In classical physics, a potential V(x) acts on a particle at a location x regardless of its velocity which is taken to be constant within dx (even though the particle accelerates). Thus writing potential energy V(x) at a given x is like taking a snapshot of the particle when it is at x. The snapshot is static and does not indicate the dynamics of the problem as the particle may be moving either to the right or left, which is irrelevant in terms of the exact potential energy and even kinetic energy at point x. A quantum particle moving with constant velocity is described by a two-dimensional dynamic probability exp(ipx), with cos(px) representing a static density and sin(px) a time-shifted density to capture one directional motion such that exp(ipx)exp(-ipx)=1, we argue. V(x), however, only acts on the static density (linked to cos(px) and not sin(px)). Thus it seems that bound state quantum mechanics needs to remove the dynamic pieces of free particle wavefunctions exp(ipx) through interference. One might argue that for wavelengths of the order of the system length one does not know if one has exp(ipx) or exp(-ipx) at an x point, but for high energy levels, wavelengths become small with respect to the system length and yet interference is still present. Thus it seems that for the calculation of energy one requires a static average snapshot of the bound particle. Physically, we argue that the dynamic piece is still present i.e.. if one knocks out a bound particle and it is found in the p momentum state, it is represented by exp(ipx). Thus the purpose of the bound state wavefunction is to facilitate the calculation of average energy which is associated with a static snapshot type of density and not necessarily represent all of the dynamical information in the system i.e. the two dimensionality of each exp(ipx). As a result, the bound wavefunction does not show motion to the right and then to the left in the high energy limit.
In classical physics, a potential V(x) acts on a particle at a location x regardless of its velocity which is taken to be constant within dx (even though the particle accelerates). Thus writing potential energy V(x) at a given x is like taking a snapshot of the particle when it is at x. The snapshot is static and does not indicate the dynamics of the problem as the particle may be moving either to the right or left, which is irrelevant in terms of the exact potential energy and even kinetic energy at point x. A quantum particle moving with constant velocity is described by a two-dimensional dynamic probability exp(ipx), with cos(px) representing a static density and sin(px) a time-shifted density to capture one directional motion such that exp(ipx)exp(-ipx)=1, we argue. V(x), however, only acts on the static density (linked to cos(px) and not sin(px)). Thus it seems that bound state quantum mechanics needs to remove the dynamic pieces of free particle wavefunctions exp(ipx) through interference. One might argue that for wavelengths of the order of the system length one does not know if one has exp(ipx) or exp(-ipx) at an x point, but for high energy levels, wavelengths become small with respect to the system length and yet interference is still present. Thus it seems that for the calculation of energy one requires a static average snapshot of the bound particle. Physically, we argue that the dynamic piece is still present i.e.. if one knocks out a bound particle and it is found in the p momentum state, it is represented by exp(ipx). Thus the purpose of the bound state wavefunction is to facilitate the calculation of average energy which is associated with a static snapshot type of density and not necessarily represent all of the dynamical information in the system i.e. the two dimensionality of each exp(ipx). As a result, the bound wavefunction does not show motion to the right and then to the left in the high energy limit.
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).
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.
In a previous note (1) we argued that the translational generator d/dx being associated with momentum (px) already appears in Fermat’s principle for both reflection and refraction. We suggested that d/dx is linked to an invariant spatial variable which in turn is associated with momentum conservation in the same direction. This may be written as: d/dx (px)x = (px). Alternatively (px)x = -i ln(exp(i px x) ) with exp(i px x) being a directional probability distribution (i.e. values for p and -p differ) and ln(exp(i px x)) being of the form of information in information theory. One may then ask: How does one link average momentum in the case of a bound state with a potential V(x)? In general, such a bound system is solved using the time-independent Schrodinger equation i.e. conservation of energy using d/dx twice and a p(rms) not a p average obtain from -id/dx ln(W) = p(x) where W(x) is the wavefunction and pave= W dW/dx. After all, the form exp(i (px) x) may be obtained from considerations of d/dx and p, not energy conservation (even though E=pc for light). We argue that exp(ipx) is a directional probability. If both p and -p are present then: p exp(ipx) -p exp(-ipx) is not zero because the spatial probability distinguishes between the two. Only by multiplying exp(ipx) and exp(-ipx) does one remove motion. If p and -p are present, however, as in an OR situation, i.e. exp(ipx) + exp(-ipx), then it is as if the particle is bound or localized and P(p,x)+P(-p,x) = C cos(px). We consider the two components of exp(ipx) in: Integral dx cos(px) sin(px). Why does this equal 0 from a momentum perspective? sin(px) = 1/p d/dx cos(px). Thus the integral is: Integral dx { P(p)+P(-p)} momentum(P(p)+P(-p)) } or the average momentum at x. Overall the particle is bound and so there can be no momentum globally so the integral must be zero and hence cos(px) and sin(px) must be orthogonal over a period even though they represent the same p value. This same idea holds for a linear combination of {P(p,x)+P(-p,x)} sums for different p’s. (We consider overall positive or negative parity.) Thus: Integral dx W dW/dx = 0 holds for even-odd reasons, but also for the physical overall zero momentum. Even with a potential present: -i WW d/dx ln(W) = Pave(x). In other words, bound state quantum mechanics is linked to W dW/dx integrating to 0 even though there are positive and negative pockets at various x values. This follows from the orthogonality of cos(px) and sin(px) over a wavelength from a free particle which conserves momentum. This result holds for any V(x) even though V(x) fixes the coefficients of the linear combination of Sum over p a(p) {P(p,x)+P(-p,x)} using energy considerations. Zero average momentum follows from momentum conservation found in exp(i (px) x). In fact we have argued that the unusual probability wave form exp(i (px) x) exists to ensure constant momentum (momentum conservation) for a free particle. We argue that considering average momentum equal to 0 overall leads to ideas of zero point energy, orthogonal energy states and a crest/trough pattern without considering the time-independent Schrodinger equation which is based on energy conservation. Thus we argue that ideas related to d/dx and momentum carry over from Fermat’s principle into a bound quantum state and its average momentum which is often not considered.
