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 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.
We have argued in Parts 1 and 2, that the free particle quantum form exp(-iEt+ipx), which we call a complex probability, arises from special relativity in order to allow for conservation of energy and momentum (in an elastic collision). One may see from single particle bound states how exp(-iEt) plays the part of a binding energy. We have noted in Parts 1 and 2, that exp(-iEt+ipx) follows from the Lorentz invariant -Et+px, but that one does not consider the deterministic x=vt equation which applies to the center-of-mass, but rather treats t and x as independent variables which force the independence of E and p (two different interaction variables) and also force separate E and p conservation. The question becomes: How is it possible to have x,t play two such different roles in special relativity? Here we try to answer this question. We argue that special relativity is based on determinism, i.e. x=vt. One may create a Lorentz transformation for a particle with rest mass mo at x=0 with t, a running variable. As seen from a frame moving with -v, one has x’=vt’. In other words, a separate x’,t’ is associated with each v. In Newtonian mechanics, it is possible to have mov1 and mov2 in the same frame (i.e. a frame with a ruler and clock), but in special relativity, each mo in the rest frame maps to a different x1’,t’ and x2’,t2’. This, however, seems to lead to a contradiction. If one considers two particles p1,E1 and p2,E2 (one dimensional motion) which elastically collide, then before the collision, they may be thought of as a composite particle (for more than a single x,t point) with energy E1+E2 and momentum p1+p2. Given the validity of special relativity, we argue that one may associate a Lorentz invariant -(E1+E2)t + (p1+p2) x with this composite particle. Furthermore, any outcome Ei+Ej = E1+E2 and pi+pj = p1+p2 would satisfy this same invariant. At this point there is no contradiction. The contradiction arises if one considers a single t and x for a composite particle, while internally there exist t1’,x1’ and t2’, x2’. The composite particle, in terms of special relativity, is not compatible with each -E’t’+p’x. The energies and momenta, however, add for each particle (in conservation equations) and so we argue that one should have a relativistically compatible approach which allows one to link each particle to a composite particle and see conservation of E and p. We note that it is precisely x=vt for each particle which leads to x1’,t1’ and x2’,t2’ and so argue that if one gives up this notion of determinism for x and t and introduces independent x, t values (fluctuations), then one can indeed retain the -Et+px form for each particle and use it to create composite particles. This seems to give rise to an AND situation which does not exist for both p,x and E,t in the deterministic relativistic approach. In that approach, one can only think of p’s and E’s adding separately, without any linking of x’s and t’s. In the fluctuation approach, it is possible to have the same x and t fluctuation points for each particle. We argue that it is this kind of thinking which leads to the probabilistic form exp(-iEt+ipx) which still respects the Lorentz invariant form and at the same time allows for conservation of E,p with the same t,x fluctuation points allowed. x=vt does not allow one to combine x1’,t1’ and x2’,t2’ for a composite particle because these may coincide for only one x,t point and one wishes to describe a composite particle with an x,t which exists over more than one x,t. This is why we suggest the physical fluctuation scenario which exists together with x=vt. x=vt describes the deterministic motion of the center-of-mass, while exp(-iEt+ipx) describes probabilistic hits in x and energy in t. It is special relativity which forces one to associate time fluctuations with E and spatial ones with p. As a result, we argue that special relativity gives rise to both probabilistic and deterministic versions, both of which respect the form -Et+px. The deterministic version is linked with x=vt center-of-mass motion, while exp(-iEt+ipx) represents probabilistic appearances of E in t and p in x, i.e. there is a spatial probability for impulse hit delivery.
