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Quantum Free Particle exp(-iEt+ip dot r) and State Probability Francesco R. Ruggeri Hanwell, N.B. Nov. 15, 2025 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. Classical Conservation Probability Newtonian mechanics is deterministic by nature. Nevertheless, it is standard procedure to be unable to deterministically know the result of a two body elastic collision. In particular, one considers any solution which conserves energy and momentum and this solution presumably has the same probability to occur as any other which follows conservation. One may note that a free particle is independent before it collides and so there cannot be any real value weight to the probability. As a result, we propose a unit modulus complex number (as done in previous notes): exp(iC1 E) and exp(iC2 p) ((1)) Here p is the magnitude of the momentum vector together with its sign, i.e. the x-axis is taken along the direction of motion. ((1)) changes under a Lorentz transformation, but conservation of energy and momentum are preserved. In particular: exp(iC1 E1) exp(iC1 E2) = exp(iC1 E3)exp(iC1 E4) and exp(iC2 p1)exp(iC2 p2) = exp(iC2 p3)exp(iC2 p4) ((2)) After the Lorentz transformation: p’ = g(v)p + g(v)vE and E’= g(v)vp + g(v)E ((2)) Then: E1’+E2’ = g (E1+E2) + gv (p1+p2) and E3’+E4’= g(E3+E4) + gv(p3+p4) ((3)) Thus: E1’+E2’= E3’+E4’ ((4)) A similar argument holds for: p1’+p2’ = p3’+p4’ ((5)) Even though energy and momentum values change under a Lorentz transformation, conservation of both is not lost. We suggest there is a problem with the use of ((1)). In previous notes, we are argued that for momentum one requires time reversal invariance of probability, i.e.
Under p→ -p and x→-x the probability should remain unchanged, which does not follow for exp(iC2 p). We thus proposed exp(ipx) as a possible solution. Although we think time reversal symmetry should hold, it might seem a little arbitrary to simply insert x without further arguments. Here we try to consider some possible ones. We first note that conservation of energy and momentum should apply to an interaction and classically this occurs at a given x and t which applies to all particles. For example, one might have particles which do not interact and are separated by a great distance. In such a case, there is no reason to write a product exp(iC1 E1)exp(iC1 E2), but there is nothing stopping one from doing this mathematically and equating it to another set of values even though no interaction occurs between E1 and E2. In other words, one wishes to enforce conservation of momentum and energy for a two body elastic collision and so it seems that one must introduce x and t into the probabilities ((1)), so that conservation only occurs if all particles have the same t and x. As a first step, one might do this in the following manner: exp(i E (f(x)+g(t)) exp(i p (f(x)+g(t)) ((6)) Here f(x) and g(t) are unknown functions. If one takes the notion of spacetime seriously, one would set f(x)=g(ct). Now, x and t are measured from zero points. One may have x1 and x2 and t1 and t2 and also x1+x2 and t1+t2. We then suggest that: exp(i E (f(x1)+g(t1)) means that one has a particle with E and p which has moved from x=0, t=0 to x1 and t1. For x2,t2 one has a similar interpretation. Mathematically, one may multiply these two probabilities together. Given that the position of origins is arbitrary, one might consider that x2 and t2 are measured from x1 and t1 instead of from x=0, t=0. In such a case, one might suggest that: exp(i E (f(x1)+g(t1)) exp(i E (f(x2)+g(t2)) = exp(iE (f(x1+x2)+ g(t1+t2)) ((7)) This suggests a simple solution of: exp(iE (x+ct)) ((8)) and an identical result for the p case. At this point, one may realize that instead of dealing with p and E, ((8)) deals with p,E,x and t, i.e. the entire classical state description of a free particle. By focusing on energy and momentum conserving interactions, one must introduce space and time to describe the point of interaction and in doing so, the probability becomes a state one. As a result, probability is now a property of the state, like particle number, and we argue that it should be invariant under a Lorentz transformation. This conclusion could have been reached without the considerations of ((6)) and ((7)) and one could enforce Lorentz invariance immediately, leading to the notion of a Lorentz invariant, but we consider ((6)) and ((7)) useful. Thus, we suggest that there exists a state probability: exp(-iEt+ip dot r) ((8))
