A classical free particle has a specific momentum p and kinetic energy pp/2m. In an ideal gas with no potential P(p)=P(-p) so p average=0 at each x. A quantum free particle has a wavefunction exp(ipx), a specific momentum of p and a kinetic energy of pp/2m. A special feature seems to be that exp(ipx) is associated with a length proportional to 1/p which repeats itself. We argue that this length is associated with the motion p in one dimension together with the two dimensionality. In other words, motion is not a simple translation. If a particle is placed in an infinite well of 1 dimension and length L, then low bound states have a special length (wavelength) of the order of the box. In other words one does not know at which point the particle is located or whether it has already reflected. Thus we argue that from a probabilistic point of view one must include p and -p together. These have different probabilities exp(ipx) and exp(-ipx) because this is a dynamic probability. The interference of the two removes one dimension exposing the periodic cos(px) like nature. If one calculates pp/2m average using -1/2m d/dx d/dx W / W (where W is the wavefunction) one obtains pp/2m (the classical result) because both p and -p have the same magnitude. The dynamic probability, however, does not have the same value for each. Thus p average = -i dW/dx / W is not zero and this result still holds if one multiplies by W*W. In other words there at each x there seems to be more weight to p or to -p. This suggests that an average acceleration may appear in fact: d/dx { -dW/dx / W ) = d/dx d/dx W / W - (dW/dx / W)(dW/dx / W). The quantum bound scenario is a statistical one with an inherent physical (we argue) kind of motion linked to a wavelength. If a particle is knocked out of bound state it has a wavefunction exp(ipx) and a momentum p. It is no longer in a hybrid inteferring statistical state. The modulus of exp(ipx) is 1 at each x as well. Furthermore P(p)=P(-p) = a*(p)a(p)=a*(-p)a(-p) which is not the same result as exp(ipx) versus exp(-ipx) for the dynamical probability. Given P(p)=P(-p) and no knowledge of x one might guess that one has a constant spatial density classical scenario, but this is not the case. As a result a special type of measurement is needed to discern W*(x)W(x) it seems, but this picture of humps and low probability regions is also in keeping with an average acceleration we argue.
A classical free particle has a specific momentum p and kinetic energy pp/2m. In an ideal gas with no potential P(p)=P(-p) so p average=0 at each x. A quantum free particle has a wavefunction exp(ipx), a specific momentum of p and a kinetic energy of pp/2m. A special feature seems to be that exp(ipx) is associated with a length proportional to 1/p which repeats itself. We argue that this length is associated with the motion p in one dimension together with the two dimensionality. In other words, motion is not a simple translation. If a particle is placed in an infinite well of 1 dimension and length L, then low bound states have a special length (wavelength) of the order of the box. In other words one does not know at which point the particle is located or whether it has already reflected. Thus we argue that from a probabilistic point of view one must include p and -p together. These have different probabilities exp(ipx) and exp(-ipx) because this is a dynamic probability. The interference of the two removes one dimension exposing the periodic cos(px) like nature. If one calculates pp/2m average using -1/2m d/dx d/dx W / W (where W is the wavefunction) one obtains pp/2m (the classical result) because both p and -p have the same magnitude. The dynamic probability, however, does not have the same value for each. Thus p average = -i dW/dx / W is not zero and this result still holds if one multiplies by WW. In other words there at each x there seems to be more weight to p or to -p. This suggests that an average acceleration may appear in fact: d/dx { -dW/dx / W ) = d/dx d/dx W / W - (dW/dx / W)(dW/dx / W). The quantum bound scenario is a statistical one with an inherent physical (we argue) kind of motion linked to a wavelength. If a particle is knocked out of bound state it has a wavefunction exp(ipx) and a momentum p. It is no longer in a hybrid inteferring statistical state. The modulus of exp(ipx) is 1 at each x as well. Furthermore P(p)=P(-p) = a(p)a(p)=a*(-p)a(-p) which is not the same result as exp(ipx) versus exp(-ipx) for the dynamical probability. Given P(p)=P(-p) and no knowledge of x one might guess that one has a constant spatial density classical scenario, but this is not the case. As a result a special type of measurement is needed to discern W*(x)W(x) it seems, but this picture of humps and low probability regions is also in keeping with an average acceleration we argue.
