In Part I we noted that the quantum oscillator result E= hbar w (n+.5) implies that for exp(-iEt) there are special times proportional to 1/E which in turn are proportional to 1/w which is proportional to the classical cycling time of the oscillator. In other words for oscillators with different w there are different 1/w times such that -Et (which is part of a free particle action A= -Et+px) is constant i.e. one cannot distinguish as in statistical mechanical equilibrium. In this note, we consider the free particle action in more detail. From A=-Et+px there are special times (1/E) and lengths 1/p proportional to wavelength such that different E and p values when multiplied by these times and length have the same value. Furthermore A=0 for this values. We argue that it is this feature of px in exp(ipx) which is responsible for the energy of an oscillator being proportional to w, the classical angular frequency. Thus the idea of a special length (wavelength) such that different p’s when multiplied by 1/p produce the same value leads to a quantum energy such that 1/E is proportional to a physical cycling time in the classical problem. Thus different w’s lead to E/E being constant just as p/p is constant.
In Part I we noted that the quantum oscillator result E= hbar w (n+.5) implies that for exp(-iEt) there are special times proportional to 1/E which in turn are proportional to 1/w which is proportional to the classical cycling time of the oscillator. In other words for oscillators with different w there are different 1/w times such that -Et (which is part of a free particle action A= -Et+px) is constant i.e. one cannot distinguish as in statistical mechanical equilibrium. In this note, we consider the free particle action in more detail. From A=-Et+px there are special times (1/E) and lengths 1/p proportional to wavelength such that different E and p values when multiplied by these times and length have the same value. Furthermore A=0 for this values. We argue that it is this feature of px in exp(ipx) which is responsible for the energy of an oscillator being proportional to w, the classical angular frequency. Thus the idea of a special length (wavelength) such that different p’s when multiplied by 1/p produce the same value leads to a quantum energy such that 1/E is proportional to a physical cycling time in the classical problem. Thus different w’s lead to E/E being constant just as p/p is constant.
We take as our starting point the free particle classical action which may be written as A= -Et+px if t and x are treated as independent variables. It is possible to find two relationships linking v (velocity) to x and t from dA=0. The first is the classical wave result vp = E while the second is p=Ev where E=mo/sqrt(1-vv) (c=1). Furthermore one may write dA/dx partial = p ((1a)) and dA/dt partial = -E. ((1b)) This may be generalized to two eigenvalue differential equations, but this immediately begs the question: Why do this? We argue that Lorentz invariant form -Et+px represents scaling of both t and x. For example, p scales x and we argue there should be physical relevance to this. This physical relevance results in the wavelength proportional to 1/p. In other words, -Et+px represents more than velocity. The treatment of the scaling for a quantum wave and classical wave differ, we argue, in that the former keeps E,t and p,x separate while a classical wave does not. In other words pv=E applies to a classical wave which is described by cos(-Et+px). In other words, one takes ((1a)) and ((1b)) which separate E,t and p,x into two independent equations and takes account of scaling by combining the two (to include both x and t) i.e. creating the classical wave equation: 1/vv d/dt d/dt partial W(x,t) = d/dx d/dx partial W(x,t). We argue that quantum mechanics on the other hand keeps ((1a)) and ((1b)) separate. The velocity (relativistic) p=Ev is not a wave velocity. A solution to ((1a)) is exp(ipx) and to ((1b)) exp(-iEt). These are usually combined for a free particle to give exp(-iEt+ipx) and a wave is described with a phase velocity pv(phase)=E. We prefer to keep exp(-iEt) and exp(ipx) separate . For example, for two slit interference one does not make use of phase velocity or exp(-iEt). For a bound state one uses p=Ev, but a single free particle is represented by exp(ipx) in W(x)=wavefunction = Sum over p a(p)exp(ipx). exp(-iEn t) appears after one has found En, using the exp(ipx)’s in the sense that one does not include an exp(- i pp/2m t) for each exp(ipx). We argue that by separating E,t and p,x, the latter leads to a scheme for which there is “no time”. It is true that the particle moves at a speed x/t, but p=m1v1=m2v2 etc so p does not uniquely define velocity, rather it defines an impulse hit 2p = Integral F dt . Thus the scaling px which leads to an x, p only wave exp(ipx) is linked to an interaction. If the particle is moving freely and not interacting, then one has classical determinism i.e. v=constant = x/t. As soon as an interaction occurs, one changes from variables x,t which are linked to x, P conditional (x/p) and there is uncertainty in the position of x within a wavelength. The phase in exp(ipx) is only dependent on x not on the time the particle reaches x. exp(ipx) thus represents a probability and one must consider a sum of different possible outcomes (as in a two slit experiment) in order to describe an interaction. This is very different from the local interaction of classical mechanics. Nevertheless for a bound state one makes use of the two formalisms i.e. KE(x)= Sum over p pp/2m a(p)exp(ipx) / W(x) = KEclassical(x). We try to discuss some of 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).
