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Speculation on the Inter-relation between Special Relativity and Quantum Mechanics Francesco R. Ruggeri Hanwell, N.B. Dec. 8, 2025 Special relativity seems to be associated with linking physical relations as seen in one frame with those as seen in a frame moving at a speed relative to such a frame. A constant speed is often used for the moving frame. This suggests that classically there is a special frame for a particle with rest mass called a rest frame. In such a case, one simply introduces a parameter for rest mass and does not worry about the complicated motion and interactions which actually create the rest mass. It seems that even quantum mechanics, which considers a particle to be moving as one does not know its location, one uses the idea of a potential V(x) which in principle is associated with a very large rest mass, presumably at rest as no variables of such a rest mass appear in the Klein-Gordon or Dirac equations. It seems that one uses a preferred frame when solving relativistic quantum mechanical equations. Furthermore, given V(x) and a bound state with exp(-i E bound t), E applies to the overall bound system, not to a possible exp(ipx) constituent. The special relativistic invariant equation seems to exist at the system level, like with the rest mass and one may apparently calculate E in the special rest frame of the large mass and hence find a rest mass of the bound system, i..e large rest mass*cc + mo cc + E, mo is the rest mass of the single bound particle. As a result of the use of the special rest mass frame for the V(x) producing large rest mass, it seems that one does not have to worry about special relativistic constraints on the interaction with V(x), as only one frame is involved. In other words, a bound calculation is about finding a new rest mass for an overall new system. Then this system may be used in Lorentz boosts. One simplifies by working in the rest frame of the mass creating V(x). Perhaps surprisingly then, we note that exp(-iEt+ipx) for a free particle is Lorentz invariant and in fact must be Lorentz invariant for the following reason. Imagine that one wishes to create a probability for free particles which is linked with conservation of momentum. Any p must have the same real value weight because for free particles there is no bias for one p or the other. There is, however, a bias for a p1 and p2. It should have the same probability as a p3 and p4, if p1+p2 = p3+p4. (We use one dimension for simplicity.) As a result, exp(ip) seems to be a probability which respects these conditions. The problem is that one may have p=m1v1=m2v2 hence (E1,p) and (E2,p). With respect to momentum, it is irrelevant if one particle has E1 and the other E2, but not if one does a Lorentz boost. Then, in one case: p→ g(v) p + g(v) v E1 and in the second, p→ g(v) p + g(v)vE2. Here g(v) =1/sqrt(1-vv/cc). In other words, p is the same for each particle in one frame and not in the other which violates conservation of momentum because in frame 1 both particles can complete the same conservation of momentum equation, but in the second frame this is not the case, because each has a different momentum. Thus, exp(ip) cannot hold. One requires a linear form in p which is Lorentz invariant, i.e. exp(-iEt+ipx), we argue (and have done so in previous notes). We see now that momentum probability is linked with space in exp(ipx). The idea seems to be that for free colliding particles there is no preferred frame and the probabilities used must reflect this Lorentz invariance.
As a result, we argue that a free particle probability in E,p arises because of Lorentz invariance, but then apply it to a V(x) problem and no longer worry about Lorentz invariance at the interaction level. V(x) with no associated rest mass variables assumes a rest frame for the V(x) producing mass. A Lorentz boost which then invalidates exp(ip), would only apply to the total rest mass of the bound particle system which has no overall momentum. It seems that one does apply the Lorentz transformation to the interaction of exp(ipx)’s with V(x). One is only interested in the overall bound system as a single entity for special relativity. As a result, one may only impose momentum considerations as V(x) is a spatial function and one must link it with exp(ipx). In fact, one may write V(x) = Sum over k Vk exp(ikx) to associate it with momentum hits linked with Vk energies, which already moves away from the idea of special relativity as Vk and k are not linked as a 4-vector. The idea of exp(ip1x)exp(i k1x) = exp(ip2x)exp(ik2x), however, seems to still hold (as it does for free particles), linking these two values (i.e. Vk1 and Vk2) with (p1+k)(p1+k) exp(i (p1+k)x). Lorentz invariance for a particle with rest mass seems to be linked with the idea that one ultimately has a rest frame for the particle. One then establishes a relation between the rest mass in one frame and its new E and p as seen by another frame. One does not, however, need to impose Lorentz invariance conditions on the interaction which creates the rest mass it seems. For example, exp(-iEt+ipx) is the Lorentz form used to describe a free particle, but exp(ipx) is the only portion used to describe its interaction with V(x) in the rest frame of the large mass creating V(x). Furthermore, exp(-iEt+ipx) ensures conservation of energy and p for a free particle, but in an interaction case, that does not happen as the potential may absorb energy. Special Relativistic Equation and Quantum Mechanics Special relativity yields the Lorentz invariant for a free particle: -EE + p dot p cc = -momocccc ((1)) This becomes the Klein-Gordon equation through: E→ id/dt partial and p→ -i grad (hbar=1) ((2)). If a potential exists, one may use: E→ E-V(x). ((3)) The point here, however, is that having a single E for a bound state and the presence of V(x) with no variables of the rest mass, assumes that one is in a preferred frame of the large rest mass creating V(x). E is then binding energy and one has a system at rest with energy: M1cc +mocc + E, where M1 is the rest mass of the V(x) mass and mo, of the bound particle ((4)) The form ((3)) tries to have E-V(x) represent kinetic energy of a free particle at x, but one clearly has an interacting particle. As we note, special relativistic frames are linked with ((4)). E in ((3)) is an average E and is linked with a prms(x). These are not 4-vectors because they do not