In a classical bound state, conservation of energy: p(x)p(x)/2m + V(x) = E holds for all x. Examined in a statistical sense, one may consider each constant dx to have a probability density proportional to dt(x) where v(x)=velocity =dx/ dt(x). Thus the particle is followed in time. In quantum mechanics, even for a free particle exp(ipx), there exists a wavelength hbar/p. Within this region one cannot follow the particle in time. In fact, there are two probability distributions cos(px) and the shifted sin(px). The two dimensional nature of exp(ipx) is required to describe the direction of motion. There exists a mathematical abstraction or average which holds for any x within the wavelength and yields the classical x= p/m t. Thus there is a duality, but within a wavelength one cannot follow the particle exactly using x=p/m t. Rather there is no time and one thinks of probabilities to be at one x or another. The operator -id/dx yields the abstracted or average momentum at any x and (-id/dx)(-id/dx) average pp if one divides by probability. In other words: (-id/dx) exp(ipx)/ exp(ipx) = p and (-id/dx) (-id/dx) exp(ipx) / exp(ipx) = pp. These then apply to any x point. For a bound state, exp(ipx) is replaced with W(x)= Sum over p a(p)exp(ipx). Thus p and -p are included together. From a statistical point of view this is fine because a quantum bound state is described in terms of energy conservation (time-independent Schrodinger equation). Unlike the classical case, the probabilities of p and -p differ, i.e. exp(ipx) and exp(-ipx). The sum exp(ipx)+exp(-ipx) yields cos(px) which is the resolution probability distribution. Taking -d/dx d/dx of W(x) and then dividing by W(x) yields the average kinetic energy at any x within the wavelength. In other words this is the average kinetic energy which matches the classical result for any energy level. As a result, prms(x)prms(x)/2m + V(x) = En holds for any quantum level, but prms(x) is an rms average value which bypasses wave structure. That is why the exp(ipx) for p,-p and -id/dx must be used to calculate average quantities within a wavelength. Many different p,-p combinations are needed so that at each x within a wavelength the correct average kinetic energy value appears. Now consider a high energy level and choose a fixed dx length to represent a tiny region in the system, a cutoff for which there is no further resolution. p(x) is constant within this region. If hbar/ prms(x) <= dx, the wave structure of quantum mechanics does not play a role. (Note that for an oscillator, W(x) is a set of crests and troughs with different heights and widths.) As a result, there is no need for d/dx acting on W(x) to describe average kinetic energy within dx. The average kinetic energy is prms(x)prms(x)/2m and dx is treated like a point. Only if exp(ipx) is used does -id/dx yield the proper classical result. (An alternative scenario is that d/dx acting on a function divided by the function yields the same result as acting on exp(ipx) divided by exp(ipx) e.g. -d/dx d/dx cos(px)/cos(px) = -d/dx d/dx exp(ipx) / exp(ipx). If one has the quantum kinetic energy expression -1/2m d/dx dW/dx / W ((1)) and considers one momentum, namely prms(x), and uses W(x)= cos( prms(x) x), then with prms(x) constant at x, ((1)) yields prms(x) prms(x)/ 2m. Thus d/dx has not broken down even though it does not describe motion within dx which is now a point. A problem arises, however, if one considers Wn(x)= Wn1(x)Wn2(x) as this yields an equation containing dW1/dx dW2/dx/ (W1W2). Then the above procedure yields: dW1/dx/W1 (prms(x) sin(prms(x) x) / cos(prms(x) x). This does not match the classical value. Furthermore, dW2/dx/ W2 contains +prms(x) and -prms(x) with different weights while a classical scenario is -pP(-p,x) + pP(p,x) = 0 because P(-p,x)=P(p,x) while quantum mechanically it doesn’t. We analyze this situation and conclude that d/dx breaks down/ does not apply in the classical limit because it is used to calculate average values within a wavelength only. It only “works” if the effect of the derivative(s) on a function divided by the function is the same as the effect on exp(ipx) divided by exp(ipx). If a wavelength fits within dx, there is no need to calculate values at any lower x level and d/dx is not needed. In fact, there is a single prms(x) value in dx and the average time links different dx’s so that one follows the particle (on average in the sense that one ignores structure within each x) by x(t). As a practical consequence, using -id/dx together with the wavefunction W(x) can lead to problems in the classical limit because d/dx does apply. In short, using -id/dx operating on Wn(x) divided by Wn(x) leads to issues unless it produces the same effect as acting on exp(ipx) divided by exp(ipx).
