Gibb’s Paradox is associated with introducing a 1/N! factor in the partition function= 1/N! {Sum over i exp(-ei/T)} power N (1). It is often stated that N! is required to deal with identical particles. It is further stated that in classical physics, particles have an identity, but in quantum mechanics, they are identical and so the N! is really a quantum mechanical feature. In (2), we argued that dE=TdS - PdV is entirely focused on interactions and so if S = -k ln(number of states), one should consider interaction states, not identity states. We argued that a green particle differs in colour from a red one, but this has nothing to do with interactions and one would completely ignore colour when calculating the number of states. The same reasoning applies to two identifiably different particles with the same e,p being at x1, and x2 versus x2, x1. In terms of identity states, there are two, but they both represent the same physical interaction and are counted once. Even though classical particles have an identity, and quantum ones also because they have a center-of-mass, from the point of view of an interaction, a particle at (x,y,z) with a certain momentum and energy is all that one needs to know for interaction purposes. It is irrelevant which particle is at this (x,y,z) and so N! Is introduced for this reason, as argued in (2). Here we consider the quantum mechanical case. We have argued in previous notes that quantum mechanics follows from special relativity. In particular, a particle still has a center-of-mass and so an identity. If it moves with constant speed, one has the classical equation x=vt for the center-of-mass. This begs the question: Where are the quantum mechanical features? We argue they are in the interaction. The fact that the center-of-mass is at some x,t does not mean that a p (momentum) impulse hit is rendered at that point. In fact, the Lorentz invariant -Et+px, which is more general than x=vt, may be considered to be a kind of free particle potential with x,t and x+hbar/p, t+hbar/E having the same potential values and rendering the same hit, regardless of whether x=vt. -Et+px is already created using the notion x=vt for the center-of-mass and so -Et+px holds for more x, t points than simply x=vt. As a result, the quantum free particle wavefunction exp(-iEt+ipx) represents a probability to interact, not a center-of-mass identity of the particle. In other words, quantum mechanics is a statistical theory of interaction and as such the probability bypasses the actual position of the center-of-mass which identifies the particle. In a 2-slit experiment, a particle has a specific center-of-mass and moves as x=vt, but at the 2-slit apparatus (with slits about hbar/p apart), it may probabilistically interact with both slits. This is a probabilistic interaction calculation which has little to do with the center-of-mass position and hence any position identity of the particle. As a result, a more general quantum wavefunction W(x)= Sum over p a(p)exp(ipx) (bound state) is also an interaction probability. Thus, quantum mechanics automatically only considers interactions and this leads some to argue that there is no particle identity in quantum mechanics. We argue that just like classical physics, there is a particle identity (center-of-mass), but the interaction is the main focus and does not depend on this identity position. In other words, one only deals with interaction probabilities, just like classical statistical mechanical problems should only deal with interaction states (as argued in (2)), and not with any physical state which may be identified, but does not change the interaction. The difference between quantum mechanics and classical mechanics, is that quantum mechanics is formulated on the notion of interaction (so the particle identity is removed from the calculation right away), while in the classical case one retains the identity and has to manually correct for identity states in order to convert the number to the number of interaction states.
