In a previous note (1), we argued that the Maxwell-Boltzmann (MB) distribution exp(-p*p/2mT) and similar forms such as exp(-e1/T) and exp(-V(x)/T) were based on reactions of the form P(p1)P(p3)=P(p2)P(p4) or P(p1)V(1 to 2) = P(p2)V(2 to 1) ((1)) where p1, momentum, is being scattered into p2. Taking the natural log of this expression and comparing it to a conservation of kinetic energy (or energy levels in other cases) leads to the form exp(-p*p/2mT). This approach was applied to static statistical mechanics with a potential V(x), statistical mechanics with time independent flow, quantum mechanics with a temperature and fixed energy levels and the T=0 ground state quantum oscillator. It was argued that the MB type distribution breaks down for other bound state cases because there are two velocities in such cases. The first is the vrms(x) velocity where .5mvrms(x)*vrms(x) is the classical kinetic energy and u(x)=(1/W(x)) d/dx W(x) where W(x) is the bound state wavefunction. In this note, we try to give a specific calculation involving velocity (or flux) for the statistical mechanical case to show how it complements a reaction balance ((1)) as long as the flux velocity obeys Newtonian mechanics. For the ground state quantum oscillator, this is the case for the flux 1/W d/dx W, but not for other levels or potentials. Thus, this lack of Newton’s law in the flux 1/W d/dx W (which is responsible for d/dx d(x), where d(x) is the spatial density) seems to lead to the breakdown of the reaction balance ((1)) and the MB distribution in quantum mechanical bound states, leading to a new kind of statistical requirement for quantum bound states.
In a previous note (1), we argued that the Maxwell-Boltzmann (MB) distribution exp(-pp/2mT) and similar forms such as exp(-e1/T) and exp(-V(x)/T) were based on reactions of the form P(p1)P(p3)=P(p2)P(p4) or P(p1)V(1 to 2) = P(p2)V(2 to 1) ((1)) where p1, momentum, is being scattered into p2. Taking the natural log of this expression and comparing it to a conservation of kinetic energy (or energy levels in other cases) leads to the form exp(-pp/2mT). This approach was applied to static statistical mechanics with a potential V(x), statistical mechanics with time independent flow, quantum mechanics with a temperature and fixed energy levels and the T=0 ground state quantum oscillator. It was argued that the MB type distribution breaks down for other bound state cases because there are two velocities in such cases. The first is the vrms(x) velocity where .5mvrms(x)*vrms(x) is the classical kinetic energy and u(x)=(1/W(x)) d/dx W(x) where W(x) is the bound state wavefunction. In this note, we try to give a specific calculation involving velocity (or flux) for the statistical mechanical case to show how it complements a reaction balance ((1)) as long as the flux velocity obeys Newtonian mechanics. For the ground state quantum oscillator, this is the case for the flux 1/W d/dx W, but not for other levels or potentials. Thus, this lack of Newton’s law in the flux 1/W d/dx W (which is responsible for d/dx d(x), where d(x) is the spatial density) seems to lead to the breakdown of the reaction balance ((1)) and the MB distribution in quantum mechanical bound states, leading to a new kind of statistical requirement for quantum bound states.
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.
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.
