Full text
Quantum Free Particle exp(-iEt+ip dot r) and State Probability Part 4 Francesco R. Ruggeri Hanwell, N.B. Nov. 19, 2025 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). Two Body Elastic Scattering In Part 3 we started with the notion of Newtonian elastic two-body scattering and sought a probability in p,E such that any (ei,ej) (pi,pj) (momentum vector) outcome should have the same weight. This automatically forces one to consider E and p, but not x and t. We first postulated: exp(iC1 E) and exp(iC2 p) where p lies along the x-axis ((1)) The main point of Part 2 was to argue that E and p are defined by viewing a particle at rest from a frame moving with constant speed -v. At the same, x and t of the rest frame must be transformed into x’,t’. We noted that one may have two frames with v and -v and that to conserve probability one must consider not only p,E, but also x, t to create a Lorentz invariant. Otherwise, for probabilities exp(iC2 p) and exp(iC2 (-p)) one may mix frames and create AND situations which do not correspond to probabilities in the rest frame. Here we wish to start with no consideration of p and E, but only of the classical probability equation: P(x)dx = dx/L where L is an arbitrary length ((2)) P(x)dx=dx/L P(x)dx=dx/L is a uniform distribution for particles at rest. One may, however, view this from two moving frames -v and v (both constant speeds) to obtain: P(x,t, v) and P(x,t, -v) ((3)) There is no notion of p in ((3)). We argue that one may always apply an AND situation to probabilities and so: P(x,t,v) P(x,t-v) = P(x) ((4)) As a result, motion seems to create a kind of square root probability of P(x)dx =dx/L.
There is, however, an immediate paradox in ((4)) because there is no reason why one cannot consider: P(x,t,v=0) which should also give the information of P(x) ((5)) How can one have two P(x,t,v) act as a square root probability in ((4)) and a full probability in ((5))? The only way it seems is for P(x,t,v) =1, i.e. to have no velocity dependence. This suggests that probability for moving particles is exactly the same as that for one at rest, but we argue that this does not make sense from the point of view of special relativity. Special Relativistic Considerations We reject the notion of P(x,t,v) =1 and search for a probability which is Lorentz invariant and also contains dynamics, i.e information about v. In special relativity v is obtained through E and p which are part of a 4-vector and so we propose the Lorentz invariant probability: exp(-iEt+ipx) (or exp(-iEt+i p dot r) ((5)) One may notice immediately that this introduces E and p based units in time and space, overriding the infinite resolution associated with Newtonian rulers and clocks: dt = hbar/E and dx = hbar/p ((6)) Thus, there is an inherent physicality, namely ((6)), associated with probability linked to motion. One does not simply have 1, but the modulus is 1. One may note that setting p=0 in ((5)) yields: exp(-iEt) = exp(-i mocc t) ((7)) As a result, the x portion of probability may seem to be identical to P(x)=1/L because hbar/p goes to infinite, but there is still the time resolution. Thus, special relativity is consistent with a dynamical probability which is not that same as its modulus of 1 and introduces new physics. This new physics is presumably linked with interactions because ((5)) contains p which is physically linked with impulse. This begs the question: Why should one then even consider an equation like P(x)dx=dx/L if there is no infinite resolution for x? We suggest that the answer is linked to coarse graining to remove the effects of interaction. As an example, one may consider two-slit interference or even a bound state: W(x)=Sum over p a(p)exp(ipx) ((8)) One may note that W(x) is both positive and negative, but counting particles in spatial bins cannot yield a negative number. The negative number in ((8)) is fine for interaction calculations because it shows probability being removed from one place and added to another. Furthermore,
one cannot simply take the absolute value of ((8)) because this is not a continuous function. One requires an approach which removes the interactional feature associated with hbar/p. To find this we return to ((4)), but now use the Lorentz invariant form: exp(iEt-ipx) exp(-iEt+ipx) = 1 ((9)) exp(-iEt+ipx) shows the Lorentz invariant probability needed for probabilistic interaction calculations, but this is a dynamic kind of probability. The modulus ((9)) is really a subset of the information which may be all that is needed if one is simply counting particles in bins in space. There is no quantum interaction assumed to be occurring in this counting approach and so the coarse grained approach of ((9)), i.e. using the modulus instead of the entire probability is used. Formally, this has the form of multiplying a probability for motion to the right with motion to the left. Hence for a bound state, one considers spatial density to be: P(x) = W*(x)W(x) ((10)) We note (1) also considers W(x) to be a probability and defines W*(x)W(x) a priori without reference to special relativity. We suggest that special relativity is key for the quantum free particle wavefunction. Conclusion In conclusion, in Part 3 we started with a probability based on p and E which describes equal weights for any (ei,ej) (pi,pj) (momentum vector) outcome set in a Newtonian elastic 2-body scattering interaction. This led to a guess of exp(iC1 E) and exp(iC2 p) as probabilities, but we argued that it is special relativity (Lorentz transform) which defines E and p and this same transform must act on x and t. In order to have conserved probability, we argued in Part 2 that one must include x and t in a P(p)P(E) expression. This led to exp(-iEt+ipx), i.e. a Lorentz invariant form. Here we start with P(x)dx=dx/L, the classical spatial result and no p or E appears as this applies to stationary particles. One may view this set from a frame moving with constant v or -v and define P(x,t,v) and P(x,t,-v). Formally, one would expect that P(x,t,v)P((x,t-v) = 1, i.e. the P(x) result, but this is a paradox because P(x,t,v=0) should also yield P(x). To resolve this paradox, we argue that if one considers frames, then one should use special relativity and create a Lorentz invariant probability. We reject the idea of a probability that is 1 for a particle in motion and see that special relativity allows for the solution exp(-iEt+ipx). This automatically introduces resolution in time (hbar/E) and space (hbar/p) which is not found in Newtonian mechanics. Newtonian mechanics, however, is an approximation. As argued in Part 2, the full values of p and E follow from special relativity and so we wish to consider Lorentz invariant probabilities. We note that hbar/E and hbar/p have consequences for interactions and that exp(-iEt+ipx) should be used when describing interactions. We also note that the modulus of exp(-iEt+ipx) is 1 and if one is not interested in an interaction, the modulus squared information is good enough. In other words, one may formally remove the interaction information by using
exp(iEt-ipx)exp(-iEt+ipx) leaving only coarse-grained information, we argue. This even holds in the rest frame in which one has exp(-i mocc t). References 1. Antonakos, Charalampos The Quantum Wavefunction as a Complex Probability Distribution (2025) https://arxiv.org/html/2502.10523v3
