In classical physics, one states in words the existence of conservation of momentum and energy. An example would be an elastic collision of two particles. In particular, this conservation is imposed, it does not mathematically appear unless one describes the process by say a Kronecker delta, i.e. delta( momentum vector before, momentum vector after) or delta(energy before, energy after). Even in such a case, the Kronecker delta is isolated and so one still has to manually ensure that energies and momenta chosen in a calculation ensure conservation. A Kronecker delta is mathematically equivalent to the dot product of two orthonormal states in linear algebra. This implies an operator which may obtain the particular value in question, momentum vector or energy, when operating on the state. A Hermitian matrix operator is guaranteed to have real eigenvalues and orthogonal eigenvectors, which is required. Thus, mathematically, one may postulate energy and momentum states and corresponding operators because this automatically creates the Kronecker delta which describes conservation in classical processes. This approach holds both nonrelativistically and relativistically. The question becomes: How does one specifically describe the operator and its state? For both a relativistic and nonrelativistic free particle classical action, which describes motion, one may write: A= -Et+px where E,p are the relativistic free particle values in one case and E=p dot p/2mo and p=mov in the other. One may note that E and p are values which may be conserved in interactions. The action shows that the operators -d/dt partial and d/dx partial yield E and p respectively, but one does not have an eigenstate. This may be obtained using id/dt partial exp(-iEt+ipx) = E exp(-iEt+ipx) and -id/dx partial exp(-iEt+ipx) = p exp(-iEt+ipx). Because one uses partial derivatives, t and x are treated as independent relative to the time and space translation operators d/dt partial and d/dx partial. As a result, we argue that by mathematically linking a Kronecker delta associated with classical conservation with linear algebra, i.e. momentum and energy states and finding a corresponding operator by examining the classical action leads to the quantum free particle wavefunction exp(-iEt+ipx). One may apply this to an elastic two particle collision to see that the Kronecker delta and its associated state/operator form describe conservation of both energy and momentum.One may ask whether exp(ip dot r) has any physical relevance. Two-slit interference experiments using particles with rest mass shows that it does.
In classical physics, one states in words the existence of conservation of momentum and energy. An example would be an elastic collision of two particles. In particular, this conservation is imposed, it does not mathematically appear unless one describes the process by say a Kronecker delta, i.e. delta( momentum vector before, momentum vector after) or delta(energy before, energy after). Even in such a case, the Kronecker delta is isolated and so one still has to manually ensure that energies and momenta chosen in a calculation ensure conservation. A Kronecker delta is mathematically equivalent to the dot product of two orthonormal states in linear algebra. This implies an operator which may obtain the particular value in question, momentum vector or energy, when operating on the state. A Hermitian matrix operator is guaranteed to have real eigenvalues and orthogonal eigenvectors, which is required. Thus, mathematically, one may postulate energy and momentum states and corresponding operators because this automatically creates the Kronecker delta which describes conservation in classical processes. This approach holds both nonrelativistically and relativistically. The question becomes: How does one specifically describe the operator and its state? For both a relativistic and nonrelativistic free particle classical action, which describes motion, one may write: A= -Et+px where E,p are the relativistic free particle values in one case and E=p dot p/2mo and p=mov in the other. One may note that E and p are values which may be conserved in interactions. The action shows that the operators -d/dt partial and d/dx partial yield E and p respectively, but one does not have an eigenstate. This may be obtained using id/dt partial exp(-iEt+ipx) = E exp(-iEt+ipx) and -id/dx partial exp(-iEt+ipx) = p exp(-iEt+ipx). Because one uses partial derivatives, t and x are treated as independent relative to the time and space translation operators d/dt partial and d/dx partial. As a result, we argue that by mathematically linking a Kronecker delta associated with classical conservation with linear algebra, i.e. momentum and energy states and finding a corresponding operator by examining the classical action leads to the quantum free particle wavefunction exp(-iEt+ipx). One may apply this to an elastic two particle collision to see that the Kronecker delta and its associated state/operator form describe conservation of both energy and momentum.One may ask whether exp(ip dot r) has any physical relevance. Two-slit interference experiments using particles with rest mass shows that it does.
