In a previous note (1), we argued that even though photons and particles with rest mass both have the same quantum free particle probability (wavefunction) exp(-iEt+ipx), they have different spins, spin 1 (photon) and spin ½ (fermion like nucleons, electron). We argued that one may find equations for spin matrices by identifying an inherent force associated with the object. For a particle with rest mass, we argued that one chooses rest mass, mo, which is proportional to the force, inertia. For a photon, one chooses the electric field. The key is then to find an equation which is linear in the inherent force and contains p, E (for special relativistic effects and geometry). For a particle with rest mass, the Lorentz invariant -EE + cc p dot p = -momo cccc is linearized as Dirac showed. For a photon, one seeks an equation linear in the electric field. One may linearize d/dt partial (.5eo El dot El + .5/uo B dot B) + 1/uo grad dot (El x B) = 0, where El=electric field and B=magnetic field. For a particle with rest mass, one must retain the identity of the rest mass when forces interact. If one has two identical mo’s with the same center-of-mass, one must somehow distinguish between the two to deliver an impulse hit to only one. This may be done by using a magnetic field and having one particle have spin up, and the other, spin down. This leads to the Pauli exclusion principle. Using the Pauli exclusion principle in a two body elastic scattering problem, one may easily establish time reversal balance by: p(e1)p(e2) {1-p(e3)}{1-p(e4)} = p(e3)p(e4) {1-p(e1)}{1-p(e2)} ((1)). From this one may obtain the Fermi-Dirac distribution. The question becomes: What does one do for photons? We argue that photons do not interact in the same manner as two particles with rest mass which elastically scatter. Instead we consider photons interacting with electrons in atoms. We note that in the case of particles with rest mass, there is a grouping of two energies, i.e. e1,e2 on the LHS side of ((1)), and e3,e4, on the RHS. We keep this notion of grouping, but note that e1, and e2 are not being converted into e3, e4 photons. Rather, all four types of photons are present and may interact with the electrons in atoms. Thus, we consider the sum of the probabilities for the e1,e2 grouping, namely (e1,e2) + (e1,e2) e3 + (e1,e2) e4 + (e1,e2) e3, e4. We argue that the sum of probabilities for this set of groupings should equal the sum of probabilities for the grouping of (e3,e4), i.e. (e3,e4)+ (e3,e4) e1 + (e3,e4) e2 + (e3,e4) e1,e2. This is equivalent to: p(e1)p(e2) {1+p(e3)}{1+p(e4)} = p(e3)p(e4) {1+p(e1)}{1+p(e2)}. From this, the Bose-Einstein distribution follows.
In a previous note (1), we argued that even though photons and particles with rest mass both have the same quantum free particle probability (wavefunction) exp(-iEt+ipx), they have different spins, spin 1 (photon) and spin ½ (fermion like nucleons, electron). We argued that one may find equations for spin matrices by identifying an inherent force associated with the object. For a particle with rest mass, we argued that one chooses rest mass, mo, which is proportional to the force, inertia. For a photon, one chooses the electric field. The key is then to find an equation which is linear in the inherent force and contains p, E (for special relativistic effects and geometry). For a particle with rest mass, the Lorentz invariant -EE + cc p dot p = -momo cccc is linearized as Dirac showed. For a photon, one seeks an equation linear in the electric field. One may linearize d/dt partial (.5eo El dot El + .5/uo B dot B) + 1/uo grad dot (El x B) = 0, where El=electric field and B=magnetic field. For a particle with rest mass, one must retain the identity of the rest mass when forces interact. If one has two identical mo’s with the same center-of-mass, one must somehow distinguish between the two to deliver an impulse hit to only one. This may be done by using a magnetic field and having one particle have spin up, and the other, spin down. This leads to the Pauli exclusion principle. Using the Pauli exclusion principle in a two body elastic scattering problem, one may easily establish time reversal balance by: p(e1)p(e2) {1-p(e3)}{1-p(e4)} = p(e3)p(e4) {1-p(e1)}{1-p(e2)} ((1)). From this one may obtain the Fermi-Dirac distribution. The question becomes: What does one do for photons? We argue that photons do not interact in the same manner as two particles with rest mass which elastically scatter. Instead we consider photons interacting with electrons in atoms. We note that in the case of particles with rest mass, there is a grouping of two energies, i.e. e1,e2 on the LHS side of ((1)), and e3,e4, on the RHS. We keep this notion of grouping, but note that e1, and e2 are not being converted into e3, e4 photons. Rather, all four types of photons are present and may interact with the electrons in atoms. Thus, we consider the sum of the probabilities for the e1,e2 grouping, namely (e1,e2) + (e1,e2) e3 + (e1,e2) e4 + (e1,e2) e3, e4. We argue that the sum of probabilities for this set of groupings should equal the sum of probabilities for the grouping of (e3,e4), i.e. (e3,e4)+ (e3,e4) e1 + (e3,e4) e2 + (e3,e4) e1,e2. This is equivalent to: p(e1)p(e2) {1+p(e3)}{1+p(e4)} = p(e3)p(e4) {1+p(e1)}{1+p(e2)}. From this, the Bose-Einstein distribution follows.
