approach of (1) is somehow linked to a change in direction, i.e. collision, which ultimately is associated with a conserved m, i.e. E (energy). In previous notes, we suggested that one may obtain the free particle quantum wavefunction exp(-iEt+ipx) by suggesting that a probability exists in Newtonian two-body elastic scattering. We argued that given initial (e1,e2) and (p1,p2) (energy, momentum) particles, any (ei,ej) (pi,pj) vectors has the same product probability (equal to the product probability of the (e1,e2) (p1,p2) case) has the same value. In (2), we showed that one cannot use exp(ip) for the momentum probability because it is not unique and violates time reversal symmetry. We suggested using a Lorentz invariant result exp(-iEt+ipx). The argument we make in this note is that the approach of (1) and (2) seem to be the same. At first this seems to be surprising. (1) obtains exp(-iEt) strictly through a “special” anti-Bernouilli somewhat-random walk approach. There is no concept of energy or momentum, hence no concept of their conservation and no concept of special relativity. (2) on the other hand is based on all of these ideas. We suggest that given that probability is introduced in (2), the notion of physical paths and changes is common to both approaches. Furthermore, we suggest that both are based on time reversal symmetry. We have already noted that the number m in (1) is equivalent to energy. Thus, energy is linked with the anti-Bernouilli probability motion in time (and p with motion in x). This, in a sense, is the same as a conservation of energy and momentum approach, we argue. If one considers that energy and momentum follow from special relativity by Lorentz boosting a rest mass, then one really has the concept of conservation of the rest mass which is the same as conservation of a certain kind of probability. Thus, we argue that (1) is an interesting approach to flushing out the probabilistic nature of the Dirac equation which we argue is also directly linked to introducing a dynamic directional probability which describes energy and momentum conserving interactions.
In (1), the notion of a stochastic classical action is introduced and a probability is linked to exp( constant * action). The objective of (1) is not to derive quantum mechanics, but to derive stochastic dynamics based on maximum entropy in action space. We have argued in a number of previous notes (2) that one may use the classical action to create a quantum free particle probability. This classical action, however, is not chosen a priori, but rather arises due to other considerations. We suggest that stochasticity is present in Newtonian mechanics at some time scale because dt is finite in dp/dt= Force (Newton’s second law) and so postulate that force delivery at some x,t is probabilistic within some dt and dx. A priori, dt and dx are not known, but if there is stochasticity, there should be a free particle probability, i.e. a probability(x,t) where x,t represent an x, t interaction point. Furthermore, the notion of dx and dt should arise naturally. We note that the form exp(i function(x,t,p,E) may be introduced because free particles have no distributed variable (like energy in the Maxwell-Boltzmann case). exp(i function(x,t)) immediately introduces periodicity, but we have argued that stochasticity can only exist within tiny, fixed dx, dt linked to dp/dt=F. In particular, one classical Newtonian scales this stochasticity is not seen. The periodicity should be directly in x and t space, suggesting exp( i A1 t + i A2 x). This expression, however, should be Lorentz invariant, and A1 and A2 should be constants, at least for a fixed mo and v. The reason that the one must have periodicity is that the dx and dt ranges of stochasticity in dp/dt=Fr must appear naturally. At this point, one is led to exp(-iEt+ipx) with -Et+px being a Lorentz invariant which satisfies the above criteria. This Lorentz invariant happens to be the classical action in both the relativistic and nonrelativistic cases, hence giving rise to action as being a stochastic variable (of (1)) because it is a function of the two scholastic variables x and t. Alternatively, one may choose the exp(iA1x+iA2t) due to its conservation properties. One may then argue that A1 and A2 could also be conserved for x,t held constant, i.e. A1 and A2 could be linked to Newtonian energy and momentum in elastic collisions. This allows one to solve for the special relativistic g(v) =1/sqrt(1-vv/cc). Given the idea of a conserved quantity form for a probability, one may argue that exp(i Action) satisfies a maximization of Shannon’s entropy subject to a conserved constraint (which is again a link with (1).) The question then becomes: Why should the Lorentz invariant A= -Et+px be the classical action.One may note that x,t usually refer to trajectory values and so for a free particle, one has x=vt. This means that one may factor out t, i.e. A= t( -E+pv), but this is automatically the form of the Lagrangian with E being the Hamiltonian. This holds relativistically and non-relativistically with the form: dL/dv partial = p. The remaining question is: Why should the classical action be Lorentz invariant? We consider E(v)-p(v) v = -L such that dE/dv - v dp/dv -p = -dL/dv partial. This means that dE/dv - v dpd/v = 0 ((1)). A Lorentz transformation, however, changes mo in a rest frame into p=mov/sqrt(1-vv/cc) and E=mocc/sqrt(1-vv/cc). One may see that this solution may be obtained from ((1)) showing a direct link between the classical action approach and special relativity. The connection may be carried further as Lt = -Et+px is a Lorentz invariant, but also the classical action (in both the relativistic and nonrelativistic cases). We argue that the use of the action in (1) as a stochastic variable is directly related to special relativity. In summary, from finite dt in dp/dt=F, t and x are stochastic variables within some naturally appearing dx and dt. One may write a linear combination of x and t and to create a stochastic Lorent invariant variable which should appear in exp(i Lorentz invariant) and create the dx and dt units. It may then be shown that this Lorentz invariant in E,p,x,t should be equivalent to the classical action.