In a previous note (1) we argued that both the relativistic A= -mo t sqrt(1-vv) and nonrelativistic A=t .5mvv classical free particle actions may be written with v=x/t and x and t varied independently to yield: dA/dx partial = p and dA/dt partial = -E. Thus A is equivalent to -Et + px for a free particle with x and t being treated as independent. We argue in this note that such an assumption is equivalent to a statistical picture for a free particle even at the classical level because x and t are treated as independent. Thus a constant p may occur at any x because no time is linked to x. Given that all space points are the same for constant p one expects a constant probability at all x points. Knowing p exactly means not knowing anything about which x one has. Secondly, one may create an ensemble of free particle p classical actions for a situation containing a potential V(x). Given that x and t are independent, there cannot be any notion of single p accelerating, although a vrms from an ensemble may. At a given x (x known exactly) one does not know which p is present. Thus the ideas of uncertainty of p and x follow immediately from this picture. If one considers a single particle and an ensemble of free particles in the x-t independent scenario, then each free particle action p is linked to A=-Et+px where E=pp/2m. This ensemble would arguably be stirred up by a potential V(x). If one wishes to consider an equilibrium scenario for which there is an overall En for the bound system one would expect an overall Ent for the system and for -pp/2m t pieces to disappear. Thus we postulate that pp/2m +{p part} G(p’s k’s) = En for all p where {p part}G(p’s, k’s) is an unknown function and V(x)=Sum over k Vk exp(ikx). Next one may consider the spatial situation for which V(x) is the classical potential for a single particle. The kinetic energy follows from an ensemble of free particles with the overall -Et removed. From classical considerations one would expect to have a KE(x). Thus one expects: KE(x) = {Sum over p a(p) f(px) } / {Sum over a(p)f(px) } with f(px), an orthonormal basis set, because the action contains px. (It is orthonormal because one wishes the identity of p to exist if not at a point, then over all of x.) We try to argue in this note that f(px)=exp(ipx), the momentum eigenstate which follows from A=-Et+px. We find G(p’s, k’s) which implies entanglement or mixing of a given p state with other p and k states from the potential V(x)=Sum over k Vk exp(ikx). Thus one expects an entropy linked to this momentum entanglement. Given the form exp(ipx), finding a set of ap’s which satisfy pp/2m + G(p’s, k’s)=En leads to a spatial distribution W(x)=Sum over p a(p)exp(ipx) so there should be an entropy contribution from this spatial part as well. In classical statistical mechanics one has both momentum and spatial contributions to entropy. We argue for S=Sp+Sx, but note that even though px has x, p symmetry as does the Heisenberg uncertainty principle var x var p >= hbar/2, Sp and Sx may not be symmetric in terms of important parameters such as n, energy level. In other words, Sp the entanglement entropy with V(x) may be the driver entropy sensitive to energy level changes while Sx the resulting spatial density may not be as sensitive. Furthermore the overall entropy Sp+Sx gives information about uncertainties in x and p as is already known in the literature where bound state Sp+Sx bounds are given.
In a previous note (1) we argued that both the relativistic A= -mo t sqrt(1-vv) and nonrelativistic A=t .5mvv classical free particle actions may be written with v=x/t and x and t varied independently to yield: dA/dx partial = p and dA/dt partial = -E. Thus A is equivalent to -Et + px for a free particle with x and t being treated as independent. We argue in this note that such an assumption is equivalent to a statistical picture for a free particle even at the classical level because x and t are treated as independent. Thus a constant p may occur at any x because no time is linked to x. Given that all space points are the same for constant p one expects a constant probability at all x points. Knowing p exactly means not knowing anything about which x one has. Secondly, one may create an ensemble of free particle p classical actions for a situation containing a potential V(x). Given that x and t are independent, there cannot be any notion of single p accelerating, although a vrms from an ensemble may. At a given x (x known exactly) one does not know which p is present. Thus the ideas of uncertainty of p and x follow immediately from this picture. If one considers a single particle and an ensemble of free particles in the x-t independent scenario, then each free particle action p is linked to A=-Et+px where E=pp/2m. This ensemble would arguably be stirred up by a potential V(x). If one wishes to consider an equilibrium scenario for which there is an overall En for the bound system one would expect an overall Ent for the system and for -pp/2m t pieces to disappear. Thus we postulate that pp/2m +{p part} G(p’s k’s) = En for all p where {p part}G(p’s, k’s) is an unknown function and V(x)=Sum over k Vk exp(ikx). Next one may consider the spatial situation for which V(x) is the classical potential for a single particle. The kinetic energy follows from an ensemble of free particles with the overall -Et removed. From classical considerations one would expect to have a KE(x). Thus one expects: KE(x) = {Sum over p a(p) f(px) } / {Sum over a(p)f(px) } with f(px), an orthonormal basis set, because the action contains px. (It is orthonormal because one wishes the identity of p to exist if not at a point, then over all of x.) We try to argue in this note that f(px)=exp(ipx), the momentum eigenstate which follows from A=-Et+px. We find G(p’s, k’s) which implies entanglement or mixing of a given p state with other p and k states from the potential V(x)=Sum over k Vk exp(ikx). Thus one expects an entropy linked to this momentum entanglement. Given the form exp(ipx), finding a set of ap’s which satisfy pp/2m + G(p’s, k’s)=En leads to a spatial distribution W(x)=Sum over p a(p)exp(ipx) so there should be an entropy contribution from this spatial part as well. In classical statistical mechanics one has both momentum and spatial contributions to entropy. We argue for S=Sp+Sx, but note that even though px has x, p symmetry as does the Heisenberg uncertainty principle var x var p >= hbar/2, Sp and Sx may not be symmetric in terms of important parameters such as n, energy level. In other words, Sp the entanglement entropy with V(x) may be the driver entropy sensitive to energy level changes while Sx the resulting spatial density may not be as sensitive. Furthermore the overall entropy Sp+Sx gives information about uncertainties in x and p as is already known in the literature where bound state Sp+Sx bounds are given.
