Given a single particle bound wavefunction, W(x)=Sum over p a(p)exp(ipx), it is traditionally stated that W(x)W(x) is the probability to find a particle at x. We have argued that W(x)W(x) is actually a resolution profile or a probability for interaction and does not indicate whether a particle is at x or not. (We discuss this briefly in the note.) It is linked to an observer trying to find the particle, on average at x, not if the particle is present there and interacting with V(x). If one normalizes Integral dx W(x)W(x)=1, then it follows that Sum over p a(p)a(p) = 1. Thus, a(p)a(p) has the appearance of a probability as well and one might argue that it is the probability to have a momentum p. The catch is that with W(x)W(x) one considers spatial resolution/interaction without knowing p and with a(p)a(p), one knows nothing about x. Given exp(ipx), the probability for a free particle impulse hit at x, one knows both p and x and may use this probability to calculate KEave(x) = {Sum over p a(p)pp/2m exp(ipx)} / W(x). We note that this probability does not have 0 points when W(x)=0. It shows the particle does not avoid W(x)=0 points in internal interactions with V(x), we argue. a(p)exp(ipx) / W(x) is a third probability. Given that it holds at a fixed x and ranges over p (it is sometimes called P(p/x)) it should be linked to a(p)a(p), but it considers spatial issues. We suggest that it should be combined with W(x)W(x) as this demonstrates resolution or measuring interaction hits and is linked to an external observer. The product probability, however, imposes the 0 points of W(x). This simply means, we argue, that a person making measurements does not interact with the particle at these W(x)=0 points, not that the particle does not exist there. Using this probability, as we have shown before, yields an average of the Shannon’s entropies created using W(x)W(x) and a(p)a(p) as probabilities. Here we try to understand the relationship between these three probabilities more closely without using the usual P(x)=W(x)W(x), P(p) = a(p)a(p) and P(p/x) = a(p)exp(ipx)/W.
Given a single particle bound wavefunction, W(x)=Sum over p a(p)exp(ipx), it is traditionally stated that W(x)W(x) is the probability to find a particle at x. We have argued that W(x)W(x) is actually a resolution profile or a probability for interaction and does not indicate whether a particle is at x or not. (We discuss this briefly in the note.) It is linked to an observer trying to find the particle, on average at x, not if the particle is present there and interacting with V(x). If one normalizes Integral dx W(x)W(x)=1, then it follows that Sum over p a(p)a(p) = 1. Thus, a(p)a(p) has the appearance of a probability as well and one might argue that it is the probability to have a momentum p. The catch is that with W(x)W(x) one considers spatial resolution/interaction without knowing p and with a(p)a(p), one knows nothing about x. Given exp(ipx), the probability for a free particle impulse hit at x, one knows both p and x and may use this probability to calculate KEave(x) = {Sum over p a(p)pp/2m exp(ipx)} / W(x). We note that this probability does not have 0 points when W(x)=0. It shows the particle does not avoid W(x)=0 points in internal interactions with V(x), we argue. a(p)exp(ipx) / W(x) is a third probability. Given that it holds at a fixed x and ranges over p (it is sometimes called P(p/x)) it should be linked to a(p)a(p), but it considers spatial issues. We suggest that it should be combined with W(x)W(x) as this demonstrates resolution or measuring interaction hits and is linked to an external observer. The product probability, however, imposes the 0 points of W(x). This simply means, we argue, that a person making measurements does not interact with the particle at these W(x)=0 points, not that the particle does not exist there. Using this probability, as we have shown before, yields an average of the Shannon’s entropies created using W(x)W(x) and a(p)a(p) as probabilities. Here we try to understand the relationship between these three probabilities more closely without using the usual P(x)=W(x)W(x), P(p) = a(p)a(p) and P(p/x) = a(p)exp(ipx)/W.
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.
