Challenges
Datasets
Workspaces
Papers
Discussions
Leaderboard
Log inSign up
Challenges
Datasets
Workspaces
Papers
Discussions
Leaderboard
Blog
Job Board
Q3AS

© 2026 Aqora Quantum S.A.S.

TermsPrivacyLegal Notice

Wavefunction and the Identity of a Momentum State in a Quantum Bound State Part II

Francesco R. Ruggeri

Published Jul 22, 2021• 0 citations
Publisher

Abstract

In Part I of this note, we argued that in classical statistical mechanics, the probability P(p), where p is momentum, is well defined because there is a balance of two-body elastic collisions and their time-reversed reactions. In particular, this may be applied to a point x where one imagines a tiny box with many particles. In quantum bound states, collisions are with a potential and one does not have a balance of reactions and their time reversed cases. Thus, the notion of p at x does not seem to be clearly defined. A particle may be a x with momentum p, but at the same time may be reacting with V(x) and transforming into another pj. Thus, there is an ambiguity which does not exist in classical statistical mechanics. There, a particle may be colliding and so a p appears to disappear, but a new p is created by a time reversed reaction. P(p), we argue, must hold over time in an equilibrium. Due to this issue, it appears that the best one may do is to define P(p) for the whole x region in quantum mechanics. This P(p) is measurable by taking an ensemble of identical single particles, knocking them out and measuring their momentum. As a result, knowing “p” exactly means one does not know “x”. As a result, a different view must be taken to dealing with probabilities in a quantum bound state. We use the idea of invariance in x and argue that a wave-like behaviour in probability may be used to describe results. This is unusual because it appears as if particles are appearing or disappearing at different x points. If one has periodic motion, a question which arises is: What is the spatial frequency? We argue that the answer follows from classical physics. There: delta p = delta t (-dV/dx). If p initial =0 and delta t is constant, one sees that a large p final corresponds to a large change in V(x) in a tiny dx or that there is very good resolution in x. Thus a large p (linear) implies high resolution in x. Thus, we use px as the argument of the periodic probability function.

DOI
10.5281/zenodo.5122939

Related papers

Wavefunction and the Identity of a Momentum State in a Quantum Bound State

Francesco R. Ruggeri

Addendum July 19, 2021 The Fermi-Dirac and Bose-Einstein statistical cases, like the Maxwell-Boltzmann case, may be formulated in terms of a reaction balance between a regular two body elastic collision and its time reversed reaction: e1+e2 = e3 + e4 and f(e1)[1-/+f(e3)]f(e2)[1-/+f(e4)] = f(e3)[1-/+ f(e1)] f(e4)[1-/+ f(e2)]. Both the Fermi-Dirac and Bose-Einstein cases reduce to the Maxwell-Boltzmann case under certain conditions. An unusual feature of bound state quantum mechanics is that there is a spatial density (or probability to be at x = P(x)), just as in classical statistical mechanics, but also a wavefunction W(x) such that P(x)=W(x)W(x). In other words, W(x) for the bound state is like a square root of probability. Furthermore, there is a P(p)=a(p)a(p) where W(x)=Sum over p a(p)exp(ipx). In this note, we try to argue that P(p), which is based on the measurement of momenta of particles knocked out of an ensemble of identical single particle bound states, does not have physical meaning within the bound state, unlike P(p,x) which has physical meaning for a Maxwell-Boltzmann gas both inside the gas and for particles released from a small hole in the gas wall. This forces one to create another function to “link” the two physically observed probabilities P(x) and P(p).

