A quantum single particle bound state involves a single particle, like a classical bound state, but it also involves spatial densities like a Maxwell-Boltzmann gas. For instance, a single particle has potential energy V(x) at x, while a density(x) has potential energy V(x) density(x). As a result, there seems to be a dichotomy. How does one solve the problem of single particle bound state? Does one use the concept of a single particle or of a density(x)? We argue that one focuses on a single particle, but with an added piece of physics. In classical physics, a particle with constant momentum (speed) moves in a deterministic way described by x=vt. On average, a quantum single particle creates its own ruler gradation and a sense of a direction in space through exp(ipx) which maps to 1 via exp(-ipx) exp(ipx). A single quantum bound particle which gives the semblance of an entity moving in a straight line back and forth (in a distance of the same order of hbar/p ) needs to create a semblance of acceleration as an average single particle. As a result, one may think of an ensemble i.e. W(x)=Sum over p a(p)exp(ipx). The difference with a free particle is that one has interference (+p’s and -p’s) which readily display a spatial W(x) which may ultimately be linked to a spatial density through W*(x)W(x). Thus, one now has the idea of spatial density which exists in a Maxwell-Boltzmann gas, This, however, is a consequence of the single particle approach. We argue that the single particle viewpoint leads to the spatial density one and not vice versa because of the way KE(x)+V(x) = E is treated. If one writes V(x)= Sum over k Vk exp(ikx), one does not write: KE(x) = Sum over k bk exp(ikx). We argue the reason is that such a KE(x) is not normalized so as to create the sense of a single particle. We argue that such a normalization leads to: KE(x) = {Sum over p a(p)pp/2m exp(ipx)} / W(x) meaning that on average one has a moving entity. V(x), on the other hand, is not a moving entity.
A quantum single particle bound state involves a single particle, like a classical bound state, but it also involves spatial densities like a Maxwell-Boltzmann gas. For instance, a single particle has potential energy V(x) at x, while a density(x) has potential energy V(x) density(x). As a result, there seems to be a dichotomy. How does one solve the problem of single particle bound state? Does one use the concept of a single particle or of a density(x)? We argue that one focuses on a single particle, but with an added piece of physics. In classical physics, a particle with constant momentum (speed) moves in a deterministic way described by x=vt. On average, a quantum single particle creates its own ruler gradation and a sense of a direction in space through exp(ipx) which maps to 1 via exp(-ipx) exp(ipx). A single quantum bound particle which gives the semblance of an entity moving in a straight line back and forth (in a distance of the same order of hbar/p ) needs to create a semblance of acceleration as an average single particle. As a result, one may think of an ensemble i.e. W(x)=Sum over p a(p)exp(ipx). The difference with a free particle is that one has interference (+p’s and -p’s) which readily display a spatial W(x) which may ultimately be linked to a spatial density through W*(x)W(x). Thus, one now has the idea of spatial density which exists in a Maxwell-Boltzmann gas, This, however, is a consequence of the single particle approach. We argue that the single particle viewpoint leads to the spatial density one and not vice versa because of the way KE(x)+V(x) = E is treated. If one writes V(x)= Sum over k Vk exp(ikx), one does not write: KE(x) = Sum over k bk exp(ikx). We argue the reason is that such a KE(x) is not normalized so as to create the sense of a single particle. We argue that such a normalization leads to: KE(x) = {Sum over p a(p)pp/2m exp(ipx)} / W(x) meaning that on average one has a moving entity. V(x), on the other hand, is not a moving entity.
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.
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.
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.
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).
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 previous notes, we argued one may formulate bound state quantum mechanics by assuming a stochastic potential which delivers “hits” of momentum k V(k) exp(ikx), while having an average form of V(x). A main constraint seems to be that the different k V(k) exp(ikx) hits may occur at any x point and that V(x) be of the form Sum over k b(kx) so that -d/dx b(kx) = -k b(kx). In other words, conservation of energy exists at the level of each hit as well as for the overall average. To force all of these constraints, a solution of b(kx)=V(k) exp(ikx) emerges which introduces periodicity and complex numbers. It should be noted that V(x)= Sum V(k) exp(ikx) so each term is actually a “piece of potential energy”. Furthermore, we argue that V(x) is decomposed for physical reasons i.e. V(x) is a low resolution result which incorporates all of the “pieces”. In a quantum mechanical situation, a “piece” V(k)exp(ikx) may act at t1 and V(k2)exp(i k2x) at t2. We argue that exp(ikx) should have physical meaning and leads to sin(kx) terms for the “k” and “-k” pieces. Thus, there is forward and backward motion which “cancel” i.e. have nothing to do with V(x). For a particle, however, in this region, there may be forward and backward motion, both of which have kinetic energy which should not be part of the average kinetic energy at x. The various physical (resolved) pieces of the potential lead to momentum distribution at each x point with P(p/x)= a(p) exp(ipx)/ W(x) (W(x)=wavefunction). In this case, one has a probability, not “pieces” of KE(x). At first, one may be alarmed by complex probabilities, but they are necessary to allow for cancellation of kinetic energy pieces due to the positive/negative values combined V(k)exp(ikx) pieces. Thus, even though KE(x) (average) matches classical results, the actual motion is more complicated. The potential seems to be the driver of the ensemble picture and should be linked to the microscopic physical nature of the potential. In this note, we wish to examine features and consequences of such a stochastic potential, including the idea of tunneling and infinite potential walls.