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 Part 1, we noted that W*(x)W(x) represents the average number of interaction hits at x for an equilibrium (p, -p) scenario. This follows from exp(ipx) representing the probability for a p hit at x. If -p is also present with the same weight, then exp(ipx)+exp(-ipx) is proportional to cos(px), showing that some x points have low or no probability for a p hit. We argued, however, that these hit points are not at the center-of-mass so the center-of-mass could be at a 0 point. More importantly, average kinetic energy exists at these low or no hit points. In this note, we wish to examine the average kinetic energy more closely. It is built upon both classical Newtonian (nonrelativistic) ideas and the notion of a probabilistic x point for p hits, given by the probability exp(ipx). We note that both x and V(x) are continuous, but suggest that a probabilistic picture is compatible with non-continuous impulse hits. In particular, lengths are of the order of 10 power -10 m (atomic) or 10 power -15 m (nuclear) and particle speeds, on the order of 10 power 7 m/s. This suggests that V(x) and exp(ipx) are associated with ensembles. In the classical case with very slow speeds and large lengths compared to the above numbers, a tiny dx seems to represent an ensemble. Each dx represents an average result p(x)p(x)/2m + V(x) = E. In the quantum case, one needs to use an average based on exp(ipx) (as argued in Part 1) because pp/2m is seen to exist and change at an interaction point. Thus, {Sum over p a(p)pp/2m exp(ipx) }/{Sum over p a(p)exp(ipx)} represents an average KE at an x interaction point which may be linked to the V(x) at this same point. One must have kinetic energy at an interaction point for it to be compatible with V(x)
In previous notes, we suggested that quantum bound states may be analysed in terms of the conditional probability P(p/x)= a(p)exp(ipx)/W(x) (where W(x) is the wavefunction) which allows for the calculation of average kinetic energy at each point x i.e. [Sum over p pp/2m P(p/x)]/ W(x). This added to V(x) yields E (average energy) at each x and yields the time-independent Schrodinger equation). We argued that probability is handled at the level of W(x)=Sum over p a(p) exp(ipx). Measurements (even internal) may knock a particle from one p1 state to another so another set of probabilities (P(x) and P(p)) arise, P(x) being based on: a(p1)exp(i p1x) exp(i p2x). Averages of f(p,x) follow from P(x) f(p,x) P(p/x). Thus, there are two sets of probabilities in quantum mechanics. Entropic dynamics (1) appears to begin with the probability P(x) and considers a transition probability P(x1 |x) with x1-x= an infinitesimal quantity. A product of transition probabilities is then taken to allow for finite x1-x differences. It is stated in (1) that there exists information about x but not p. The maximization of entropy density may be used to determine the form of P(x1 | x). A “drift velocity” is also included (which is later shown to be related to the phase of the wavefunction). Ultimately, the Schrodinger equation is also obtained. We argue one may use P(p/x) to also calculate P(x1 | x) and compare this result with that obtained through the maximization of entropy density. A type of drift velocity appears naturally in this approach. We argue that the form a(p)exp(ipx) already incorporates statistical ideas (i.e. maximum entropy) in that the modulus of exp(ipx) is identical for all x, endowing all x points with a “kind” of equivalent status. exp(ipx) is a function of p and x because one must calculate average kinetic energy at each x and this must change with x i.e. KEave(x) + V(x) = E. Thus, we think there are strong similarities between entropic dynamics and the approach of using P(p/x). There are also some differences in that the coefficient A of delta(x)delta(x) in P(x1 | x) = 1/Z exp{ A delta(x)delta(x) + B d Phi/dx delta(x)} is treated as independent of x in entropic dynamics, but is a specific function of x (related to d ln(W)/dx d ln(W)/dx + 1/W d/dx d/dx W) in the P(p/x) approach.
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.
In the literature, Shannon’s entropy with spatial density as probability i.e. density(x) ln(density(x)) appears frequently. It seems, however, that this calculation of entropy assumes no knowledge of momentum, yet density=W(x)W(x) (for a bound state) where W(x) = Sum over p a(p) exp(ipx). In other words, the wavefunction W(x) contains information about the momentum distribution associated with P(p/x)= a(p) exp(ipx)/W(x). In classical statistical mechanics, entropy is calculated using exp[ - (p*p/2m + V(x))/T) which includes both momentum and spatial considerations. In a previous note (1), we calculated a possible expression for entropy using P(p/x)P(x). This includes Shannon’s spatial entropy and other terms even if one sums over p after forming Shannon’s entropy with P(p/x)P(x). In this note, we try to consider the interpretation of these two forms of entropy.