Challenges
Datasets
Workspaces
Papers
Discussions
Leaderboard
Log inSign up
Challenges
Datasets
Workspaces
Papers
Discussions
Leaderboard
Blog
Job Board
Q3AS

© 2026 Aqora Quantum S.A.S.

TermsPrivacyLegal Notice

Momentum and Wavelength in Quantum Bound States

Francesco R. Ruggeri

Published Jul 15, 2021• 0 citations
Publisher

Abstract

Early experiments linked to bound state quantum mechanics noticed discrete energy E levels. (Ignoring the issue of radiation from accelerating charges), classical physics predicts continuous E levels. If one tries a statistical approach i.e. thinks in terms of a “perceived” density dt=C dx/v(x) where v(x) is velocity, one again has continuous energy levels. Thus, experimental results may suggest a non-classical condition which affects both acceptable densities and energy levels. Bypassing Newtonian mechanics which traces a particle through x(t), we note that even classically a potential interacts with p and not v i.e. pp/2m - popo/2m = -V(x)+V(xo). This equation holds for p= mo v = (.5mo) 2v etc. Thus, energy flow i.e. p is of interest and if one does not trace the particle as x(t) or consider acceleration (which is not considered statistically), one may view energy flux p as being “spread” over the x axis, it seems. The length 1/p then corresponds to one unit of energy flux for the case of p. In other words, each p is associated with a length 1/p. If this length may be physically resolved, one immediate trial result suggests a spatial probability which varies over 1/p. In other words one may have a periodic function such as cos(px). This physical scenario seems to arise only when the quantum particle interacts with a potential. Calling 2*3.14/p = wavelength, this length is not associated with the velocity of the particle, but with energy flow. Each particle has an identical type of cycling within its own wavelength i.e. cos(px) behaviour as each wavelength represents one unit of energy flux. The wavelength limits resolution and leads to a probability scheme. In a bound quantum system dx associated with V(x) is usually much smaller than the larger wavelengths and so classical ideas such as p and pp/2m still apply, but are modified i.e. multiplied by probabilities which may be positive or negative. Having a spatial probability linked to density which also creates a probability scheme used to calculate average kinetic energy at each x based on 1/p type wavelengths is independent of classical physics. Trying to force average kinetic energy at x in the wavelength scheme to match KE(x) classical leads to discrete E values.

DOI
10.5281/zenodo.5108364

Related papers

Classical Limit of Quantum Mechanics and the Breakdown of -id/dx

Francesco R. Ruggeri

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).

•Dec 1, 2022•0 citations

Speculation on the Quantum Wavefunction in the Classical Limit

Francesco R. Ruggeri

Newtonian mechanics often employs a potential V(x) such that .5m v(x)v(x) + V(x) =E ((1)) where v(x) is velocity. Mathematically V(x) is defined at each x, but in reality there should be a tiny cutoff i.e. finite dx. Newtonian mechanics makes use of the idea of small dx lengths which tend to 0 mathematically. To first order v(x) is constant within a dx range, but the overall mathematics is continuous instead of discrete. Thus there is a continuous equation ((1)) for conservation of energy. A photon may be described by an eclectic field varying as cos(-Et+px). In terms of wavelengths exp(ipx) is often used as a mathematical convenience and noncontinous behaviour is not an issue. For example in refraction, momentum = hbar/wavelength abruptly changes and one straight line path is joined to another with a different slope. exp(ipx) is associated with repeatable wavelength patterns with a single wavelength representing a spatial distribution/ (kind of probability) region. A particle with rest mass may also be described by exp(ipx) which suggests a characteristic length hbar/p. Within this length (i.e at each x point which exist within hbar/p) the particle has p and pp/2m. Thus for a square well potential there is conservation of energy at each x point even though there is a wavelength in the problem. This is fine for a constant momentum, but what about a potential V(x) which implies p must change at each dx region? exp(ipx) is an eigenfunction of -id/dx which is linked to spatial invariance. It manifests itself physically by interference (e.g. two slits) which again involves non-smooth motion (i.e. one compares rays with completely different slopes). The characteristic length hbar/p is fixed and associated with a single p, yet momentum is supposed to change from one x point to the next. hbar/p does not fit within a single dx and this is a problem. Quantum mechanics remedies this issue by allowing for a superposition of exp(ipx)s i.e. W(x) = wavefunction = Sum over p a(p) exp(ipx) such that energy conservation is forced at each x. Thus one combines the idea of wavelength (which is physical) with energy conservation which is also physical. A consequence of this, however, is that -id/W/dx / W leads to a strange average momentum which is not the square root of -1/2m d/dx dW/dx/ W as well as discrete energies. A mathematical benefit is that one has a continuous differential equation, namely the time-independent Schrodinger equation. The question we ask is: What happens in the so-called classical limit? In the nonrelativstic case this leads to Newtonian mechanics in which there is no wavelength. How does the wavelength disappear so to speak? We suggest (as has been suggested often in the literature and textbooks) that the wavelengths in W(x) may become extremely tiny so as to not be detected easily in classical length scales. Mathematically the time-independent Schrodinger equation holds and there is a smooth W(x) at all points, but we suggest that this smoothness may not be necessary because it is not found in refracting light. If the wavelength is tiny compared to small dx regions in the classical limit, then many wavelengths of a given p fit into dx. Thus the idea of spatial invariance approximately holds within dx and there is no issue of the wavelength spilling out of the dx region as it does when it is comparable to the size of the system as in quantum bound problems. Thus within dx a single p with many (or at least several) wavelengths holds and there is a single p. The kinetic energy is pp/2m and the potential V(x) where x is in the center of dx. One may think of p changing abruptly (nonsmoothly) at the boundary of one dx with another such that conservation of energy holds. Thus one has p wavelengths which do not spill out of dx and give an approximate sense of spatial invariance together with energy conservation. The only loss is that of a smooth wavefunction, but it is not clear that such smoothness is a requirement. V(x) is physically associated with a limiting dx length (at which the concept of V(x) does not hold, but rather there are stochastic hits) and if one expands this length i.e. chooses a much bigger dx in keeping with classical scales, it seems that there should not be a problem. It as if V(x) delivers an impulse hit at each dx junction (like a series of different media in refraction).. This approach leads to the decoherence of wavelengths in the classical limit.

