Quantum Free Particle exp(-iEt+ip dot r) and State Probability Part 3

Quantum Free Particle exp(-iEt+ip dot r) and State Probability Part 3 Francesco R. Ruggeri Hanwell, N.B. Nov. 18, 2025 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. Newtonian Mechanics and a Single Frame Newtonian mechanics, which is nonrelativistic, uses a single x,t frame and introduces the ideas: p = momentum = mov (in one direction as it is a vector) ((1a)) Kinetic energy = .5mo vv ((1b)) mo is rest mass It also introduces various equations such as the second law: Force = dp/dt (vector equation) ((2)) The point we make is that one has a single x,t frame and applies different forces to rest masses mo in order to create different p and ke values. If one speaks of p and E, it is automatically assumed that they are associated with a single x,t frame and there are no issues of mixed frames. Defining Momentum and Energy Using Frames Moving at Constant Speeds There is nothing preventing one from viewing a particle at rest at x=0 with rest mass mo from a frame moving with constant -v. From the point of view of a person in such a frame, the particle is moving with a speed v and if the person knows Newtonian mechanics, he or she would assume that that particle has a kinetic energy and momentum. If on the other hand, one wishes to see a -p, one creates a frame moving with v. An immediate problem arises with a person assuming that one has Newtonian energy and momentum and we argue that this is linked to the notion of a fixed x-t set as described in the first paragraph. As we have shown in previous notes, the x,t measuring system which applies to a person in the rest frame cannot possibly apply for the person traveling in either the v or - v frame for the following reason. If a person in the rest frame sees a particle at x=0 at t and x=0,t=0 coincide for all frames, then: Frame moving with v, particle moves with -v and x’= - g(v) v t and t’ = g(v)t ((3a)) Frame moving with -v, particle moves with v x’ = g(v)vt t’=g(v)t ((3b)) If x=x’=0 and t=t’=0 for all frames, then at a later time in the rest frame one may have x=0, t, but not in the moving frame. The moving frame must show motion and x must change. As a result, the notion of a fixed x,t system does not hold if one uses frames moving at constant speed. This begs the question: Why does one wish to use frames moving at constant speeds relative to each other? We argue that these frames are important because they define momentum and energy. In the first paragraph, we simply listed formulas for Newtonian momentum and kinetic energy without giving any derivation. In fact, there is no direct derivation of momentum in Newtonian mechanics. It is given a priori. Such is not the case when using moving frames. In particular: Mo at x=0, t in a rest frame maps to: E’, p’, x’, t’ with x’/t’ = v if the frame moves with -v ((4)) For a particle with rest mass, one might argue that p’/E’ = v as well, in keeping with Newtonian mechanics for p, but having E be linked to mo. This suggests a matrix transformation: | g(v) v/c g(v) | ((5)) | v/cg(v) g(v)| If one wishes to have a link between a rest frame mo and E’,p’ and yet have a kind of modulus equation, then one is lead to: -E’E’ + cc p’p’ = momocccc g(v)g(v) - cc momo vv g(v)g(v) = - momo ccccc ((6)) All of this is well known. The point we make is that ((5)) applies to both E.,p and x,t transformations, not to one set alone. This is very different from Newtonian mechanics in which one has a single x,t set and uses the notion of force to create different E, p values. Here we note that: E= mocc/ sqrt(1-vv/cc) —> mocc + .5movv for v<<c ((7)) Newton’s kinetic energy appears in the v<<c limit, and p=mov in the same limit. As a result, the constant moving frame approach yields more general results than Newtonian mechanics. We argue that there are two key ideas: ((7a)) The definition of energy (including kinetic energy) and momentum formally and accurately follows from the constant frame approach ((7b)) The constant frame approach involves transforming both E,p and t,x using ((5)), not just the E,p set. We suggest that both ((7a)) and ((7b)) must be respected in a formalism and argue that this ultimately has consequences for free particle quantum mechanics. Momentum and Energy Based Probabilities In some previous notes, we suggested that one may consider two body elastic scattering in Newtonian mechanics and argue that there exist probabilities P(ke) and P(p) which respect conservation of momentum and energy and allow for equal product probabilities for any (ei,ej) , (pi,pj) (momentum vectors) outcome. In other words, any outcome which conserves momentum and energy is