In previous notes, we introduced a complex probability into Newtonian two-body elastic scattering to ensure that for a given (e1,e2) and (p1, p2) (momentum vectors), any (ei, ej) (pi, pj) have the same probability to occur. This ultimately led to the Lorentz invariant probability exp(-iEt+i p dot r). Here we begin with the delta functions delta( pinitial - pfinal) and delta( Einital - Final) and see that one may write these as Integral da C exp(-i pfinal + i p initial) and a similar expression for E. Here a is simply a math variable. (Note: This is only one math form for a delta function, but this particular form will prove useful from physical considerations later.) The delta function may be used in an expression Integral delta(pinitial - pfinal) dpfinal and only pfinal values which satisfy conservation of momentum yield a nonzero integral. Furthermore, this integral may be multiplied by various functions and still maintain conservation of momentum. At this point, the above is simply a calculation formalism. The question we ask is whether one may make the argument that a particle itself is responsible for conservation of momentum in an interaction. We suggest this may be the case based on Newton’s notion of action-reaction. It is not simply that one applies a force to a particle and it changes. The particle itself resists and this is the idea of inertia as well as action-reaction. The particle itself is strongly involved in a reaction and we suggest that a particle carries a property of ensuring conservation of momentum and energy itself. It is not something that is imposed by the interaction, otherwise there would be no need for the notion of action-reaction. The particle reacts and the manner in which it reacts leads to conservation of momentum and energy. We suggest that there seem to be grounds for considering a particle to actually be the driver of energy and momentum conservation and that this actually extends to further features, namely probabilistic results in interactions which do not appear in Newtonian mechanics. This property might then be taken as exp(i p a) and exp(i E b), where a and b are math variables used to establish the delta function. One, however, cannot really establish a property of a particle if math running variables a and b are present. The entire expression should consist of particle state variables. This seems to suggest the consideration of Lorentz invariance because E and p change when the particle is seen from a frame moving with -v. One may create a state based exp(-iEt+ipx) energy-momentum conserving property of a particle which is Lorentz invariant. We next note that p and E are linked to interactions, but ones that differ. In particular, p is linked with an impulse which has nothing to do with velocity , i.e. E. On the other hand, E = .5movv in the nonrelativistic limit and is dependent on v, i.e. a m1v1= m2v2 have the same impulse, but different kinetic energies. Thus, we suggest that exp(-iEt) and exp(ipx) be used separately depending on the type of interaction one considers. In a quantum single particle bound state, one is interested in impulse hits with the potential we argue and so does not need to consider exp(-iEt) as one cannot measure the particle moving through the potential in a classical manner. We argue that a constraint (momentum and energy conservation) may actually be due to properties of the particle and that this ultimately follows from Newton’s action-reaction. In (1), we argued that exp(-iEt+ipx) follows from Lorentz invariance which is based on the Newtonian idea of work increasing energy, i.e. 1/sqrt(1-vv/cc) and A= -Et+px have minus signs for this very Newtonian reason. We argued that dx = hbar/p and dt=hbar/E follow as fluctuations which leave A unchanged due to the presence of the minus sign.Here we again argue that a Newtonian idea is responsible for exp(-iEt+ipx), namely action-reaction. We show that the two approaches are identical.
The classical Lagrangian L leads to Newton’s second law which is equivalent to an energy-momentum conservation equation. The classical action: A = Integral dt1 L(t1) for a constant velocity v=X/T may be varied with respect X and T i.e dA/dT = -E and dA/dX= p. Thus, one may create a differential equation in (dA/dT)(dA/dt) and dA/dx dA/dx for the EE=pp+momo equation. Alternatively, one may create other differential equations. Quantum mechanical differential equations, however, are quite different with exp(i Action) being a solution. Even this may be written as the solution of a non-Schrodinger equation i.e. id/dt exp(iAction) = -1/2m (d/dx exp(iAction)) (d/dx exp(iAction)) for a free particle. For a constant velocity, exp(iAction) is equivalent to exp(ipx-iEt). We argue that this solution is statistical because it separates x and t (as independent variables) and holds for all x and t. Thus, quantum mechanics seems to be a statistical approach using free solutions exp(ipx-iEt) to create an ensemble as in classical statistical mechanics. A difference, however, is that exp(ipx-iEt) “interferes” even if one only takes the real part cos(px-Et). We consider the case of a particle in a box with infinite potential walls to argue that the periodicity is warranted i.e. positive and negative values (interference). Even though the modulus is 1, exp(ipx-iEt) may be operated on by various generators e.g. d/dx the generator of translations. This is analogous to operations on vectors in linear algebra. As noted above, quantum equations are not only energy-momentum conservation equations. They yield a wavefunction W(x) which is an unusual momentum distribution W(x)=Sum over p a(p) exp(ipx). exp(-iEt) holds overall because average E is the same at each x. Thus, one needs a W(x) which leads to a momentum-energy conservation equation, but also satisfies certain boundary conditions which become linked to spatial density= W*W. This leads to discrete energy levels and also cases of humps and troughs in W(x). We attempt to investigate some of these ideas in this note.
