In (1), it is suggested that a Bohmian quantum potential for a wavefunction A(x,t) exp(i S(x,t)) is given by ( -1/2m d/dx d/dx A)/A. Here we wish to describe how the notion of such a potential might arise based on viewing the free particle wavefunction as a complex momentum conserving probability. In a number of previous notes, we argued that one may derive the free particle wavefunction exp(ipx) by seeking a free particle probability which conserves energy and momentum in Newtonian 2-body (or more) elastic scattering. We argued that for a given initial e1,e2 (energies) and p1,p2 (momentum vectors), any ei,ej and pi, pj (vectors) which conserve energy and momentum have equal likelihood of occurring as final outcomes after the collision. This leads to the exp(i C1 E) and exp(i C2 p) (p along the x-axis) as a solution, but for p=m1v1=m2v2, the same p satisfies a conservation of momentum equation. In a boosted frame , the p’s are different and so do not satisfy the same momentum conserving equation which is a problem, so one requires a Lorentz invariant argument, i.e. exp(-iEt+ipx). If one only considers momentum, then one has exp(ipx). In other words, a particle with momentum p and kinetic energy pp/2m (nonrelativistic) is linked with a momentum conserving probability exp(ipx) which is impulse-centric. Even if one considers kinetic energy pp/2m, in the probabilistic view of exp(ipx), one cannot separate energy from impulse hits as one does in Newtonian mechanics. For example, pp/2m + V(x) = E is strictly about energy. A quantum scenario cannot be so. It must combine both ideas and it this notion of including a picture of impulse hits which gives rise to a kinetic energy term for a bound state (i.e. the Bohmian quantum potential -1/2m d/dx d/dx A), which is a statistical average, but at the same time has the appearance of a “potential”. In this view, the potential V(x) may be written as Sum over k Vk exp(ikx) with exp(ikx) also representing momentum impulse hits, while at the same time Vk represents energy.
In (1), it is suggested that a Bohmian quantum potential for a wavefunction A(x,t) exp(i S(x,t)) is given by ( -1/2m d/dx d/dx A)/A. Here we wish to describe how the notion of such a potential might arise based on viewing the free particle wavefunction as a complex momentum conserving probability. In a number of previous notes, we argued that one may derive the free particle wavefunction exp(ipx) by seeking a free particle probability which conserves energy and momentum in Newtonian 2-body (or more) elastic scattering. We argued that for a given initial e1,e2 (energies) and p1,p2 (momentum vectors), any ei,ej and pi, pj (vectors) which conserve energy and momentum have equal likelihood of occurring as final outcomes after the collision. This leads to the exp(i C1 E) and exp(i C2 p) (p along the x-axis) as a solution, but for p=m1v1=m2v2, the same p satisfies a conservation of momentum equation. In a boosted frame , the p’s are different and so do not satisfy the same momentum conserving equation which is a problem, so one requires a Lorentz invariant argument, i.e. exp(-iEt+ipx). If one only considers momentum, then one has exp(ipx). In other words, a particle with momentum p and kinetic energy pp/2m (nonrelativistic) is linked with a momentum conserving probability exp(ipx) which is impulse-centric. Even if one considers kinetic energy pp/2m, in the probabilistic view of exp(ipx), one cannot separate energy from impulse hits as one does in Newtonian mechanics. For example, pp/2m + V(x) = E is strictly about energy. A quantum scenario cannot be so. It must combine both ideas and it this notion of including a picture of impulse hits which gives rise to a kinetic energy term for a bound state (i.e. the Bohmian quantum potential -1/2m d/dx d/dx A), which is a statistical average, but at the same time has the appearance of a “potential”. In this view, the potential V(x) may be written as Sum over k Vk exp(ikx) with exp(ikx) also representing momentum impulse hits, while at the same time Vk represents energy.
