A classical bound state does indicate the direction of motion if one only considers kinetic energy pp/2m. It is the vector quantities, p and x, which indicate direction. In quantum mechanics, a free particle interaction probability (for impulse hits) is directional, i.e. exp(ipx) not= exp(-ipx). Unless one has a single value of pp/2m, KE+= sum over p>0 a(p)exp(ipx) pp/2m / Sum over p>0 a(p)exp(ipx) is not equal to KE- even if a(p)=a(-p). In order to use conservation of energy in quantum mechanics, one is forced to consider both a(p)exp(ipx) +a(p)exp(-ipx) with a(p)=a(-p) which is the correct statistical weighting for a bound state. One may note that exp(ipx) is a probability for interaction and is separate from x=p/m t even though the notion of this velocity is tied to the fact that nonrelativistic kinetic energy is pp/2m. As a result, one must use an average scenario for calculating kinetic energy which seems to be tied to an ensemble average. Such a notion should hold for all energy levels. If one considers the classical limit, then one has p(x)p(x)/2m in a tiny dx. If one were to simply use 1/sqrt(p)exp(ip(x) x)/sqrt(dx) in dx, then Integral W*W dx in dx = C/p. The relative weight of two adjacent dx regions is: p1/ p2 = p1 / (p1+dp). This is equivalent to probabilities proportional to dt= dx/v, where p=mv. Thus, the area represented by W*W captures the number of possible interactions in dx if attempted hits have a fixed period. We argue, however, that given the bound state conservation recipe, one must use a(p) (exp(ipx)+exp(-ipx)) to describe a dx (with p=p(x)), even in the high energy limit. exp(ipx) is simply an interaction pattern (and is a probability) and so if the particle is bound, the correct statistical thing to do for an ensemble seems to be to combine exp(ipx) with exp(-ipx). We then show that quantum mechanics lowers the number of interactions in an area with increased speed more than the classical case. In particular, we find instead of p/(p+dp), 1- dp/p - 2dp/(pp).
A classical bound state does indicate the direction of motion if one only considers kinetic energy pp/2m. It is the vector quantities, p and x, which indicate direction. In quantum mechanics, a free particle interaction probability (for impulse hits) is directional, i.e. exp(ipx) not= exp(-ipx). Unless one has a single value of pp/2m, KE+= sum over p>0 a(p)exp(ipx) pp/2m / Sum over p>0 a(p)exp(ipx) is not equal to KE- even if a(p)=a(-p). In order to use conservation of energy in quantum mechanics, one is forced to consider both a(p)exp(ipx) +a(p)exp(-ipx) with a(p)=a(-p) which is the correct statistical weighting for a bound state. One may note that exp(ipx) is a probability for interaction and is separate from x=p/m t even though the notion of this velocity is tied to the fact that nonrelativistic kinetic energy is pp/2m. As a result, one must use an average scenario for calculating kinetic energy which seems to be tied to an ensemble average. Such a notion should hold for all energy levels. If one considers the classical limit, then one has p(x)p(x)/2m in a tiny dx. If one were to simply use 1/sqrt(p)exp(ip(x) x)/sqrt(dx) in dx, then Integral WW dx in dx = C/p. The relative weight of two adjacent dx regions is: p1/ p2 = p1 / (p1+dp). This is equivalent to probabilities proportional to dt= dx/v, where p=mv. Thus, the area represented by WW captures the number of possible interactions in dx if attempted hits have a fixed period. We argue, however, that given the bound state conservation recipe, one must use a(p) (exp(ipx)+exp(-ipx)) to describe a dx (with p=p(x)), even in the high energy limit. exp(ipx) is simply an interaction pattern (and is a probability) and so if the particle is bound, the correct statistical thing to do for an ensemble seems to be to combine exp(ipx) with exp(-ipx). We then show that quantum mechanics lowers the number of interactions in an area with increased speed more than the classical case. In particular, we find instead of p/(p+dp), 1- dp/p - 2dp/(pp).
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).
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).