We have argued in previous notes that special relativity is often developed from the point of view of a particle with rest mass, leading to the Lorentz invariant -EE + cc p dot p = -momo cccc ((1)). Then, as a second step, one may obtain the equation E= [p] c by setting mo=0. This appears to be a straightforward math step to obtain a photon relevant result, but we have suggested in (1) that mo represents an inherent force, called inertia (Newton). Thus, special relativity for a particle with rest mass is really linked to this inherent force. In fact, we have suggested in (1), that spin ½ is obtained by finding an equation linear in a particle’s inherent force (rest mass) which is linked to E and p from special relativity and geometry. This, we suggest, is why Dirac linearized ((1)) to find spin ½. The reason for linearization is to have an equation linear in the inherent force, because force appears in F=dp/dt in a linear fashion. The rest mass case seems to be largely based on mo, inertia, which we call the inherent force. We further argued in previous notes that free particle quantum probability arises from -Et+px = Lorentz invariant ((2)) because x,t and x+hbar/p, t+hbar/E give the same result and we consider -Et+px to act somewhat like a free particle potential. This suggests that a photon should also have a probability exp(-iEt+ipx), which is verified as both a photon and particle with rest mass, with the same p, give rise to the same 2-slit interference. The problem in the photon case is that mo=0 and so no inherent force appears in special relativity. We argue that physically, there must be an inherent force as interactions (as in the 2-slit case) depend on it. In (1) we argued that the inherent force of a photon is the electric field, but this does not appear in ((1)) or ((2)), but emerges from Maxwell’s electromagnetic equations. This is why the photon has spin 1 and not spin ½ even though it has the probability exp(-iEt+ipx). (See (1) for details on the specific equation which gives rise to spin 1.) Here, we argue that for a particle with spin ½ (i.e. rest mass), exp(-iEt+ipx), which is a probability which conserves energy and momentum, the momentum and energy are proportional to rest mass (inertia or inherent force) and so exp(ipx) is really a probability distribution for inherent force (augmented to E,p due to the Lorentz boost). In the case of the photon, one still has exp(ipx), but the force is due to the electric field. This begs the question: What is the spatial distribution of the electric field? Is it free to be any function? We argue that given that exp(-iEt+ipx) is the energy/momentum conserving probability from special relativity, the inherent force cannot contradict this distribution. The electric field, which is a real valued function, must be cos(-Et+px). We note that one may formally obtain a classical wave equation from Maxwell’s EM equations to find El proportional to cos(-Et+px), but this does not explain the complex probability exp(-iEt+ipx) which holds for both a photon and particle with rest mass. Our objective is to argue that given exp(-iEt+ipx) for a photon, the electric field must behave as cos(-Et+px). Interestingly, the photon momentum density 1/uo Electric field x Magnetic field is proportional to cos(-Et+px) * cos(-Et+px). This does not follow Real(exp(-iEt+ipx), but it is the electric field and not momentum in the photon case which is linked to force. We argue that ultimately it is force which governs interactions, even though momentum (not momentum density) is conserved.