•Nov 3, 2022•0 citations

Quantum Bound States and Conditional Probability

Francesco R. Ruggeri

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).

•May 29, 2026•0 citations

Speculation on Kinetic Energy in Quantum Bound States

Francesco R. Ruggeri

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.

•Jul 16, 2022•0 citations

Momentum/Spatial Bound Quantum States and Particle-Wave Duality?

Francesco R. Ruggeri

Quantum mechanics is often compared with classical mechanics, but it seems that the momentum/spatial treatment of bound quantum mechanics provides for a particle-wave duality with both features having an experimental basis. In bound state quantum mechanics, one has a density pattern often similar to a two slit pattern i.e. with humps, yet at the same time one may knock a particle out of the bound state and obtain a probability distribution a(p)a(p) of momentum states. Describing a particle simply by its momentum even in the bound state seems to be classical as does using pp/2m to describe the kinetic energy of a plane wave in the decomposition of W(x), the wavefunction. Wave aspects are linked to classical particle features like momentum and do not change the particle energy pp/2m. The wave, we argue, may be related to some type of periodic cyclic forward backward motion which creates a spatial probability needed to estables resolution on the order of a wavelength proportional to 1/p. It is often not seen unless there is a potential or other probability waves which may interfere with it In this note, we consider the scenario that momentum (i.e. mov in the nonrelativistic case) is directly linked to a kind of forward-backward cyclic motion of a quantum particle regardless of its spin. The fact that spin exists suggests some kind of internal motion. In the case of a photon, it may be directly linked to angular motion of energy, although for spin ½ matters are more complicated. This motion again does not change classical energy pp/2m unless spin interacts with a magnetic field etc. We argue that a bound state is stochastic and so various free particle momentum states exist without being restricted to an x position. This is the particle scenario. The internal cycling motion, however, gives rise to a resolution of the order of 1/p suggesting a spatial distribution to account for such a resolution. Furthermore, the internal cycling (i.e. forward and backwards motion) even for a particle traveling to the right requires two dimensions to describe overall forwards and backwards motion. The spatial density of the cycling motion may be associated with internal changes in velocity, but overall there is a particle with mass mo moving with v (nonrelativistic case) and kinetic energy pp/2m. This holds for a free particle, but also for a particle in an infinite potential well with a single average |p|. An issue occurs, however, if one wishes to form an average of different p values at the same point x. In such a case the spatial probability density exp(i pj x) differs for each p at x and so the average [p1 exp(i p1 xo) + p2 exp(i p2 xo) ] / [ exp(i p1 xo) + exp(i p2 xo)] becomes relevant. In other words, one takes the average momentum, say p1, and links it to the spatial probability i.e. multiplies two averages and then divides by spatial probability to find a momentum flow at a specific point within a wavelength region. For the particle aspect, such a resolution is non-issue. The particle is the same at each x point in space.

