Physics is based on experimental observables. Some of these observables are functions of other observables. For example, mass is an observable associated with inertia or gravity and velocity is classically measured, but the product mov yields nonrelativistic momentum which may be measured independently through an impulse hit with a target. A further issue with measurement is the case of vectors. For a vector pointing in one direction, one may orient the z axis to lie along this direction. If one always does this, momentum appears as a scalar, but one has to continuously rotate the co-ordinate system to account for different orientations. In this note, we are interested in spin which is linked to a kind of rotation and so we focus on the momentum vector, but only consider a function of momentum which yields a linear E, free particle energy. Thus, for the Lorentz invariant: -EE + p dot p = -momo, we consider the nonrelativistic limit: E = p dot p /2mo. In quantum mechanics, a free particle is represented by the complex probability (wavefunction) exp(ip dot r).This function has physical implications as it manifests itself in n-slit interference experiments so it is physical (not a mathematical convenience), but at the same time is responsible for conservation of momentum in x,y,z because Integral dx exp(ip1x) exp(ip2x) = 0 if p1 is not equal to p2. A physical function (property) through orthogonality creates an important relationship, namely conservation of momentum. We suggest that a similar idea may be at work in a relationship linking a linear E (free particle energy) to a function of momentum. We start with a linear E related to a function of p and only focus on the p vector because if it is treated as a number, the co-ordinate axis must rotate to account for its orientation. We only consider a situation in which p(x), p(y) and p(z) behave in an independent manner in the function of p (i.e. in the same manner as p itself) and suggest that instead of using vector analysis, there should exist a physical co-ordinate axis for each projection p(x), p(y) and p(z) if they are independent. We argue that p dot p = p(x)p(x) + p(y)p(y)+p(z)p(z) does not contain cross terms such as p(x)p(y). This is a kind of orthogonality which we associate with a rotational kind of operator linked with the p(i) operator -id/dx(i). In other words, we seek: p vector = generalized number = p(x)sx+p(y)sy+p(z)sz, where s(i) are generalized numbers or matrices, so that when this multiplied by itself, it yields p dot p. It turns out that the s(i) are the 2x2 Pauli matrices. In the case of a photon, there is no nonrelativistic limit and a linear E equation related to p is: E= |p|c. This equation is not independent in p(x), p(y), p(z) i.e. one cannot write p(x)sx+p(y)sy+p(z)sz an obtain E = {p(x)sx+p(y)sy+p(z)sz} c. Thus we look for another equation containing a linear E and a function of p. In this case, we use energy and momentum densities as given by the continuity equation: d/dt partial energy density + d grad dot ElxB =0 where d is a constant and El and B electric and magnetic 3-vectors. As in the above analysis, one wishes to have a momentum operator linked with a kind of rotational operator which allows one to eliminate vector math and use generalized numbers. In such a case, grad in the continuity equation becomes the momentum vector and eijk (Levi-Civita symbol in the cross product) becomes the spin matrix. Thus in both cases, spin governs the form of the function containing momentum which yields a linear E. In other words, it is not momentum alone which yields the relationship between a linear E (free particle energy) and momentum.
