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).
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).
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 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.
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.
A quantum free particle is linked with a complex probability for impulse strikes at x of exp(ipx), but the momentum and kinetic energy are p and pp/2m everywhere. This suggests that Newtonian ideas of work are associated with p. Conservation of energy, however, is not only linked with pp/2m (nonrelativistically), but with interaction. If one has statistical inteferfence, i.e. an OR probability scenario, then it seems that average kinetic energy for particular p values does not match classical expectations because translational symmtery is broken. There is a shift in interaction probability in x (no longer a unit modulus) and kinetic energy linked with interaction must shift also. Given the root exp(ipx), this shift must occur in additions and subtractions of interaction in space and hence kinetic energy linked with probabilistic interaction. A simple case is that of a single particle bound state with positive parity. In such a case, pp/2m {a(p)exp(ipx) + a(-p)exp(-ipx}) (with a(p)=a(-p) yields a(p) 2 cos(px) pp/2m. Classically, one expects an average kinetic energy from p and -p of pp/2m, but the quantum result may be 0 any fraction less than 1 of pp/2m or the negative of this. It is only the overal KEave(x) that, when added to V(x), yields E. At the single p value one has unusual results. These seem to be due to interference and so we consider two other cases. We next examine exp(ipx) + exp(i2px). Classically, one would expect an average kinetic energy of ½ ( pp/2m + 4pp/2m) = 2.5 pp/2m, but for px= 3.14, one has 3 (pp/2m) which exceeds the classical average. As a third example, one may consider an equilibrium of a sin(pave x) state in an infinite well potential with a sin(2pave x) state (maintained by a photon bath at a certain temperature. If exp(-pp/2mT) is set to .5, then for pave x= 3.14 + 3.14/4, one has an average kinetic energy of .52 pave* pave/2m which is much less than the minimum classical average of pave*pave/2m. Thus, quantum interference seems to have unusual effects on average kinetic energy with respect to classical expectations.
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.
In Part 1, we argued that the consideration of all possible outcomes is a defining idea of a probabilistic calculation. We then argued that in free particle quantum mechanics there were two sets of outcomes, one driving the other. In particular, a single particle moves as x=vt, but at each x,t point, one has interaction uncertainty/fluctuations of dt and dx linked with exp(-iEt) and exp(ipx). As a result, in a 2-slit interference problem, exp(ipx) (a probability linked with possible positional fluctuations about x) leads to a further probability of interacting with both slits. The two sets of outcomes must be considered, i.e. exp(i p dot r1) + exp(i p dot r2), where r1 is a vector from the center of the first slit to a point on a screen far away, and r2, a vector from the center of the second slit to the same screen point. We also considered the case of reflection-refraction from an n1-n2 index of refraction junction. The point is that these two cases are clearly probabilistic, i.e. not deterministic at all. Here we consider two different scenarios which have the appearance of a deterministic result, but are actually probabilistic and so linked with all possible outcomes, i.e. an average over these which must be understood as an average (as in classical probability). The first is a single particle bound state and the second, transmission through a potential wall. In the bound state case, the analogue is a Newtonian bound state and in fact between the classical turning points KEave(x) + V(x) = E in the probabilistic case which exactly matches the classical one. One is, however, dealing with an average KEave(x) = -1/2m d/dx d/dx W/ W, where W=Sum over p a(p)exp(ipx), it involves all possible outcomes and is not a physical state. Given the similarity to a deterministic problem there is a danger of not considering the quantum bound calculation as a probabilistic one. In the quantum single particle case, one may be surprised by tunneling, but the quantum probabilistic scenario is one in which all possible interaction outcomes are considered together (as is required by probability) and so one cannot possibly ignore interactions linked to a particle entering or leaving the potential region. Furthermore, it is only KEave(x) which equals the classical result and so E which is single value in the classical case, is linked to an average value in the quantum case, i.e. a particle may have a p such that pp/2m is not zero at the classical turning point. Only KEave(x) =0 at this point. An interesting feature is that a probabilistic calculation applies to a single particle run (which only produces one outcome). One, however, cannot measure the full probability associated with a problem without making a very large number of measurements.