In the literature (1), the Schrodinger equation: {p1p1/2m + p2p2/2m + V(x1-x2)} W(x1,x2,t) = id/dt partial W(x1,x2,t) may be transformed using P=p1+p2 and (p=m1p2-m2p1)/M and x=x1=x2 where M=m1+m2 ((1)). This leads to a bound Schrodinger equation in x and a “center of mass” equation” in P i..e Hcm =PP/2M which depends on m1 and m2, but not binding energy. . In a previous note, we argued that a free particle’s classical action A (relativistic or nonrelativistic) may be obtained by setting v=x/t and treating x and t as independent (2). This leads to dA/dx partial=p and dA/dt partial = -E. One may write eigenvalue equations: -id/dx partial exp(iA) = pexp(iA) and id/dt partial exp(iA) = E exp(iA). A solution is A=-Et+px with E,t,p and x independent. -Et+px is a Lorentz invariant and this form is used in both relativistic and nonrelativistic quantum mechanics (although E and p have different values). We argue that a free particle rest mass mo may actually involve internal quantum physics, yet the “system” is replaced with a parameter mo and exp(-iEt+px) used to describe it. One may try to apply this approach to a bound system, but exp(-iEt+px) is a statistical probability like object. Thus one needs a statistical theory. We have suggested using an ensemble of exp(-iEt+px) with E=En for all p to indicate a resonance. Then: { Sum over p pp/2m a(p)exp(ipx) } / {Sum over p a(p)exp(ipx) } + V(x) = En ((2)) This is the Schrodinger bound equation. We argue that W=Sum over p a(p)exp(ipx) should be real and this implies that Sum over p a(p) p exp(ipx) = 0 i.e. average momentum for this type of average is 0. As a result, one may create a 4-vector using { 0, mcc + En} and use the Lorentz boost approach to find the momentum and energy in a moving frame. One may then take the nonrelativistic limit to find kinetic energy i.e. the full center of mass kinetic energy. We argue that this is equivalent to considering exp(-iEt+ipx) for the overall bound system which is now translating in space. We argue that ((2)) is a different approach to ((1)) in that binding energy, which may be significant with respect to mcc, should be part of the overall energy of the translating bound system which is called the “center of mass” system in the literature.
In previous notes (1) we argued that the free particle action A (either nonrelativistic or relativistic) may have v (velocity) written as x/t and x and t varied independently to yield: dA/dx partial = p and dA/t partial = -E with A=-Et+px, the Lorentz invariant classical Action = t Lagrangian. We further argued that one may write eigenvalue equations i.e. -idexp(iA)/dx partial = p exp(iA) and idexp(iA)/dt partial = Eexp(iA) and call exp(iA) a type of probability. We noted that if a potential is present one may use P(p/x)= a(p)exp(ipx)/W(x) [W(x)=wavefunction = Sum over p a(p)exp(ipx)] and establish a conservation of energy equation KE(x) + V(x) = E such that there exists the same constant E at each x. We described this in two ways. First E represents binding energy so at each x one has the same binding energy i.e. each x point maps to the same E as part of the mass of an overall moving system. Secondly, we argued that E was a parameter indicating that equilibrium existed at each x point, much like temperature T. In this note we wish to consider a different idea suggested in (2), namely that if one has a quantum state: W(x)= Sum over n a(n,t) Wn(x) this may be written as: <i |H| j> a*(i,t)a(j,t) = <W H W>.. If Hamilton’s equations for complex variables are applied to these then: i d/dt partial a*(i,t) = d/da(i,t) partial { <W H W> } ((A)) (2) argues this is the time-dependent Schrodinger equation. In other words, probability weights of the Schrodinger equations (which are complex) obey Hamilton’s classical equation. In this note, we point out that the formalism dA/dx partial = p already introduces operators into classical physics, so it follows that H is an operator which may act on states. We consider the idea of ((A)) as applied to a free quantum particle moving. Next, we apply ((A)) to a quantum bound state and argue that a classical Hamilton’s equation applied to d/dt a(p) partial may describe equilibrium. Finally ((A)) may be applied to a sum of eigenstates, but this is already shown in (2).
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.