equally likely. We argued that as a first guess, this leads to: exp(iC1 ke) and exp(i C2 p) where C1 and C2 are constants used to fix units and p is along the x-axis ((8)) In Part 2, we argued for a probability P(x,t,p,E) based on state variables and argued for Lorentz invariance, but here we do not do this. We try to argue that one may consider the physical example of elastic two body scattering. Furthermore, we do not wish to use the argument of time reversal symmetry, i.e. complain about exp(iC2 p) and exp(-iC2 p) having different values. We then ask: What is wrong with ((8))? We suggest that to see the problem with ((8)) one must go to the root of the definition of E and p, which is special relativity. E and p emerge from a Lorentz transformation ((5)) of a particle at rest, and such a transformation must also apply to x and t if one considers p and E as existing because they are seen from different frames. In Newtonian mechanics, p and E are seen because a force acts on a particle, but momentum mov is introduced a priori and one does obtain the full correct form. Thus, we choose to consider E and p being created due to frames and ((7b)) suggests that one cannot then ignore the variable x,t. We begin by assuming that one may ignore these variables and simply use ((8)) which displays only p and E. Given this, we argue one may use mixed frames to create different E and p values. For example: E,p may be created using a frame moving with -v relative to the rest frame ((9a)) E,-p may be created using a frame moving with v relative to the rest frame ((9a)) If one uses ((8)) which only displays E and p values, one may mix frames (v and -v) in order to create E,p and E,-p. These frames, however, are associated with different x,t schemes i.e. x=0 t in the rest frame is: x’= -v g(v)t t’=g(v)t for a frame with v ((10a)) and x’= v g(v)t t’=g(v) t for a frame with -v ((10b)) In other words, one uses different p’s in ((8)) without any consideration of the fact that they are associated with different x’,t’ scales. Instead, one would employ a single x,t scale as in Newtonian mechanics which does not consider different constant moving frames. We suggest, however, that one has a problem if one uses ((8)). If one considers a product probability: exp(iC1 E) exp(iC1 E) exp(iC2 p) exp(iC2 (-p)) = exp(i2C1 E) ((11)) Clearly, one no longer has the notion of a state probability as introduced in Part 2 because ((11)) does not equal: exp(iC1 2 mocc) ((12)) We suggest that one may use the notion of frames and when doing so, one should conserve probability. As a result, it is not acceptable to have ((11)) and ((12)) differing and the only way to remedy this is to bring the variable x’,t’ which differ for p and -p, i.e. exp(i2C1 E g(v) t) exp(iC2 p g(v)t) exp(iC2 (-p) (-g(v)t) For C1=C2=1 one has: exp(i2 mocc) ((13)) and so there is no issue. Thus, we argue that given that ((7a)) and ((7b)), i.e. the momentum and energy are defined by special relativity and that the matrix which creates E and p also creates x’ and t’, one cannot ignore that x’ and t’ change. Thus, a probability cannot only contain E and p as ((8)) does, but must be a state probability P(x,t,E,p) introduced in Part 2. This in turn leads to free particle quantum mechanics which is discussed in detail in Part 2 and so is not repeated here. Conclusion In conclusion, we note that the quantum free particle wavefunction exp(-iEt+ipx) (one dimension), which we call a probability, includes E,p,x and t. Here we try to eliminate t and x, arguing that one may only need probabilities containing E and p to ensure conservation of momentum and equal probabilities for any momentum and energy conserving outcomes of elastic scattering. We argue, however, that the very definitions of energy and momentum follow from special relativity, i.e. the considerations of frames moving at constant speed with respect to each other, and that this implies that not only E and p transform, but also x and t. In Newtonian mechanics, there is a single x,t scale and different E and p values arise due to force. In special relativity, one may mix frames (v, -v etc) to create different p values and we argue that in order to conserve probability one is forced to include not only E and p which change, but also x and t which also change. This leads to free particle quantum mechanics, we argue. We note that even if one does not want to call exp(-iEt+ipx) a probability, but only a math function, the same considerations apply because one must explain how such a function would have one set of values in one rest frame and a different set in a different rest frame of two systems moving at constant speeds relative to each other.