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 (1) the Wigner transform of an oscillator is calculated and the high n energy level limit taken using the WKB approximation yielding a result proportional to delta(p - p(x)) + delta(p+p(x)) where p(x) = sqrt(2m(E-.5kxx)). The Wigner transform involves quantum wavefunctions which for bound systems include +/- p components which interfere i.e. a(p)(exp(ipx) + exp(-ipx) for a(p)=a(-p). Classically, a bound particle moves to the right during one time interval and to the left during the other so there is no interference of +/-p. The quantum wavefunction retains +/-p interference for all energy levels and so does not portray the full physical information of the bound system i.e. leads to the two delta functions. For high n (energy level), crests and troughs are very narrow compared to the length of the system, approximating dx units of length. Thus, uncertainty in time is only within these tiny crests/troughs, otherwise there is a distinct time position relationship as in classical physics, but the wavefunction W(x) does not show it. How then may the wavefunction be interpreted? We argue the time independent wavefunction, say for a bound state, represents an averaged steady stream picture which is often not the physical picture. For example, a bound particle is not a steady stream average nor is a finite number of photons hitting an interface of media with different indices of refraction/reflection. Nevertheless, the steady stream approach with time removed may be used to solve the problem i.e. ultimately find a one dimensional spatial density (which averages out the original two dimensional probability of exp(ipx), energy levels etc. For instance, the energy levels of a bound system do not depend on following the particle in time. We raise this issue because we think it may have implications on the calculation of entropy in bound single particle quantum systems. In general, a spatial Shannon’s entropy is defined as: - W(x)W(x) ln[W(x)W(x)] (one dimensional problem). W(x) (wavefunction) is real because p/-p pieces interfere in a steady stream picture. A single bound particle, however, is not a steady stream picture. exp(ipx) is a dynamic probability equal to cos(px)+i sin(px) to indicate the direction of motion. Simply because exp(ipx)+exp(-px) removes the complex portion (i.e. for a steady state picture with a(p)=a(-p)) does not mean that this represents the full physical picture especially as energy levels become high and crests/troughs become narrower with respect to the overall system length it seems there is no justification for the steady stream interference picture.. In other words, using a real W(x) in a Shannon’s entropy expression treats one dimensional probability W(x) = bound wavefunction, but this is not the full probability in the picture, we argue. Full probability is two dimensional and demonstrates direction for many energy levels (especially high ones). For a free particle exp(ipx), cos(px) and sin(px) are essentially the same distribution. Thus cos(px)cos(px) describes the probability distribution in x. For W1(x)=Sum p>0 a(p)exp(ipx), Real(W1(x)) is not equal to Imaginary(W1(x)) because one does not have a simple shift of the real distribution to obtain the imaginary one. The real distribution is chosen to represent probability because the interference of exp(ipx), exp(-ipx) terms removes the imaginary piece, but it is removed only on average. If physically exp(ipx) and exp(-ipx) should not interfere, then one may need to take some kind of average of Real(W1(x))Real(W1(x)) and Imaginary(W1(x))Imaginary(W1(x)) and use this in Shannon’s entropy.
The classical limit of quantum mechanics is often considered in terms of the time dependent Schrodinger equation. In this note we consider the bound case solution i.e. a time dependence of exp(-iEt) and as in previous notes argue that time has been removed from this statistical equilibrium picture as a the quantum particle tries to establish a kind of impulse potential field based on superpositions of exp(ipx) terms. Such a potential field is linked with a bound state and does not involve following the particle about in time as it undergoes stochastic impulse hits with exp(ikx) terms from V(x)=Sum over k Vk exp(ikx). On average a classical energy balance equation KE(x)+V(x) = En holds for every energy En, but the average is created by a spread of momenta i.e. W(x)=wavefunction= Sum over p a(p)exp(ipx). In this note we consider the case of a particle in an infinite potential well and note how the spatial resolution becomes very high at the same time that the momentum spread becomes very narrow in the case of high n (energy level). Thus as n becomes approaches infinite, it is not just the vrms(x) which is linked with x, but the statistical scenario of quantum mechanics approaches a situation in which x is being resolved to a very high level and p is almost constant. Furthermore quantum entropy becomes a constant (independent of n) as n→ infinite suggesting a theory which is nonstatistical in this limit.This would represent a classical picture except for the fact that time is still removed from the quantum bound state i.e. forward and backward exp(ipx), exp(-ipx) still interfere. Given that x and p are well resolved locally for high n it seems one may follow the high n quantum particle in space and time and manually remove the forward and backward portions to create a theory of a particle moving in one direction under V(x) i.e. classical physics. This bypasses the idea of a wavepacket whose x-spread increases in time and allows one to base classical physics on a bound state solution of a quantum particle in an infinite well.