In a previous note (1) we considered a free particle entropy based on the real part of the wavefunction exp(ipx) i.e. |cos(px)|. |cos(px|) may be normalized over a certain length, just as exp(ipx) is. The idea is that there exists an uncertainty in space within a wavelength which should be reflected in an entropy expression. As p becomes very large, cos(px) becomes very narrow and has many peaks in L. Thus probability →0 so p ln(p)--> 0 using l’Hopital’s rule. This coincides with a free classical particle having zero entropy. In this note, we revisit arguments presented previously, suggesting that A= -Et+px, a Lorentz invariant and also the free particle classical relativistic and nonrelativistic action for v=x/t We suggested in the past that A is responsible for both the wavelength hbar/p and the time period hbar/E. Here we suggest uncertainty within a time period should be considered together with uncertainty within a wavelength. For instance, recent experiments (2) have shown that one may have quantum interference in time as well as space. Thus there is uncertainty for a free particle in both space and time. We add the entropies for these two. Next, we consider the case of a bound system. Such a system creates a crest-trough pattern (with different heights and widths) for its spatial wavefunction, but E, average energy is constant and there is the free particle type exp(-iEnt). We suggest that the system is characterized by uncertainty in space and time with En being a known constant at each x. This suggests that various free particle p (momentum) values combine to create the wavefunction such that -1/2m d/dx d/dx W/W is average kinetic energy which in turn adds to V(x) to create a constant En. We are thus concerned with uncertainty in time and space and not in momentum. In the limit of high energy values, a crest fits within a tiny dx and prms is essentially the classical p value which is considered certain, even though in quantum mechanics the range of a(p) weights hold for all n (W(x)=Sum over p a(p)exp(ipx)). Thus we treat a high n limit as being completely certain and not having entropy. This differs from the approach of using C/v(x), where v(x) is velocity given by .5mv(x)v(x)+V(x)=En, as a probability and calculating Shannon’s entropy using this.
The classical Lagrangian L leads to Newton’s second law which is equivalent to an energy-momentum conservation equation. The classical action: A = Integral dt1 L(t1) for a constant velocity v=X/T may be varied with respect X and T i.e dA/dT = -E and dA/dX= p. Thus, one may create a differential equation in (dA/dT)(dA/dt) and dA/dx dA/dx for the EE=pp+momo equation. Alternatively, one may create other differential equations. Quantum mechanical differential equations, however, are quite different with exp(i Action) being a solution. Even this may be written as the solution of a non-Schrodinger equation i.e. id/dt exp(iAction) = -1/2m (d/dx exp(iAction)) (d/dx exp(iAction)) for a free particle. For a constant velocity, exp(iAction) is equivalent to exp(ipx-iEt). We argue that this solution is statistical because it separates x and t (as independent variables) and holds for all x and t. Thus, quantum mechanics seems to be a statistical approach using free solutions exp(ipx-iEt) to create an ensemble as in classical statistical mechanics. A difference, however, is that exp(ipx-iEt) “interferes” even if one only takes the real part cos(px-Et). We consider the case of a particle in a box with infinite potential walls to argue that the periodicity is warranted i.e. positive and negative values (interference). Even though the modulus is 1, exp(ipx-iEt) may be operated on by various generators e.g. d/dx the generator of translations. This is analogous to operations on vectors in linear algebra. As noted above, quantum equations are not only energy-momentum conservation equations. They yield a wavefunction W(x) which is an unusual momentum distribution W(x)=Sum over p a(p) exp(ipx). exp(-iEt) holds overall because average E is the same at each x. Thus, one needs a W(x) which leads to a momentum-energy conservation equation, but also satisfies certain boundary conditions which become linked to spatial density= W*W. This leads to discrete energy levels and also cases of humps and troughs in W(x). We attempt to investigate some of these ideas in this note.
