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Joint Phase Space (x-p) Probability and Free Particle Quantum Mechanics Part 3 Francesco R. Ruggeri Hanwell, N.B. Nov. 23, 2025 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. exp(-iEt+ipx) Emerging from A Priori Probability In Part 1, we argued that one may obtain the quantum free particle wavefunction (which we call a probability) exp(-iEt+ipx) (or exp(-iEt+ip dot r)) by introducing an a priori probability into Newtonian elastic scattering. In other words, probability exists because we introduce it into this physical scattering problem by stating that one does not really know the outcome and arguing that any ei,ej (energy), pi,pj (momentum vector) sets which conserve energy and momentum have equal probability of occurring. We then argue that one requires a Lorentz invariant probability and arrive at exp(-iEt+ipx) which is complex because every free particle has the same real value weight, i.e. the same modulus of 1. Extra details of the probability are then forced to exist in the phase of the complex number. We point out, as we did in Part 2, that one might object to the introduction of a priori probability and argue that in a deterministic problem would actually know the exact outcome of a specific elastic scattering case and not need to resort to probabilistic arguments. exp(-iEt+ipx) from the Classical Action -Et+px In Part 2, we ignored a priori probability and considered the solution: x=vt ((1)) ((1)) shows no signs of probability whatsoever which is a direct contradiction to the physical intervals: dx = hbar/p and dt = hbar/E ((2)) predicted by exp(-iEt+ipx). We then noted that x=vt is ultimately a solution of: d/dt dL/dv partial - dL/dx = 0 ((3)) where L is the Lagrangian and is given by -mo sqrt(1-vv/cc) in the free particle relativistic case. We then noted that the Lagrangian is linked with momentum p through dL/dv partial and with energy E=Hamiltonian = pv - L. In other words, x,t,p,E are the variables of interest, not just x and t in ((1)). We suggested that ((3)) does not represent the full information present and argued that the Lorentz invariant (also the action Lt)
-Et + px ((4)) shows dx=hbar/p and dt=hbar/E. In other words, these intervals also appear if one considers deterministic physics as long as one examines the full action, Lt, and not simply ((1)). We next consider deterministic elastic two body scattering. Deterministic Elastic Two-Body Scattering In Part 1, we introduced probability a priori into two body Newtonian elastic scattering. Here we suggest that if Newtonian mechanics is really deterministic, one should conclude that one could know the outcome of an e1, e2 p1, p2 (energy and momentum vectors) elastic collision, i.e. there would be a single e3,e4, p3,p4 outcome and all probability would seemingly disappear. This would then nullify the arguments made in Part 1. Thus, we consider this example in more detail. If one goes into the center-of-mass frame for the e1,e2, p1,p2 collision, then one sees a p and -p. At some instant of time the two particles will be at x,t and the overall momentum will be 0. This is like having a particle at rest which is breaking apart to yield a different p1, -p1. During this change from p,-p to p1,-p1 action-reaction forces occur and momentum and energy are conserved as Newton already noted for an elastic collision. In Newtonian mechanics, an interaction occurs in dx, dt with force = -dV(x)/dx, where V(x) is the potential (if in fact it depends only on x). Newtonian mechanics allows for dx→0 and dt→0. We, however, consider special relativity and note that for a particle at rest at x=0 at t, there is a shift in x. In particular, x=0, t=0 and x’=0, t’=0 apply, but when the particle is seen to move due to a frame moving with -v, one has: x’ = g(v) v t and t’= g(v) t where g(v) = 1/sqrt(1-vv/cc) ((5)) Thus, even though x’ is considered to initially be at x=0 at t=0, when t arrives, x=0, but x’= g(v)vt. Thus, there is a finite change in dx’, if dt’ is finite. One may, however, argue for t being arbitrarily small. The point we make is that this shift in x’ is linked with a velocity measurement made by a person in the moving frame and so is technically not 0. We next note that for a moving particle: A = -E’t’ + p’ x’ ((6)) If one considers x’=0, t’=0 and an interaction occurring in a tiny interval tending to 0, then A=0 for x’=0,t’=0, but also for dt’ and dx’ =0. This bypasses, however, the notion of a shift in x’ as argued above. We note that there is in fact a second solution in ((6)) dt = hbar/E and dx= hbar/p ((7)) This allows A=0, but forces a minimal value for dt and dx based on E and p and does not allow for minimal dx→0 and dt→0. As a result, if one accepts dx=0 and dt=0, one would have to discard a legitimate solution of A=0 which seems to be more in keeping with special relativity
than dx’=0, dt’=0. After all, the person in the frame moving with -v must measure the velocity of the particle and this should not occur in dx’=0, dt’=0. It is possible that ((7)) are very tiny values which approximate 0, as in the usual classical world scenario, but there may be length scales (atoms etc) for which ((7)) are not close to 0. We suggest that even in the case of a deterministic two body elastic collision, one has initial uncertainty in dx and dt given by ((7)) for each particle. This uncertainty does not simply vanish, but must propagate, even though on average one has x=vt. This uncertainty must be directly linked to probability and we suggest once again that: exp(-iEt+ipx) ((8)) Is the relevant probability which appears due to special relativistic constraints even in a deterministic problem. Conclusion In conclusion, we note that in Part 1, we considered elastic two-body Newtonian scattering in terms of a priori probability. In particular, we argued that given an e1,e2 (energies) and p1,p2 (momentum vectors), any outcome ei,ej, pi,pj has the same product probability. This led to a probability exp(-iEt+ipx) which is Lorentz invariant only because one assumed a probabilistic problem in the first place. One may argue that a particular two body elastic interaction is deterministic and state that a specific e3,e4, p3,p4 is the outcome. Then, all probability seems to disappear. We argue here, however, that it does not due to special relativity. We consider the deterministic case in the center-of-mass frame. There, one sees a p, -p change into a p1, -p1 due to an interaction. In Newtonian mechanics, an interaction may occur in dx→0, dt→0 as is the usual treatment which applies to a force = -dV(x)/dx where V(x) is a potential. If one considers the special relativistic invariant A = -Et’+px’, which is also the classical action, then x’=0, t’=0 yields A=0 as does dx’=0, dt’=0. There is, however, another solution, dx=hbar/p and dt=hbar/E which also yields A=0 and we argue that this cannot be discarded. We also note that it takes time for a person in a moving frame to measure that a particle is moving and so suggest that dt→0 cannot really be 0. It is possible for hbar/p and hbar/E to be tiny as in the classical case in which dx→0 and dt→0 then seem to apply. For other scales (atomic etc), however, this does not hold and special relativity limits one to the actual values for dx’ and dt’ that one may use. We note that hbar/p and hbar/E do not depend on the actual interaction, but are uncertainties associated with initial motion. These uncertainties cannot simply disappear and so are present together with x=vt, the deterministic result, in the form of exp(-iEt+ipx), we argue. Thus, we suggest that it is special relativity which ultimately breaks Newtonian determinism, forcing uncertainty regions in x and t and yielding exp((-iEt+ipx), i.e free particle quantum mechanics.