The classical action (relativistic or nonrelativistic) may be written with v=x/t and x and t varied independently to yield dA/dx partial = p and dA/dt partial = -E. Thus A = -Et+px is a solution. This expression also appears in classical wave forms and in the electromagnetic description of a photon (from Maxwell’s laws). -Et+px is a Lorentz invariant and so is directly linked to special relativity and different frames moving with constant velocity. In (1) we argued that x and t should be treated as independent in A. This suggests that p is a scaling in space and E in time. Thus there should be physical lengths proportional to 1/p and physical time intervals proportional to 1/E. Special relativity considers different frames (moving with constant velocity) and 4-vectors so it seems to be quite different from a scenario with t and x separated and scaling of x by p. We argue that although -Et+px is a Lorentz invariant and as such may be used to find E(v) and p(v), quantum mechanics is usually performed in a special frame, namey one in which V(x) or V(x,v) the potential is at rest. (For electromagnetism one treats (A, phi) as a 4-vector, but this is not the norm.) As a result, energy En found in a rest frame of the potential also represents “rest energy” so En is binding energy which is added to rest mass energy. In other words motion in the rest frame of a potential creates rest energy = rest mass. We argue that V(x) may be written as a sum over k Vk exp(ikx) and so no time related terms appear, only momentum linked ones. Thus momentum is the key feature for the interaction, even though velocity and time are still in the picture (just sometimes not directly in the interaction scheme). For example, a two slit interference or single slit diffraction does not depend on time or speed although these are certainly in the problem. For a bound state v and t are also in the problem as one calculates average kinetic energy using pp/2m = .5 p v which is rate dependent, but there is no acceleration which is a time-dependent phenomenon. Rather one deals with an ensemble of exp(ipx)’s because these interact with the potential in a wavelike manner to create the effect of acceleration. exp(ipx) a momentum wave linked to the scaling of x by p in px is not only involved in an interaction with V(x), but represents extra information from the classical mechanical picture. exp(ipx) for a known rest mass mo means one knows v=x/t=dx/dt, yet there is an extra piece of information namely the phase px which is involved in even “free” interactions such as two idendical particle with different phases arriving at the same x at the same t or the average kinetic energy in a bound state. Velocity v is converted into a momentum expression through p/mo in the norelativistic case and so exp(ipx) governs the whole scenario. Acceleration then becomes a mathematical abstraction of the motion of vrms(x) for any quantum bound energy level with the real physics associated with exp(ipx) being ignored.
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 the literature (1), the Schrodinger equation: {p1p1/2m + p2p2/2m + V(x1-x2)} W(x1,x2,t) = id/dt partial W(x1,x2,t) may be transformed using P=p1+p2 and (p=m1p2-m2p1)/M and x=x1=x2 where M=m1+m2 ((1)). This leads to a bound Schrodinger equation in x and a “center of mass” equation” in P i..e Hcm =PP/2M which depends on m1 and m2, but not binding energy. . In a previous note, we argued that a free particle’s classical action A (relativistic or nonrelativistic) may be obtained by setting v=x/t and treating x and t as independent (2). This leads to dA/dx partial=p and dA/dt partial = -E. One may write eigenvalue equations: -id/dx partial exp(iA) = pexp(iA) and id/dt partial exp(iA) = E exp(iA). A solution is A=-Et+px with E,t,p and x independent. -Et+px is a Lorentz invariant and this form is used in both relativistic and nonrelativistic quantum mechanics (although E and p have different values). We argue that a free particle rest mass mo may actually involve internal quantum physics, yet the “system” is replaced with a parameter mo and exp(-iEt+px) used to describe it. One may try to apply this approach to a bound system, but exp(-iEt+px) is a statistical probability like object. Thus one needs a statistical theory. We have suggested using an ensemble of exp(-iEt+px) with E=En for all p to indicate a resonance. Then: { Sum over p pp/2m a(p)exp(ipx) } / {Sum over p a(p)exp(ipx) } + V(x) = En ((2)) This is the Schrodinger bound equation. We argue that W=Sum over p a(p)exp(ipx) should be real and this implies that Sum over p a(p) p exp(ipx) = 0 i.e. average momentum for this type of average is 0. As a result, one may create a 4-vector using { 0, mcc + En} and use the Lorentz boost approach to find the momentum and energy in a moving frame. One may then take the nonrelativistic limit to find kinetic energy i.e. the full center of mass kinetic energy. We argue that this is equivalent to considering exp(-iEt+ipx) for the overall bound system which is now translating in space. We argue that ((2)) is a different approach to ((1)) in that binding energy, which may be significant with respect to mcc, should be part of the overall energy of the translating bound system which is called the “center of mass” system in the literature.
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.
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.