Newton’s second law dp/dt = F(x)=-dV/dx may be written in terms of x alone i.e. dp/dx v(x) = -dV/dx or 1/(2m) d/dx pp = -dV/dx leading to the idea of kinetic energy pp/2m and potential energy V(x). In (1) we argued that there is velocity, rest mass mo, and momentum in quantum mechanics. A free particle, however, does not undergo smooth acceleration as in classical mechanics, and we argued there exists a momentum space wave exp(ipx) accompanying the moving particle. In such a scenario, force is considered to be probabilistic momentum hits of the form Vk exp(ikx) leading to a free particle momentum distribution W(x)= Sum over p a(p) exp(ipx). Both of these expressions involve interference due to momentum waves which do not track a particle in time. A mass flux mov or energy flux p=Ev relativistically seems to be the main consideration. Nevertheless, summing Vkexp(ikx) over k to obtain V(x) yields a function varying in x alone. This is a kind of average based on interference and Sum k Vk exp(ikx) is an average linked to momentum hits. We speculate that average momentum hits i.e. -d/dx V(x) are linked to momentum flux changes. In other words, even though an impulse is based on p and does not distinguish v i.e. p=m1v1=m2v2, momentum flux is velocity dependent and appears for an average momentum hit i.e. -dV/dx. In such a case (pave+dpave)(vave + dv ave) = 2 pave dv(ave) . This expression, however, contains both forward and backward motion which is part of quantum mechanics. Moving to an average picture seems to introduce time and direction i.e one uses pave d v(ave). This is already a change from a quantum to a classical scenario albeit for averages (which have not been specifically defined). Thus we consider pave dv(ave). If this is the case, then: d/dx pave 1/m space = -dV/dx or 1/(2m) pave pave = -d/dx V(x). Instead of treating pave pave as two separate averages one considers pave=prms or pavepave=<pp>. Thus the quantum bound picture (at any energy level) becomes a classical type of scenario which includes time and velocity. Note that for a two slit interference pattern, one does not need velocity and time because the interaction with the potentials from both slits is based on exp(ipx). One does not even need the form of the potential. Interference is the same for p=m1v1=m2v2 (even though the process takes place during different time intervals, the pattern is not affected). In other words, we argue that Newton’s second law might follow from a picture of momentum hits in which a flux i.e. p=mov is treated as a quantity which in an average scenario leads to a second flux i.e. an average momentum flux which reintroduces direction, time and velocity.. This is somewhat like a hydrodynamics scenario.
A classical bound system may be said to have a spatial density proportional to the amount of time dt a particle spends in a constant dx region i.e. C/v(x) (1) We stress that in this scenario there is a well defined and constant dx so that “dt” may be linked to velocity v(x). We also note that there is no spatial representation of action-reaction here even though this exists in Newton’s theory. In a classical Maxwell-Boltzmann gas there are many particles which each act dynamically according to .5mvv+V(x) = E (where v and E are particular to the particle when it is not colliding). At the same time there is a spatial density which “represents” macroscopic static chunks of matter which are balanced through pressure differences (P=spatial density(x) RT) and -dV/dx density where V(x) is the potential. From the pressure equation, T is proportional to energy which in turn is proportional to RMS speed squared. RMS speed may then be linked to a dt if a fixed dx is understood. In the classical statistical case, however, spatial density is directly linked to the number of particles in a region. Time enters in the pressure expression through a term proportional to the average kinetic energy i.e. pv where p is like an impulse and v a flux. What happens in a single particle quantum bound state? In a sense it is like a classical bound state as the equation: prms(x)prms(x)/2m + V(x) = En holds for any energy level. Can one then take vrms(x), choose a fixed dx and argue that spatial density should be equal to dt or C/vrms(x)? We argue that one cannot due to the lack of defined “dx” quantity. Why is dx problematic to define? We argue that it is because quantum mechanics is based on action-reaction x-probaiblity distributions linked to impulse. For example, exp(ipx) for a free particle captures only p or impulse (not velocity) and cos(px) and sin(px) are its action-reaction regions.These lead ultimately to a density is given by the well-known W*(x)W(x) where W(x)=Sum over p a(p)exp(ipx), although kinetic energy with pp/2m = .5pv is used to find W(x) as well. . W*(x)W(x) has crests and troughs similar to cos(px) in exp(ipx), except that these now have varying heights and widths. We try to argue that even though each x point maps to a different KEave(x), the hump physically represents action-reaction in the system (just as the humps in cos(px) do except for a single p) while KE(x) represents an average of p’s. This action-reaction is linked with impulses. In general, if one takes d/dx of the time-independent Schrodinger equation one finds an average impulse term, -dV/dx and KE(x) dW/dx / W where dW/dx / W is the average impulse. At peaks, dW/dx =0 and so an average force through time*impulse term balances -dV/dx=Newtonian force. These peaks are also associated with W*(x at peak)W(x at peak) which is the spatial density. We interpret this as representing the average time the single particle spends near x at peak. This value, however, follows from ideas of impulse and not from speed as there is no well defined dx even though there is a well defined vrms(x). Thus the ratio of W*(x)W(x) at two peaks is not the inverse of the ratio of vrms(x) except apparently when the energy level n becomes very high. We investigate these ideas.
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).