Physics is based on experimental observables. Some of these observables are functions of other observables. For example, mass is an observable associated with inertia or gravity and velocity is classically measured, but the product mov yields nonrelativistic momentum which may be measured independently through an impulse hit with a target. A further issue with measurement is the case of vectors. For a vector pointing in one direction, one may orient the z axis to lie along this direction. If one always does this, momentum appears as a scalar, but one has to continuously rotate the co-ordinate system to account for different orientations. In this note, we are interested in spin which is linked to a kind of rotation and so we focus on the momentum vector, but only consider a function of momentum which yields a linear E, free particle energy. Thus, for the Lorentz invariant: -EE + p dot p = -momo, we consider the nonrelativistic limit: E = p dot p /2mo. In quantum mechanics, a free particle is represented by the complex probability (wavefunction) exp(ip dot r).This function has physical implications as it manifests itself in n-slit interference experiments so it is physical (not a mathematical convenience), but at the same time is responsible for conservation of momentum in x,y,z because Integral dx exp(ip1x) exp(ip2x) = 0 if p1 is not equal to p2. A physical function (property) through orthogonality creates an important relationship, namely conservation of momentum. We suggest that a similar idea may be at work in a relationship linking a linear E (free particle energy) to a function of momentum. We start with a linear E related to a function of p and only focus on the p vector because if it is treated as a number, the co-ordinate axis must rotate to account for its orientation. We only consider a situation in which p(x), p(y) and p(z) behave in an independent manner in the function of p (i.e. in the same manner as p itself) and suggest that instead of using vector analysis, there should exist a physical co-ordinate axis for each projection p(x), p(y) and p(z) if they are independent. We argue that p dot p = p(x)p(x) + p(y)p(y)+p(z)p(z) does not contain cross terms such as p(x)p(y). This is a kind of orthogonality which we associate with a rotational kind of operator linked with the p(i) operator -id/dx(i). In other words, we seek: p vector = generalized number = p(x)sx+p(y)sy+p(z)sz, where s(i) are generalized numbers or matrices, so that when this multiplied by itself, it yields p dot p. It turns out that the s(i) are the 2x2 Pauli matrices. In the case of a photon, there is no nonrelativistic limit and a linear E equation related to p is: E= |p|c. This equation is not independent in p(x), p(y), p(z) i.e. one cannot write p(x)sx+p(y)sy+p(z)sz an obtain E = {p(x)sx+p(y)sy+p(z)sz} c. Thus we look for another equation containing a linear E and a function of p. In this case, we use energy and momentum densities as given by the continuity equation: d/dt partial energy density + d grad dot ElxB =0 where d is a constant and El and B electric and magnetic 3-vectors. As in the above analysis, one wishes to have a momentum operator linked with a kind of rotational operator which allows one to eliminate vector math and use generalized numbers. In such a case, grad in the continuity equation becomes the momentum vector and eijk (Levi-Civita symbol in the cross product) becomes the spin matrix. Thus in both cases, spin governs the form of the function containing momentum which yields a linear E. In other words, it is not momentum alone which yields the relationship between a linear E (free particle energy) and momentum.
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 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.
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.
In two previous notes (Parts 1 and 2), we argued thta spin matrices arise when one attempts to separate a quadratic equation (continuity equation with the Poynting vector for a photon, Klein-Gordon equation for a spin .5 particle) into two factors. These appear as generalized numbers (i.e. matrices) which may be multiplied together without considering vector operations such as the cross or dot product. Here we try to examine the problem in an alternative way. We note first that in the case of the Klein-Gordon equation one has a 4-vector containing (p vector, E) or (r vector, t) to cite two examples. These transform under the unitary Lorentz transformation, i.e. via a unitary 4x4 matrix. A subset, however, namely p(x), p(y),p(z) or x,y,z of this 4-vector transforms physically under rotations in 3-space, i.e. via a 3x3 unitary matrix. Furthermore, there is an invariance under this 3x3 matrix transformation. In particular, objects such as the dot product or cross product do not change. Classically, one has no issue with vector operators in an equation. In quantum mechanics, however, a free particle is described by the wavefunction exp(-iEt+ip dot r) and the operator i d/dt partial and -i grad return E and p (vector). exp(-iEt+ip dot r), however, describes a translational invariance in space and time, i.e. there is a periodicity. This invariance is clearly seen if the complex conjugate of the free particle wavefunction is multiplied by the wavefunction to yield 1 everywhere.Thus one starts with a quadratic equation W*W and linearizes it to obtain the function W(x,t). The operators which act on W are the generators related to the invariance in W*W. For the quantum free particle, these are id/dt and -i grad.. We suggest that if such a scheme exists for this translational invariance in t and r, one should treat the rotational invariance in the Klein-Gordon equation in a similar manner. In other words, there should exist operators and a function associated with rotational invariance. In this case, the operator is a matrix (Pauli matrices) and the function a spinor.