In previous notes, we presented a simple argument for obtaining the quantum free particle wavefunction exp(-iEt+ipx). We suggested that there are situations for which one cannot predict the outcome of an elastic collision and that given initial e1,e2 (energies) and p1,p2 (momentum vectors in 1-direction for simplicity), any ei,ej and pi,pj which conserve energy and momentum have the same product probability. Even in Newtonian elastic collisions of free particles, one sees calculations which allow any outcome which conserves momentum and energy. If the probability has the same real value for any free particle as no variable is being distributed and one wishes to impose Lorentz invariance, then exp(-iEt) and exp(ipx) represent the probabilities. This automatically gives rise to units in time and space, hbar/E and hbar/p (the hbar comes from considerations linked to a photon). One may ask, however: What is the real physics behind dx and dt? We argued in (1) that this physics is associated with force, in particular an intrinsic force. We considered a particle with rest mass mo and argued that this rest mass represents intrinsic force as it is proportional to Newton’s inertia and the force needed to set it in motion. If viewed from a moving frame, (0, mocc) maps to (E,p) with E containing the intrinsic inertial force as well as enhancements due to motion and p containing the intrinsic mo as well as the velocity vector. For a particle with rest mass, we suggest that a key idea is this notion of intrinsic force. We argued that one may consider the Lorentz invariant -Et+px = -mocc t1 and introduce dx=hbar/E, dt=hbar/p, dt1=hbar/(mocc) and dx1-> infinite as p->0. This leads to -E dt + p dx =0 or -E/dx + p /dt = 0. We then argued in (1) that p/dt is like an average force and from this we introduced exp(-iEt+ipx). Thus, dt and dx are linked with the notion of an average force. We suggest that this notion of intrinsic force is the reason behind linearizing -EE + p dot p = -momo cccc ((1)). -Edt+pdx=0 is too simple to show the geometry linked with rotational invariance etc in ((1)), yet at the same time ((1)) is not linear in the intrinsic force mo. This is the physical reason why ((1)) should be linearized via Dirac’s process, giving rise to spin ½. As a result, we postulate that a particle has an intrinsic force, in this case rest mass, which is directly linked with the probability exp(-iEt+ipx) and that one should be able to write an equation containing operators and the intrinsic force (in linear form) which also shows geometry. It is this equation which gives rise to spin. At this point, this is a postulate, i.e. speculation. It would be good if one could find an example in which these ideas arise naturally, i.e. a case in which one did not have to linearize an equation like ((1)) or make special arguments in order to obtain exp(-iEt+ipx). In that case, an equation linear in the intrinsic force would exist a priori and a direct link to exp(-iEt+ipx) would also naturally exist. We suggest that such an example does exist, namely for a photon. One may use the special relativistic equations developed for a particle with rest mass and set mo=0 to obtain E=[p] c for a photon. This is a fine equation, but unlike the rest mass case mo (proportional to the intrinsic force), there is no rest mass and no intrinsic force present for the photon. Certainly p (momentum) is linked with a force as both a particle with rest mass and a photon impart an impulse on a target, but momentum is not an intrinsic force, it is a consequential one. In the case of a rest mass, it arises because the rest mass (the intrinsic force) is seen to be moving. For a photon, one would then expect the existence of an intrinsic force which gives rise to momentum. We suggest that the electric field (together with the magnetic) represents this intrinsic force and that Maxwell’s equations allow one to directly write a linear equation in El (electric field) (no linearization required) which shows the geometry in the system. One may also explicitly show that El behaves as exp(-iEt+ipx) if one extends cos(-Et+ipx) to a more general complex number. In other words, the postulates made for the particle with rest mass mo seem to naturally arise in the case of the photon. This, we argue, is why the procedures for finding spin for a spin ½ particle with rest mass mo and a photon are so different even though both a photon and rest mass particle satisfy special relativity and are associated with exp(-iEt+ipx). It is the difference in the intrinsic force which is the key to finding spin, we speculate.
