Statistical mechanics seems to be based on the idea of distributing a fixed amount of energy among a number of particles. (Usually this number is very large.) Alternatively, one may think of a single particle changing its state (e.g. momentum) during many time intervals. In both cases, one takes an average of the energy, but this means one needs weights for each single particle state. These weights seem to be based on the idea that each ‘vector’ of states has the same probability of occurrence. Thus, if one has N particles and an energy eN, one vector involves one particle having eN and the rest 0. Many vectors, however, may be created with N particles having energy near e. Thus, the weight for a single particle to have energy e is much higher than for it to have energy E. The weight is ultimately taken to be the Boltzmann factor. In quantum mechanics, it seems a very different assumption is made. Unlike classical physics where one follows a particle in space and time, knowing its acceleration, position and velocity at each x, in quantum mechanics, collisions with the potential seem to be granular, and the main assumption seems to be that a quantum particle may be in any momentum plane wave state at a given time. Like statistical mechanics, there is a set of weights, but one no longer uses the scenario of distributing energy among different states to determine these weights. Instead, the weights are obtained by noting that a quantum average at a point x is taken using fp exp(ipx)/ (Sum over fp exp(ipx)). The weights must ensure that: -1/2m (Sum over p p2/2m fp exp(ipx)) / (Sum over fp exp(ipx)) = E-V(x) at each x point. Thus, the plane wave behavior of quantum particles becomes a dominant factor. For the case of V(x)=0, the plane wave nature of the possible single particle momentum states leads to an overall (Sum over fp exp(ipx)) which has the appearance to two plane waves. This is very different from the classical statistical mechanical case. In this note, we attempt to examine the difference of these two statistical theories, and understand why in one case, the ground state of an oscillator, the two become the same.
Statistical mechanics seems to be based on the idea of distributing a fixed amount of energy among a number of particles. (Usually this number is very large.) Alternatively, one may think of a single particle changing its state (e.g. momentum) during many time intervals. In both cases, one takes an average of the energy, but this means one needs weights for each single particle state. These weights seem to be based on the idea that each ‘vector’ of states has the same probability of occurrence. Thus, if one has N particles and an energy eN, one vector involves one particle having eN and the rest 0. Many vectors, however, may be created with N particles having energy near e. Thus, the weight for a single particle to have energy e is much higher than for it to have energy E. The weight is ultimately taken to be the Boltzmann factor. In quantum mechanics, it seems a very different assumption is made. Unlike classical physics where one follows a particle in space and time, knowing its acceleration, position and velocity at each x, in quantum mechanics, collisions with the potential seem to be granular, and the main assumption seems to be that a quantum particle may be in any momentum plane wave state at a given time. Like statistical mechanics, there is a set of weights, but one no longer uses the scenario of distributing energy among different states to determine these weights. Instead, the weights are obtained by noting that a quantum average at a point x is taken using fp exp(ipx)/ (Sum over fp exp(ipx)). The weights must ensure that: -1/2m (Sum over p p2/2m fp exp(ipx)) / (Sum over fp exp(ipx)) = E-V(x) at each x point. Thus, the plane wave behavior of quantum particles becomes a dominant factor. For the case of V(x)=0, the plane wave nature of the possible single particle momentum states leads to an overall (Sum over fp exp(ipx)) which has the appearance to two plane waves. This is very different from the classical statistical mechanical case. In this note, we attempt to examine the difference of these two statistical theories, and understand why in one case, the ground state of an oscillator, the two become the same.
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.
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.
In classical statistical mechanics, C exp[ -( mv*v/2 + V(x))/T) represents density i.e. the probability for a particle to have a velocity v and be at x. If applies this probability to Shannon’s entropy formula, one obtains C/T ( mv*v/2 + V(x)) exp[ -( mv*v/2 + V(x))/T) + exp[ -( mv*v/2 + V(x))/T) ln(C). Thus, it seems energy times density + density times ln(C) is the entropy density, a well known result in classical statistical mechanics, although sometimes not explicitly stated. The first term integrated over x is essentially the kinetic energy, while the second integrated over velocity yields the potential energy term. Thus, the first term partial density seems to be a purely momentum dependent kinetic energy piece times probability and the second a purely x dependent potential energy piece. (There is still an integral over velocity for the first term and one over x for the second to obtain the full entropy.) In quantum mechanics, we argue that a(p1)exp(ip1 x) a(p2) exp(ip2 x) serves as the probability where W(x)=Sum over p a(p) exp(ipx) = wavefunction. This leads to an expression for entropy density with two terms, one of which may be immediately integrated over momentum and the other over space. If one wishes to make a link to the classical statistical mechanical case, one may ask what requirement is necessary for the momentum related piece to represent average kinetic energy and the spatial part, average potential energy. The requirement, we find, yields the wavefunction of the ground state harmonic oscillator. For other cases, the partial density is not equivalent to average energy. Thus, we try to establish a link between entropy density and energy density, but require the kinetic energy related term to be strictly a function of momentum and the potential related term strictly one of x. In quantum mechanics, one may form averages at a point x. For example, KEave(x)= [Sum over p p*p/2m a(p) exp(ipx)]/W(x). If one examines a p/-p pair with a(p)=a(-p) (for example), cos(px) terms arise which are positive at some x and negative at others. We argued in previous notes that this represents stochastic flow in the opposite direction from that expected by -dV/dx. In other words, -dV/dx is an average which consists of stochastic force addends which may give momentum hits to the right or left, but which add to yield -dV/dx. As a result, one may have negative contributions to the kinetic energy (and density as well). In this note, we argue one may also have negative contributions to the entropy density due to this phenomenon.