In Part 1, we suggested that exp(ipx) represents a spatial probability for the rendering of momentum hits. A single particle has a center-of-mass which undergoes x=vt, but momentum hits are not rendered from this point, but are rather delivered in x according to the complex probability exp(ipx) which is linked with the resolution delta x = hbar/p. This accounts for the fact that a particle can probabilistically interact with each slit of a 2-slit apparatus if they are about hbar/apart. If one is only interested in energy, pp/2m, the classical result holds for both exp(-ipx) and exp(ipx). In a classical bound state, one considers only energy p(x)p(x)/2m + V(x)=E because each p(x) exists at x in dx->0. One cannot distinguish any probabilistic hit distribution. exp(ipx) and exp(-ipx) differ even though they are linked with the same kinetic energy pp/2m (nonrelativistic case). We argued in Part 1, that the Newtonian conservation of energy equation actually includes a force profile linked with V(x). The catch is that for V(x) one has a force at each x, dx->0. Such is not the case for exp(ipx). We argued in Part 1 that one must consider not only pp/2m, but the associated spatial probability distribution for hits (i.e. energy together with force). Here, we note that this idea gives rise to an equilibrium calculation. This is somewhat startling because classically, and even quantum mechanically for higher energy levels, a particle moves to the right part of the time and then to the left for other times. As a result, a calculation which respects time would not average over exp(ipx) and exp(-ipx). Only an equilibrium calculation in time would do so. We suggest that one must consider the Newtonian conservation of energy equation as applying to both p(x) and -p(x) and so from an energy point of view one may consider it to be an average. The catch is that each term p(x) and -p(x) yields the same kinetic energy profile in the Newtonian case and the same V(x). For a quantum particle the kinetic energy with probabilistic force distribution is not the same for p and -p. As a result, to use V(x) which applies to forward and backwards motion, one is forced to consider the average and this leads to an equilibrium scenario. We show that the average kinetic with probabilistic force hit profile in space is not the same for +p and -p. Measurements, however, take time and may be averages of much data and so the notion of an average or equilibrium seems to make sense. We note, however, that an average is very different from the Newtonian forwards-backwards motion picture. Only in the high energy limit does the energy pp/2m with probabilistic force profile become the same for both p and -p. Thus, the main point of this note is to stress the notion of a single particle bound state as an average and to point out the consequences of this.
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.
Quantum mechanics utilizes a wavefunction W(x)=Sum over p a(p)exp(ipx) (in time independent cases) with exp(ipx) being what we call a free particle complex probability. In such a case, W(x) is an OR scenario of these various probabilities. We have argued in previous notes that exp(ipx) is the probability for a p impulse hit at x, even if the center-of-mass of the particle is not at x. This means that impulse hits may occur probabilistically in space as in the 2-slit example which requires adding two probabilities as an OR case for each slit. The point then seems to be that W(x) (or even its momentum frozen form W*(x)W(x) gives a profile in space of the number of interactions (various p impulse hits) at each x. Classically, the kinetic energy (nonrelativistic) is pp/2m. In a classical bound problem this kinetic energy is considered to exist (in calculations) at the center-of-mass, but forces are also said to be rendered at this point so pp/2m, the center-of-mass and the force hit point all coincide. In quantum mechanics, a p impulse hit and the center-of-mass do not coincide. If p is said to hit at a probabilistic x, then it seems that pp/2m should also be seen to exist at this same hit point. A physical interaction point verifies in a sense the location of pp/2m in a probabilistic manner, i.e. a(p) exp(ipx)/W(x) is the probability used to calculate KEaverage and this is based on impulse hit locations. By normalizing using W(x), one removes the relative number of average hits at each x and obtains a pure energy value at x. (This may then be used in a KEave(x)+V(x)=E equation.) On average W(x) has zero points, which means that if one considers W*(x)W(x) to represent the motion-frozen-out spatial distribution of hits, then no hits are rendered at certain x points (on average). One might argue then that the particle cannot exist at x, but the center-of-mass can exist anywhere. One might next argue that there is no kinetic energy at x, but that cannot be true because 1/2m d/dx d/dx W has the same 0 points as W and the two divide. Thus, there is average kinetic energy at all bound points and this represents an average based on impulse hits (exp(ipx)). Classically, if one has many particles in a potential, one may write density(x)V(x) to find the potential energy of the particles. This implies that the particles interact with V(x), e.g. neutrinos usually do not interact. Density(x) in a classical sense means that the center-of-mass