In Part 1, we suggested that exp(ipx) represents a spatial probability for the rendering of momentum hits. A single particle has a center-of-mass which undergoes x=vt, but momentum hits are not rendered from this point, but are rather delivered in x according to the complex probability exp(ipx) which is linked with the resolution delta x = hbar/p. This accounts for the fact that a particle can probabilistically interact with each slit of a 2-slit apparatus if they are about hbar/apart. If one is only interested in energy, pp/2m, the classical result holds for both exp(-ipx) and exp(ipx). In a classical bound state, one considers only energy p(x)p(x)/2m + V(x)=E because each p(x) exists at x in dx->0. One cannot distinguish any probabilistic hit distribution. exp(ipx) and exp(-ipx) differ even though they are linked with the same kinetic energy pp/2m (nonrelativistic case). We argued in Part 1, that the Newtonian conservation of energy equation actually includes a force profile linked with V(x). The catch is that for V(x) one has a force at each x, dx->0. Such is not the case for exp(ipx). We argued in Part 1 that one must consider not only pp/2m, but the associated spatial probability distribution for hits (i.e. energy together with force). Here, we note that this idea gives rise to an equilibrium calculation. This is somewhat startling because classically, and even quantum mechanically for higher energy levels, a particle moves to the right part of the time and then to the left for other times. As a result, a calculation which respects time would not average over exp(ipx) and exp(-ipx). Only an equilibrium calculation in time would do so. We suggest that one must consider the Newtonian conservation of energy equation as applying to both p(x) and -p(x) and so from an energy point of view one may consider it to be an average. The catch is that each term p(x) and -p(x) yields the same kinetic energy profile in the Newtonian case and the same V(x). For a quantum particle the kinetic energy with probabilistic force distribution is not the same for p and -p. As a result, to use V(x) which applies to forward and backwards motion, one is forced to consider the average and this leads to an equilibrium scenario. We show that the average kinetic with probabilistic force hit profile in space is not the same for +p and -p. Measurements, however, take time and may be averages of much data and so the notion of an average or equilibrium seems to make sense. We note, however, that an average is very different from the Newtonian forwards-backwards motion picture. Only in the high energy limit does the energy pp/2m with probabilistic force profile become the same for both p and -p. Thus, the main point of this note is to stress the notion of a single particle bound state as an average and to point out the consequences of this.
In Part 1, we suggested that exp(ipx) represents a spatial probability for the rendering of momentum hits. A single particle has a center-of-mass which undergoes x=vt, but momentum hits are not rendered from this point, but are rather delivered in x according to the complex probability exp(ipx) which is linked with the resolution delta x = hbar/p. This accounts for the fact that a particle can probabilistically interact with each slit of a 2-slit apparatus if they are about hbar/apart. If one is only interested in energy, pp/2m, the classical result holds for both exp(-ipx) and exp(ipx). In a classical bound state, one considers only energy p(x)p(x)/2m + V(x)=E because each p(x) exists at x in dx->0. One cannot distinguish any probabilistic hit distribution. exp(ipx) and exp(-ipx) differ even though they are linked with the same kinetic energy pp/2m (nonrelativistic case). We argued in Part 1, that the Newtonian conservation of energy equation actually includes a force profile linked with V(x). The catch is that for V(x) one has a force at each x, dx->0. Such is not the case for exp(ipx). We argued in Part 1 that one must consider not only pp/2m, but the associated spatial probability distribution for hits (i.e. energy together with force). Here, we note that this idea gives rise to an equilibrium calculation. This is somewhat startling because classically, and even quantum mechanically for higher energy levels, a particle moves to the right part of the time and then to the left for other times. As a result, a calculation which respects time would not average over exp(ipx) and exp(-ipx). Only an equilibrium calculation in time would do so. We suggest that one must consider the Newtonian conservation of energy equation as applying to both p(x) and -p(x) and so from an energy point of view one may consider it to be an average. The catch is that each term p(x) and -p(x) yields the same kinetic energy profile in the Newtonian case and the same V(x). For a quantum particle the kinetic energy with probabilistic force distribution is not the same for p and -p. As a result, to use V(x) which applies to forward and backwards motion, one is forced to consider the average and this leads to an equilibrium scenario. We show that the average kinetic with probabilistic force hit profile in space is not the same for +p and -p. Measurements, however, take time and may be averages of much data and so the notion of an average or equilibrium seems to make sense. We note, however, that an average is very different from the Newtonian forwards-backwards motion picture. Only in the high energy limit does the energy pp/2m with probabilistic force profile become the same for both p and -p. Thus, the main point of this note is to stress the notion of a single particle bound state as an average and to point out the consequences of this.