Gibb’s Paradox is associated with introducing a 1/N! factor in the partition function= 1/N! {Sum over i exp(-ei/T)} power N (1). It is often stated that N! is required to deal with identical particles. It is further stated that in classical physics, particles have an identity, but in quantum mechanics, they are identical and so the N! is really a quantum mechanical feature. In (2), we argued that dE=TdS - PdV is entirely focused on interactions and so if S = -k ln(number of states), one should consider interaction states, not identity states. We argued that a green particle differs in colour from a red one, but this has nothing to do with interactions and one would completely ignore colour when calculating the number of states. The same reasoning applies to two identifiably different particles with the same e,p being at x1, and x2 versus x2, x1. In terms of identity states, there are two, but they both represent the same physical interaction and are counted once. Even though classical particles have an identity, and quantum ones also because they have a center-of-mass, from the point of view of an interaction, a particle at (x,y,z) with a certain momentum and energy is all that one needs to know for interaction purposes. It is irrelevant which particle is at this (x,y,z) and so N! Is introduced for this reason, as argued in (2). Here we consider the quantum mechanical case. We have argued in previous notes that quantum mechanics follows from special relativity. In particular, a particle still has a center-of-mass and so an identity. If it moves with constant speed, one has the classical equation x=vt for the center-of-mass. This begs the question: Where are the quantum mechanical features? We argue they are in the interaction. The fact that the center-of-mass is at some x,t does not mean that a p (momentum) impulse hit is rendered at that point. In fact, the Lorentz invariant -Et+px, which is more general than x=vt, may be considered to be a kind of free particle potential with x,t and x+hbar/p, t+hbar/E having the same potential values and rendering the same hit, regardless of whether x=vt. -Et+px is already created using the notion x=vt for the center-of-mass and so -Et+px holds for more x, t points than simply x=vt. As a result, the quantum free particle wavefunction exp(-iEt+ipx) represents a probability to interact, not a center-of-mass identity of the particle. In other words, quantum mechanics is a statistical theory of interaction and as such the probability bypasses the actual position of the center-of-mass which identifies the particle. In a 2-slit experiment, a particle has a specific center-of-mass and moves as x=vt, but at the 2-slit apparatus (with slits about hbar/p apart), it may probabilistically interact with both slits. This is a probabilistic interaction calculation which has little to do with the center-of-mass position and hence any position identity of the particle. As a result, a more general quantum wavefunction W(x)= Sum over p a(p)exp(ipx) (bound state) is also an interaction probability. Thus, quantum mechanics automatically only considers interactions and this leads some to argue that there is no particle identity in quantum mechanics. We argue that just like classical physics, there is a particle identity (center-of-mass), but the interaction is the main focus and does not depend on this identity position. In other words, one only deals with interaction probabilities, just like classical statistical mechanical problems should only deal with interaction states (as argued in (2)), and not with any physical state which may be identified, but does not change the interaction. The difference between quantum mechanics and classical mechanics, is that quantum mechanics is formulated on the notion of interaction (so the particle identity is removed from the calculation right away), while in the classical case one retains the identity and has to manually correct for identity states in order to convert the number to the number of interaction states.
We have argued in previous notes that both Newtonian mechanics and free particle quantum mechanics arise from special relativity. Thus, special relativity gives an expression for free particle energy and momentum and these are related to concepts of work. The Lorentz invariant -Et+px shows that x,t and x+hbar/p, t+hbar/E give the same solution to this potential-like term showing that there is stochasticity in the position and time delivery of interaction hits. In other words, even though the center-of-mass moves as x=vt, a momentum hit (proportional to rest mass mo) does not necessarily act at the center-of-mass. This gives the impression of shifts in probability, kinetic energy etc in the case of interference, i.e. interactions. For a single free particle, there is no interaction and -1/2m d/dx d/dx exp(ipx) = pp/2m exp(ipx). We argue that exp(ipx) is a probability and that in general one may add independent probabilities in an OR situation as in a 2-slit interference case. A bound state, however, is different because even though one adds weighted exp(ipx)s, they are part of a resonance or equilibrium and this is no longer an independent OR situation. We suggest that one uses a conditional probability when computing a Newtonian nonrelativistic conservation of energy equation to try to remove as much of the quantum features from this equation. As a result, one has: KEclassical(x) + V(x) = E = Sum over p pp/2m P(p/x) + V(x). One cannot bypass the fact that a quantum particle interacts differently from a classical one and that this involves a probabilistic granularity (wavelength), but classical notions of energy and energy conservation still exist. As a result, the bound state wavefunction Wn(x) = Sum over p an(p) exp(ipx) is not a usual OR sum, but is rather 1 over a normalization factor for an equilibrium picture. As a result, one cannot use Wn1(x)Wn2(x) = W(n1+n2) (x) as one would for independent probabilities. Furthermore, given that one has Wn(x) and Wn1(x)+Wn2(x) which are both of the form Sum over p a(p)exp(ipx), and both represent average energies, one needs a way to distinguish a resonance/equilibrium from a sum of resonance/equilibrium normalization factors. We suggest that the orthogonality of resonance Wn(x)’s provides this feature. The standard textbook explanation is based on Hermitian operators, but in the nonrelativistic case, we argue that one may obtain the result based on integration by parts based on the form of the p operator = -grad. Thus, it is ultimately the form exp(ipx) which leads to orthogonal bound state Wn(x)’s, together with the boundary condition that Wn(x)-->0 of x-> +infinite and -infinite.