and force hit point are both at x. In quantum mechanics, exp(ipx) represents the probability for a force hit point, W(x) represents an average and W*(x)W(x) represents a motion-frozen average hit point. W*(x)W(x)V(x) seems to represent the average potential energy of the particle at x because W*(x)W(x) is linked with interaction. Even if a particle is not present, however, there is energy V(x) in space. It seems that if W*(x)W(x)=0 at some x, then there is no average particle interaction at this point, but there is still average kinetic energy Sum over p a(p)pp/2m exp(ipx) / W(x) based on the probability for average probability hits. The key idea is that one normalizes probability by dividing by W(x) in the average kinetic energy calculation. W*(x)W(x) is normalized not at x, but integrating over dx and setting this value to 1. One has relative hit at x values represented by W*(x)W(x). For exp(-ipx)exp(ipx)=1, each dx has the same average hit value (with one not being able to distinguish details in hbar/p, i.e. this is a motion frozen result. The question then becomes: Why would one write: W*(x)W(x) KEave(x)? One might argue that this term occurs if one takes the quantum mechanical equation: KEave(x) +V(x) = En and multiplies by W*(x)W(x). One may ask: Does V(x)W*(x)W(x) have physical meaning? V(x) is the potential energy at x. The average interaction probability with V(x) at x should be W(x)V(x) and not W*(x)W(x)V(x). The latter is a classical mechanical idealization. If W*(x)W(x) is 3 at x1 and 2 at x2, then it is as if the particle interacts 1.5 times more at x1 and so if V(x1)=V(x2), then this approach assumes 1.5 the interaction energy at x1. KEave(x), however, is simply free particle energy (average) at x and one does not need to take this average value and multiply it by the relative number of interactions at x. W(x) has zero points meaning that there is no interaction at such a point on average. KEave(x) on the other hand is the average energy at x based on the particle kinetic energy presence pp/2m being at the point of an impulse hit. This does not mean that an interaction is occurring with V(x) at this point in this formalism. Even a free particle has an exp(ipx) probability when it is not interacting. The sum Sum over p a(p)exp(ipx) means that all of the interactions linked with x have occurred and one is making an OR calculation. KEave(x) is free particle kinetic energy at x and V(x) is potential energy (not interaction potential energy with the particle). Multiplying by W*(x)W(x) is a classical minded approach to creating a classical equation: density(x) (KEave(x)+V(x)) = E density(x) ((1)) Here KEave(x) would represent the average kinetic energy of a single particle and density(x) would be the average number of particles at x. density(x)KE(x) has a clear meaning. W*(x)W(x) KEave(x) on the other hand is different. KEave(x) is actually the classical kinetic energy value at x and so it is not clear why one would multiply it by W*(x)W(x). W*(x)W(x) represents the probability for an impulse hit at x in a p-frozen out picture. The notion of exp(ipx) (i.e. probability for a p hit at x) is already buried within the quantum calculation of KEave(x) (which yields the classical value). In order to find the physical meaning of W*(x)W(x) multiplied by KEave(x) or V(x) it seems that one has broken dx into equally sized segments which correspond to different detection hits of the particle. This is then somehow equated with the presence of the particle per unit time in dx which is then somehow rendered as the average energy at x multiplied by the detected presence at x through a steady rate of measurement/time. Thus, multiplying by W*(x)W(x) is not multiplying by a physical density.
In a previous note (1) we argued that a potential V(x), which exists in Newtonian mechanics, may be written as Sum over k Vk exp(ikx) (i.e. in a Fourier series). In another note (2) we suggested that the free particle wavefunction form exp(ipx) may be obtained from special relativity i.e. from the Lorentz invariant -Et+px. Thus in (1) we argued that the potential delivers impulse hits in a quantum mechanical scenario. In this note we revisit the emergence of exp(ipx) from special relativity, which like quantum mechanics, does not include acceleration i.e. a frame moves at a constant speed. (In Newtonian mechanics a particle moving within a constant dx is also said to have a constant speed, but this speed is differs in the next dx.) Given exp(ipx) in the context of special relativity, we consider the possibility of a system of impulse hits which may change p to p1 etc i.e. a set of exp(ipx)s. This leads to an average <pp> = {Sum over p a(p)exp(ipx) pp } / {Sum over p a(p) exp(ipx)}. A free particle satisfies: EE = pp + momo (c=1). If one wishes to retain a constant E (energy), one may replace pp with <pp> and imagine a function V(x) such that -(E-V(x)) (E-V(x)) + <pp> - momo. With p→-id/dx one obtains the Klein-Gordon equation. Taking a nonrelativistic limit yields the Schrodinger equation. We suggest the notion of potential may appear directly from special relativity together with the notion of a px dependent dynamic probability, namely exp(ipx) which is a quantum mechanical function. We further argue that the Newtonian approach ignores the information contained in -Et+px and uses a smoothing procedure.