It is sometimes stated in the literature (1) that quantum mechanics is based on conditional probabilities. If this is the case, one would expect such conditional probabilities to appear in a single particle bound state. A free quantum particle is associated with a probability to interact of exp(ipx) (linked to 2-slit interference, 1-D reflection-refraction at an n1-n2 junction etc), but the kinetic energy and momentum are pp/2m (nonrelativistic) and p at all x even though each x has a different phase according to exp(ipx). The momentum is p at x, but the probability for an impulse hit at x is not the same even though its modulus is. When different probabilities exp(ipx) add in an OR situation, this leads to a spatial pattern which no longer has a modulus of 1, i.e. translational invariance is broken. Such a pattern also occurs for a single particle bound state W(x)=Sum over p a(p)exp(ipx). W(x) can have 0 points, but the average kinetic energy -1/2m d/dx dW/dx / W is not 0 at such a point. One has, on average (a statistical calculation), no average interaction at such an x point. One might ask why this should have anything to do with average kinetic energy? If one writes a conservation law: KEave(x) + V(x) = 0, then one considers kinetic energy interacting with potential energy. If there is no interaction at some x point for a given p (not W(x) on average), then it should not be part of the conservation of energy at this point. One is only interested in kinetic energy which interacts with potential energy at each x point. Thus, a(p) exp(ipx) + a(-p) exp(-ipx) (with a(p)=a(-p)) may have a 0 point at some x and so it does not have an average interaction at such a point, i.e. pp/2m (a(p)exp(ipx) + a(-p)exp(-ipx)) = 0. As a result, in a bound state, one considers kinetic energy per p, -p multiplied by the average probability (p, -p). Different a(p)exp(ipx) + a(-p)exp(-ipx) (a(p)=a(-p)) weights in this average picture indicate the extent to which the particle interacts. This idea holds for a single p,-p pair, but one must normalize over all such pairs, leading to the idea of a conditional probability. Given that hbar/p must be “created physically” in space so to speak, one must have interaction probability which creates this physical situation even in cos(px)+ i sin(px). This leads to the idea of interaction probability being 0, positive, but <1, or even negative and it is the interaction which allows one to consider conservation of energy with a potential V(x) in the first place. As a result, one cannot include a full pp/2m if there is negative, 0 or only particle interaction. Thus, pp/2m is multiplied by 2 a(p)cos(px) /W(x) for a(p)=a(-p).
Newtonian force is based on a change in momentum in time, but momentum is a vector and so may change in magnitude and direction. In many cases, the change is of one form or the other. For example, for a bound particle the change is in kinetic energy i.e. in magnitude, while for a particle moving at constant speed in a circle or pressure in a Maxwell-Boltzmann gas, the change is in direction. For an MB gas, the probability for p and -p are the same so the average momentum at any point x is 0 and thus chunks of the gas do not move, even though a single particle in the gas moves. The pressure, on the other hand, is the average of 2pv (2p for impulse, v for flux) and so pressure for p and -p do not cancel. A quantum free particle is based on exp(ipx) which does not contain time and is a dynamical probability, This is both an eigenfunction of the momentum operator -id/dx and the kinetic energy operator -1/2m d/dx d/dx, yet as pointed out in Part II, there are two pieces (cos(px),sin(px)) with a spatial density and hence a kind of pressure/force. These, however, combine to give the sense of constant motion in one direction. We argue this is due to a dynamical probability, unlike a static balance of pressures/forces found in an MB gas or Newtonian statics. There is a density with pressure distribution cos(px), but this is continuously changing because the particle moves in one direction with constant p. Thus the two pressure distributions “balance” so to speak to create a constant p. Interference of exp(ipx) and say exp(-ipx) may “bring out” a “static” spatial density i.e. 2cos(px), but behind the scene the dynamical probabilities still exist. It is these which “hold” the cos(px) together which appears strange from a static point of view. For example, a quantum bound state has a static W*(x)W(x) spatial density at the same time that the single particle has an average classical kinetic energy KE(x)=En-V(x). Interference also gives rise to the notion of an impulse-pressure in quantum as one ignores the isin(px) and i sin(-px) terms. Creating a pressure scenario with p and -p leads to two considerations. First, there is an impulse-pressure (defined without flux) at each x point and secondly, this impulse-pressure changes with x. The time-independent Schrodinger equation is formulated in terms of energy (i.e. a scalar): KE(x)+V(x)=En where KE(x)= {-1/2m d/dx dW/dx} \ W and W=Sum over p a(p)exp(ipx). Thus, as in Newtonian mechanics, one thinks of -dV/dx solely accelerating a particle. The quantum particle however, is described by an ensemble W(x) which has pressure. This pressure is due to a combination of the form exp(ipx) and V(x), so V(x) does not simply change root mean square speed. exp(ipx), by itself, creates pressure when interference occurs as in a two slit scenario. We try to examine these ideas in this note.