•Jul 13, 2021•0 citations

Quantum Bound Density and Time Through Impulse

Francesco R. Ruggeri

A classical bound system may be said to have a spatial density proportional to the amount of time dt a particle spends in a constant dx region i.e. C/v(x) (1) We stress that in this scenario there is a well defined and constant dx so that “dt” may be linked to velocity v(x). We also note that there is no spatial representation of action-reaction here even though this exists in Newton’s theory. In a classical Maxwell-Boltzmann gas there are many particles which each act dynamically according to .5mvv+V(x) = E (where v and E are particular to the particle when it is not colliding). At the same time there is a spatial density which “represents” macroscopic static chunks of matter which are balanced through pressure differences (P=spatial density(x) RT) and -dV/dx density where V(x) is the potential. From the pressure equation, T is proportional to energy which in turn is proportional to RMS speed squared. RMS speed may then be linked to a dt if a fixed dx is understood. In the classical statistical case, however, spatial density is directly linked to the number of particles in a region. Time enters in the pressure expression through a term proportional to the average kinetic energy i.e. pv where p is like an impulse and v a flux. What happens in a single particle quantum bound state? In a sense it is like a classical bound state as the equation: prms(x)prms(x)/2m + V(x) = En holds for any energy level. Can one then take vrms(x), choose a fixed dx and argue that spatial density should be equal to dt or C/vrms(x)? We argue that one cannot due to the lack of defined “dx” quantity. Why is dx problematic to define? We argue that it is because quantum mechanics is based on action-reaction x-probaiblity distributions linked to impulse. For example, exp(ipx) for a free particle captures only p or impulse (not velocity) and cos(px) and sin(px) are its action-reaction regions.These lead ultimately to a density is given by the well-known W*(x)W(x) where W(x)=Sum over p a(p)exp(ipx), although kinetic energy with pp/2m = .5pv is used to find W(x) as well. . W*(x)W(x) has crests and troughs similar to cos(px) in exp(ipx), except that these now have varying heights and widths. We try to argue that even though each x point maps to a different KEave(x), the hump physically represents action-reaction in the system (just as the humps in cos(px) do except for a single p) while KE(x) represents an average of p’s. This action-reaction is linked with impulses. In general, if one takes d/dx of the time-independent Schrodinger equation one finds an average impulse term, -dV/dx and KE(x) dW/dx / W where dW/dx / W is the average impulse. At peaks, dW/dx =0 and so an average force through time*impulse term balances -dV/dx=Newtonian force. These peaks are also associated with W*(x at peak)W(x at peak) which is the spatial density. We interpret this as representing the average time the single particle spends near x at peak. This value, however, follows from ideas of impulse and not from speed as there is no well defined dx even though there is a well defined vrms(x). Thus the ratio of W*(x)W(x) at two peaks is not the inverse of the ratio of vrms(x) except apparently when the energy level n becomes very high. We investigate these ideas.

