Quantum mechanics is largely based on a wavefunction, which we consider to be a probability. The theory is, however, not solely probabilistic because if an interaction does not occur, x=vt holds as does the classical notion of impulse hit and kinetic energy = pp/2m (nonrelativistic case). The probability exp(-iEt+ipx) describes probabilistic interactions in time and space (we argue) and so one does not think of x and t as in x=vt. As a result, there is a probability for a quantum particle to interact with both slits of a two slit apparatus if they are about hbar/p apart. This leads us to the notion that W*(x)W(x) (W(x)=Sum over p a(p)exp(ipx)=wavefunction) is not spatial density, but rather an interaction profile. A particle can still exist at W(x)=0 points, it just has very little to no probability of interacting at these points on average, with the average considering both exp(ipx) and exp(-ipx) in a bound case scenario. Given that E=kinetic energy = pp/2m (nonrelativistic case only) and W*(x)W(x) not= spatial density, we suggest that one must be careful with interpreting d/dt partial W*(x)W(x) = constant grad dot Im (W* (-id/dx) W) (from (1)) as a continuity equation which represents hydrodynamic motion. We suggest that the quantum features are governed by exp(-iEt+ipx) and that any derivatives d/dx and d/dt are related to changes in phase and not hydrodynamic flow (in a classical sense). We note that for a free particle d/dt partial density = 0 = constant grad dot Im (W* (-id/dx) W). The same zero equation holds for a bound state (even though there is physical motion to the right and then left) , but if one creates a superposition of bound states (which is a probability statement), then the same equation is not 0. We note that for a classical bound state there exists a vrms(x), but this is an average quantity. It does not show any resolution pattern which is represented by W(x) and so one must be careful when interpreting it, we argue.
Quantum mechanics is largely based on a wavefunction, which we consider to be a probability. The theory is, however, not solely probabilistic because if an interaction does not occur, x=vt holds as does the classical notion of impulse hit and kinetic energy = pp/2m (nonrelativistic case). The probability exp(-iEt+ipx) describes probabilistic interactions in time and space (we argue) and so one does not think of x and t as in x=vt. As a result, there is a probability for a quantum particle to interact with both slits of a two slit apparatus if they are about hbar/p apart. This leads us to the notion that W*(x)W(x) (W(x)=Sum over p a(p)exp(ipx)=wavefunction) is not spatial density, but rather an interaction profile. A particle can still exist at W(x)=0 points, it just has very little to no probability of interacting at these points on average, with the average considering both exp(ipx) and exp(-ipx) in a bound case scenario. Given that E=kinetic energy = pp/2m (nonrelativistic case only) and W*(x)W(x) not= spatial density, we suggest that one must be careful with interpreting d/dt partial W*(x)W(x) = constant grad dot Im (W* (-id/dx) W) (from (1)) as a continuity equation which represents hydrodynamic motion. We suggest that the quantum features are governed by exp(-iEt+ipx) and that any derivatives d/dx and d/dt are related to changes in phase and not hydrodynamic flow (in a classical sense). We note that for a free particle d/dt partial density = 0 = constant grad dot Im (W* (-id/dx) W). The same zero equation holds for a bound state (even though there is physical motion to the right and then left) , but if one creates a superposition of bound states (which is a probability statement), then the same equation is not 0. We note that for a classical bound state there exists a vrms(x), but this is an average quantity. It does not show any resolution pattern which is represented by W(x) and so one must be careful when interpreting it, we argue.
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.