•Jul 19, 2021•0 citations

Quantum Mechanics and Orthonormal Basis Functions

Francesco R. Ruggeri

For quantum bound states, classical style probabilities P(p)=a(p)a(p) and P(x)=W(x)W(x) [where W(x)= the wavefunction = Sum over p a(p) exp(ipx)] exist. These are real and normalized to one, but unusually, they separate p and x completely. For example, P(p) is the probability to find a momentum p anywhere in the system - no information about x is known. This seems to suggest that the P(p/x) is not defined in a classical sense. (Similar arguments hold for P(x).) We have argued in a previous note (1), that due to a lack of time reversal balance, there is no clear classical notion of P(p/x) in quantum mechanics while in classical statistical mechanics, there is a time reversal balance of two-body collisions and so P(p/x) is well-defined. In particular, a momentum p collides with a pj to form another momentum pk. There is ambiguity if p is changing, but if at the same time a pk is changing (colliding) back into a p at x, then the ambiguity is removed. In quantum mechanics, the ambiguity is not removed at x, but one may discern one p state from another throughout the entire x space. This suggests orthonormal probabilities e.g. say exp(ip1 x) and exp(i p2 x). These exp(i p x) probabilities comprise the wavefunction which is linked to P(p/x) = a(p) exp(ipx)/W(x), but although Sum over p P(p/x)=1, P(p/x) is complex with the real and complex portions taking on positive and negative values. Nevertheless, it is W(x) which is the solution of the Schrodinger (and other) equations. It is known that solutions Wn(x) of the Schrodinger equation have discrete En (energy) values. It is possible to have [H, Operator]=0 and possibly to have H written in terms of this operator. In such a case, Wn(x) might be both an eigenfunction of H and the Operator. Alternatively, a linear combination of Operator eigenfunctions may be used. If En is discrete, one may expect the operator eigenfunctions to also be discrete, leading to quantization of the Operator eigenvalues. In some cases, one may have two operators which commute with H, e.g. LL and Lz (related to angular momentum). Lz eigenfunctions might not be linked to H so one might ask why Lz should be quantized. Another H, however, may distinguish between Lz eigenfunctions and so they also have discrete eigenvalues. In this note, we try to consider why one should have orthonormal eigenfunctions of these commuting operators which make up the wavefunction and try to interpret their physical meaning.

•Aug 3, 2021•0 citations

Time Reversal and Wavefunction*(x,t) Wavefunction(x,t)?

Francesco R. Ruggeri

In a previous note (1) we argued that the free particle classical action A (relativistic or nonrelativistic) may be written in terms of v=x/t and then varied in x and t independently. One finds that dA/dx partial = p and dA/dt partial = -E implying A= -Et+px with x and t independent. This is the opposite of the classical mechanical idea of x(t). Creating an eigenvalue equation yields -id/dx partial exp(iA) = pexp(iA) and id/dt partial exp(iA) = Eexp(iA). exp(iA) plays the role of an unusual type of probability we argue which describes two dimensional periodicity (cos(px) and sin(px) ). We have argued that the quantum particle is the medium for a two dimensional wavelike motion of an energy-momentum wave. A wave type structure is needed to describe dynamics (like a classical wave), but unlike a classical wave two dimensions are required because there is no preference of any x point for a free particle. Both cos(px) and sin(px) both treat x in a preferential manner as their values vary with x. For a classical wave this is fine because the creation point of the wave imposes boundary conditions, but there is no such boundary condition imposed on a free particle. exp(ipx-iEt0 must somehow then link to P(x)=constant. One may note that exp(-iA)exp(iA) = 1 for the free particle. This may be argued to be linked to spatial invariance i.e. P(x)=constant where P(x) is not exp(ipx). Here we argue that exp(-iA)exp(iA) represents a product of two unusual probabilities , one for motion in a certain direction and the other for the time reversed scenario. If the quantum particle is the medium of an energy-momentum wave exp(-iEt+ipx), then the product of the regular and time reversed unusual probabilities cancel or undo each other. In other words one does not see any change in space as long as one has a noninteracting exp(-iEt+ipx) . Given that -id/dx is a linear operator this product yields a momentum of 0. We consider this idea and its implications on a momentum distribution Sum over p a(p)exp(ipx) which may represent a bound state or other states. In particular, it seems that P(x)=W*(x,t)W(x,t) which yields a classical type of probability (P(x) real, positive and normalized to 1) is linked to a product (merging) of forward and time reversed situations so that any spatial density which remain due to the quantum particle acting as a medium for an energy-momentum wave (as well as the time reversed part) would represent an actual change in spatial density. exp(ipx) already contains two periodic spatial functions (cos(px) and sin(px)), but does not manifest it unless there is an interaction.