In classical physics, a potential V(x) acts on a particle at a location x regardless of its velocity which is taken to be constant within dx (even though the particle accelerates). Thus writing potential energy V(x) at a given x is like taking a snapshot of the particle when it is at x. The snapshot is static and does not indicate the dynamics of the problem as the particle may be moving either to the right or left, which is irrelevant in terms of the exact potential energy and even kinetic energy at point x. A quantum particle moving with constant velocity is described by a two-dimensional dynamic probability exp(ipx), with cos(px) representing a static density and sin(px) a time-shifted density to capture one directional motion such that exp(ipx)exp(-ipx)=1, we argue. V(x), however, only acts on the static density (linked to cos(px) and not sin(px)). Thus it seems that bound state quantum mechanics needs to remove the dynamic pieces of free particle wavefunctions exp(ipx) through interference. One might argue that for wavelengths of the order of the system length one does not know if one has exp(ipx) or exp(-ipx) at an x point, but for high energy levels, wavelengths become small with respect to the system length and yet interference is still present. Thus it seems that for the calculation of energy one requires a static average snapshot of the bound particle. Physically, we argue that the dynamic piece is still present i.e.. if one knocks out a bound particle and it is found in the p momentum state, it is represented by exp(ipx). Thus the purpose of the bound state wavefunction is to facilitate the calculation of average energy which is associated with a static snapshot type of density and not necessarily represent all of the dynamical information in the system i.e. the two dimensionality of each exp(ipx). As a result, the bound wavefunction does not show motion to the right and then to the left in the high energy limit.
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 Part III we considered the quantum free particle wavefunction exp(ipx) as consisting of two parts cos(px) and sin(px) treated as two different dimensions. We also argued that each, is a probability distribution in space and so associated with an internal force. One may notice that exp(ipx) does not contain time, but still predicts the direction of motion. Given that one cannot consider x(t) within a wavelength hbar/p even though on average the particle moves as x=p/m t, there is no real sense of time within a wavelength. Extending this idea, one repeats cos(px) humps. Given that the particle is moving, one must somehow juxtapose two spatial probability densities. The derivative of the distribution is proportional to average momentum and so one may see the notion of action-reaction occurring between cos(px) and sin(px) and within cos(px) and sin(px) accounting for the sign changes with x. We argue that d/dx cos(px) regions which change sign imply that the impulse direction changes. Furthermore, cos(px) has regions near the endpoints of the wavelength which have very low values (close to zero) which is fine if one has two juxtapositioned distributions. If, however, one creates a single particle bound state, exp(ipx) and exp(-ipx) for a(p)=a(-p) (a specific parity) give the appearance of a cancelation leaving the impression of 2cos(px) i.e a single “static” distribution. This occurs because there is no internal time in the picture and so p and -p probabilities add. The so-called canceled pieces have not really disappeared. as will be argued, they simply do not contribute to an average impulse which is what experiments apparently detect. Thus “negative” quantum probability is linked to impulse in the opposite direction, we argue. Thus interference is really linked to the notion of probabilistic impulses i.e. p exp(ipx) even though exp(ipx) is used as a kind of probability. exp(ipx) contains internal action-reaction impulse hits which have experimental consequences as we argue is shown in a reflection/refraction problem, but are also linked with impulse hits associated with p. Thus there are two sets of forces to deal with. This, we argue, explains why one writes the energy conservation equation AA/c1=BB/c1+CC/c2 for reflection/refraction where A,B,C represent the incident, reflected and refracted rays and AA simply shows that intensity/flux is positive as two equations: A+B=C and Ap - pB = p2C with E=pc. The two equations,, needed to solve for B,C in terms of A,, describe the internal force in exp(ipx) and the probabilistic impulse p exp(ipx). For a single particle bound state, one considers a linear combination of free particle states indicating that impulse hits represent the underlying physics. Given the apparent spatial invariance of a free particle exp(ipx), one may wonder how a kinetic energy profile appears. We argue that different p values yield different peak points for maximum impulse hits and these are out of synch leading to a spatial distribution of KE(x) or d/dx KE(x) which is equivalent to -dV/dx =Force where V(x) is the potential.