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 classical physics, a particle traveling to the right is the mirror image of one traveling to the left. In terms of a spatial density (the time spent in each constant dx), the value is the same, as is the energy. A similar notion of such a spatial invariance is in the classical conservation of energy equation for a bound particle: p(x)p(x)/2m + V(x) = E. Quantum mechanics seems to be different because exp(ipx) and exp(-ipx) are distinctly different spatial probabilities, i.e. they are dynamic. It is not just that case that a particle moves to the right for exp(ipx) and to the left for exp(-ipx). The impulse hit probabilities (represented by exp(ipx) and exp(-ipx)) are distinctly different. If one wishes to create an OR situation and write: W(x) = Sum over p a(p)exp(ipx) and use this in a conservation of energy equation, then simply including only positive or negative p values in W(x) leads to a complex average kinetic energy -1/2m d/dx dW/dx / W which cannot be used with V(x) and E. Furthermore, W_ (only negative p) and W+ (only positive p) values do not yield the same average kinetic energy. Fortunately, given that this is a statistical average for a localized object (somewhat analogous to a gas), one may combine W+ + W- and obtain a real kinetic energy. We have argued in previous notes that there exists a quantum-classical duality. For example, for a free particle one has two independent notions. One is x=vt which describes the center-of-mass of the particle. We take this to represent a state description which treats v and -v as mirror images. exp(ipx), on the other hand, represents probabilistic impulse hits and does not define the state by itself, we argue, only its interactions. Interactions are physical and define what happens in an interacting system. One may, however, measure a system, i.e interact with exp(ipx), and this should define a measurement state ( a second state ensemble average). For example, exp(ipx)exp(-ipx) = 1 which is equivalent to the constant dt in each dx associated with x=vt. Thus, the two states coincide, but exp(ipx)exp(-ipx) means that one may place the 2-slit apparatus at any x point and is not about the center-of-mass, as x=vt is. In some cases, the two states are quite different. W*(x)W(x) represents an interaction state, but vrms(x) from .5mvrms(x)vrms(x) = -1/2m d/dx dW/dx / W represents a center-of-mass state description. (Note: in these cases a state means an average state, i.e. an average over many ensemble entries.) If one takes W*(x)W(x) and integrates over x, then interference terms vanish and one obtains a classical counting of distinct p value states with their weight a*(p)a(p).This, we argue, leads to the notion of orthogonality of probabilities. In classical probability, there is no such notion, but there is the notion of AND (multiply) probabilities. Given exp(ipx) and exp(-ipx), if one multiplies one removes p, i.e. the interactions and only sees the state left. W*(x)W(x) yields a set of peaks which represent a pattern of probabilistic hits in space which define a state which one would measure in space. The interference pattern, however, must be one such that if one integrates over x, one only obtains a classical sum of probabilities a*(p)a(p)’s as mentioned above. This leads to the notion of orthogonal exp(ipx)’s (which follows mathematically anyway). If one considers different bound En states, then these two are orthogonal because W*n1(x)Wn2(x) mean n1 and n2 are associated with different energies and integrating over x cannot show a result which looks like a state result, where W*n(x)Wn(x) shows a state result. In order to achieve this orthogonality (linked directly to the notion of a state) one must have that Wn(2) usually has 0 points and is positive and negative with the possible exception of a ground state. In other words, if Wn1*(x)Wn2(x) integrated over x is not 0, one violates conservation of energy. As a result, we argue an average probability for a bound state must statistically include +p and -p contributions and must create a function which has positive and negative peaks (except possibly for the ground state), i.e. W(x) must have 0 points. This W(x) is linked to a measurement state W*(x)W(x), but this is strictly considering exp(ipx)s. We argue that there are two pieces of information for a free particle, x=vt and exp(ipx). There must also be an average state linked to x=vt and this given by KEave(x) = .5m vrms(x)vrms(x) + V(x) = E which shows that the particle may exist and have average kinetic energy even at W(x)=0 points. When measuring a system, one interacts with the exp(ipx)s and so the W*(x)W(x) state is the one which is traditionally presented, but whether one measures the center-of-mass motion or not, one also has the notion of the x=vt, i.e. vrms(x) state (average) as well, we argue.