•Apr 1, 2022•0 citations

Quantum Bound and Classical Statistical Interactions and a Constant "Energy" in Space

Francesco R. Ruggeri

Both quantum bound state equilibrium and classical statistical mechanical equilibrium are characterized by d/dt partial density(x,t)=0. For a quantum particle, W(x,t)=exp(-iEt)W(x) and d(x,t)=W*W. Density is linked to fluid flow and arguments of pressure balance such that the overall average fluid flow is zero are made. This pressure-force balance is common in classical statistical mechanics and used to calculate the variation of density with x i.e. obtain a differential equation with d d(x,t) /dt partial. A similar balance equation is used in (1) as the authors try to derive the Schrodinger equation using a fluid dynamical approach. In this note, we try to argue that one may think of the two equilibria in terms of an energy quantity which remains constant throughout space. This energy quantity depends directly on the type of interactions in the system at any point x. For a classical ideal gas, balanced two body time reversal scattering represents the interaction. V(x), a potential, does not change this fundamental manner of interacting at any given point, although it does affect density. As a result the energy related constant at each x is T, related to average kinetic energy. We argue that this already points to the dominant type of interaction creating the statistical interaction i.e. one that depends only on p in interactions which conserve kinetic energy. This “happens” to be equivalent to maximizing Shannon’s entropy, but we argue this is particular to this scenario in the sense that not all interacting systems will have this equivalence. In other words, we argue that one must be careful with regards to the maximization of entropy as a fundamental rule. In particular, we consider the single particle bound quantum state. The interaction exists between the particle and the potential V(x), but one may consider a distribution of free particle momentum probabilities i.e W(x)=wavefunction = Sum over p a(p)exp(ipx). The form exp(ipx) comes from separate arguments as is discussed in the note, but the point is that there are no two body scattering reactions occurring at a given x, neither is there acceleration from on x to another (except for a vrms mathematical quantity). Nevertheless the “free particle” exp(ipx) interacts at each x with V(x). As a result of this interaction, one has free particles exp(ipx) which are no longer associated with pp/2m, but rather with an Eave which is a constant for every p. In other words the identity of exp(ipx) is unusual because it is linked to exp(ipx) and pp/2m, but also to Vk1 a(p-k1), Vk2 a(p-k2) etc where V(x)=Sum over k Vk exp(ikx). Thus each x point is associated with a constant Eave and this defines the equilibrium. (It is also equivalent to conservation of energy at each x.) In such a case, one does not maximize entropy to find a(p) values. In the two cases, the common point seems to be considering the interactions at each x and finding a constant “energy” type quantity which does not vary with x. This we argue appears to define equilibrium for both the classical statistical and quantum bound states. Instead of considering density which does not change in time at each x, there is constant energy which not only does not change in time, but has a constant value at each x which seems to be an alternative rule.

•Feb 14, 2022•0 citations

Implications of Average Momentum On Quantum Bound State Behaviour

Francesco R. Ruggeri

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.

•Nov 26, 2022•0 citations

Conditional Probability and Time in Quantum Bound States

Francesco R. Ruggeri

We argue that in a quantum bound state with one particle, the particle receives stochastic hits from a potential. As a result, if one examines a point x, the particle may be present at ti with momentum pi (in both the forward and backward directions) leading to a momentum distribution with a conditional probability P(p/x) = a(p) exp(ipx)/W(x) where W(x) is the wavefunction. Thus, one does not follow a stochastic particle in time, but rather describes it by a probability distribution at each x. One may next ask whether this conditional probability at x changes in time? This is not the actual time related to stochastic motion of the particle. To answer this question, we argue one must consider interactions which occur at x which are part of an energy equation. Even in classical physics, the energy equation at x indicates how a particle’s motion changes. Thus, we argue one does not simply take measurements at x of ti and pi, but that at each ti, a particle with pi is undergoing dynamic changes and these may be considered as well. If one describes the quantum picture by W(x) i.e. a momentum distribution, one may find a collective interaction given by -1/2m d/dx d/dx W + VW. The coupling between V(x)=Sum over k V(x)exp(ikx) and the particle conditional probabilities in W is apparent in the VW term. This energy term should describe whether the momentum distribution changes in time i.e should be proportional to dW/dt (partial derivative), but should also equal the average energy. Thus, a link exists between ensemble energy and time. In this note, we argue that these ideas may be applied to P(p/x) in a bound state resonance because the same E applies to each p.