Quantum mechanics utilizes a wavefunction W(x)=Sum over p a(p)exp(ipx) (in time independent cases) with exp(ipx) being what we call a free particle complex probability. In such a case, W(x) is an OR scenario of these various probabilities. We have argued in previous notes that exp(ipx) is the probability for a p impulse hit at x, even if the center-of-mass of the particle is not at x. This means that impulse hits may occur probabilistically in space as in the 2-slit example which requires adding two probabilities as an OR case for each slit. The point then seems to be that W(x) (or even its momentum frozen form W*(x)W(x) gives a profile in space of the number of interactions (various p impulse hits) at each x. Classically, the kinetic energy (nonrelativistic) is pp/2m. In a classical bound problem this kinetic energy is considered to exist (in calculations) at the center-of-mass, but forces are also said to be rendered at this point so pp/2m, the center-of-mass and the force hit point all coincide. In quantum mechanics, a p impulse hit and the center-of-mass do not coincide. If p is said to hit at a probabilistic x, then it seems that pp/2m should also be seen to exist at this same hit point. A physical interaction point verifies in a sense the location of pp/2m in a probabilistic manner, i.e. a(p) exp(ipx)/W(x) is the probability used to calculate KEaverage and this is based on impulse hit locations. By normalizing using W(x), one removes the relative number of average hits at each x and obtains a pure energy value at x. (This may then be used in a KEave(x)+V(x)=E equation.) On average W(x) has zero points, which means that if one considers W*(x)W(x) to represent the motion-frozen-out spatial distribution of hits, then no hits are rendered at certain x points (on average). One might argue then that the particle cannot exist at x, but the center-of-mass can exist anywhere. One might next argue that there is no kinetic energy at x, but that cannot be true because 1/2m d/dx d/dx W has the same 0 points as W and the two divide. Thus, there is average kinetic energy at all bound points and this represents an average based on impulse hits (exp(ipx)). Classically, if one has many particles in a potential, one may write density(x)V(x) to find the potential energy of the particles. This implies that the particles interact with V(x), e.g. neutrinos usually do not interact. Density(x) in a classical sense means that the center-of-mass and force hit point are both at x. In quantum mechanics, exp(ipx) represents the probability for a force hit point, W(x) represents an average and W*(x)W(x) represents a motion-frozen average hit point. W*(x)W(x)V(x) seems to represent the average potential energy of the particle at x because W*(x)W(x) is linked with interaction. Even if a particle is not present, however, there is energy V(x) in space. It seems that if W*(x)W(x)=0 at some x, then there is no average particle interaction at this point, but there is still average kinetic energy Sum over p a(p)pp/2m exp(ipx) / W(x) based on the probability for average probability hits. The key idea is that one normalizes probability by dividing by W(x) in the average kinetic energy calculation. W*(x)W(x) is normalized not at x, but integrating over dx and setting this value to 1. One has relative hit at x values represented by W*(x)W(x). For exp(-ipx)exp(ipx)=1, each dx has the same average hit value (with one not being able to distinguish details in hbar/p, i.e. this is a motion frozen result. The question then becomes: Why would one write: W*(x)W(x) KEave(x)? One might argue that this term occurs if one takes the quantum mechanical equation: KEave(x) +V(x) = En and multiplies by W*(x)W(x). One may ask: Does V(x)W*(x)W(x) have physical meaning? V(x) is the potential energy at x. The average interaction probability with V(x) at x should be W(x)V(x) and not W*(x)W(x)V(x). The latter is a classical mechanical idealization. If W*(x)W(x) is 3 at x1 and 2 at x2, then it is as if the particle interacts 1.5 times more at x1 and so if V(x1)=V(x2), then this approach assumes 1.5 the interaction energy at x1. KEave(x), however, is simply free particle energy (average) at x and one does not need to take this average value and multiply it by the relative number of interactions at x. W(x) has zero points meaning that there is no interaction at such a point on average. KEave(x) on the other hand is the average energy at x based on the particle kinetic energy presence pp/2m being at the point of an impulse hit. This does not mean that an interaction is occurring with V(x) at this point in this formalism. Even a free particle has an exp(ipx) probability when it is not interacting. The sum Sum over p a(p)exp(ipx) means that all of the interactions linked with x have occurred and one is making an OR calculation. KEave(x) is free particle kinetic energy at x and V(x) is potential energy (not interaction potential energy with the particle). Multiplying by W*(x)W(x) is a classical minded approach to creating a classical equation: density(x) (KEave(x)+V(x)) = E density(x) ((1)) Here KEave(x) would represent the average kinetic energy of a single particle and density(x) would be the average number of particles at x. density(x)KE(x) has a clear meaning. W*(x)W(x) KEave(x) on the other hand is different. KEave(x) is actually the classical kinetic energy value at x and so it is not clear why one would multiply it by W*(x)W(x). W*(x)W(x) represents the probability for an impulse hit at x in a p-frozen out picture. The notion of exp(ipx) (i.e. probability for a p hit at x) is already buried within the quantum calculation of KEave(x) (which yields the classical value). In order to find the physical meaning of W*(x)W(x) multiplied by KEave(x) or V(x) it seems that one has broken dx into equally sized segments which correspond to different detection hits of the particle. This is then somehow equated with the presence of the particle per unit time in dx which is then somehow rendered as the average energy at x multiplied by the detected presence at x through a steady rate of measurement/time. Thus, multiplying by W*(x)W(x) is not multiplying by a physical density.