In classical physics, a particle traveling to the right is the mirror image of one traveling to the left. In terms of a spatial density (the time spent in each constant dx), the value is the same, as is the energy. A similar notion of such a spatial invariance is in the classical conservation of energy equation for a bound particle: p(x)p(x)/2m + V(x) = E. Quantum mechanics seems to be different because exp(ipx) and exp(-ipx) are distinctly different spatial probabilities, i.e. they are dynamic. It is not just that case that a particle moves to the right for exp(ipx) and to the left for exp(-ipx). The impulse hit probabilities (represented by exp(ipx) and exp(-ipx)) are distinctly different. If one wishes to create an OR situation and write: W(x) = Sum over p a(p)exp(ipx) and use this in a conservation of energy equation, then simply including only positive or negative p values in W(x) leads to a complex average kinetic energy -1/2m d/dx dW/dx / W which cannot be used with V(x) and E. Furthermore, W_ (only negative p) and W+ (only positive p) values do not yield the same average kinetic energy. Fortunately, given that this is a statistical average for a localized object (somewhat analogous to a gas), one may combine W+ + W- and obtain a real kinetic energy. We have argued in previous notes that there exists a quantum-classical duality. For example, for a free particle one has two independent notions. One is x=vt which describes the center-of-mass of the particle. We take this to represent a state description which treats v and -v as mirror images. exp(ipx), on the other hand, represents probabilistic impulse hits and does not define the state by itself, we argue, only its interactions. Interactions are physical and define what happens in an interacting system. One may, however, measure a system, i.e interact with exp(ipx), and this should define a measurement state ( a second state ensemble average). For example, exp(ipx)exp(-ipx) = 1 which is equivalent to the constant dt in each dx associated with x=vt. Thus, the two states coincide, but exp(ipx)exp(-ipx) means that one may place the 2-slit apparatus at any x point and is not about the center-of-mass, as x=vt is. In some cases, the two states are quite different. W*(x)W(x) represents an interaction state, but vrms(x) from .5mvrms(x)vrms(x) = -1/2m d/dx dW/dx / W represents a center-of-mass state description. (Note: in these cases a state means an average state, i.e. an average over many ensemble entries.) If one takes W*(x)W(x) and integrates over x, then interference terms vanish and one obtains a classical counting of distinct p value states with their weight a*(p)a(p).This, we argue, leads to the notion of orthogonality of probabilities. In classical probability, there is no such notion, but there is the notion of AND (multiply) probabilities. Given exp(ipx) and exp(-ipx), if one multiplies one removes p, i.e. the interactions and only sees the state left. W*(x)W(x) yields a set of peaks which represent a pattern of probabilistic hits in space which define a state which one would measure in space. The interference pattern, however, must be one such that if one integrates over x, one only obtains a classical sum of probabilities a*(p)a(p)’s as mentioned above. This leads to the notion of orthogonal exp(ipx)’s (which follows mathematically anyway). If one considers different bound En states, then these two are orthogonal because W*n1(x)Wn2(x) mean n1 and n2 are associated with different energies and integrating over x cannot show a result which looks like a state result, where W*n(x)Wn(x) shows a state result. In order to achieve this orthogonality (linked directly to the notion of a state) one must have that Wn(2) usually has 0 points and is positive and negative with the possible exception of a ground state. In other words, if Wn1*(x)Wn2(x) integrated over x is not 0, one violates conservation of energy. As a result, we argue an average probability for a bound state must statistically include +p and -p contributions and must create a function which has positive and negative peaks (except possibly for the ground state), i.e. W(x) must have 0 points. This W(x) is linked to a measurement state W*(x)W(x), but this is strictly considering exp(ipx)s. We argue that there are two pieces of information for a free particle, x=vt and exp(ipx). There must also be an average state linked to x=vt and this given by KEave(x) = .5m vrms(x)vrms(x) + V(x) = E which shows that the particle may exist and have average kinetic energy even at W(x)=0 points. When measuring a system, one interacts with the exp(ipx)s and so the W*(x)W(x) state is the one which is traditionally presented, but whether one measures the center-of-mass motion or not, one also has the notion of the x=vt, i.e. vrms(x) state (average) as well, we argue.