We try to consider free particle and single bound state quantum mechanics from the viewpoint of classical physics. We first assume that a free particle moves as x=vt. Secondly, we assume that dx and dt cannot be 0 as in Newtonian mechanics, but must be finite. This begs the question: How does one measure dx and dt? If one can measure a particle at a given x and t, then dx→0, dt→0. Thus, finite dx, dt means that a given p=constant and pp/2m=kinetic energy (nonrelativistic) interact with a force over a finite dx. In other words, one does not know where the interaction occurs within dx. As long as the free particle moves without interacting, then x=vt and there is no interest in any dx or interactions. For example, one may measure very large x and t ranges. Even if there are dx and dt effects at the start and end points, these are small compared to the large x and t values. As a result, one has a F(x,p) probability within dx, which repeats itself for the next dx (i.e. is periodic), where F is a probability for a p impulse hit at x within dx. At the same time, the particle moves with x=vt it seems within dx. In general, one might then think that F(x,p) and x=vt are decoupled. In previous notes, we have argued that one way to couple F(x,p) and x=vt is through special relativity, i.e. A = -Et+px. X and t values used would presumably satisfy x=vt, but for any x,t one has x+hbar/p and t+hbar/E as solutions yielding the same A value and so might consider dx= hbar/p and dt=hbar/E. The question we ask here is whether one may arrive at this conclusion without using special relativity. The answer seems to be that one can if one suggests that F(x,p) must contain information making it consistent with x=vt, i.e. with every x and t point having the same weight. Clearly, however, each x,t do not have the same weight within dx and dt, but are rather governed by F(x,p). This means, however, that F(x,-p) also represents a probability distribution and so if one has F(x,p)F(x,-p) there is no momentum at a given x point. In such a case, the distribution within dx must disappear and one must have a uniform distribution. This holds even if F(x,p) has the same dx value for any p. Thus, F(x,p)F(x,-p) = constant and must be periodic in x. This implies that F(x,p) = 1/F(x,-p) for constant=1. Given that F(x,p) =0 at the initial and final points of dx, then F(x,-p) = infinite at these points and that is not allowable. Thus, one is forced to use complex variables and try exp(i g(p) x). If g(p) is a function of p (i.e. not a constant), then conservation of momentum implies that exp(i g(p1) x) exp(i g(p2) x) = exp(i g(p1+p2) x), suggesting exp(ipx). What, however, if g(p) = constant? In such a case, one has issues creating an average kinetic energy: ( Sum over p a(p) exp(i constant x) pp/2m) / (Sum over p a(p) exp(i constant x) ((1)) This leads to a kinetic energy which does not depend on x and cannot be used with V(x) in a conservation of energy equation. Thus, we choose exp(ipx). Note: there is no reason to choose a(p,x) because the physics of finite dx and dt is restricted to exp(i constant x). Finally, we try to explain why the same dx=hbar/p and dt=hbar/E together with x=vt follow from special relativity.
We have argued in previous notes that both Newtonian mechanics and free particle quantum mechanics arise from special relativity. Thus, special relativity gives an expression for free particle energy and momentum and these are related to concepts of work. The Lorentz invariant -Et+px shows that x,t and x+hbar/p, t+hbar/E give the same solution to this potential-like term showing that there is stochasticity in the position and time delivery of interaction hits. In other words, even though the center-of-mass moves as x=vt, a momentum hit (proportional to rest mass mo) does not necessarily act at the center-of-mass. This gives the impression of shifts in probability, kinetic energy etc in the case of interference, i.e. interactions. For a single free particle, there is no interaction and -1/2m d/dx d/dx exp(ipx) = pp/2m exp(ipx). We argue that exp(ipx) is a probability and that in general one may add independent probabilities in an OR situation as in a 2-slit interference case. A bound state, however, is different because even though one adds weighted exp(ipx)s, they are part of a resonance or equilibrium and this is no longer an independent OR situation. We suggest that one uses a conditional probability when computing a Newtonian nonrelativistic conservation of energy equation to try to remove as much of the quantum features from this equation. As a result, one has: KEclassical(x) + V(x) = E = Sum over p pp/2m P(p/x) + V(x). One cannot bypass the fact that a quantum particle interacts differently from a classical one and that this involves a probabilistic granularity (wavelength), but classical notions of energy and energy conservation still exist. As a result, the bound state wavefunction Wn(x) = Sum over p an(p) exp(ipx) is not a usual OR sum, but is rather 1 over a normalization factor for an equilibrium picture. As a result, one cannot use Wn1(x)Wn2(x) = W(n1+n2) (x) as one would for independent probabilities. Furthermore, given that one has Wn(x) and Wn1(x)+Wn2(x) which are both of the form Sum over p a(p)exp(ipx), and both represent average energies, one needs a way to distinguish a resonance/equilibrium from a sum of resonance/equilibrium normalization factors. We suggest that the orthogonality of resonance Wn(x)’s provides this feature. The standard textbook explanation is based on Hermitian operators, but in the nonrelativistic case, we argue that one may obtain the result based on integration by parts based on the form of the p operator = -grad. Thus, it is ultimately the form exp(ipx) which leads to orthogonal bound state Wn(x)’s, together with the boundary condition that Wn(x)-->0 of x-> +infinite and -infinite.