In Part 1, we considered Newtonian elastic scattering and proposed a probability exp(-iEt+ip dot r) to allow for equal product probability weights for any (ei, ej) (energy) and (pi, pj) (momentum vector) outcome pair. We insisted on the probability being a Lorentz scalar and argued that all free particles have the same real value weight, hence the modulus of 1. This notion is based on probability as it is applied to possible outcomes. This means that one does not have a fully deterministic situation because otherwise one would know the exact ei,ej, pi,pj outcome and would not require probabilities in the first place. In Part 2, we argued that a deterministic approach based on x=vt for a free particle contains no notion of probability and so this seems to contradict exp(-iEt+ipx) with its physical intervals: delta t = hbar/E and delta x = hbar/p. We then noted that these intervals and hence probability are found in the Action= Lagrangian * t = -Et+px with v=x/t. We concluded that x=vt, a solution of d/dt dL/dv partial - dL/dv = 0 loses some of the information present and that examining the action reveals more information because one is forced to deal with x,t, E and p at the same time. Here, we return to Newtonian elastic two-body scattering and assume that one has the following deterministic case: e1,e2, p1, p2 → e3,e4, p3,p4. One would then assume that no probability of any kind is present. We suggest that momentum is conserved due to action-reaction, something that Newton already proposed without any notion of probability required. Energy is conserved because it cannot be ultimately created or destroyed and we consider special elastic collisions. We then consider the Newtonian idea of a particle being acted upon by a force -dV(x)/dx, where V(x) is a potential. The interaction occurs for a dt, dx, but Newtonian physics assumes that dx→0, dt→0. Thus, Newtonian mechanics allows for a minimum dx->0 for conservation of momentum. By going into the center-of-mass frame of two particles which collide elastically, one has p and -p in this frame. When the two particles come together at x,t, for a tiny time one has a joint particle with no net momentum which is interacting and breaking apart. Then, in the 1D case, one particle moves out with p1 and the other with -p1 in the cm-frame. If one considers A= -Et+px for one particle, then Newton’s dt->0, dx->0 as well as x=0, t=0 yield A=0. It is known from special relativity, however, that there is a jump in x for a particle viewed from a moving frame. In particular, if the particle is at x=0 at t in the rest frame (cm frame), then in the moving frame: x’ = g(v) v t where g(v)=1/sqrt(1-vv/cc).This is linked to the person in the moving frame (-v) measuring the particle which seems to move to him/her. This measurement takes some minimum amount of time. Thus, it seems that one must reject dx→0 on the grounds of special relativity. The question then becomes: What is the smallest dx and dt one may have based on special relativity? We suggest that given the A=0 case, one may have delta x =hbar/p and delta t = hbar/E. Even in a deterministic elastic interaction, the notion that dt and dx cannot be 0 leads to minimal values for these which are probabilistic, i.e. they are based on special relativity and not on the actual interaction. If one has initial uncertainty in x and t, even if the particle then moves with x=vt, this uncertainty should also be propagated along as well, because the dx, dt uncertainty cannot simply vanish. We suggest that the notion of exp(-iEt+ipx) emerges even if one removes the initial probabilistic assumption of elastic scattering with unknown outcomes. Known outcomes still lead to uncertain dx=hbar/p and dt =hbar/E in keeping with the uncertainty contained in Lt= -Et+px, which is used for deterministic calculations. Thus, it is special relativity which places a constraint on how small dx and dt may be in a physical reaction and one cannot simply use the Newtonian dx→0, dt→0.
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.
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).
In a previous note (1) we argued that the free particle classical action A (relativistic or nonrelativistic) may be written in terms of v=x/t and then varied in x and t independently. One finds that dA/dx partial = p and dA/dt partial = -E implying A= -Et+px with x and t independent. This is the opposite of the classical mechanical idea of x(t). Creating an eigenvalue equation yields -id/dx partial exp(iA) = pexp(iA) and id/dt partial exp(iA) = Eexp(iA). exp(iA) plays the role of an unusual type of probability we argue which describes two dimensional periodicity (cos(px) and sin(px) ). We have argued that the quantum particle is the medium for a two dimensional wavelike motion of an energy-momentum wave. A wave type structure is needed to describe dynamics (like a classical wave), but unlike a classical wave two dimensions are required because there is no preference of any x point for a free particle. Both cos(px) and sin(px) both treat x in a preferential manner as their values vary with x. For a classical wave this is fine because the creation point of the wave imposes boundary conditions, but there is no such boundary condition imposed on a free particle. exp(ipx-iEt0 must somehow then link to P(x)=constant. One may note that exp(-iA)exp(iA) = 1 for the free particle. This may be argued to be linked to spatial invariance i.e. P(x)=constant where P(x) is not exp(ipx). Here we argue that exp(-iA)exp(iA) represents a product of two unusual probabilities , one for motion in a certain direction and the other for the time reversed scenario. If the quantum particle is the medium of an energy-momentum wave exp(-iEt+ipx), then the product of the regular and time reversed unusual probabilities cancel or undo each other. In other words one does not see any change in space as long as one has a noninteracting exp(-iEt+ipx) . Given that -id/dx is a linear operator this product yields a momentum of 0. We consider this idea and its implications on a momentum distribution Sum over p a(p)exp(ipx) which may represent a bound state or other states. In particular, it seems that P(x)=W*(x,t)W(x,t) which yields a classical type of probability (P(x) real, positive and normalized to 1) is linked to a product (merging) of forward and time reversed situations so that any spatial density which remain due to the quantum particle acting as a medium for an energy-momentum wave (as well as the time reversed part) would represent an actual change in spatial density. exp(ipx) already contains two periodic spatial functions (cos(px) and sin(px)), but does not manifest it unless there is an interaction.