represent a free particle. ((4)) represents the rest mass of a free particle, so it, with ptotal =0 represents a 4-vector. In other words, one begins with a Lorentz invariant equation ((1)) describing a free particle with rest mass mo. The transformation is linked to the special rest mass frame represented by momo cccc. A single particle bound system is again a rest system with an overall new mass given by ((4)) and one may associate this with a 4-vector. The particle interacting with V(x) is not really a free particle at all at any instant in the classical sense. In the quantum mechanical case, it is a superposition of part of the free particle probability, i.e. W(x) = Sum over p a(p)exp(ipx). As a result, we suggest that even in the use of quantum relativistic equations, one does not need to worry about an interacting particle being Lorentz invariant, i.e one may only exp(ipx) and not a full exp(-iEt+ipx) which is required for Lorentz invariant. For a bound situation, it is the overall bound system which is now the entity, not the interacting particle, it seems. To see this another way, consider a particle at rest at x=0 at t. A frame moving at constant -v would see: X’ = g(v)v t and t’ = g(v) t ((4)) The person in the moving frame sees the particle move, but that requires a measurement and a delta t and delta x. If the particle is accelerating, it is changing as this measurement is being done. One would require extremely tiny dt and dx values, but quantum mechanics places a bound on these sizes and so we argue that one must be careful with expecting Lorentz invariance for an interacting particle. This begs the question: Why does Lorentz invariance appear in exp(-iEt+ipx)? Lorentz Invariance in exp(-iEt+ipx) Consider creating a probability which is linked to conservation of momentum. In principle, for a free particle, any p value should have the same real value weight or probability. One is not favored over another. Such is not the case for AND combinations. For instance, if: P1+p2 = p3+p4 ((5)) (one dimension for simplicity) Then P(p1)P(p2) = P(p3)P(p4) not= P(p5)P(p6) if p1+p2 not= p5+p6 ((5)) If P(p)=1, however, ((5)) does not hold. Thus, we first propose: P(p) = exp(ip) ((6)) ((6)) is fine in one frame, but p=m1v1=m2v2, i.e (E1,p) and (E2,p). If one is interested in conservation of momentum, one does not care that E1 not= E2. If one Lorentz boosts, however, one has a problem. Either p with E1 and p with E2 satisfy a conservation of momentum equation in frame 1. In frame 2: P with E1 → g(v) p + gvE1 while p with E2 → g(v)p+gvE2 ((7))
In frame 2 both particles cannot satisfy the same conservation of momentum equation and this is unphysical. One must thus resort to a Lorentz invariant: exp(-iEt+ipx) ((8)) ((8)), however, imposes bounds on minimal dt and dx for a given E and p, i.e. dt =hbar/E and dx = hbar/p ((9)) This is a major physical issue, because a particle’s position is uncertain in x. In Newtonian mechanics, one may consider a tiny dx and argue that one has v(x) in such a tiny dx. One may then perform a Lorentz boost. One cannot do that with exp(-iEt+ipx). There is no tiny dx, unless hbar/p is tiny compared with the length of the system which is not the quantum case. Thus, for a V(x) one does not follow the particle in a trajectory and does not need to impose a Lorentz boost on the interaction. One may choose V(x) as representing a rest frame and use exp(ipx). For free particles, however, colliding, one needs ((8)) which is linked with conservation of energy and momentum (i.e. elastic collisions). Electromagnetism One may note that in electromagnetism, an electrostatic potential is the 4-th component of a 4-vector. Thus, one may solve the quantum problem with this potential, i.e. in the preferred rest frame of the large mass creating the potential as is done in the literature, even in the Dirac equation approach which yields more details. Nevertheless, one has a rest frame. One may Lorentz boost the entire system, but then the electrostatic potential transforms to a new electrostatic one and a magnetic vector potential and both are functions of x-vt, i.e one no longer has V(x), but V(x-vt). Similarly, if the mass creating V(x) is moving, V(x)-->V(x,t). Conclusion In conclusion, we argue that if one wishes to introduce a probability which is linked with conservation of momentum, then exp(ip) (one dimension) works in one frame, but one may have p=m1v1=m2v2. Thus, one has E1,p and E2, p and both p’s (which are the same) satisfy a conservation of momentum equation. In a Lorentz boosted frame, p,E1 → p =gp+gvE1 and p,E2 → pg + pgE2. The new momenta are not the same and cannot satisfy the same conservation equation which is a physical issue. Thus, one requires a Lorentz invariant probability exp(-iEt+ipx) linked with both energy and momentum conservation. This is fine for free particles which are an entity by themselves. There is no preferred frame for free particles which collide and so one must have a probability scheme which holds in all frames. An interacting particle with V(x) implies a rest frame as there are no variables linked with the large rest mass creating V(x). There is a preferred frame in this case because the bound state is creating a new overall rest mass. This is very different from colliding particles.
In such a case, the interacting exp(ipx)s are not free particles and one has constant E on average. One may only use exp(ipx) and not really worry about internal Lorentz invariance as all one requires is that one finds an E such that M1cc+mocc+E is the rest mass of the overall system (and part of a 4-vector). exp(ipx)s are used because they are associated with physical impulse hits in the free particle case and this must carry over to interactions with a potential. hbar/p does not simply turn off because a potential is present. There is no need to worry about exp(-Et+ipx) for the interaction of the single particle with V(x). exp(ipx)s suffice and one does not consider different frames. The rest mass is the rest mass, just as in special relativity. For instance, a particle at rest may have a rest mass created by complicated interactions. This same rest mass is used in the moving frame, but the overall energy changes to mocc/sqrt(1-vv/cc). A person in a moving frame does not perform a calculation of the constituent interaction creating mo. They simply take mo and link it with 1/sqrt(1-vv/cc) which is associated with movement of the entire mo.