Even though this is a complex valued probability, it is a state probability which should be conserved which is associated with it being used in AND (multiply) and OR (add) situations, particularly the latter. This has important consequences because we argue that one may apply ((8)) a priori in a one dimensional n1-n2 index of refraction reflection-refraction problem. One Dimensional Reflection-Refraction We have discussed one dimensional reflection-refraction at x=0 at an n1-n2 index of refraction junction from a quantum mechanical point of view. This involves writing: Aexp(ipx)+Bexp(-px) = Cexp(ip2 x) at x=0 ((8a)) and Ap exp(ipx) -Bp exp(-ipx) = Cp2 exp(ip2x) at x=0 ((8b)) These two equations solve the problem, but there may be some uneasiness about using exp(ipx) as a kind quantity as is done in ((8a)). Alternatively, one may think in terms of math continuity, but we argue that there must be an accompanying physical interpretation to ((8a)) and ((8b)). Traditionally, this problem is solved using continuity of the electric field of a photon (1), but in the literature (and in some of our notes), we have argued for the wavefunction of a photon as being given by: El + i B where El is the electric field and B, the magnetic ((9)) Thus, one is really dealing with a quantum mechanical situation and there should be an a priori reason for using ((8)) as a quantity and this, we argue, follows from it being a state probability. We argue that the same reasoning that justifies it being Lorentz invariant is linked to it acting like a “quantity” in ((8a)), even though its weight A turns out to be the square root of flux, which is not an obvious physical value. It is also the state nature of mixing E,p,x,t with the first two being constants and the latter two running variables, which yields a structure in x and t due to p and E, i.e. the wavelength and frequency. These ideas then lead to the consequences of free particle quantum mechanics. Conclusion In conclusion, we argue (as we have in previous notes) that probability exists in Newtonian deterministic physics, namely in the prediction of the outcomes of elastic two body scattering. Traditionally one accepts any solution which conserves momentum and energy. Thus, any such solution has the same weight or probability as another. One may try to quantify this idea by writing a probability expression which demonstrates this conservation, i.e. using exp(iC1 E) and exp(iC2 p). Here p is a positive or negative number as the x-axis is chosen to lie along the p vector and we consider one dimensional scattering for simplicity. One may note that under a Lorentz transformation, E and p change, but conservation of momentum and energy in product probabilities remains.
The problem with exp(iC1 E) and exp(iC2 p) is that they are introduced to describe an interaction. One wishes to have the two interacting particles at the same x and t. This might seem obvious, but it is not mathematically the case given exp(iC1 E) and exp(iC2 p). One may have two particles which are far apart and not interacting and multiply their probabilities and then argue for conservation which follows from the math even though no interaction has occurred. The math should enforce conservation only for the particle (2 incoming, 2 outgoing) being at the same x and t. This seems to imply that one must introduce x and t into the probabilities exp(iC1 E) and exp(iC2 p). We suggest doing this through exp(i E (f(x)+g(t)) and exp(ip (f(x)+g(t)). We provide arguments above to suggest that f(x)+g(t)= x+ct, but then point out that instead of only dealing with E and p, one now uses E,p,x,t in a probability function for a single particle. These four variables describe the entire state of the free particle and so we argue that one has a state probability. As a consequence this should be Lorentz invariant, i.e. one should use exp(-iEt+ip dot r). Even more importantly, this probability now may be used in OR (add) situations and behaves like a “quantity”. This we argue allows one to write a priori Aexp(ipx) + Bexp(-ipx) = Cexp(ip2x) at x=0 in an n1-n2 (Index of refraction) 1-D reflection-refraction problem. (One may also write an equation using p.) We suggest that given a bonafide state probability, one may use it in such a manner as one does any probability in classical probability theory. The consequences of E,p,x,t in the same formula are, however, large as E and p create units or cycles in time and space. The spatial one is called wavelength and given by hbar/p. Thus, it is the state probability feature, i.e. the requirement of all variables which describe the state (E,p,x,t) which ultimately lead to the quantum free particle description. References 1. https://en.wikipedia.org/wiki/Reflection_phase_change
