Quantum spatial probability is given by W*(x)W(x) where W(x)=Sum over p a(p)exp(ipx) is the wavefunction. If one knows about a particle’s location through W*(x)W(x) then momentum is uncertain. On the other hand, a*(p)a(p) is the probability for a particle to have a specific p free particle momentum. In that case, there is no knowledge of the location of the particle. In (1) it is argued that for an infinite potential one dimensional box, the wavefunction is sin(n3.14/L) and so W(x)W(x) may have zero values within the box. It is stated that this is unphysical as there is no reason a particle cannot be at such an x point. (The resolution in (1) is to consider a linear superposition of energy eigenfunctions based on certain rules.) In this note, we consider that a(p) values suggest there are probabilities for a particle to have a constant p momentum. This momentum means the particle may move through all points x because even in the quantum free particle case exp(ipx), a two dimensional dynamical probability implies: exp(ipx)exp(-ipx)=1. Thus a measurement of a p value (i.e. a constant value) even with no known corresponding x value suggests the particle may be anywhere in the system, even at zero points of the wavefunction. The bound wavefunction is real and so is an average which removes the dynamism of the system i.e. the two A particle does not exist in all p states at once. The zeros of W(x) appear only because exp(ipx) and exp(-ipx) interfere revealing one dimension of the two dimensional probability. Thus we argue that W(xa)=0 does not mean that a particle cannot be at xa.
Quantum spatial probability is given by W*(x)W(x) where W(x)=Sum over p a(p)exp(ipx) is the wavefunction. If one knows about a particle’s location through W*(x)W(x) then momentum is uncertain. On the other hand, a*(p)a(p) is the probability for a particle to have a specific p free particle momentum. In that case, there is no knowledge of the location of the particle. In (1) it is argued that for an infinite potential one dimensional box, the wavefunction is sin(n3.14/L) and so W(x)W(x) may have zero values within the box. It is stated that this is unphysical as there is no reason a particle cannot be at such an x point. (The resolution in (1) is to consider a linear superposition of energy eigenfunctions based on certain rules.) In this note, we consider that a(p) values suggest there are probabilities for a particle to have a constant p momentum. This momentum means the particle may move through all points x because even in the quantum free particle case exp(ipx), a two dimensional dynamical probability implies: exp(ipx)exp(-ipx)=1. Thus a measurement of a p value (i.e. a constant value) even with no known corresponding x value suggests the particle may be anywhere in the system, even at zero points of the wavefunction. The bound wavefunction is real and so is an average which removes the dynamism of the system i.e. the two A particle does not exist in all p states at once. The zeros of W(x) appear only because exp(ipx) and exp(-ipx) interfere revealing one dimension of the two dimensional probability. Thus we argue that W(xa)=0 does not mean that a particle cannot be at xa.
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 this note, we try to understand why tunneling should be part of a quantum mechanical treatment of a one-dimensional bound state problem. We suggest that the ideas behind tunneling begin with the directional probability exp(ipx) used to describe conservation of momentum in Newtonian 2-body scattering. Given the notion of a wavelength in space, there must be a probability scheme in space which “physically” creates the wavelength, as we have mentioned before. This means one cannot have uniform probability in x and so some directional probability (linked to exp(ipx)) must be removed from one x region and moved to another. Thus, one cannot point to a particular x1 point, for which cos(px1)=0 and say that the particle has disappeared. Probability is simply shifted from one point to another and this leads to the idea of positive and negative values linked to this probability as one sees from the negative and positive values of sin(px) and cos(px). As mentioned in Part 4, one may create various averages of functions of p including the function 1 which is then linked to particle presence (i.e. spatial density although it does not directly equal this). Given the periodicity of exp(ipx), W(x) =Sum over a(p)exp(ipx) usually tends to be periodic as well, even if one considers a(p)=a(-p) and a real W(x). It is possible, however, to have smooth valued functions for {Sum over p f(p)exp(ipx) a(p)} / W(x). This might lead one to think that one has a classical scenario with no wavelength effects present, but we argue here that this is not the case. Furthermore, we argue that this type of argument seems to explain why tunneling is linked with a one-dimensional bound state solution. We noted in Part 4, that one may solve an equation like the time-independent Schrodinger one to obtain a linear combination solution of directional probabilities exp(ipx)s, i.e. W(x). In Part 4, we were concerned with what to do with W(x) and argued for imposing boundary conditions- constraints -continuity on W(x). Here we consider the case of classical turning point, for which KEave(x)=0. For a classical particle, delta x being tiny is huge compared with the wavelength of exp(ipx). Thus, p→0 leads to wavelength = hbar/p, but for a tiny p, a large wavelength is not a problem as it is still not greater than delta x. Thus, one may have a p (momentum) become very small at a classical turning point (and even argue that it is 0.) In the quantum region, one cannot do this because a delta x near the classical turning