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 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).
Newtonian mechanics often employs a potential V(x) such that .5m v(x)v(x) + V(x) =E ((1)) where v(x) is velocity. Mathematically V(x) is defined at each x, but in reality there should be a tiny cutoff i.e. finite dx. Newtonian mechanics makes use of the idea of small dx lengths which tend to 0 mathematically. To first order v(x) is constant within a dx range, but the overall mathematics is continuous instead of discrete. Thus there is a continuous equation ((1)) for conservation of energy. A photon may be described by an eclectic field varying as cos(-Et+px). In terms of wavelengths exp(ipx) is often used as a mathematical convenience and noncontinous behaviour is not an issue. For example in refraction, momentum = hbar/wavelength abruptly changes and one straight line path is joined to another with a different slope. exp(ipx) is associated with repeatable wavelength patterns with a single wavelength representing a spatial distribution/ (kind of probability) region. A particle with rest mass may also be described by exp(ipx) which suggests a characteristic length hbar/p. Within this length (i.e at each x point which exist within hbar/p) the particle has p and pp/2m. Thus for a square well potential there is conservation of energy at each x point even though there is a wavelength in the problem. This is fine for a constant momentum, but what about a potential V(x) which implies p must change at each dx region? exp(ipx) is an eigenfunction of -id/dx which is linked to spatial invariance. It manifests itself physically by interference (e.g. two slits) which again involves non-smooth motion (i.e. one compares rays with completely different slopes). The characteristic length hbar/p is fixed and associated with a single p, yet momentum is supposed to change from one x point to the next. hbar/p does not fit within a single dx and this is a problem. Quantum mechanics remedies this issue by allowing for a superposition of exp(ipx)s i.e. W(x) = wavefunction = Sum over p a(p) exp(ipx) such that energy conservation is forced at each x. Thus one combines the idea of wavelength (which is physical) with energy conservation which is also physical. A consequence of this, however, is that -id/W/dx / W leads to a strange average momentum which is not the square root of -1/2m d/dx dW/dx/ W as well as discrete energies. A mathematical benefit is that one has a continuous differential equation, namely the time-independent Schrodinger equation. The question we ask is: What happens in the so-called classical limit? In the nonrelativstic case this leads to Newtonian mechanics in which there is no wavelength. How does the wavelength disappear so to speak? We suggest (as has been suggested often in the literature and textbooks) that the wavelengths in W(x) may become extremely tiny so as to not be detected easily in classical length scales. Mathematically the time-independent Schrodinger equation holds and there is a smooth W(x) at all points, but we suggest that this smoothness may not be necessary because it is not found in refracting light. If the wavelength is tiny compared to small dx regions in the classical limit, then many wavelengths of a given p fit into dx. Thus the idea of spatial invariance approximately holds within dx and there is no issue of the wavelength spilling out of the dx region as it does when it is comparable to the size of the system as in quantum bound problems. Thus within dx a single p with many (or at least several) wavelengths holds and there is a single p. The kinetic energy is pp/2m and the potential V(x) where x is in the center of dx. One may think of p changing abruptly (nonsmoothly) at the boundary of one dx with another such that conservation of energy holds. Thus one has p wavelengths which do not spill out of dx and give an approximate sense of spatial invariance together with energy conservation. The only loss is that of a smooth wavefunction, but it is not clear that such smoothness is a requirement. V(x) is physically associated with a limiting dx length (at which the concept of V(x) does not hold, but rather there are stochastic hits) and if one expands this length i.e. chooses a much bigger dx in keeping with classical scales, it seems that there should not be a problem. It as if V(x) delivers an impulse hit at each dx junction (like a series of different media in refraction).. This approach leads to the decoherence of wavelengths in the classical limit.
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.
Quantum mechanics is concerned with measurement and we suggest that in a bound problem impulses due to momentum are the form of measurement which occurs in x space. The Lorentz invariant A= -Et+px, which is also the free particle relativistic or nonrelativistic action with v=x/t, associates energy = pp/2m with time and p with x. Creating eigenvalue equations yields id/dt exp(-iEt) = E exp(-iEt) and -id/dx exp(ipx) = p exp(ipx). P (momentum) is associated with physical impulses and with its own two dimensional probability distribution exp(ipx). E is associated with t and has its own distribution in time exp(-iEt) which implies that the same E holds for each x point without any notion of x probability associated with it. In classical physics, a momentum p exists in a tiny dx (dx→0). For exp(ipx), a p impulse may occur within hbar/p with different probabilities, so one cannot restrict the measurement of p to a tiny dx unless dx is of the order of hbar/p. Nevertheless energy exists at all x points. Given that p measurements occur through specific impulses at different x points with exp(ipx) probabilities, one must calculate an average energy using {Sum over p pp/2m a(p) exp(ipx)} / {Sum over p a(p)exp(ipx)} using the probability associated with the measurement of p in space i.e. exp(ipx). This adds to V(x) to yield a constant E at each x point. We suggest that weighted exp(ipx)s are used to calculate average energy at each x, because it is through impulse measurements that one determines p at different x values and hence the average energy. One may note that one does not calculate {Sum over p a(p)exp(ipx) p } / {Sum over p a(p)exp(ipx)}. An impulse hit is associated with a certain probability and this measurement in x yields information about the corresponding energy pp/2m. Each impulse hit also yields information about a time probability exp(-i ppt/2m), but only if the particle is knocked out of the bound system i.e. only if pp/2m is the only energy present, which is not the case for a single particle bound system.
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 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.
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).