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.
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.
Linking quantum mechanics to classical mechanics seems to have been an early goal in quantum theory as represented by the correspondence principle. More recently there have been attempts to link the entropy of both a single bound quantum state and a thermal statistical system of quantum states to classical entropies (1),(2). For the former, the entropy is proportional to Integral dx density(x) ln[density(x)] where density(x)= delta t = 1 / v(x) with v(x) being velocity. Thus, density, which is a probability in Shannon’s entropy equation, is a measure of the change in the value of delta t(x) in this classical entropy. Bound state quantum mechanics is said to be time-independent, but by the correspondence principle (1) the probability of the envelope of the wavefunction i.e. Envelope(W(x))* Envelope(W(x)) = delta t(x) so quantum probability (albeit linked to the envelope) is related to classical time intervals which adjust with x to yield the proper ration v(x)=delta x / delta t(x) with delta x being constant. In this note we argue that the expectation value: Integral W(x) d/dx dW/dx = Integral WW KEclassical(x) for a bound state is, in the high energy limit Integral dt KEclassical(x) i.e. a portion of the action, just as in the classical case (i.e. the correspondence principle). This leads to the association of the higher energy limit of the envelope of W times itself with delta t(x).
In the literature (e.g. (1)), the expression - density(x) ln(density(x)) is used as Shannon’s spatial entropy for a quantum system (bound or not). For the bound case, the correspondence principle is said to hold for high energy levels (2). In that case, quantum density = W(x)W(x) (W(x)=real wavefunction) is said to be equivalent to the classical density C/v(x) where v(x) the velocity is given by .5mv(x)v(x) + V(x) = E. Classical density yields a curve which passes through peak regions of quantum density humps. In the classical case, particle motion is completely determinate. C/v(x) as a density (probability) is proportional to dt, the amount of time a particle spends passing through a fixed length dx. If the particle moves quickly, it spends little time, and hence there is less time for a measurement (i.e one may miss the particle due to its speed). We argue in this note that such a scenario does not seem to be in keeping with the ideas of Shannon’s entropy where one has probabilities for certain mutually exclusive events at each ti. For example at each ti, one may toss a coin. If it is not “heads”, this means it is tails. For the classical particle in the correspondence principle, each dx region has its own probability proportional to dt = dx/v(x). Thus, each dx is a region of its own in terms of probability. Thus, if Ci measurements of the presence of the particle are made within dx of xi out of N trials, the probability that a measurement is made in dx at xi is Ci/N and the probability that it is not is 1-Ci/N. Thus, unlike Shannon’s picture, various arrangements do not have the same weight or probability. We investigate this scenario in more detail and argue that it does not follow the approach of Shannon’s entropy. We also try to connect with the quantum situation.
In bound state quantum mechanics, one has statistical observables (based on P(x) where P is probability) such as spatial density d(x), potential energy density V(x)d(x) and kinetic energy density .5mv(x)v(x) d(x), where v(x) is the classical velocity obtained from .5mv(x)v(x) + V(x) = E. (Kinetic energy density can also be written as -1/2m W(x) d/dx d/dx W(x) where W(x) is the wavefunction). The energy of a state can be obtained by integrating these statistical observables i.e. Integral dx d(x) V(x) + Integral dx d(x) .5mv(x)v(x) = E. If one performs this integration for two different states, one may calculate the energy difference. This begs the question: why does quantum mechanics deal with conditional probabilities such as P(p/x) directly related to W(x)? The issue perhaps begins with the appearance of V(x) and E in the time independent Schrodinger equation. Kinetic energy (not kinetic energy density) should appear and be somehow linked to some motion. A question arises as to why the energy density is proportional to the spatial density i.e. Energy density = E d(x)? This is directly linked to the time-independent Schrodinger equation, but one may try to argue that classical energy conservation .5mv(x)v(x) + V(x) = E at each point x should hold in quantum mechanics with .5mv(x)v(x) equal to Sum over p p*p/2m P(p/x) where P(p/x) = fp exp(ipx)/W(x). If one, however, considers excited states written as Wn(x)= Wo(x)Hn(x) where Wo(x) is the ground state wavefunction, one finds that: -1/2m d/dx d/dx Hn(x) - 1/m [d/dx Wo(x)]/Wo(x) [d/dx Hn(x)] = E1 Hn(x) ((1)) In other words, an eigenvalue equation appears again with no V(x) present. If one assumes Hn(x) has physical meaning (i.e. is related to a phonon or photon), then ((1)) may suggest there is a physical principle in place, i.e. the idea that one must have a “constant” energy at each point x. Here E1 is strictly kinetic energy (although not all of it). This “principle” has consequences as it restricts the values of E1 (i.e. discrete eigenvalues) and leads often to a periodic Hn(x) which has direct effect on kinetic energy density and potential energy density. Thus, having Hn(x) interact with Wo(x) to produce a constant energy E1 times Hn(x) does affect kinetic and energy densities at each x point, just not kinetic energy itself (it is constant i.e. E1). The purpose of this note is to examine this possible principle behind quantum mechanics.