•Feb 16, 2023•0 citations
Early experiments linked to bound state quantum mechanics noticed discrete energy E levels. (Ignoring the issue of radiation from accelerating charges), classical physics predicts continuous E levels. If one tries a statistical approach i.e. thinks in terms of a “perceived” density dt=C dx/v(x) where v(x) is velocity, one again has continuous energy levels. Thus, experimental results may suggest a non-classical condition which affects both acceptable densities and energy levels. Bypassing Newtonian mechanics which traces a particle through x(t), we note that even classically a potential interacts with p and not v i.e. pp/2m - popo/2m = -V(x)+V(xo). This equation holds for p= mo v = (.5mo) 2v etc. Thus, energy flow i.e. p is of interest and if one does not trace the particle as x(t) or consider acceleration (which is not considered statistically), one may view energy flux p as being “spread” over the x axis, it seems. The length 1/p then corresponds to one unit of energy flux for the case of p. In other words, each p is associated with a length 1/p. If this length may be physically resolved, one immediate trial result suggests a spatial probability which varies over 1/p. In other words one may have a periodic function such as cos(px). This physical scenario seems to arise only when the quantum particle interacts with a potential. Calling 2*3.14/p = wavelength, this length is not associated with the velocity of the particle, but with energy flow. Each particle has an identical type of cycling within its own wavelength i.e. cos(px) behaviour as each wavelength represents one unit of energy flux. The wavelength limits resolution and leads to a probability scheme. In a bound quantum system dx associated with V(x) is usually much smaller than the larger wavelengths and so classical ideas such as p and pp/2m still apply, but are modified i.e. multiplied by probabilities which may be positive or negative. Having a spatial probability linked to density which also creates a probability scheme used to calculate average kinetic energy at each x based on 1/p type wavelengths is independent of classical physics. Trying to force average kinetic energy at x in the wavelength scheme to match KE(x) classical leads to discrete E values.
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).
Newtonian mechanics often employs a potential V(x) such that .5m v(x)v(x) + V(x) =E ((1)) where v(x) is velocity. Mathematically V(x) is defined at each x, but in reality there should be a tiny cutoff i.e. finite dx. Newtonian mechanics makes use of the idea of small dx lengths which tend to 0 mathematically. To first order v(x) is constant within a dx range, but the overall mathematics is continuous instead of discrete. Thus there is a continuous equation ((1)) for conservation of energy. A photon may be described by an eclectic field varying as cos(-Et+px). In terms of wavelengths exp(ipx) is often used as a mathematical convenience and noncontinous behaviour is not an issue. For example in refraction, momentum = hbar/wavelength abruptly changes and one straight line path is joined to another with a different slope. exp(ipx) is associated with repeatable wavelength patterns with a single wavelength representing a spatial distribution/ (kind of probability) region. A particle with rest mass may also be described by exp(ipx) which suggests a characteristic length hbar/p. Within this length (i.e at each x point which exist within hbar/p) the particle has p and pp/2m. Thus for a square well potential there is conservation of energy at each x point even though there is a wavelength in the problem. This is fine for a constant momentum, but what about a potential V(x) which implies p must change at each dx region? exp(ipx) is an eigenfunction of -id/dx which is linked to spatial invariance. It manifests itself physically by interference (e.g. two slits) which again involves non-smooth motion (i.e. one compares rays with completely different slopes). The characteristic length hbar/p is fixed and associated with a single p, yet momentum is supposed to change from one x point to the next. hbar/p does not fit within a single dx and this is a problem. Quantum mechanics remedies this issue by allowing for a superposition of exp(ipx)s i.e. W(x) = wavefunction = Sum over p a(p) exp(ipx) such that energy conservation is forced at each x. Thus one combines the idea of wavelength (which is physical) with energy conservation which is also physical. A consequence of this, however, is that -id/W/dx / W leads to a strange average momentum which is not the square root of -1/2m d/dx dW/dx/ W as well as discrete energies. A mathematical benefit is that one has a continuous differential equation, namely the time-independent Schrodinger equation. The question we ask is: What happens in the so-called classical limit? In the nonrelativstic case this leads to Newtonian mechanics in which there is no wavelength. How does the wavelength disappear so to speak? We suggest (as has been suggested often in the literature and textbooks) that the wavelengths in W(x) may become extremely tiny so as to not be detected easily in classical length scales. Mathematically the time-independent Schrodinger equation holds and there is a smooth W(x) at all points, but we suggest that this smoothness may not be necessary because it is not found in refracting light. If the wavelength is tiny compared to small dx regions in the classical limit, then many wavelengths of a given p fit into dx. Thus the idea of spatial invariance approximately holds within dx and there is no issue of the wavelength spilling out of the dx region as it does when it is comparable to the size of the system as in quantum bound problems. Thus within dx a single p with many (or at least several) wavelengths holds and there is a single p. The kinetic energy is pp/2m and the potential V(x) where x is in the center of dx. One may think of p changing abruptly (nonsmoothly) at the boundary of one dx with another such that conservation of energy holds. Thus one has p wavelengths which do not spill out of dx and give an approximate sense of spatial invariance together with energy conservation. The only loss is that of a smooth wavefunction, but it is not clear that such smoothness is a requirement. V(x) is physically associated with a limiting dx length (at which the concept of V(x) does not hold, but rather there are stochastic hits) and if one expands this length i.e. chooses a much bigger dx in keeping with classical scales, it seems that there should not be a problem. It as if V(x) delivers an impulse hit at each dx junction (like a series of different media in refraction).. This approach leads to the decoherence of wavelengths in the classical limit.