In physics (including classical physics), conservation of momentum is always considered, but one must be more careful with conservation of energy as it may include forms like sound etc. As a result, there is a tendency to focus on conservation of momentum all of the time, and conservation of particle energy only in elastic collisions. In addition to this, one may think of a particle striking a target, in which case two equal momentum (but different energies due to different v speed and mo rest mass) cause the same impact. In the case of an interaction with a potential, however, it is strictly energy considerations which are important classically, i.e one uses p dot p /2m + V(x) = E. Knowing momentum without rest mass is not sufficient. One may consider it unusual that one may consider two different scenarios, one which focuses only on momentum (impulse) or conservation of momentum in collisions while only energy is the focus in classical bound states or motion in V(x). Given the classical focus on momentum, imagine that one wishes to describe all free particles with different momenta as a uniform probability distribution such that a product of two probabilities for p1,p2 is the same as for p3,p4 if p1+p2=p3+p4. (We use 1-dimension for simplicity.). One cannot use P= 1/n because p is absent and n->infinite. In one sense, all particles must have the same probability, but at the same time must contain extra information related to p for the conservation of p part of the problem. The notion of a uniform distribution is independent of conservation of momentum, hence of p momentum. Conservation of momentum requires an explicit presence of p. As a result, there seems to be a paradox. One may, however, resolve this paradox by arguing that a probability is a mathematical way of presenting a subset of the information present. If one argues for a uniform probability, then it does not need to contain p, and one would have 1 as an unnormalized probability for any free particle p. A different probability (representing a different subset of information) which respects all free particles as having the same weight, but includes p can also be constructed, namely exp(i C p). C is a constant introduced for unit issues. In this case, the modulus gives all free particle probabilities with p the same weight, but the presence of p allows one to have conservation of momentum. Given that momentum is the key variable that is conserved in classical physics, one should be satisfied with exp(i C p). One might argue that there is no need to consider energy E. We note, however, that if one has p, E1 and -p, E2 in a frame, then in a frame moving with -v, one has gp+gvE1= p1’ and p2’ = -gp + gvE2. Adding yields gv(E1+E2) and so energy automatically appears and cannot be discarded. Furthermore for p, E1 and -p, E2, exp(i C 0) is exp(i C gv(E1+E2)) as seen in the moving frame and so one is forced to consider conservation of energy together with momentum and include both p and E in the more detailed probability. In other words, special relativity automatically links p and E in a 4-vector, leading to the idea that both are important. We argue that one cannot simply consider exp(i C p) due to special relativity. Both E and p must be considered together and this is a key idea, we argue, because if one wishes to have a Lorentz invariant probability, it must be in E and p, i.e. -Et+px. Instead of having a probability for an interaction with p given by exp(i C p) , one has exp(-iEt+ipx) and we argue that this must have the same value in all frames for a particular interaction. One may note that E and p are, on the one hand, separated allowing for impulse hit versus E related interactions, but that overall the probability for interaction includes both and it is the full -Et+ipx which must be invariant. This further suggests that if one does not artificially force an interaction to occur at the center-of-mass, then x,t and x+hbar/p, t+hbar/E are equally good points for the interaction. We argue that there is no need to force x=vt because exp(-iEt+ipx) * exp(iEt+ipx) already shows that one has uniformity, not only for any p,E, but also for any x, t, as there are now four variables. Thus, there is no reason to enforce x=vt in exp(-iEt+ipx), when considering an interaction, we argue. This means that one has two possible interactions at different x,t values, i.e. x,t and x+hbar/p, t+hbar/E. If one is unable to follow the interaction in time, but may follow it in space as in the case of 2-slits about hbar/p apart or for a potential V(x), then one may use the exp(ipx) part of the interaction probability piece by itself. Thus, we argue that special relativity does not allow one to treat p independently of E. For example, for p values that are the same in one frame, but have E1, E2, will be different in another frame. One must consider E and p together and so instead of creating a probability for p alone, one must create one for E,p and then impose Lorentz invariance which leads to different x,t pairs yielding the same probability value for an interaction, i.e. free particle quantum mechanics.