•Oct 7, 2020•0 citations
In Part I of this note, we argued that in classical statistical mechanics, the probability P(p), where p is momentum, is well defined because there is a balance of two-body elastic collisions and their time-reversed reactions. In particular, this may be applied to a point x where one imagines a tiny box with many particles. In quantum bound states, collisions are with a potential and one does not have a balance of reactions and their time reversed cases. Thus, the notion of p at x does not seem to be clearly defined. A particle may be a x with momentum p, but at the same time may be reacting with V(x) and transforming into another pj. Thus, there is an ambiguity which does not exist in classical statistical mechanics. There, a particle may be colliding and so a p appears to disappear, but a new p is created by a time reversed reaction. P(p), we argue, must hold over time in an equilibrium. Due to this issue, it appears that the best one may do is to define P(p) for the whole x region in quantum mechanics. This P(p) is measurable by taking an ensemble of identical single particles, knocking them out and measuring their momentum. As a result, knowing “p” exactly means one does not know “x”. As a result, a different view must be taken to dealing with probabilities in a quantum bound state. We use the idea of invariance in x and argue that a wave-like behaviour in probability may be used to describe results. This is unusual because it appears as if particles are appearing or disappearing at different x points. If one has periodic motion, a question which arises is: What is the spatial frequency? We argue that the answer follows from classical physics. There: delta p = delta t (-dV/dx). If p initial =0 and delta t is constant, one sees that a large p final corresponds to a large change in V(x) in a tiny dx or that there is very good resolution in x. Thus a large p (linear) implies high resolution in x. Thus, we use px as the argument of the periodic probability function.
Addendum July 19, 2021 The Fermi-Dirac and Bose-Einstein statistical cases, like the Maxwell-Boltzmann case, may be formulated in terms of a reaction balance between a regular two body elastic collision and its time reversed reaction: e1+e2 = e3 + e4 and f(e1)[1-/+f(e3)]f(e2)[1-/+f(e4)] = f(e3)[1-/+ f(e1)] f(e4)[1-/+ f(e2)]. Both the Fermi-Dirac and Bose-Einstein cases reduce to the Maxwell-Boltzmann case under certain conditions. An unusual feature of bound state quantum mechanics is that there is a spatial density (or probability to be at x = P(x)), just as in classical statistical mechanics, but also a wavefunction W(x) such that P(x)=W(x)W(x). In other words, W(x) for the bound state is like a square root of probability. Furthermore, there is a P(p)=a(p)a(p) where W(x)=Sum over p a(p)exp(ipx). In this note, we try to argue that P(p), which is based on the measurement of momenta of particles knocked out of an ensemble of identical single particle bound states, does not have physical meaning within the bound state, unlike P(p,x) which has physical meaning for a Maxwell-Boltzmann gas both inside the gas and for particles released from a small hole in the gas wall. This forces one to create another function to “link” the two physically observed probabilities P(x) and P(p).
For quantum bound states, classical style probabilities P(p)=a(p)a(p) and P(x)=W(x)W(x) [where W(x)= the wavefunction = Sum over p a(p) exp(ipx)] exist. These are real and normalized to one, but unusually, they separate p and x completely. For example, P(p) is the probability to find a momentum p anywhere in the system - no information about x is known. This seems to suggest that the P(p/x) is not defined in a classical sense. (Similar arguments hold for P(x).) We have argued in a previous note (1), that due to a lack of time reversal balance, there is no clear classical notion of P(p/x) in quantum mechanics while in classical statistical mechanics, there is a time reversal balance of two-body collisions and so P(p/x) is well-defined. In particular, a momentum p collides with a pj to form another momentum pk. There is ambiguity if p is changing, but if at the same time a pk is changing (colliding) back into a p at x, then the ambiguity is removed. In quantum mechanics, the ambiguity is not removed at x, but one may discern one p state from another throughout the entire x space. This suggests orthonormal probabilities e.g. say exp(ip1 x) and