In previous notes, we argued that one could describe elastic two particle collisions, even in Newtonian mechanics, in a classical statistical manner. In particular, we argued that one could introduce a probability which depends on E,p, but at the same time has the same weight (in a sense) for any free particle. Product probabilities would be linked with a uniform distribution in the sense that exp(-iE1)exp(-iE2)= exp(-iE3)exp(-iE4) if E1+E2=E3+E4 and a similar expression for momentum. If one respects the special relativistic invariant -Et+px, then a possible solution is exp(-iEt+ipx), but even though this yields the free particle wavefunction (probability), this does not explain quantum mechanics. These are simply ideas of classical probability theory with the enforced special relativistic invariant -Et+px. Here, we argue that the real feature leading to quantum mechanics (the idea of hbar/p and hbar/E) arises from the notion of inertia (which we have argued in previous notes also gives rise to spin). We argue that a single force (inertia) in a rest frame becomes two independent interaction notions, E and p as well as creating x’ as a running variable. Thus, mo,t and E,t’ and p,x’ are intertwined. We suggest that there must be a physical mechanism which keeps these two concepts E,p independent and so one cannot simply rely on x=vt. Given the Lorentz invariant and the classical notion of probability exp(-iEt+ipx), one must go one step further and postulate that there are independent fluctuations in t’ and x’ in such a calculation such that hbar/E and hbar/p emerge. We argue that this is a key feature of quantum mechanics and that it does not immediately arise from exp(-iEt+ipx) if one only considers x=vt. One is forced to allow x,t to fluctuate independently which leads to a fluctuation in mocc in the rest frame as well. We argue that this feature is actually associated with special relativity as it allows a single force inertia to manifest itself as two independent interaction quantities E,p. As a result, we suggest that one may obtain exp(-iEt+ipx) without any notion of probability, but only by postulating the existence of a physical mechanism (fluctuation of x’,t’ or t in the rest frame) which keeps E,p independent because -Et+px does not ensure their independence if one only considers x=vt. One may then interpret exp(-iEt+ipx) as having the form of a classical probability which respects special relativity.
In Part 1, we tried to understand why spin arises. The basic conclusion was that it is linked with an intrinsic force which in turn is associated with E,p (energy momentum) and geometry (rotational symmetry and other symmetries). In practical terms, this means finding an equation linear in intrinsic force (as force appears linearly in nature), but still tied to E,p and geometry. Interestingly, this leads to two very different results for particles with rest mass (spin ½) and photons, although both satisfy the same special relativistic equations (albeit the photon with rest mass mo=0) and both have exp(-iEt+ipx) free wavefunctions. In the photon case, the Electric field El is the intrinsic force (together with the magnetic B) which gives rise to both energy and momentum density. Specifically, energy density= .5eo El dot El + .5/uo B dot B and momentum density = 1/uo El x B. In the photon case, however, there is no rest frame and so the classical vector El is directly associated with E (energy) and p (momentum) without one needing to consider special relativity. Instead one directly uses Maxwell’s electromagnetic equations in order to find a linear equation describing the classical vector El. This equation involves d/dt partial and grad partial which are linked to the quantum operators for E and p (id/dt, -igrad) and El itself is proportional on exp(-iEt+ipx) (actually cos(-wt+kx)), also from Maxwell’s equations. Given this scenario, one has a classical vector El which is perpendicular to B and both lie in a plane perpendicular to motion. For circular polarization, El moves in a circle and given that it gives rise to energy, this energy circulates like classical angular momentum. For the 3-vector for El, one says that one has spin 1, but for circular polarized light, this is classical angular momentum. As a result, spin in the photon case follows from the behaviour of the intrinsic force (a linear equation in this force as shown in Part 1), but one must also have a clear link between the intrinsic force and the consequential energy and force, because it is the energy which actually has to rotate. It is not enough to have El rotate. In the case of a particle with rest mass, we argued in Part 1 that mo (Newton’s inertia) is the intrinsic force which together with velocity gives rise to the consequential energy E and momentum p. The situation is that one actually has two physical frames, the rest frame with the intrinsic force mo and the moving frame in which one has E,p. In order to link the intrinsic force mo to E, p, one must use a Lorentz invariant. We suggested in Part 1 that -Et+px = -mocc t1 is associated with -E dt + p dx = -mocc dt1 + (p=0) * (dx1->inifite) =0. Then dx=hbar/p, dt=hbar/E and E/dx and p/dt are average forces as is mocc/dt1. Tthe idea of force is very much the cornerstone of exp(-iEt+ipx) we argue. In order to see spin arise, one needs an equation linear in mo (the intrinsic force), but linked to E,p and showing geometrical invariances, according to the recipe of Part 