Classical statistical equilibrium appears to be based on a no-flow situation with a balance of pressure (including -dV/dx in the case of a potential present). It may be noted, however, that like quantum mechanics, a single particle has different probabilities of being at one x versus another. Furthermore, the Maxwell-Boltzmann, Fermi-Dirac and Bose-Einstein distributions may be obtained using reaction balance which is linked to conservation of energy i.e. g(f(ei))g(f(ej)) = g(f(el))g(f(ek)) where ei+ej=el+ei=e. Taking ln of the former expression and linking to the latter yields f(ei), the probability distribution. g=1 for the MB case, f(e)/(1-f(ei)) for the FD and f(e)/(1+f(ei)) for the BE. In classical mechanics, one has the conservation equation: p(x)*p(x)/2m + V(x) = E. In quantum bound systems one has: KE(ave)(x) + V(x) = E which appears to be very similar, although it is based on ensemble averages and there is a momentum distribution P(p/x) at each x. A question which arises is: What is the “conservation of energy” equation for a particular p of the P(p/x) distribution? In this note, we argue the “potential” term associated with a component p*p/2m is related to the part of the potential which contributes to the “creation” of p. In particular, we consider V(x)= Sum over k V(k) exp(ikx) where each k term renders a kick of momentum k. One may create a momentum p using k and p-k for each value of k. We further argue that: p*p/2m P(p/x) + contributing portion of V(x) = E P(p/x) ((1)) for all p. Thus, in quantum equilibrium, which involves a flow, there is a “constant E” type of equation i.e. ((1)), different in form but similar in idea, to that of classical statistical mechanical equilibrium. In previous notes, we referred to this as cycling, but in this note we wish to consider it in terms of dynamics. We further note that in both classical statistical mechanics and quantum mechanics, the energy-probability p equation makes use of the overall average energy which is bT for the classical case and E for the quantum. (b is constant)
In bound state quantum mechanics, high energy eigenvalues are said to lead to a density W(x)W(x) where W(x) is the wavefunction, whose envelope approaches the classical value 1/v(x), where v(x) is the classical velocity given by sqrt[2m(E-V(x)]. For the quantum oscillator, the ground state mimics classical statistical mechanics, while the high energy states mimic classical mechanics. For both classical statistical mechanics and classical physics, we argue in this note that one may use a reaction picture to derive probabilities. Traditionally, the classical density 1/v(x) is obtained from dx = v dt or dx/v = dt with “dt” being proportional to probability. In classical statistical mechanics, one has two particle scattering and imposes a balance using P(p1)P(p2)=P(p3)P(p4). In the case of V(x), one may also write a reaction equation (1) showing particles leaving and entering a little box. In this note, it we argue it would be good to have a consistent picture when dealing with densities, namely a reaction picture, with “breakup or decay” balancing “synthesis”. We then try to argue that such a reaction picture may actually also apply to high energy eigenvalue quantum mechanical bound states.
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.
In a previous note (1), it was shown that for a ground state oscillator wavefunction, one could analyse the problem in momentum space as a free gas of momentum particles and obtain exp(-p*p/2mT) as fp*fp, fp being the momentum wavefunction. A free energy equal to zero was obtained, with kinetic energy being the energy term in E-TS. This suggests quantum particles scatter (with conservation of momentum) like classical point particles in momentum space, but differ from classical point particles in position space. In particular, one takes a Fourier transform of fp as it seems that each momentum wave carries an exp(ipx) function which is a sort of “baggage” term, possibly due to zitterbewegung. This seems to explain why a free gas in momentum space for the ground state oscillator yields a Gaussian density with W(x)=exp(-ax*x) (W(x)=wavefunction). An oscillator potential is then needed to keep this density with its momentum space pressure in balance. Thus, quantum mechanics and classical statistical mechanics become very similar for this scenario. This begs the question: What about other potentials or oscillator states higher than the ground state? In such a case, it still seems that one may consider “particles colliding in momentum space” with each carrying an exp(ipx) “baggage” term, but one does not have a “classical statistical mechanical” equilibrium scenario any longer. Rather, it seems that one has a net force created by -dV/dx (V(x)=potential) and a difference in pressure d/dx [Average(p*p)d(x)] where d(x) is density. There also seems to be an average velocity (1/W)dW/dx. In momentum space such a velocity would not exist for f(p)=f(-p), but because of the baggage of exp(ipx) it is present i.e. p exp(ipx) - p exp(-ipx). Newton’s law, with force acting on this velocity appears in the equation d/dx [Average(p*p)d(x)]=F(x) d(x). This same equation is equivalent to the gradient of the time-independent Schrodinger equation. For quantum mechanics, one has one particle rather than many as in classical statistical mechanics. Thus, it seems that this “Newton’s law” balance is based on a long term time average. It seems, however, that there should be a period T during which the average can establish itself. After this, one would have periodic time motion where the system keeps scattering back into itself in this characteristic time. This period can be deduced, it seems, from the time-independent Schrodinger equation and is found to be E, the energy.