In previous notes, we argued that one could describe elastic two particle collisions, even in Newtonian mechanics, in a classical statistical manner. In particular, we argued that one could introduce a probability which depends on E,p, but at the same time has the same weight (in a sense) for any free particle. Product probabilities would be linked with a uniform distribution in the sense that exp(-iE1)exp(-iE2)= exp(-iE3)exp(-iE4) if E1+E2=E3+E4 and a similar expression for momentum. If one respects the special relativistic invariant -Et+px, then a possible solution is exp(-iEt+ipx), but even though this yields the free particle wavefunction (probability), this does not explain quantum mechanics. These are simply ideas of classical probability theory with the enforced special relativistic invariant -Et+px. Here, we argue that the real feature leading to quantum mechanics (the idea of hbar/p and hbar/E) arises from the notion of inertia (which we have argued in previous notes also gives rise to spin). We argue that a single force (inertia) in a rest frame becomes two independent interaction notions, E and p as well as creating x’ as a running variable. Thus, mo,t and E,t’ and p,x’ are intertwined. We suggest that there must be a physical mechanism which keeps these two concepts E,p independent and so one cannot simply rely on x=vt. Given the Lorentz invariant and the classical notion of probability exp(-iEt+ipx), one must go one step further and postulate that there are independent fluctuations in t’ and x’ in such a calculation such that hbar/E and hbar/p emerge. We argue that this is a key feature of quantum mechanics and that it does not immediately arise from exp(-iEt+ipx) if one only considers x=vt. One is forced to allow x,t to fluctuate independently which leads to a fluctuation in mocc in the rest frame as well. We argue that this feature is actually associated with special relativity as it allows a single force inertia to manifest itself as two independent interaction quantities E,p. As a result, we suggest that one may obtain exp(-iEt+ipx) without any notion of probability, but only by postulating the existence of a physical mechanism (fluctuation of x’,t’ or t in the rest frame) which keeps E,p independent because -Et+px does not ensure their independence if one only considers x=vt. One may then interpret exp(-iEt+ipx) as having the form of a classical probability which respects special relativity.