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).
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.
Quantum mechanics is often compared with classical mechanics, but it seems that the momentum/spatial treatment of bound quantum mechanics provides for a particle-wave duality with both features having an experimental basis. In bound state quantum mechanics, one has a density pattern often similar to a two slit pattern i.e. with humps, yet at the same time one may knock a particle out of the bound state and obtain a probability distribution a(p)a(p) of momentum states. Describing a particle simply by its momentum even in the bound state seems to be classical as does using pp/2m to describe the kinetic energy of a plane wave in the decomposition of W(x), the wavefunction. Wave aspects are linked to classical particle features like momentum and do not change the particle energy pp/2m. The wave, we argue, may be related to some type of periodic cyclic forward backward motion which creates a spatial probability needed to estables resolution on the order of a wavelength proportional to 1/p. It is often not seen unless there is a potential or other probability waves which may interfere with it In this note, we consider the scenario that momentum (i.e. mov in the nonrelativistic case) is directly linked to a kind of forward-backward cyclic motion of a quantum particle regardless of its spin. The fact that spin exists suggests some kind of internal motion. In the case of a photon, it may be directly linked to angular motion of energy, although for spin ½ matters are more complicated. This motion again does not change classical energy pp/2m unless spin interacts with a magnetic field etc. We argue that a bound state is stochastic and so various free particle momentum states exist without being restricted to an x position. This is the particle scenario. The internal cycling motion, however, gives rise to a resolution of the order of 1/p suggesting a spatial distribution to account for such a resolution. Furthermore, the internal cycling (i.e. forward and backwards motion) even for a particle traveling to the right requires two dimensions to describe overall forwards and backwards motion. The spatial density of the cycling motion may be associated with internal changes in velocity, but overall there is a particle with mass mo moving with v (nonrelativistic case) and kinetic energy pp/2m. This holds for a free particle, but also for a particle in an infinite potential well with a single average |p|. An issue occurs, however, if one wishes to form an average of different p values at the same point x. In such a case the spatial probability density exp(i pj x) differs for each p at x and so the average [p1 exp(i p1 xo) + p2 exp(i p2 xo) ] / [ exp(i p1 xo) + exp(i p2 xo)] becomes relevant. In other words, one takes the average momentum, say p1, and links it to the spatial probability i.e. multiplies two averages and then divides by spatial probability to find a momentum flow at a specific point within a wavelength region. For the particle aspect, such a resolution is non-issue. The particle is the same at each x point in space.
A classical bound system may be said to have a spatial density proportional to the amount of time dt a particle spends in a constant dx region i.e. C/v(x) (1) We stress that in this scenario there is a well defined and constant dx so that “dt” may be linked to velocity v(x). We also note that there is no spatial representation of action-reaction here even though this exists in Newton’s theory. In a classical Maxwell-Boltzmann gas there are many particles which each act dynamically according to .5mvv+V(x) = E (where v and E are particular to the particle when it is not colliding). At the same time there is a spatial density which “represents” macroscopic static chunks of matter which are balanced through pressure differences (P=spatial density(x) RT) and -dV/dx density where V(x) is the potential. From the pressure equation, T is proportional to energy which in turn is proportional to RMS speed squared. RMS speed may then be linked to a dt if a fixed dx is understood. In the classical statistical case, however, spatial density is directly linked to the number of particles in a region. Time enters in the pressure expression through a term proportional to the average kinetic energy i.e. pv where p is like an impulse and v a flux. What happens in a single particle quantum bound state? In a sense it is like a classical bound state as the equation: prms(x)prms(x)/2m + V(x) = En holds for any energy level. Can one then take vrms(x), choose a fixed dx and argue that spatial density should be equal to dt or C/vrms(x)? We argue that one cannot due to the lack of defined “dx” quantity. Why is dx problematic to define? We argue that it is because quantum mechanics is based on action-reaction x-probaiblity distributions linked to impulse. For example, exp(ipx) for a free particle captures only p or impulse (not velocity) and cos(px) and sin(px) are its action-reaction regions.These lead ultimately to a density is given by the well-known W*(x)W(x) where W(x)=Sum over p a(p)exp(ipx), although kinetic energy with pp/2m = .5pv is used to find W(x) as well. . W*(x)W(x) has crests and troughs similar to cos(px) in exp(ipx), except that these now have varying heights and widths. We try to argue that even though each x point maps to a different KEave(x), the hump physically represents action-reaction in the system (just as the humps in cos(px) do except for a single p) while KE(x) represents an average of p’s. This action-reaction is linked with impulses. In general, if one takes d/dx of the time-independent Schrodinger equation one finds an average impulse term, -dV/dx and KE(x) dW/dx / W where dW/dx / W is the average impulse. At peaks, dW/dx =0 and so an average force through time*impulse term balances -dV/dx=Newtonian force. These peaks are also associated with W*(x at peak)W(x at peak) which is the spatial density. We interpret this as representing the average time the single particle spends near x at peak. This value, however, follows from ideas of impulse and not from speed as there is no well defined dx even though there is a well defined vrms(x). Thus the ratio of W*(x)W(x) at two peaks is not the inverse of the ratio of vrms(x) except apparently when the energy level n becomes very high. We investigate these ideas.