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.
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 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 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 Part 3, we started with the notion of a free particle probability linked to energy and momentum conservation in two body Newtonian elastic scattering. In particular, we argued that one needed a probability which shows an equal weight for any outcome (ei,ej) and (pi,pj) (momentum vector) set which conserves energy and momentum. This led to a first guess of exp(iC1 E) and exp(iC2 p), but in Part 3, we argued the momentum and energy are defined through a Lorentz transformation of mo at rest and such a transformation must also affect x,t → x’,t’. We concluded one cannot create any probability based on p and E without also considering x and t which transform according to the same transformation. This then allows one to create a Lorentz invariant probability. Here we take an opposite approach, i.e. one which does not involve p and E a priori. It is well known from classical physics that P(x)dx=dx/L, where L is an arbitrary length, describes a uniform distribution for particles at rest. One may, however, view these particles from two frames, one moving with constant v and the other with constant -v. This would seemingly introduce probabilities P(x1’,t1’,v) and P(x2’,t2’,-v) with no notion of energy or momentum present. It would seem that the product of the two for an x1’=x2’ and t1’=t2’ should equal the rest frame result. This suggests an exp(f(x,t,v)) solution, but there is no reason to have an ever increasing or decreasing exponential and so exp(i f(x,t,v)) seems like a solution. There is, however, a problem, because given P(x,t,v=0), and such a solution is then not P(x)= 1/L unless P(x,t,v)=1, which we reject as it assumes a moving particle is identical to one at rest in terms of probability.. In other words, there are two probabilities for the rest frame, P(x)=1/L and P(x,t,v=0) = exp(i f(x,t,v=0). Given that exp(i f(x,t,v) is based on Lorentz frames, one would also expect probability to be conserved and not be created or destroyed simply because a particle with rest mass is viewed from a rest frame or one moving at constant speed. This seems to force one to use relativistic variables linked to a Lorentz invariant, namely E,p as well as x,t. As a result, in Part 3 we argued that one cannot consider a probability of p,E without considering x,t as well and here we suggest that one cannot even consider a probability of x which is conserved in different constant moving frames without introducing t,p and E. Thus, P(x,t,v=0) which seemed like a “P(x)” type of probability in the rest frame is really a different probability, namely one in time exp(i f(x, v=0, t)), i.e. . for this to be Lorentz invariant, one has exp(-iEt+ipx) = exp(-iEt) for any x. In a sense, this is like P(x)=1/L in that each x carries the same weight, but there is still the time variable and the probability actually changes in time unlike P(x)=1/L. This seems to lead to a paradox because the given exp(-imocct) one may formally argue that exp(-px)=1 for p=0 which is the x part of the probability, yet above we argued that in order to obtain this value one should use P(x1’,t1’v)P(x2’,t2’,-v) at x1’=x2’ and t1’=t2’. The difference between P(x)=1/L is that the time measurement has infinite resolution, while exp(-imocc t) implies that it does not because otherwise information about x is the same. Thus, special relativistic considerations lead to the notion of uncertainty in time (hbar/E) and space hbar/p, but in the Newtonian scenario, these do not exist. We argue that the question then becomes: Why would one wish to change from a special relativistically related probability exp(-iEt+ipx) which shows physical uncertainty units of time and space, to one which shows infinite resolution for x and t, through exp(-iEt+ipx)exp(iEt+ipx)? We suggest that exp(-iEt+ipx) or exp(ipx) in time-independent problems is a dynamic probability needed to describe interactions. One may see that cos(px) and sin(px) may have positive and negative values, implying the removal of probability from one place and its addition somewhere else. One the interaction result has been obtained, one wishes to measure spatial results, i.e.count particles in bins in space. There is no need for the interaction information and it may be formally removed using exp(iEt-ipx) exp(-iEt+ipx).
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.
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 Part I of this note, we argued that the exp(ipx) portion of a free quantum particle’s wavefunction follows from d/dx A(x,t) = p extended to an eigenvalue equation: -id/dx exp(iA) = p exp(iAc) “p” appears as a scalar because it is the momentum magnitude, but one may consider flow in 3-space i.e. px,py,pz. In such a case, one might think of using basis vectors ei such that ei dot ei =1 (i=1,2,3). Given that one has an operator equation, one uses basis operators or matrices instead leading to the Pauli matrices or 4x4 matrices as found by Dirac if one imposes the restriction that upon “squaring” the operators disappear because one only has pp. As a result, we argued that intrinsic spin is linked to 3-space recognition for a free particle. The resulting spin, however, is spin ½ so what about other spins say 0 or 1? This seems to imply that free recognition of 3-space does not apply to other spin cases as it does to spin ½. The Stern-Gerlach experiment already shows that spin ½ is interesting in that one may have spin up or down project along any 3-space axis. In Part I, we showed that a wave-equation developed for the photon from d/dt (partial) b (ElEl + BB) + grad dot ElxB = 0 contains an eijk (-i d/dxi) term with eijk being the Levi-Civita symbol as well as the spin matrix. This is not the Pauli matrix (or matrix containing it) and suggests a different geometry or basis scheme. Thus, there is physical angular momentum (spin) in a plane perpendicular to the momentum and the (1,0) spin component is “hidden”, but associated with angular momentum in another spatial direction. As a result, 3-space is not treated in an independent way as in the spin ½ case. In this note, we investigate this idea further by considering the deuteron, He4 and He3 and try to argue that intrinsic spin still seems to have a connection to 3-space recognition, but that the fully independent 3-space recognition corresponds to spin ½. We only consider spin 0, ½ and 1 in this note.