We try to consider free particle and single bound state quantum mechanics from the viewpoint of classical physics. We first assume that a free particle moves as x=vt. Secondly, we assume that dx and dt cannot be 0 as in Newtonian mechanics, but must be finite. This begs the question: How does one measure dx and dt? If one can measure a particle at a given x and t, then dx→0, dt→0. Thus, finite dx, dt means that a given p=constant and pp/2m=kinetic energy (nonrelativistic) interact with a force over a finite dx. In other words, one does not know where the interaction occurs within dx. As long as the free particle moves without interacting, then x=vt and there is no interest in any dx or interactions. For example, one may measure very large x and t ranges. Even if there are dx and dt effects at the start and end points, these are small compared to the large x and t values. As a result, one has a F(x,p) probability within dx, which repeats itself for the next dx (i.e. is periodic), where F is a probability for a p impulse hit at x within dx. At the same time, the particle moves with x=vt it seems within dx. In general, one might then think that F(x,p) and x=vt are decoupled. In previous notes, we have argued that one way to couple F(x,p) and x=vt is through special relativity, i.e. A = -Et+px. X and t values used would presumably satisfy x=vt, but for any x,t one has x+hbar/p and t+hbar/E as solutions yielding the same A value and so might consider dx= hbar/p and dt=hbar/E. The question we ask here is whether one may arrive at this conclusion without using special relativity. The answer seems to be that one can if one suggests that F(x,p) must contain information making it consistent with x=vt, i.e. with every x and t point having the same weight. Clearly, however, each x,t do not have the same weight within dx and dt, but are rather governed by F(x,p). This means, however, that F(x,-p) also represents a probability distribution and so if one has F(x,p)F(x,-p) there is no momentum at a given x point. In such a case, the distribution within dx must disappear and one must have a uniform distribution. This holds even if F(x,p) has the same dx value for any p. Thus, F(x,p)F(x,-p) = constant and must be periodic in x. This implies that F(x,p) = 1/F(x,-p) for constant=1. Given that F(x,p) =0 at the initial and final points of dx, then F(x,-p) = infinite at these points and that is not allowable. Thus, one is forced to use complex variables and try exp(i g(p) x). If g(p) is a function of p (i.e. not a constant), then conservation of momentum implies that exp(i g(p1) x) exp(i g(p2) x) = exp(i g(p1+p2) x), suggesting exp(ipx). What, however, if g(p) = constant? In such a case, one has issues creating an average kinetic energy: ( Sum over p a(p) exp(i constant x) pp/2m) / (Sum over p a(p) exp(i constant x) ((1)) This leads to a kinetic energy which does not depend on x and cannot be used with V(x) in a conservation of energy equation. Thus, we choose exp(ipx). Note: there is no reason to choose a(p,x) because the physics of finite dx and dt is restricted to exp(i constant x). Finally, we try to explain why the same dx=hbar/p and dt=hbar/E together with x=vt follow from special relativity.
Newtonian mechanics presents a simultaneous description of dynamics x(t) and interactions (F=dp/dt) because it assumes infinite resolution, i.e. dx->0, dt->0. We have argued in a previous note (1), that quantum mechanics follows from free particle quantum mechanics, which in turn follows from special relativity. Special relativity uses the notion of x=vt and provides formulas for E=mocc/sqrt(1-vv/cc) and p=mov/sqrt(1-vv/cc) even with no free particle quantum mechanical notions. We have also argued in previous notes, however, that these classical concepts are present in the Lorentz invariant -Et+px which we have argued (2) acts as a kind of free particle potential. This potential, however, has the same value for x, t and x+hbar/p, t+hbar/E and so we have suggested that free particle quantum mechanics is comprised of two parts. The first is a complex probability exp(-iEt+ipx) which describes possible interactions with a free particle and the second is x=vt. In other words, interactions and dynamics are split, but the interaction probability depends on classical notions such as x=vt, p and E. In the limit of hbar/p and hbar/E <<1, the interaction resolution becomes the Newtonian dx->0, dt.->0. Thus, as in (1), we argue that quantum and classical mechanics are intertwined. In (3), entitled Unifying Quantum and Classical Dynamics, a similar argument for the intertwining of quantum mechanics and classical mechanics is presented, but in a different manner. The basic argument of (3) seems to be that quantum mechanics, written in the Heisenberg approach, yields operator equations which are identical in form to classical equations, in particular, d/dt X(t) = P(t) and d/dt d/dt X(t) = -d/dx V( X(t)). Here X(t) and P(t) are operators given by X(t) = exp(iH t) X(0) exp(-iHt) (and a similar expression for P(t). We try to show that given that free particle quantum mechanics is directly linked to special relativity and hence to the acceptance of x=vt, that in the free particle case, any operator description of d/dt X(t) = P(t) cannot yield any result other than a classical one because no interaction is occurring. In other words, quantum mechanics is directly linked to classical mechanics at the free particle level. Later, one takes a superposition (OR probability addition) of exp(ipx)s, as in the case of bound states. In such a case, one deals with average interactions and so one must be careful when interpreting d/dt d/dt X(t) = -d/dx V( X(t)) as the system description is now an average one.