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.
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 timeimpulse 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.
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 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).
Both quantum bound state equilibrium and classical statistical mechanical equilibrium are characterized by d/dt partial density(x,t)=0. For a quantum particle, W(x,t)=exp(-iEt)W(x) and d(x,t)=W*W. Density is linked to fluid flow and arguments of pressure balance such that the overall average fluid flow is zero are made. This pressure-force balance is common in classical statistical mechanics and used to calculate the variation of density with x i.e. obtain a differential equation with d d(x,t) /dt partial. A similar balance equation is used in (1) as the authors try to derive the Schrodinger equation using a fluid dynamical approach. In this note, we try to argue that one may think of the two equilibria in terms of an energy quantity which remains constant throughout space. This energy quantity depends directly on the type of interactions in the system at any point x. For a classical ideal gas, balanced two body time reversal scattering represents the interaction. V(x), a potential, does not change this fundamental manner of interacting at any given point, although it does affect density. As a result the energy related constant at each x is T, related to average kinetic energy. We argue that this already points to the dominant type of interaction creating the statistical interaction i.e. one that depends only on p in interactions which conserve kinetic energy. This “happens” to be equivalent to maximizing Shannon’s entropy, but we argue this is particular to this scenario in the sense that not all interacting systems will have this equivalence. In other words, we argue that one must be careful with regards to the maximization of entropy as a fundamental rule. In particular, we consider the single particle bound quantum state. The interaction exists between the particle and the potential V(x), but one may consider a distribution of free particle momentum probabilities i.e W(x)=wavefunction = Sum over p a(p)exp(ipx). The form exp(ipx) comes from separate arguments as is discussed in the note, but the point is that there are no two body scattering reactions occurring at a given x, neither is there acceleration from on x to another (except for a vrms mathematical quantity). Nevertheless the “free particle” exp(ipx) interacts at each x with V(x). As a result of this interaction, one has free particles exp(ipx) which are no longer associated with pp/2m, but rather with an Eave which is a constant for every p. In other words the identity of exp(ipx) is unusual because it is linked to exp(ipx) and pp/2m, but also to Vk1 a(p-k1), Vk2 a(p-k2) etc where V(x)=Sum over k Vk exp(ikx). Thus each x point is associated with a constant Eave and this defines the equilibrium. (It is also equivalent to conservation of energy at each x.) In such a case, one does not maximize entropy to find a(p) values. In the two cases, the common point seems to be considering the interactions at each x and finding a constant “energy” type quantity which does not vary with x. This we argue appears to define equilibrium for both the classical statistical and quantum bound states. Instead of considering density which does not change in time at each x, there is constant energy which not only does not change in time, but has a constant value at each x which seems to be an alternative rule.
In Part I, we considered the ideal gas law P= density(x) RT of a Maxwell-Botlzmann gas, which follows from the notion of impulse 2p multiplied by v (flux) and then averaged,and examined how it balanced (on two sides of a box) a force -dV(x)/dx multiplied by density. We argued that in a quantum bound state there is no notion of internal time and hence no flux and so considered an average impulse 2p multiplied by an overall internal time in the system, namely hbar/E. We found that this scenario balanced with -dVd/x matches the quantum oscillator ground state. In this note we ask: What about other potential V(x) situations? Quantum mechanics seems to be a mixture of statistical mechanics and Newtonian dynamics i.e. the time-independent Schrodinger equation is, on average, an energy conservation equation for all energy levels i.e. KE(x)+V(x) = En where KE = {Sum over p a(p)pp/2m exp(kpx)}/{Sum over p a(p)exp(ipx)}. In (1) we took d/dx to bring out a force term -dV/dx and showed if is linked to <ppp/2m> and <pp/2m> , but analyzed these in terms of fluxes. The notion of flux makes sense using external time, but internally there is no time and so here we try to examine the situation in terms of various impulses.
In a previous note (1) we argued that both the relativistic A= -mo t sqrt(1-vv) and nonrelativistic A=t .5mvv classical free particle actions may be written with v=x/t and x and t varied independently to yield: dA/dx partial = p and dA/dt partial = -E. Thus A is equivalent to -Et + px for a free particle with x and t being treated as independent. We argue in this note that such an assumption is equivalent to a statistical picture for a free particle even at the classical level because x and t are treated as independent. Thus a constant p may occur at any x because no time is linked to x. Given that all space points are the same for constant p one expects a constant probability at all x points. Knowing p exactly means not knowing anything about which x one has. Secondly, one may create an ensemble of free particle p classical actions for a situation containing a potential V(x). Given that x and t are independent, there cannot be any notion of single p accelerating, although a vrms from an ensemble may. At a given x (x known exactly) one does not know which p is present. Thus the ideas of uncertainty of p and x follow immediately from this picture. If one considers a single particle and an ensemble of free particles in the x-t independent scenario, then each free particle action p is linked to A=-Et+px where E=pp/2m. This ensemble would arguably be stirred up by a potential V(x). If one wishes to consider an equilibrium scenario for which there is an overall En for the bound system one would expect an overall Ent for the system and for -pp/2m t pieces to disappear. Thus we postulate that pp/2m +{p part} G(p’s k’s) = En for all p where {p part}G(p’s, k’s) is an unknown function and V(x)=Sum over k Vk exp(ikx). Next one may consider the spatial situation for which V(x) is the classical potential for a single particle. The kinetic energy follows from an ensemble of free particles with the overall -Et removed. From classical considerations one would expect to have a KE(x). Thus one expects: KE(x) = {Sum over p a(p) f(px) } / {Sum over a(p)f(px) } with f(px), an orthonormal basis set, because the action contains px. (It is orthonormal because one wishes the identity of p to exist if not at a point, then over all of x.) We try to argue in this note that f(px)=exp(ipx), the momentum eigenstate which follows from A=-Et+px. We find G(p’s, k’s) which implies entanglement or mixing of a given p state with other p and k states from the potential V(x)=Sum over k Vk exp(ikx). Thus one expects an entropy linked to this momentum entanglement. Given the form exp(ipx), finding a set of ap’s which satisfy pp/2m + G(p’s, k’s)=En leads to a spatial distribution W(x)=Sum over p a(p)exp(ipx) so there should be an entropy contribution from this spatial part as well. In classical statistical mechanics one has both momentum and spatial contributions to entropy. We argue for S=Sp+Sx, but note that even though px has x, p symmetry as does the Heisenberg uncertainty principle var x var p >= hbar/2, Sp and Sx may not be symmetric in terms of important parameters such as n, energy level. In other words, Sp the entanglement entropy with V(x) may be the driver entropy sensitive to energy level changes while Sx the resulting spatial density may not be as sensitive. Furthermore the overall entropy Sp+Sx gives information about uncertainties in x and p as is already known in the literature where bound state Sp+Sx bounds are given.