Quantum mechanics is often compared with classical mechanics because both may deal with a single particle subject to a potential. Classical mechanics is said to be deterministic describing x(t) and its derivatives i.e showing how a particle moves and accelerates while bound state quantum mechanics produces a real wavefunction W(x) such that W(x)W(x) yields a spatial density profile. The comparison is carried further with the correspondence principle which states that at high energy levels quantum mechanics should mimic classical mechanics. Classical mechanics does not simply describe x(t) and its derivatives i.e. deterministic motion. It also includes two body scattering, force and torque balances for a stationary object etc. In the case of the classical statistical mechanics of an equilibrium ideal gas, the Maxwell-Boltzmann (MB) distribution in the absence of a potential is C1 exp(- pp/(2mT) ). If a potential V(x) is added, one may find P(x) two different ways using Newtonian ideas. The first is to consider the gas as a collective entity like a tiny macroscopic classical object at x. Then one may argue that equilibrium means the system is stationary and apply Newtonian statics i.e. a balance of pressures and -dV/dx to obtain exp(-V(x)/T) as is often done in textbooks. Alternatively, one may consider a single particle and apply Newtonian mechanics differently, namely use a conservation of energy equation: p1p1/2m + V(x1) = p2p2/2m + V(x2). Again one obtains exp(-V(x)/T). The spatial part of probability may be based on Newtonian mechanics indicating that it is not stochastic. The stochastic function is C1 exp(-pp/2mT). Interestingly, however, the two Newtonian approaches are different with respect to a single particle. A collective nonmoving set of particles acts as a stationary object, yet within it each particle seems to be “bouncing” off of an imaginary wall creating pressure. Thus one has an impulse free particle picture yet obtains the correct result exp(-V(x)/T). Alternatively one may think of a single particle accelerating which it does if it is not colliding with other particles. This is a very different picture of the single particle than of one bouncing back and forth hitting an imaginary wall. In fact, there is usually a wall holding the gas together and pressure and bouncing particles are the real picture there, but within the gas one would normally think of a single particle accelerating. Nevertheless there are two scenarios which yield the same result. The idea of pressure balance is more in keeping with an overall statistical picture of the gas. In other words, one may actually ignore the idea of following a single particle as it accelerates. (That is not to say that single particle acceleration does not exist - it does.) Quantum mechanics we argue is a statistical theory. In the case of a single bound particle one has a momentum distribution W(x)=wavefunction = Sum over p a(p)exp(ipx). One may apply conservation of average energy at each x to obtain the time-independent Schrodinger equation. In such a picture, like with the statistical mechanical gas, we argue that one may imagine impulse hits with the potential instead of considering acceleration. The quantity vrms(x) from KEclassical= .5m vrms(x)vrms(x) = -1/2m d/dx dW/dx / W does display acceleration, but this is a mathematical average. As energy levels become higher the correspondence principle is said to apply, but a classical particle exhibits acceleration. The quantum bound state does not, but like in the case of the statistical mechanical gas it may be that two scenarios lead to the same result. In other words, the quantum bound W(x) still has crests and troughs, but with low resolution one may only see an envelope roughly touching W(x)W(x) which matches a classical C2/v(x) where v(x) is classical velocity. Given that quantum mechanics is a statistical theory it may not be necessary to have it explicitly display acceleration.
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.
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.