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).
In physics (including classical physics), conservation of momentum is always considered, but one must be more careful with conservation of energy as it may include forms like sound etc. As a result, there is a tendency to focus on conservation of momentum all of the time, and conservation of particle energy only in elastic collisions. In addition to this, one may think of a particle striking a target, in which case two equal momentum (but different energies due to different v speed and mo rest mass) cause the same impact. In the case of an interaction with a potential, however, it is strictly energy considerations which are important classically, i.e one uses p dot p /2m + V(x) = E. Knowing momentum without rest mass is not sufficient. One may consider it unusual that one may consider two different scenarios, one which focuses only on momentum (impulse) or conservation of momentum in collisions while only energy is the focus in classical bound states or motion in V(x). Given the classical focus on momentum, imagine that one wishes to describe all free particles with different momenta as a uniform probability distribution such that a product of two probabilities for p1,p2 is the same as for p3,p4 if p1+p2=p3+p4. (We use 1-dimension for simplicity.). One cannot use P= 1/n because p is absent and n->infinite. In one sense, all particles must have the same probability, but at the same time must contain extra information related to p for the conservation of p part of the problem. The notion of a uniform distribution is independent of conservation of momentum, hence of p momentum. Conservation of momentum requires an explicit presence of p. As a result, there seems to be a paradox. One may, however, resolve this paradox by arguing that a probability is a mathematical way of presenting a subset of the information present. If one argues for a uniform probability, then it does not need to contain p, and one would have 1 as an unnormalized probability for any free particle p. A different probability (representing a different subset of information) which respects all free particles as having the same weight, but includes p can also be constructed, namely exp(i C p). C is a constant introduced for unit issues. In this case, the modulus gives all free particle probabilities with p the same weight, but the presence of p allows one to have conservation of momentum. Given that momentum is the key variable that is conserved in classical physics, one should be satisfied with exp(i C p). One might argue that there is no need to consider energy E. We note, however, that if one has p, E1 and -p, E2 in a frame, then in a frame moving with -v, one has gp+gvE1= p1’ and p2’ = -gp + gvE2. Adding yields gv(E1+E2) and so energy automatically appears and cannot be discarded. Furthermore for p, E1 and -p, E2, exp(i C 0) is exp(i C gv(E1+E2)) as seen in the moving frame and so one is forced to consider conservation of energy together with momentum and include both p and E in the more detailed probability. In other words, special relativity automatically links p and E in a 4-vector, leading to the idea that both are important. We argue that one cannot simply consider exp(i C p) due to special relativity. Both E and p must be considered together and this is a key idea, we argue, because if one wishes to have a Lorentz invariant probability, it must be in E and p, i.e. -Et+px. Instead of having a probability for an interaction with p given by exp(i C p) , one has exp(-iEt+ipx) and we argue that this must have the same value in all frames for a particular interaction. One may note that E and p are, on the one hand, separated allowing for impulse hit versus E related interactions, but that overall the probability for interaction includes both and it is the full -Et+ipx which must be invariant. This further suggests that if one does not artificially force an interaction to occur at the center-of-mass, then x,t and x+hbar/p, t+hbar/E are equally good points for the interaction. We argue that there is no need to force x=vt because exp(-iEt+ipx) * exp(iEt+ipx) already shows that one has uniformity, not only for any p,E, but also for any x, t, as there are now four variables. Thus, there is no reason to enforce x=vt in exp(-iEt+ipx), when considering an interaction, we argue. This means that one has two possible interactions at different x,t values, i.e. x,t and x+hbar/p, t+hbar/E. If one is unable to follow the interaction in time, but may follow it in space as in the case of 2-slits about hbar/p apart or for a potential V(x), then one may use the exp(ipx) part of the interaction probability piece by itself. Thus, we argue that special relativity does not allow one to treat p independently of E. For example, for p values that are the same in one frame, but have E1, E2, will be different in another frame. One must consider E and p together and so instead of creating a probability for p alone, one must create one for E,p and then impose Lorentz invariance which leads to different x,t pairs yielding the same probability value for an interaction, i.e. free particle quantum mechanics.
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.