exp(i p2 x). These exp(i p x) probabilities comprise the wavefunction which is linked to P(p/x) = a(p) exp(ipx)/W(x), but although Sum over p P(p/x)=1, P(p/x) is complex with the real and complex portions taking on positive and negative values. Nevertheless, it is W(x) which is the solution of the Schrodinger (and other) equations. It is known that solutions Wn(x) of the Schrodinger equation have discrete En (energy) values. It is possible to have [H, Operator]=0 and possibly to have H written in terms of this operator. In such a case, Wn(x) might be both an eigenfunction of H and the Operator. Alternatively, a linear combination of Operator eigenfunctions may be used. If En is discrete, one may expect the operator eigenfunctions to also be discrete, leading to quantization of the Operator eigenvalues. In some cases, one may have two operators which commute with H, e.g. LL and Lz (related to angular momentum). Lz eigenfunctions might not be linked to H so one might ask why Lz should be quantized. Another H, however, may distinguish between Lz eigenfunctions and so they also have discrete eigenvalues. In this note, we try to consider why one should have orthonormal eigenfunctions of these commuting operators which make up the wavefunction and try to interpret their physical meaning.
In a previous note (1) we argued that the free particle classical action A (relativistic or nonrelativistic) may be written in terms of v=x/t and then varied in x and t independently. One finds that dA/dx partial = p and dA/dt partial = -E implying A= -Et+px with x and t independent. This is the opposite of the classical mechanical idea of x(t). Creating an eigenvalue equation yields -id/dx partial exp(iA) = pexp(iA) and id/dt partial exp(iA) = Eexp(iA). exp(iA) plays the role of an unusual type of probability we argue which describes two dimensional periodicity (cos(px) and sin(px) ). We have argued that the quantum particle is the medium for a two dimensional wavelike motion of an energy-momentum wave. A wave type structure is needed to describe dynamics (like a classical wave), but unlike a classical wave two dimensions are required because there is no preference of any x point for a free particle. Both cos(px) and sin(px) both treat x in a preferential manner as their values vary with x. For a classical wave this is fine because the creation point of the wave imposes boundary conditions, but there is no such boundary condition imposed on a free particle. exp(ipx-iEt0 must somehow then link to P(x)=constant. One may note that exp(-iA)exp(iA) = 1 for the free particle. This may be argued to be linked to spatial invariance i.e. P(x)=constant where P(x) is not exp(ipx). Here we argue that exp(-iA)exp(iA) represents a product of two unusual probabilities , one for motion in a certain direction and the other for the time reversed scenario. If the quantum particle is the medium of an energy-momentum wave exp(-iEt+ipx), then the product of the regular and time reversed unusual probabilities cancel or undo each other. In other words one does not see any change in space as long as one has a noninteracting exp(-iEt+ipx) . Given that -id/dx is a linear operator this product yields a momentum of 0. We consider this idea and its implications on a momentum distribution Sum over p a(p)exp(ipx) which may represent a bound state or other states. In particular, it seems that P(x)=W*(x,t)W(x,t) which yields a classical type of probability (P(x) real, positive and normalized to 1) is linked to a product (merging) of forward and time reversed situations so that any spatial density which remain due to the quantum particle acting as a medium for an energy-momentum wave (as well as the time reversed part) would represent an actual change in spatial density. exp(ipx) already contains two periodic spatial functions (cos(px) and sin(px)), but does not manifest it unless there is an interaction.
