Quantum Free Particle Probability and the Notion of Outcomes Part 2

Quantum Free Particle Probability and the Notion of Outcomes Part 2 Francesco R. Ruggeri Hanwell, N.B. Dec. 2, 2025 In Part 1, we argued that the consideration of all possible outcomes is a defining idea of a probabilistic calculation. We then argued that in free particle quantum mechanics there were two sets of outcomes, one driving the other. In particular, a single particle moves as x=vt, but at each x,t point, one has interaction uncertainty/fluctuations of dt and dx linked with exp(-iEt) and exp(ipx). As a result, in a 2-slit interference problem, exp(ipx) (a probability linked with possible positional fluctuations about x) leads to a further probability of interacting with both slits. The two sets of outcomes must be considered, i.e. exp(i p dot r1) + exp(i p dot r2), where r1 is a vector from the center of the first slit to a point on a screen far away, and r2, a vector from the center of the second slit to the same screen point. We also considered the case of reflection-refraction from an n1-n2 index of refraction junction. The point is that these two cases are clearly probabilistic, i.e. not deterministic at all. Here we consider two different scenarios which have the appearance of a deterministic result, but are actually probabilistic and so linked with all possible outcomes, i.e. an average over these which must be understood as an average (as in classical probability). The first is a single particle bound state and the second, transmission through a potential wall. In the bound state case, the analogue is a Newtonian bound state and in fact between the classical turning points KEave(x) + V(x) = E in the probabilistic case which exactly matches the classical one. One is, however, dealing with an average KEave(x) = -1/2m d/dx d/dx W/ W, where W=Sum over p a(p)exp(ipx), it involves all possible outcomes and is not a physical state. Given the similarity to a deterministic problem there is a danger of not considering the quantum bound calculation as a probabilistic one. In the quantum single particle case, one may be surprised by tunneling, but the quantum probabilistic scenario is one in which all possible interaction outcomes are considered together (as is required by probability) and so one cannot possibly ignore interactions linked to a particle entering or leaving the potential region. Furthermore, it is only KEave(x) which equals the classical result and so E which is single value in the classical case, is linked to an average value in the quantum case, i.e. a particle may have a p such that pp/2m is not zero at the classical turning point. Only KEave(x) =0 at this point. An interesting feature is that a probabilistic calculation applies to a single particle run (which only produces one outcome). One, however, cannot measure the full probability associated with a problem without making a very large number of measurements. Single Particle Bound State Problem In classical mechanics, a bound state problem is based on a single particle which has one value of energy, i.e. pp/2m + V(x) = E ((1)) Here there are two clear turning points at which v(x)=p(x)=0 and so the particle cannot move beyond these two points. A quantum single particle bound state uses an average equation which is of the same math form as ((1)) i.e. KEave(x) + V(x) = En ((2)) where KEave(x) = 1/2m d/dx dW/dx / W, W= Sum over p a(p)exp(ipx) ((3)) In the quantum case, it is only KEave(x turning point) = 0. KEave(x), however, is based on a single particle having an entire range of p values, even at the turning point. On average it seems as if the particle stops, but this is an average of all possible outcomes and most of these include p values which are not zero and may continue outside of the turning point. In particular, a quantum problem is based on exp(ipx) which is an interaction probability which sits upon a deterministic x=vt. One must consider all possible interactions with a potential and all possible p values which includes tunnelling in and out of the classical bound region. The only constraints are W(x) =0 at x= +/- infinite and the notion of a single average E at all x. The question may then arise as to why should ((2)) hold, i.e. why should it mimic the classical equation ((1)) within the turning points? KEave(x) is an average kinetic energy value which does not physically represent a single outcome, as in deterministic physics. Clearly, one has p values larger than 0 at the classical turning points and particles moving in and out at these points. Furthermore, KEave(x) is based on quantum interference due to exp(ipx)s i.e. KEave(x) = {Sum over p a(p)exp(ipx) pp/2m} / W(x) ((4)) This is a probabilistic average based on a quantum complex probability exp(ipx). We note that even if one takes averages at each x point, average energy must be conserved, i.e. KEave(x) + V(x) = En ((5)) We note that exp(ipx) when added in W(x) = Sum over p a(p)exp(ipx) for a(p)=a(-p) (example) may lead to W(x) being positive and negative. W(x), however, is not representative of a physical quantity. It is W*(x)W(x) which is spatial probability. W(x) is associated with continuity properties, but that does not mean that it must be positive. The question then becomes: What happens with an average created using W(x), i.e. exp(ipx)’s. Is one guaranteed to have a positive KEave(x)? If one breaks a potential into constant V values for tiny dx regions, then one has a solution of Acos(px)+Bsin(px) for each region as the overall W(x) must be