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 Newtonian mechanics, force and energy are clearly separated. For example, pp/2m = Integral F dx = Integral dp/dt dx = m/2 vv = pp/2m. The equation pp/2m + V(x) = E is considered to be strictly an energy equation. In special relativity, mocc is considered to be strictly an energy. We have suggested in previous notes that mo is proportional to inertia which is a force. Thus, mo and mocc are thought of in two very different manners simultaneouly, i.e. as an energy and as a force. We propose that it might be beneficial to think of both mo and V(x) in terms of energy and force distribution simultaneously. In particular, we consider that V(x) is not only potential energy at x, but is also linked to force or interaction at x through -d/dx V(x). The reason this duality of energy-force may be important is that it is directly linked to quantum mechanics, we argue. In general, it is argued that classical mechanics is a high energy level limit of quantum mechanics, We argue, however, that for a single particle bound state, the energy linked with a momentum p (at any energy level) is pp/2m (nonrelativistic) which is deterministic (and also follows from special relativity) and has nothing to do with quantum mechanics. The difference between classical mechanics and quantum, is that there exists an interaction (impulse hit) probability of exp(ipx) which means that instead of a hit occurring at the center-of-mass point, it may occur within an hbar/p range. Thus, one may have a particle interact and move through either of two slits if they are separated by about hbar/p. If p, however, is changed in an interaction, p1->p2, then the change in energy p2p2/2m - p1p1/2m (one dimension) is classical-deterministic. It is, however, the duality of V(x) with force that force one to consider not only energy pp/2m (deterministic), but also its associated force distribution given by exp(ipx) / W(x), where W(x) = Sum over p a(p)exp(ipx). As a result, one creates the average kinetic energy term: {Sum over p a(p) pp/2m exp(ipx) } / W(x). One may call exp(ipx)/W(x) a probability and so formally one has KEave(x), but physically exp(ipx) represents a force distribution. The balance to V(x) is a combination of a deterministic energy pp/2m with a force distribution exp(ipx)/W(x). This, we argue, is a physical statement and helps to describe why V(x) may be expanded as a series of virtual particles (e.g. photons etc). It is not just energy, but force that must be considered together. It is only in classical physics, pp/2m + V(x) = E that one solely considers the energy aspects, i.e. pp/2m and V(x) only as energy, but the more general picture, we argue, is to consider not only energy, but force distribution together. This is the quantum picture and it incorporates not only probability, but deterministic Newtonian mechanics (or special relativity) through the use of pp/2m.
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 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.
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.
We have argued in previous notes that exp(-iEt+ipx) is a modulus 1 (representing a uniform distribution under reduced information) complex probability. It represents the spatial probability for hits in space by the inherent force. Given that this complex probability may interfere, it has an effect on average kinetic energy: -1/2m d/dx d/dx W/W where W= Sum over p a(p)exp(ipx). One has various free particle interaction probabilities as the particle is interacting with a potential. Both the numerator and denominator of the average kinetic energy are related to the interaction spatial density, but by dividing the two one obtains the classical kinetic energy. Thus, the classical picture is present alongside the quantum one. The interference of exp(ipx)s in a time independent problem can give rise to interaction points with zero or very low probability. This, however, does not mean that the velocity is 0 at these points. For example, a particle in an infinite well has KEave(x)= -1/2m d/dx d/dx sin(pave x) / sin(pave x) = pave pave/2m. This result matches x=vrms t which is the motion of the particle because it is not interacting with a potential. Nevertheless, the interaction spatial density is W*W = C sin(pave x) sin(pave x).We have argued that exp(-iEt+ipx) is a probability which conserves energy and momentum in a two body elastic collision. Energy conservation should hold for a 1-dimensional single particle bound state as one creates an energy resonance when creating a bound state. This is important because we argue that transitions from one energy level to another should be linked with energy considerations, in particular < W1 V W2>. We first note that the Schrodinger equation, which is a conservation of nonrelativistic energy, establishes an energy resonance. V(x) is the classical energy which may be considered to be a number at x, one has E on the RHS of the equation and average kinetic energy: {-1/2m Sum over p a(p)pp/2m exp(ipx)} / {Sum over p a(p)exp(ipx)) on the LHS. Thus, it requires two pieces of information {-1/2m Sum over p a(p)pp/2m exp(ipx)} and V(x)W(x) to determine if one has a resonance energy state. In principle, if one only had V(x) W(x), this is a math function which may be written as Sum over i an Wn(x), where n is the energy level. We argue that given V(x)W(x), one should first calculate -1/2m d/dx dW/dx / W to see if one has an energy resonance, because this is a steady solution (if no V perturbation exists). Otherwise, V(x)W(x) may be seen as a sum of states an Wn(x) and should not depend on the energy of W(x). Given that V(x)W(x) represents an interaction between W(x) and V(x), Sum over n an Wn(x) represents the weighted bound state amplitudes and < W1 V(x) W2 > * <W2 V(x) W1> should be proportional to the transition probability and not depend on any energy of W1 or W2. One may create units of probability/per sec by dividing by 1/hbar and multiplying by density(E) which has units of 1/E. Thus, we argue that it is the single particle bound state resonance based on conservation of energy which gives rise to Fermi’s Golden rule. We note that (1) derives Fermi’s Golden Rule in a somewhat different manner. We suggest that -Et+px, a Lorentz invariant, is akin to a potential for a particle with rest mass and so exp(-iEt+ipx) is linked with a potential. This also holds for an electrostatic potential, while a photon has cos(-Et+px) apply directly to the inherent force and not a potential. A single particle Schrodinger equation, however, deals with a particle with rest mass mo and so it is the potential energy which is the key idea for quantum interactions. We suggest that quantum mechanics is inherently a potential based theory (with the exception of the photon) and this is why Fermi’s Golden Rule is based on <W1 V W2>. In other words, in quantum mechanics, force follows from a potential and not the other way around as in Newtonian mechanics. In quantum mechanics, the rest mass of a particle represents an inherent force which is linked with a probabilistic hit pattern in space given by exp(-iEt+ipx). Thus, the inherent force gives rise to a probabilistic potential which in turn gives rise to a term proportional to force -grad (-iEt+ipx) = p. This seems to explain why quantum mechanics equations deal with potential instead of force. The probabilistic wavefunction is linked with a potential related to the inherent force for particles with rest mass.