In physics (including classical physics), conservation of momentum is always considered, but one must be more careful with conservation of energy as it may include forms like sound etc. As a result, there is a tendency to focus on conservation of momentum all of the time, and conservation of particle energy only in elastic collisions. In addition to this, one may think of a particle striking a target, in which case two equal momentum (but different energies due to different v speed and mo rest mass) cause the same impact. In the case of an interaction with a potential, however, it is strictly energy considerations which are important classically, i.e one uses p dot p /2m + V(x) = E. Knowing momentum without rest mass is not sufficient. One may consider it unusual that one may consider two different scenarios, one which focuses only on momentum (impulse) or conservation of momentum in collisions while only energy is the focus in classical bound states or motion in V(x). Given the classical focus on momentum, imagine that one wishes to describe all free particles with different momenta as a uniform probability distribution such that a product of two probabilities for p1,p2 is the same as for p3,p4 if p1+p2=p3+p4. (We use 1-dimension for simplicity.). One cannot use P= 1/n because p is absent and n->infinite. In one sense, all particles must have the same probability, but at the same time must contain extra information related to p for the conservation of p part of the problem. The notion of a uniform distribution is independent of conservation of momentum, hence of p momentum. Conservation of momentum requires an explicit presence of p. As a result, there seems to be a paradox. One may, however, resolve this paradox by arguing that a probability is a mathematical way of presenting a subset of the information present. If one argues for a uniform probability, then it does not need to contain p, and one would have 1 as an unnormalized probability for any free particle p. A different probability (representing a different subset of information) which respects all free particles as having the same weight, but includes p can also be constructed, namely exp(i C p). C is a constant introduced for unit issues. In this case, the modulus gives all free particle probabilities with p the same weight, but the presence of p allows one to have conservation of momentum. Given that momentum is the key variable that is conserved in classical physics, one should be satisfied with exp(i C p). One might argue that there is no need to consider energy E. We note, however, that if one has p, E1 and -p, E2 in a frame, then in a frame moving with -v, one has gp+gvE1= p1’ and p2’ = -gp + gvE2. Adding yields gv(E1+E2) and so energy automatically appears and cannot be discarded. Furthermore for p, E1 and -p, E2, exp(i C 0) is exp(i C gv(E1+E2)) as seen in the moving frame and so one is forced to consider conservation of energy together with momentum and include both p and E in the more detailed probability. In other words, special relativity automatically links p and E in a 4-vector, leading to the idea that both are important. We argue that one cannot simply consider exp(i C p) due to special relativity. Both E and p must be considered together and this is a key idea, we argue, because if one wishes to have a Lorentz invariant probability, it must be in E and p, i.e. -Et+px. Instead of having a probability for an interaction with p given by exp(i C p) , one has exp(-iEt+ipx) and we argue that this must have the same value in all frames for a particular interaction. One may note that E and p are, on the one hand, separated allowing for impulse hit versus E related interactions, but that overall the probability for interaction includes both and it is the full -Et+ipx which must be invariant. This further suggests that if one does not artificially force an interaction to occur at the center-of-mass, then x,t and x+hbar/p, t+hbar/E are equally good points for the interaction. We argue that there is no need to force x=vt because exp(-iEt+ipx) * exp(iEt+ipx) already shows that one has uniformity, not only for any p,E, but also for any x, t, as there are now four variables. Thus, there is no reason to enforce x=vt in exp(-iEt+ipx), when considering an interaction, we argue. This means that one has two possible interactions at different x,t values, i.e. x,t and x+hbar/p, t+hbar/E. If one is unable to follow the interaction in time, but may follow it in space as in the case of 2-slits about hbar/p apart or for a potential V(x), then one may use the exp(ipx) part of the interaction probability piece by itself. Thus, we argue that special relativity does not allow one to treat p independently of E. For example, for p values that are the same in one frame, but have E1, E2, will be different in another frame. One must consider E and p together and so instead of creating a probability for p alone, one must create one for E,p and then impose Lorentz invariance which leads to different x,t pairs yielding the same probability value for an interaction, i.e. free particle quantum mechanics.
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.