In Part 2, we suggested the existence of a free particle probability P(x,t,E,p) and argued that it must be Lorentz invariant as equations and formulas should appear the same as written in terms of the variables of various rest frames moving at constant speeds relative to each other. This led to P(-Et+px) and we saw that energy and momentum conservation followed. Here, we consider the following. Imagine that one wishes to introduce a momentum and energy conserving probability and does not wish to make it Lorentz invariant. In previous notes, we argued that one would be led to exp(iC1 E) and exp(iC2 p) where C1 and C2 are constants related to units and p is along an x-axis. We argued that for time reversal invariance exp(iC2p) → exp(-iC2p) and suggested that this is unphysical. Here, we show that using exp(iC1 E) and exp(iC2 p) leads to problems if one compares results from mixed frames (v and -v) to a rest frame. Given that probability exp(iC1E) and exp(iC2 p) only contain E and p, one may formally consider such probabilities as being created using different frames. For example, one may take a rest particle and create p using a frame moving with -v (relative to the rest frame) and -p using a frame moving with v. exp(iC2 p) provides no information about a frame, it is only linked to a momentum consideration. This leads, we argue, to issues, and forces one to include x,t as well as E,p, i.e. use Lorentz invariance when dealing with such probabilities. As a result, we argue that there is no way to avoid introducing x,t into such a probability and as a result, quantum free particle features such as wavelength and frequency emerge. In other words, when one thinks only in terms of E and p, as in Newtonian mechanics, one forces a fixed x,t frame. If, however, one wishes to consider p and E as resulting from different frames and use probabilities, these must combine p,E and x,t to define the frame. Then one may actually mix probabilities linked with different frames.
In Part 2, we suggested the existence of a free particle probability P(x,t,E,p) and argued that it must be Lorentz invariant as equations and formulas should appear the same as written in terms of the variables of various rest frames moving at constant speeds relative to each other. This led to P(-Et+px) and we saw that energy and momentum conservation followed. Here, we consider the following. Imagine that one wishes to introduce a momentum and energy conserving probability and does not wish to make it Lorentz invariant. In previous notes, we argued that one would be led to exp(iC1 E) and exp(iC2 p) where C1 and C2 are constants related to units and p is along an x-axis. We argued that for time reversal invariance exp(iC2p) → exp(-iC2p) and suggested that this is unphysical. Here, we show that using exp(iC1 E) and exp(iC2 p) leads to problems if one compares results from mixed frames (v and -v) to a rest frame. Given that probability exp(iC1E) and exp(iC2 p) only contain E and p, one may formally consider such probabilities as being created using different frames. For example, one may take a rest particle and create p using a frame moving with -v (relative to the rest frame) and -p using a frame moving with v. exp(iC2 p) provides no information about a frame, it is only linked to a momentum consideration. This leads, we argue, to issues, and forces one to include x,t as well as E,p, i.e. use Lorentz invariance when dealing with such probabilities. As a result, we argue that there is no way to avoid introducing x,t into such a probability and as a result, quantum free particle features such as wavelength and frequency emerge. In other words, when one thinks only in terms of E and p, as in Newtonian mechanics, one forces a fixed x,t frame. If, however, one wishes to consider p and E as resulting from different frames and use probabilities, these must combine p,E and x,t to define the frame. Then one may actually mix probabilities linked with different frames.
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 previous notes, we argued that one may find the source of exp(-iEt+ip dot r) in Newtonian mechanics by considering two body elastic scattering. Even though Newtonian mechanics is deterministic, probability is routinely introduced in such elastic scattering events because one cannot deterministically predict the outcome in most cases. Rather, one relies on conservation of energy and momentum and any possible conforming solution has equal probability. We argued that a free particle does not have a real valued probability weight because each is independent of the other (unlike in an ideal gas). As a result, we suggested the probabilities exp(iE C1) and exp(ip C2), with the constants C1 and C2 accounting for units. We then went on to make arguments linked to time reversal symmetry for exp(ipC2) and overall Lorentz invariance to finally arrive at exp(-iEt+ip dot r). Here, we wish to examine these considerations in more detail as we think there is a more general principle present, namely that of a state probability. We first note that even though exp(iEC1) and exp(ipC2) are consistent with conservation of energy and momentum, i.e. exp(iC1 E1) exp(iC1 E2) = exp(iC1 E3) exp(iC1 E4) and a similar expression for p, one only requires such conservation if all particles are classically at the same x and t, i.e. the collision point. As a result, simply given exp(iC1 E1) and exp(iC1 E2), the particles with E1 and E2 may be far apart in space. As a result, it is not clear why one would create a product probability with them unless they were actually interacting at the same x,t classically, but mathematically there is nothing preventing this and claiming conservation even though there is no reaction. It is this idea which suggests that exp(iC1 E) and exp(iC2 p) should be linked with x and t.Then, conservation only occurs in products if x and t are the same for different particles, otherwise it does not follow. This might suggest: exp(i E * (f(x)+g(t)) ) and exp(i p * (f(x)+g(t)). Here p is taken as a one-dimensional number along the direction of momentum. It may be positive or negative.f and g are unknown functions. We next note that x is measured from an origin. Thus, x1 and x2 are distances measured from an origin as is x1+x2. If one writes exp(i E * (f(x)+g(t)), this seems to imply that x and t are both measured from zero points. If one multiplies by exp(i E * (f(x2)+g(x2)), it seems one would wish this result to be equivalent to exp(i E (f(x+x2)+g(t+t2)). This leads to f and g being linear functions in x and t. Now one may have noticed earlier that including E,p,x and t in the “conservation” probability for a free particle means that one has completely defined the state of the particle. This probability is a state probability and so even though it is complex, one may argue that this is a property which should not change through a Lorentz transformation, just like particle number. This then leads to the notion of exp(-iEt+ip dot r). Thus, one initially attempts to describe the probability to have a particular p and E such that conservation of momentum and energy holds. In such a case, there is no need to introduce x and t and a Lorentz transformation changes the probabilities, but not their conservation properties as is shown. Thus, it is the desire to enforce conservation of energy and momentum only when a classical reaction occurs with all particles at the same x and t which introduces x and t into the probability along with E and p making it a state probability which should be Lorentz invariant. We suggest that there is a further important consequence of introducing a state probability. Given that it does not change under a Lorentz boost, it is like a quantity and may be used in OR situations a priori, even though it does not represent an obvious physically conserved quantity. This allows one to immediately write: Aexp(ipx)+Bexp(-ipx)=Cexp(ip2x) at x=0 for 1-D reflection-refraction from an n1-n2 index of refraction junction. (One may write a second equation involving p as exp(ipx) represents the probability to have p at x.) This then leads to a quantum mechanical manner of dealing with an interacting free particle.
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.
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 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.
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.