point x1 is very small compared to hbar/p for p close to zero. One cannot have a single or a few small value p’s, but must rather have sizable p values. This means that the particle is in one of these p states or another and has sizable impulse and kinetic energy pp/2m. As a result, the potential barrier is not a problem- it does not hold the particle back. What is zero is the average value of KE, i..e KEave(x) = { Sum over p a(p)pp/2m exp(ipx) } / W(x). As we have argued above, for a(p)=a(-p), exp(ipx)) becomes cos(px) and even with positive a(p), one has negative and positive directional probability values which allows sizable p, pp/2m’s to have an average of zero. As noted above, however, an average kinetic energy of 0 at a single x1 does not mean that it is zero at x1+delta. Just as with cos(px), removing values from some x region means that they appear elsewhere. KEave(x)=0 is not a constraint for x outside the classical turning point region. The quantum average KEave(x)=0 describes various probabilities for sizable momenta and pp/2m and so must deal with a particle moving in and out of the so-called bound region. The question becomes: How does one make this appear like a bound state problem, i.e. one localized in space? This occurs by imposing the constraint W(x)=0 as x→ +/- infinite as noted in Part 4. It is this constraint which shows that even though one has sizable p values at the classical turning point (despite KEave(x)=0 but W(x) not=0), one must have the average W(x) drop to 0 as x→ +/- infinite. As a result, given the nature of the quantum calculation, it is physically expected that tunneling should occur for a one-dimensional bound problem because there is no constraint on a single p or a few p values becoming very small (or 0) at the classical turning point. Thus, they should keep moving past the classical turning point. This is the physical event which should occur (i.e. tunneling).
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.
Classically, it is known that the probability density P(x)=1/L for a static particle confined randomly in a length L. If one has a moving particle, one sometimes sees that argument that the particle spends delta t =constant in each constant dx for a constant speed and so P(x)=1/L as well. In such a picture, there is no quantum W(x)= exp(ipx)/sqrt(L) or W(x)= exp(-ipx)/sqrt(L), accounting for motion in both directions. On the other hand, traditional quantum mechanics uses P(x) = W*(x)W(x) so one would have 1/sqrt(L) in this case for both p and -p. The interesting point is that exp(ipx) and exp(-ipx) are both dynamic (differ for p and -p), while W*(x)W(x) does not. We first argue that by considering dt in dx, one considers an event rather than describing a particle at an instant. dt implies there is an t(initial) and t(final) and one subtracts the first from the second. This is a description of an event and not a description of the moving particle we argue. Even in classical physics, motion is state with the particle being at x at t with v(x,t). Motion to the left or right represent different physical states (described by p and -p) and there is no reason that an instantaneous probability capturing the state of motion should not depend on this direction of motion as well. Furthermore, it should depend on p as this is a variable used in special relativity. The function W*(x)W(x) = 1/L is an AND statement of probability and removes the state of motion from the probability by combining p and -p (as an AND condition). This leaves a stationary state which obeys classical probability, i.e. the probability of a static object with a uniform probability to be anywhere within L. Thus, one has the classical equation: 1/L = F(-v) * F(v), with two solutions: F(v)=constant and F(v) = exp(-iEt+ipx). We describe in detail how we obtain the second solution using invariance of probability under a Lorentz transformation. One might argue that F(p,x)=constant means that the dynamic probability (the probability of the moving particle) is the same for p and -p when in fact the particles are in different states. In such a case, one has F((p,x) = 1/sqrt(L) and all dynamic features are lost. One would expect that such a probability would be useless in trying to describe a dynamic problem. (We discuss this in more detail). The alternative is exp(-iEt+ipx) which suggests that there is periodicity in x and t based on E and p. At first, this is equally strange from a classical point of view, one might argue. We suggest, however, that for x+dx and x-dx, exp(ipx) clearly defines the direction of motion which is essential in a dynamic problem. In other words, we argue that Newtonian mechanics uses dynamic variables x(t), v(x,t) etc to solve a problem. We suggest introducing a dynamic probability which considers instantaneous probaiblies (just as Newton considers instantaneous variables). The big question, from a classical point of view, is then: Why would one use probabilities at all when Newtonian mechanics describes deterministic physics? The answer, we argue, is that P(x)=1/L is an acceptable classical probability and one is allowed products in probability theory. We suggest that there is no reason why one cannot consider a dynamic probability F(p,x) such that F(p,x)*F(-p,x) = 1/L. Unless one can refute the existence of F(p,x) in classical theory, then mathematiaclly one has F(p,x) = exp(ipx) (because F(p,x)=constant makes no sense). The question then becomes whether such an F(p,x) may be used in solving any physical problem with the answer being yes. This formalism solves the 1-dimensional reflection-refraction problem of photons, assuming the photon to be a kind of particle moving at speed c. Newtonian mechanics does not solve this problem, it seems.