1. -EE + cc p dot p = -momocc ((1)) links E,p, mo and shows invariances, but is not linear in mo. One must then linearize it. This means that one compares to Lorentz frames, the rest frame containing mo and the moving containing E,p. Both mo and p are linked to forces, but mo is the intrinsic force and p, the consequential. As a result, they must somehow be separated as they are in ((1)), in particular they are perpendicular if one considers: EE = pz pz + momo (c=1, p->px). In order to linearize without using unit vectors, one cannot use numbers, but must use matrices (as Dirac showed). The operator linked with pz must anticommute with that linked with mo. The operators associated with the consequential and intrinsic forces differ and in fact anticommute in the rest mass case. One may show that one requires at least a 2x2 matrix (the full Dirac treatment uses a 4-vector made of 2 spinors), which means that there is a two vector state. This is presumably associated with motion, if exp(ipz z) is linked to motion as well. In order to analyze this 2 vector state (spinor), we first compare the situation with the photon case. There, one had a 3-vector El which means that one knows an exact position in the plane. The second piece of information must be the direction in which this vector is moving in a circle (for circularly polarized light). There are two a priori given pieces of information which together with the link between El and energy density yields classical angular motion. In the rest mass case, one again expects rotational motion in a plane perpendicular to the z axis (because pz describes all motion along z). (1,0) and (0,1), however, describe classical rotational direction, clockwise and counterclockwise, but this spin ½ case cannot be classical rotation because the physical vector in the plane is not given (as it is by El the intrinsic force). Thus, one has only the direction of rotation. We argue that if exp(ipx) means that one has p hits at different x within hbar/p with probability exp(ipx), one may imagine different x,y positions of energy in the plane (with probabilities) as long as the rotation for (1,0) for each is in the clockwise direction (for example). This means that a physical vector may appear at some x,y in the plane and then at an x1,t1 which would be considered a classical time in the past as long as the motion is clockwise. In this way, one may have an angle of 720 degrees (instead of 360 degrees) describe the motion. In terms of interaction with a B field, one has the clear two directions (spin up clockwise and spin down counterclockwise as examples). Thus, spin must exist in the rest mass case, because one must have mo perpendicular to p if special relativity is to hold. If one does not want to have spin, then one must disregard the equivalence of frames moving at constant speeds with respect to each other and that seems unphysical. (Note: There are spin 0 objects (Higgs particle), but we do not consider that here.
We have pointed out in previous notes that the free particle wavefunction exp(-iEt+ipx) holds for both a photon and particle with rest mass, while the latter has spin ½ and the former, spin 1. We thus suggested that spin is obtained in a very different manner from exp(-iEt+ipx). Here, we start with the Lorentz invariant -EE+ p dot p cc = -momocc ((1)) and argue, as in (1), that it must be linearized to obtain a linear equation in the intrinsic force, namely mo which is linked to inertia, because force acts in a linear form in nature (e.g. F=dp/dt). At this point, however, p is a vector and there is no concept of -ihbar grad. One may mathematically linearize ((1)) using the approach of Dirac (except that now p is a vector of numbers not an operator). This leads to 4x4 matrices with off diagonal Pauli matrices (multiplied by 1 and -1). One may see that the rotational invariant p dot p may be linearized to s dot p (s = three 2x2 Pauli matrices) because these matrices anticommute. Thus, a 2x2 Pauli matrix is associated with rotational invariance. At this point, this Pauli matrix is simply a math entity. In terms of measurement, given that one has matrices, there are also commutation relations which are important. One may consider another math operator linked with rotational invariance, namely d/dphi for a rotation about a z-axis. In classical physics, one does not have such a derivative appearing as one deals with functions. If d/dphi is an operator generating rotations, then d/dx should generate translations, but this is simply math. The “catch” seems to be that if one introduces a probability f(x,p) (which is unusual because P(x)dx = dx/L, where L is an arbitrary length), then f(p1,x)f(p2,x)=f(p3,x)f(p4,x) if p1+p2=p3+p4 (we use 1-dimension for simplicity). One may, however, shift this equation to x+dx which implies that df(p1)/dx f(p2) + f(p1)df(p2)/dx = df(p3)/dx f(p4) +f(p3) df(p4)/dx. A solution which keeps p present is: exp(i C p x). We use i because we want to have equal modulus at all x points. If C=1, the one has exp(i px) with -id/dx as the operator which retrieves p. One may then apply this idea to L = r x p and find that this new operator commutes like the Pauli matrices and gives rise to d/dphi, the rotational operator postulated above. Thus, it seems that one could start with the notion of spin and obtain the notion of exp(-iEt+ipx), we argue and spin being linked to angular momentum, we argue.
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.