In Part 1, we suggested that exp(ipx) represents a spatial probability for the rendering of momentum hits. A single particle has a center-of-mass which undergoes x=vt, but momentum hits are not rendered from this point, but are rather delivered in x according to the complex probability exp(ipx) which is linked with the resolution delta x = hbar/p. This accounts for the fact that a particle can probabilistically interact with each slit of a 2-slit apparatus if they are about hbar/apart. If one is only interested in energy, pp/2m, the classical result holds for both exp(-ipx) and exp(ipx). In a classical bound state, one considers only energy p(x)p(x)/2m + V(x)=E because each p(x) exists at x in dx->0. One cannot distinguish any probabilistic hit distribution. exp(ipx) and exp(-ipx) differ even though they are linked with the same kinetic energy pp/2m (nonrelativistic case). We argued in Part 1, that the Newtonian conservation of energy equation actually includes a force profile linked with V(x). The catch is that for V(x) one has a force at each x, dx->0. Such is not the case for exp(ipx). We argued in Part 1 that one must consider not only pp/2m, but the associated spatial probability distribution for hits (i.e. energy together with force). Here, we note that this idea gives rise to an equilibrium calculation. This is somewhat startling because classically, and even quantum mechanically for higher energy levels, a particle moves to the right part of the time and then to the left for other times. As a result, a calculation which respects time would not average over exp(ipx) and exp(-ipx). Only an equilibrium calculation in time would do so. We suggest that one must consider the Newtonian conservation of energy equation as applying to both p(x) and -p(x) and so from an energy point of view one may consider it to be an average. The catch is that each term p(x) and -p(x) yields the same kinetic energy profile in the Newtonian case and the same V(x). For a quantum particle the kinetic energy with probabilistic force distribution is not the same for p and -p. As a result, to use V(x) which applies to forward and backwards motion, one is forced to consider the average and this leads to an equilibrium scenario. We show that the average kinetic with probabilistic force hit profile in space is not the same for +p and -p. Measurements, however, take time and may be averages of much data and so the notion of an average or equilibrium seems to make sense. We note, however, that an average is very different from the Newtonian forwards-backwards motion picture. Only in the high energy limit does the energy pp/2m with probabilistic force profile become the same for both p and -p. Thus, the main point of this note is to stress the notion of a single particle bound state as an average and to point out the consequences of this.
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 Part 1, we argued that the Heisenberg uncertainty principle (standard deviation p) * (standard deviation x) >= hbar/2 only applies when a measurement (i.e. interaction) has occurred. Otherwise, the notion of standard deviation does not exist. Thus, if one has a Gaussian wavepacket, it is the measuring device which has created it and constrained is standard deviations of momentum and position. A free particle, which has not been measured, is described by exp(-iEt+ipx), as argued in Part 1, and not by a wavepacket. Here we ask: How does one interpret a wavepacket calculation? In (1), we argued that exp(-iEt) is a free particle energy conserving probability and exp(ipx), a momentum conserving one. Then, a sum of exp(ipx - iEt) represents an OR situation in probability theory. There is superposition and so one obtains a weighting of possible outcomes which is the probabilistic interpretation of the scenario. If one considers two slit interference, the viewpoint is the same. exp(ipx) is a probability which allows for momentum hits of p in a range of dx=hbar/p outside the center-of-mass, which follows x=vt. This built-in stochasticity of impulse delivery leads to the notion that probabilistically, a particle may interact with both slits if their separation is about hbar/p. One writes an OR probability statement: W(x)= exp(i p dot r1) + exp(i p dot r2), where r1 is the position vector from the center of slit 1 to a point far away on a screen, and r2, a vector from the center of slit 2 to the same point. The result, W*(x)W(x) is a probabilistic expression which describes the possible outcomes probabilistically. (1). Seen in this way, the Gaussian wavepacket should be interpreted in a similar manner and not as a physical description of the particle which moves freely. A single bound state in a potential V(x) is localized by a potential and one may consider only OR cases of exp(ipx), i.e W(x) = Sum over p a(p)exp(ipx). In the measurement case, there is no quasi-permanent localization and one uses a linear superposition of exp(-iEt+ipx)s. Nevertheless for a Gaussian wavepacket, two probability distributions describing possible outcomes are created. One has W*(x)W(x)= exp(- (x-u)(x-u) / (2 sigma(t) sigma(t))) where sigma is the x standard deviation. In addition, W(x) = Sum over p a(p) exp(ipx), leading to a momentum outcome distribution of a*(p)a(p). As written, sigma(t) for x which means that the Gaussian is expanding. We called this unphysical if one considers the wavefunction as representing a description of a single physical entity (2) at times beyond t=0. Like in the 2-slit case, we argue that it represents an outcome distribution. Thus, one would not consider any more time values than t=0, at which time the measurement is completed. An exp(-iEt+ipx), describing a free particle, is measured by a specific device which creates the Gaussian form and gives it a specific standard deviation in space and momentum. The measurement is then done and W*(x,t=0)W(x, t=0) represents the possible outcome positions of the particle at t=0 (i.e. probabilities for x portions of a phase shift). The momentum distribution yields possible p outcome values. As soon as the measurement is finished, the particle moves as exp(-iEt+ipx), with p being probabilistically linked to a*(p)a(p) (a Gaussian as well), where W(x,t) = Sum over p a(p)exp(-iEt+ipx).If one has a second measuring device which creates a tiny momentum standard deviation, one would be able to carry out experiments which would determine roughly the standard deviation of p created by the first measuring device. Furthermore, one would be able to measure phase shifts as well by comparing a free particle measured by a device with a tiny p standard deviation and a large x one with free particles which were not measured.
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).