Both quantum bound state equilibrium and classical statistical mechanical equilibrium are characterized by d/dt partial density(x,t)=0. For a quantum particle, W(x,t)=exp(-iEt)W(x) and d(x,t)=W*W. Density is linked to fluid flow and arguments of pressure balance such that the overall average fluid flow is zero are made. This pressure-force balance is common in classical statistical mechanics and used to calculate the variation of density with x i.e. obtain a differential equation with d d(x,t) /dt partial. A similar balance equation is used in (1) as the authors try to derive the Schrodinger equation using a fluid dynamical approach. In this note, we try to argue that one may think of the two equilibria in terms of an energy quantity which remains constant throughout space. This energy quantity depends directly on the type of interactions in the system at any point x. For a classical ideal gas, balanced two body time reversal scattering represents the interaction. V(x), a potential, does not change this fundamental manner of interacting at any given point, although it does affect density. As a result the energy related constant at each x is T, related to average kinetic energy. We argue that this already points to the dominant type of interaction creating the statistical interaction i.e. one that depends only on p in interactions which conserve kinetic energy. This “happens” to be equivalent to maximizing Shannon’s entropy, but we argue this is particular to this scenario in the sense that not all interacting systems will have this equivalence. In other words, we argue that one must be careful with regards to the maximization of entropy as a fundamental rule. In particular, we consider the single particle bound quantum state. The interaction exists between the particle and the potential V(x), but one may consider a distribution of free particle momentum probabilities i.e W(x)=wavefunction = Sum over p a(p)exp(ipx). The form exp(ipx) comes from separate arguments as is discussed in the note, but the point is that there are no two body scattering reactions occurring at a given x, neither is there acceleration from on x to another (except for a vrms mathematical quantity). Nevertheless the “free particle” exp(ipx) interacts at each x with V(x). As a result of this interaction, one has free particles exp(ipx) which are no longer associated with pp/2m, but rather with an Eave which is a constant for every p. In other words the identity of exp(ipx) is unusual because it is linked to exp(ipx) and pp/2m, but also to Vk1 a(p-k1), Vk2 a(p-k2) etc where V(x)=Sum over k Vk exp(ikx). Thus each x point is associated with a constant Eave and this defines the equilibrium. (It is also equivalent to conservation of energy at each x.) In such a case, one does not maximize entropy to find a(p) values. In the two cases, the common point seems to be considering the interactions at each x and finding a constant “energy” type quantity which does not vary with x. This we argue appears to define equilibrium for both the classical statistical and quantum bound states. Instead of considering density which does not change in time at each x, there is constant energy which not only does not change in time, but has a constant value at each x which seems to be an alternative rule.
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.
We argue that in a quantum bound state with one particle, the particle receives stochastic hits from a potential. As a result, if one examines a point x, the particle may be present at ti with momentum pi (in both the forward and backward directions) leading to a momentum distribution with a conditional probability P(p/x) = a(p) exp(ipx)/W(x) where W(x) is the wavefunction. Thus, one does not follow a stochastic particle in time, but rather describes it by a probability distribution at each x. One may next ask whether this conditional probability at x changes in time? This is not the actual time related to stochastic motion of the particle. To answer this question, we argue one must consider interactions which occur at x which are part of an energy equation. Even in classical physics, the energy equation at x indicates how a particle’s motion changes. Thus, we argue one does not simply take measurements at x of ti and pi, but that at each ti, a particle with pi is undergoing dynamic changes and these may be considered as well. If one describes the quantum picture by W(x) i.e. a momentum distribution, one may find a collective interaction given by -1/2m d/dx d/dx W + VW. The coupling between V(x)=Sum over k V(x)exp(ikx) and the particle conditional probabilities in W is apparent in the VW term. This energy term should describe whether the momentum distribution changes in time i.e should be proportional to dW/dt (partial derivative), but should also equal the average energy. Thus, a link exists between ensemble energy and time. In this note, we argue that these ideas may be applied to P(p/x) in a bound state resonance because the same E applies to each p.