In Parts I and II we noted that a quantum oscillator time-independent Schrodinger equation may be scaled such that x→y with yy=sqrt(km)xx and E→ Esqrt(m/k). Given that the LHS side is a function strictly of y which may be integrated from -infinite to infinite, the RHS must be independent of k and m. Thus E is proportional to sqrt(k/m) which “ happens” to be the oscillator classical angular frequency. In Part II we noted that the scaling occurs in the Schrodinger equation because one uses -id/dx as the momentum operator which follows from the exp(ipx) form of a free particle wavefunction. In this note, we show that for any equation: ad/dx d/dx W + b xx W = E W scaling of the form x→y such that yy =xx sqrt(b/a) and E→ E 1/sqrt(ba) occurs. Furthermore if d/dx is associated with momentum then writing conservation of energy using m/2 vv leads to 1/a www cos(wt)cos(wt) + b sin(wt)sin(wt) = b and w is proportional to sqrt(ab). Thus E must be proportional to w, the classical frequency. (An equation of the form a df/dt df/dt + b f f = constant must have a cos(wt) solution as is already known.) We try to extend these ideas by considering a wave on a string with both ends fastened. The string produces sinusoidal waves, but sound is created if there is air. One may consider sound to come in identical units of energy proportional to the angular frequency w. This is similar to the Planck blackbody radiation assumption about light. It may be shown that the string follows an oscillator solution because d/dx dy/dx which represents the change in the sin(theta) multiplied by the constant tension from x to x+dx is set equal to y i.e Force is proportional to y which is the oscillator result. If one sets energy proportional to w as in the case of light and wishes to have an equation strictly in x describing this scenario and also linked to energy conservation then the momentum squared term should have dimensions 1/xx. This suggests d/dx d/dx as a solution because E must be proportional to w. Thus we argue that one may work “backwards” modeling phonons like photons in a classical problem to obtain the quantum operator p = -id/dx which is linked to the free particle wavefunction exp(ipx).
In Part III we argued that a Schrodinger equation of the form: a d/dx d/dx W(x) + bxx W(x) = E W(x) may be scaled x→y such that yy = xx sqrt(ba) and E→ E sqrt(a/b) so that E is proportional to sqrt(b/a). We applied this idea to the quantum oscillator and noted that classically w, the angular frequency, for the same equation is sqrt(b/a). We further argued that this implies the momentum operator in quantum mechanics should scale as 1/xx as d/dx d/dx does. In this note we use a slightly different argument to analyze this same problem. Instead of treating the classical oscillator and time-independent Schrodinger equation as separate, we link them and show that scaling must occur with E being proportional to w. Furthermore we argue that if quantum mechanically one wishes to use probabilities f(p,x) to calculate average kinetic energy at x i.e. KE(at x) = {Sum over p pp/2m f(p,x) } / {Sum over p f(p,x)}, then f(p,x) = f(px) and pp should scale as 1/xx. Thus comparison of the classical oscillator equation directly with a statistical average type of quantum equation which uses V(x), E and KE(at x) yields information about E, f(p,x) and p i.e. introduces the idea of wavelength. Furthermore the discrete quantum energy values E=hbar w (n+.5) are dependent on w not through any boundary value considerations.