In classical physics, a particle traveling to the right is the mirror image of one traveling to the left. In terms of a spatial density (the time spent in each constant dx), the value is the same, as is the energy. A similar notion of such a spatial invariance is in the classical conservation of energy equation for a bound particle: p(x)p(x)/2m + V(x) = E. Quantum mechanics seems to be different because exp(ipx) and exp(-ipx) are distinctly different spatial probabilities, i.e. they are dynamic. It is not just that case that a particle moves to the right for exp(ipx) and to the left for exp(-ipx). The impulse hit probabilities (represented by exp(ipx) and exp(-ipx)) are distinctly different. If one wishes to create an OR situation and write: W(x) = Sum over p a(p)exp(ipx) and use this in a conservation of energy equation, then simply including only positive or negative p values in W(x) leads to a complex average kinetic energy -1/2m d/dx dW/dx / W which cannot be used with V(x) and E. Furthermore, W_ (only negative p) and W+ (only positive p) values do not yield the same average kinetic energy. Fortunately, given that this is a statistical average for a localized object (somewhat analogous to a gas), one may combine W+ + W- and obtain a real kinetic energy. We have argued in previous notes that there exists a quantum-classical duality. For example, for a free particle one has two independent notions. One is x=vt which describes the center-of-mass of the particle. We take this to represent a state description which treats v and -v as mirror images. exp(ipx), on the other hand, represents probabilistic impulse hits and does not define the state by itself, we argue, only its interactions. Interactions are physical and define what happens in an interacting system. One may, however, measure a system, i.e interact with exp(ipx), and this should define a measurement state ( a second state ensemble average). For example, exp(ipx)exp(-ipx) = 1 which is equivalent to the constant dt in each dx associated with x=vt. Thus, the two states coincide, but exp(ipx)exp(-ipx) means that one may place the 2-slit apparatus at any x point and is not about the center-of-mass, as x=vt is. In some cases, the two states are quite different. W*(x)W(x) represents an interaction state, but vrms(x) from .5mvrms(x)vrms(x) = -1/2m d/dx dW/dx / W represents a center-of-mass state description. (Note: in these cases a state means an average state, i.e. an average over many ensemble entries.) If one takes W*(x)W(x) and integrates over x, then interference terms vanish and one obtains a classical counting of distinct p value states with their weight a*(p)a(p).This, we argue, leads to the notion of orthogonality of probabilities. In classical probability, there is no such notion, but there is the notion of AND (multiply) probabilities. Given exp(ipx) and exp(-ipx), if one multiplies one removes p, i.e. the interactions and only sees the state left. W*(x)W(x) yields a set of peaks which represent a pattern of probabilistic hits in space which define a state which one would measure in space. The interference pattern, however, must be one such that if one integrates over x, one only obtains a classical sum of probabilities a*(p)a(p)’s as mentioned above. This leads to the notion of orthogonal exp(ipx)’s (which follows mathematically anyway). If one considers different bound En states, then these two are orthogonal because W*n1(x)Wn2(x) mean n1 and n2 are associated with different energies and integrating over x cannot show a result which looks like a state result, where W*n(x)Wn(x) shows a state result. In order to achieve this orthogonality (linked directly to the notion of a state) one must have that Wn(2) usually has 0 points and is positive and negative with the possible exception of a ground state. In other words, if Wn1*(x)Wn2(x) integrated over x is not 0, one violates conservation of energy. As a result, we argue an average probability for a bound state must statistically include +p and -p contributions and must create a function which has positive and negative peaks (except possibly for the ground state), i.e. W(x) must have 0 points. This W(x) is linked to a measurement state W*(x)W(x), but this is strictly considering exp(ipx)s. We argue that there are two pieces of information for a free particle, x=vt and exp(ipx). There must also be an average state linked to x=vt and this given by KEave(x) = .5m vrms(x)vrms(x) + V(x) = E which shows that the particle may exist and have average kinetic energy even at W(x)=0 points. When measuring a system, one interacts with the exp(ipx)s and so the W*(x)W(x) state is the one which is traditionally presented, but whether one measures the center-of-mass motion or not, one also has the notion of the x=vt, i.e. vrms(x) state (average) as well, we argue.