In previous notes, we argued that one may introduce a probability into Newtonian elastic scattering if one assumes that for an incident (e1,e2) energies and (p1,p2) momenta vectors, any (ei,ej), (pi,pj) which satisfy conservation of momentum and energy have equal probability. We then found that for a single particle, the relevant probability is exp(-iEt+ip dot r). This is a unit modulus complex number which is Lorentz invariant, and exp(i p dot r) is time reversal invariant, i.e. p→-p and r→-r yield the same value. The question then becomes: How does one use exp(-iEt+ipx) or for time-independent problems, exp(ipx)? This is not a usual probability linked to quantity, e.g. N P(reflect) = the number of particles which reflect. exp(ipx) is a probability to have a certain momentum, but there is no reason it cannot carry a weight, i.e. a math factor. The question is: What does such a factor mean physically? Secondly, exp(ipx) depends on x (space), but exp(ipx) is not the probability to be at x, but it is somewhat related in that for any x and a given p, the modulus has the same value, suggesting equal probability to be at any x. This mapping, however, really only holds for a single exp(ipx). It seems that one can say two things about exp(ipx). First, if it is a probability for a momentum p, then one should be able to write p exp(ipx) or function(p)exp(ipx). Secondly, one should be able to write AND (multiply) and OR (add) statements. The idea of an OR or addition of exp(ipx)s is important, because one may be able to write an equation for constant kinetic energy: -1/2m grad dot grad W(x) = numberW. The general solution should be a W(x) which contains a sum of all exp(ipx) type solutions which have this kinetic energy. (In the relativistic case, one may consider d/dt d/dt W - cc grad dot grad W = -momocccc W.) The question we consider in this note is: Given that one may write W and grad W, what can only really do with these? They do not directly represent physical observables. Furthermore, if one wishes to impose continuity on W and gradW, which is reasonable mathematically, one should have a physical argument to justify it, and we suggest that this is a little complicated. Ultimately, we try to argue that one may consider continuity of W as well grad W in certain cases based on the association with a classical quantity probability equation. We also suggest one may relate W* grad W which may be generalized to W* grad W + (grad W*) W, as discussed in Part 3 and provides a link to classical probabilities.
Newton’s second law dp/dt = F(x)=-dV/dx may be written in terms of x alone i.e. dp/dx v(x) = -dV/dx or 1/(2m) d/dx pp = -dV/dx leading to the idea of kinetic energy pp/2m and potential energy V(x). In (1) we argued that there is velocity, rest mass mo, and momentum in quantum mechanics. A free particle, however, does not undergo smooth acceleration as in classical mechanics, and we argued there exists a momentum space wave exp(ipx) accompanying the moving particle. In such a scenario, force is considered to be probabilistic momentum hits of the form Vk exp(ikx) leading to a free particle momentum distribution W(x)= Sum over p a(p) exp(ipx). Both of these expressions involve interference due to momentum waves which do not track a particle in time. A mass flux mov or energy flux p=Ev relativistically seems to be the main consideration. Nevertheless, summing Vkexp(ikx) over k to obtain V(x) yields a function varying in x alone. This is a kind of average based on interference and Sum k Vk exp(ikx) is an average linked to momentum hits. We speculate that average momentum hits i.e. -d/dx V(x) are linked to momentum flux changes. In other words, even though an impulse is based on p and does not distinguish v i.e. p=m1v1=m2v2, momentum flux is velocity dependent and appears for an average momentum hit i.e. -dV/dx. In such a case (pave+dpave)(vave + dv ave) = 2 pave dv(ave) . This expression, however, contains both forward and backward motion which is part of quantum mechanics. Moving to an average picture seems to introduce time and direction i.e one uses pave d v(ave). This is already a change from a quantum to a classical scenario albeit for averages (which have not been specifically defined). Thus we consider pave dv(ave). If this is the case, then: d/dx pave 1/m space = -dV/dx or 1/(2m) pave pave = -d/dx V(x). Instead of treating pave pave as two separate averages one considers pave=prms or pavepave=<pp>. Thus the quantum bound picture (at any energy level) becomes a classical type of scenario which includes time and velocity. Note that for a two slit interference pattern, one does not need velocity and time because the interaction with the potentials from both slits is based on exp(ipx). One does not even need the form of the potential. Interference is the same for p=m1v1=m2v2 (even though the process takes place during different time intervals, the pattern is not affected). In other words, we argue that Newton’s second law might follow from a picture of momentum hits in which a flux i.e. p=mov is treated as a quantity which in an average scenario leads to a second flux i.e. an average momentum flux which reintroduces direction, time and velocity.. This is somewhat like a hydrodynamics scenario.