real for a(p)=a(-p) (i.e. a given parity). Then: -1/2m d/dx d/dx (Acos(px)+Bsin(px)) / (Acos(px)+Bsin(px)) = pp/2m >0 for all x ((6)) Thus, we suggest that one is considering all possible outcomes and creating an average scenario which has the appearance of a deterministic bound problem, but is really something quite different as it includes tunneling effects and all possible p values. The classical turning point holds for pp/2m = E, but this is only one magnitude of p in the classical problem, while the quantum problem considers all possible p values constrained by ((5)). It must be the case that some of these possible outcome p values move in and out of the turning points. This begs the question: Why should a single particle bound state be probabilistic? We argue that a quantum particle is associated with a probability exp(ipx) about a classical trajectory x. Force(x) acts at x, but the particle may be at any point in an hbar/p range and so may probabilistically receive different F(x) hits. Thus, the problem is probabilistic and one must consider all possible outcomes together as an average. Tunnelling Through a Barrier Tunneling through a barrier again has the semblance of a deterministic problem, except for the fact there is reflection at the first surface of the barrier. Inside the barrier, there is a wavefunction of exp(-K x), i.e. dropping function. A free particle, however, is described by exp(ipx), so we argue that: exp(-Kx) = Sum over p a(p) exp(ipx) within the barrier ((7)) Neither exp(-Kx) nor exp(ipx) represent the entire x range (-infinite to +infinite) and so exp(ipx) is not a true free particle state, but it shows, we argue, that exp(-Kx) represents an average of many possible outcomes, even though it might appear to be a deterministic drop in x. Outside the second surface (vertical line) of the potential wall, W(x)=D exp(ipx) where pp/2m = E is an initial condition. This seemingly deterministic looking solution, except for the initial reflection piece, is an average of all possible outcomes as it must be in a probabilistic calculation. Thus, the probability that a particle with E=pp/2m makes it through the barrier is a strictly probabilistic calculation based on solutions of -1/2m d/dx d/dx W / W +V(x)= E with E being the same for all x. In two regions (the first and third) V(x)=0, while V(x) = constant in the second. The use of d/dx d/dx presupposes a W(x)=Sum over p a(p)exp(ipx) because this is a nonrelativistic problem. A full solution is given in (1) and so is not presented here. The point we wish to make is that the average (using the quantum inherent probability exp(ipx)) seems to take on a life of its own seemingly representing a kind of deterministic process, but we suggest that one must remember that one really has an average over different possible outcomes which is not a deterministic state, but a probabilistic complete outcome scenario. The notion of average is used in the first place because a probabilistic problem considers all possible outcomes, as argued in Part 1. Conclusion In conclusion, in Part 1 we suggested that a probabilistic calculation must include all possible outcome states. We argued that a quantum free particle has an inherent interaction fluctuation probability of exp(ipx) about each x trajectory point. This means that the particle does not interact at x as in classical mechanics, but may hit at any x within the hbarh/p range, leading to a set of different outcomes, each with their own probability. There are two sets of probabilities, exp(ipx) and a second set linked to it and both have their outcomes which must be considered together as stressed in Part 1. Here we consider two additional quantum examples of total outcomes, but in these cases, the average over all outcomes has the appearance of a single quasi-deterministic solution. In other words, there is a temptation to consider an average over outcomes result (W(x)) as an actual physical state. Thus, there is surprise if a single particle bound state has tunneling because KEave(x turning)=0, just like KE(x turning) = 0 for a classical practical. KEave(x) is not the kinetic energy of a physical state, it is the average kinetic energy of all possible outcomes in a V(x) potential given a full range of p values, not a single one with p=0 at the turning point. The conditions are a single average E for all x and W(x)=0 at x=+/- infinite, in addition to the notion of conserved average energy. Tunneling through a potential wall is another example where the average of outcomes yields W(x)=exp(-Kx) which seems to represent a physical state dying in space, but is again an average of possible outcomes (represented by exp(ipx) probabilities). Thus, we caution against considering the average of outcome probabilities W(x)= Sum over p a(p)exp(ipx) as a specific physical state. It is a probabilistic calculation which yields the probabilistic average W(x) we argue. Even in classical probability, the use of an average alone is very limited as one must also consider a standard deviation and higher moments. Thus, one does not want to consider the average of outcomes as physically describing the system. After all KEave(x turning)=0 does not mean that one does not have the probability for p>0 at this point which leads to tunneling. Reference 1. https://phys.libretexts.org/Bookshelves/University_Physics/University_Physics_(OpenSta x)/University_Physics_III_-_Optics_and_Modern_Physics_(OpenStax)/07%3A_Quantum _Mechanics/7.07%3A_Quantum_Tunneling_of_Particles_through_Potential_Barriers