Quantum mechanics is based on two different probability schemes, namely W(x) =wavefunction and W*(x)W*(x)=P(x), with one being linked to the other. Why should there be two probability schemes? In (0) we argued that there are two spatial behaviour schemes because an impulse p (momentum) is automatically associated with two spatial distributions cos(px) and sin(px) in exp(ipx), yet a particle with constant p moves as x=p/m t. Thus if one is only concerned with motion, each x point has the same weight, and this is physically measurable as a photon or electron travels a certain distance in a certain time in free space. On the other hand, an impulse interaction is associated with exp(ipx) is a completely different spatial picture, and this too is measurable as in a 2-slit experiment. Thus it seems the two schemes are measured differently. We argue that this is an important issue because if there are two x-probability schemes there should be different entropies associated with each. Ttraditionally (1), one sees two quantum entropies: - W*(x)W(x) ln {W*(x)W(x)} and - a*(p)a(p) ln{ a*(p)a(p) } where W(x) = Sum over p a(p)exp(ipx). These are associated with the x=p/m t scheme or x-dependence based on motion or average motion. W*(x)W(x) represents spatial density which should be measured presumably by some sort of scattering, but we ask whether some information is lost in this approach? In particular, we note that a wavefunction may be obtained for a bound state. In the high energy limit, the envelope over the density W*(x)W(x) is said (2) to match C/v(x) from classical physics where .5mv(x)v(x) + V(x) = E. C/v(x), however, is proportional to delta t (with delta x being constant) and has nothing to do with the strength of an impulse hit. In the classical case, there is a single v(x) and so a single p(x) in each delta x thus no impulse entropy within a delta x i.e. at a point x. In quantum mechanics, however, we argue that a constant p automatically has an x probability distribution exp(ipx) associated with it and there might be an entropy linked to this.This entropy is associated with a probability to receive a p impulse at x and is so not linked to interference while the W*(x)W(x) is. In (3) we introduced the impulse or wavelength entropy (for a free particle |cos(px)|/C ln{| cos(px)|/C},, but here we wish to investigate its properties more closely.
May 1, 2026 Correction: In the example of = dW/dx / W for W= Sum over p>=0 a(p)exp(ipx), Real is finite when W(x)=0, but Im( ) tends to infinite if one uses l'Hopital's rule. In the note, only the real part of is considered, but the Im part leads to 0/0. In previous notes, we argued that the Lorentz invariant -Et+px, allows for x,t and x+hbar/p, t+hbar/p solutions. We suggested this might give rise to the quantum mechanical probability exp(-iEt+ipx). In other notes, we sought a probability which conserves energy and momentum and respects Lorentz invariance, again leading to exp(-iEt+ipx). In a couple of notes, we argued that the uncertainty in x arises from a rest mass at x=0 jumping to x’/t’ = v as seen in the moving frame. This uncertainty in x forces one to introduce probabilities. Here we combine the latter idea with the notion that special relativity yields formulas for the interaction variables E and p. We argue that one may replace the notion of a force reaction as well as the idea of frames moving at constant speed relative to each other by demanding that an interaction be represented by an AND (product) of probabilities and this must necessarily conserve energy and momentum. This probability is in the location and time of interactions. This leads to a skewed dynamic probability exp(-iEt+ipx) because exp(ipx) and ex(-ipx) are not the same. Furthermore, as we have noted before, one uses exp(ipx) as an interaction probability which leads to complex average kinetic energy (nonrelativistic case) and momentum if one only uses p>=0. The Schrodinger equation (which is based on conservation of energy) bypasses this problem by combining p>=0 and p<0. Nevertheless, we note that there can exist W(x) = Sum over p>0 a(p)exp(ipx) such that if W(x) = 0 dW/dx and d/dx dW/dx are also zero at these points even though the average kinetic energy and momentum are complex. We finally show that a complex average can be used in a physical calculation, namely one dimensional reflection-refraction.