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.
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.
Classically, probability is considered in terms of the number of possible outcomes. For example, a die toss has six possible outcomes and a coin toss, two. Probability calculations make use of probabilities and hence of the total number of possible outcomes. Thus, they are not deterministic calculations which follow “what happens” as in Newtonian mechanics. We suggest that if free particle quantum mechanics is probabilistic, it must also involve the notion of possible outcomes in an experiment and not represent a deterministic approach. We suggest, however, that something interesting applies to quantum mechanics because we have suggested that free particle quantum mechanics arises from the Lorentz invariant A = -Et+px. We note that there exist x,t trajectory points which follow x=vt. We also note, however, that one may introduce the idea of intervals dx, dt (because Newton’s dx→0, dt→0 is an idealization) and that A is unchanged for x,t and x+hbar/p , t+hbar/E. In other words, we suggest that there are intervals with probability about each x,t trajectory point. A trajectory point is associated with deterministic physics, but probability about this point is not. In other words, one has a kind of hybrid probability scheme. Thus, exp(ipx) and exp(-iEt) really represent fluctuation probabilities about a trajectory point x and t. These ranges of possible x and t fluctuations about a trajectory point x,t allow for further probability to occur in certain interactions. We consider two examples. The first is 2-slit interference which involves a deterministic particle moving along a straight line towards a 2-slit mechanism perpendicular to its path. At the 2-slit apparatus, one considers a probabilistic calculation as the uncertainty in x, i.e. dx=hbar/p allows for possible interaction with both slits. This involves considering all possible outcomes (i.e. motion through one slit or the other which is a classical situation), but exp(i p dot r1) and exp(i p dot r2) “sit on top” of these trajectories at the 2-slit apparatus and one has interference of the probabilities linked with each classical possible outcome. Thus, there are two sets of probabilities which are intertwined, the probability range hbar/p for each path and the two possible paths which have equal probability. One must consider the possible outcomes for both sets together. We suggest that a similar analysis applies to the case of 1-D reflection-refraction for x=0 at an n1-n2 index of refraction junction. Thus, we suggest that a deterministic free particle with x=vt may encounter probabilistic interaction situations, such as a two-slit apparatus or an n1-n2 index of refraction junction. These are probabilistic only because there is a probability fluctuation scheme in x associated with each deterministic x point. It is this probability which allows for a second probability to arise due to interaction with the slits or at the 1-D n1-n2 junction. In such a case, one must consider all possible outcomes which means that one does not only consider the exp(ipx) type probabilities, but also those linked with the physical interaction, i.e. two possible paths, one from each slit, or an incident, reflected and refracted photon at x=0 (n1-n2 junction). It is a combination of all possible paths which must be used in calculations. We thus suggest that even though there is a sense of determinism, i..e free particles move as x=vt and these exist in the 2-slit case and the 1-D reflection-refraction case, in a probabilistic calculation one must consider all probabilistic outcomes consistent with these. Thus, quantum mechanical calculations are true probabilistic calculations in the sense that they involve all possible outcomes in an analysis.
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.
We have suggested in previous notes that there is a tendency to consider quantum mechanics as a kind of hybrid probabilistic-deterministic theory because it is supposed to ultimately “merge’ with classical mechanics for high energy levels (tiny wavelengths). Newtonian mechanics is deterministic and so somehow it seems quantum mechanics should also be seen as deterministic, hence the sense of surprise at tunneling when KEave(x)=0, even though vrms(x)=0 does not mean that actual values creating the probability are 0. Here, we wish to argue that exp(-iEt+ipx) (or the exp(ipx) part for spatial interactions) is really a momentum probability scheme which may possibly bypass many aspects of the actual physics to describe a problem in a simpler form in terms of a momentum conserving probability. A key idea is that exp(ipx) is actually the form to use when one has randomness as we will argue for reflection-refraction from an n1-n2 index of refraction junction. We argue that this idea is already known for a particle in an infinite well for which one has sin(pave x), i.e. 1/2i (exp(i pave x) - exp(- i pave x)). In other words, one has the form exp(i pave x), even though it is linked with an average p. Given that one is not considering deterministic physics, there is nothing wrong with exp(i pave x) being composed of Sum over p a(p) exp(ipx), where exp(ipx) holds in all space, not just within the infinite well potential. exp(ipx) is a momentum conserving probability, but on average momentum must also be conserved by a free statistical entity within the well. Similarly, W(x) for a single particle bound state consists of many exp(ipx)s, but represents a fixed E ave and on average this Eave must be conserved in time, hence the time dependent exp(-i Eave t). As a result, one may see that the quantum mechanical exp(-iEt+ipx) is an attempt to describe a system in a simple manner in terms of momentum conserving probability. To see this in more detail, we consider (1) which describes reflection and refraction of a photon at an n1-n2 junction in depth. In a previous note, we have already pointed out that in n2, c/n2 simply means that the photon is interacting on average with atoms/molecules in n2, but has c in between such interactions. It is only on average that one has c/n2 and even E and p2 = p n2. Nevertheless, one may use exp(ip2 x) as a probability which describes average properties of the system, just as it does in the examples in the above paragraph. The question then becomes: Do the microscopic details of the photon physics (the actual interactions) dictate the probabilities to reflect/refract or does the simplified exp(ipx) set linked to momentum conservation do so? Based on (1), it seems that reflection is completely independent physically from “refraction”, the latter requiring re-radiation and affecting all glass molecules. As a result, it appears that it is difficult to suggest that it is the microscopic interactions which dictate probability because reflection and refraction are microscopically so different. On the other hand, reflection-refraction is not a .5-.5 probability random process as experimental results show. The question then becomes: If some kind of systematic probability driver is not supplied by the microscopic interaction, what provides it? We argued in previous notes that one may introduce the probability exp(ipx) into Newtonian scattering to account for an initial p1,p2 (momentum vector set) having equal probability to produce any pi, pj (outcome set) as long as momentum is conserved. We suggest here that it is exp(ipx) that one must use for random interaction if one insists on conserving momentum. Thus, exp(ipx) bypasses microscopic details of interaction and provides an “ any outcome pair of momenta” is equally likely if momentum is conserved. This is a major simplification of the problem, but the question is: What constraints does it impose on physical results? For a free particle, it suggests that it may interact probabilistically with both slits of 2-slit apparatus if the slits are about hbar/p apart. For randomness, one would expect equal probability for each slit and this is the case as long as one also considers exp(ipx) for each slit to a the same point on a screen. In the case of reflection-refraction at an n1-n2 index of refraction junction, we try to see how the simplification exp(ipx) can be consistent with the microscopic details of the interactions and at the same time dictate the probabilities to reflect and refract. We try to consider why an exp(ipx) scheme predicts how nature behaves while a random one with no momentum conservation considerations does not. In particular, we argue that exp(ipx) represent randomness in a two body scattering problem in which momentum is conserved and so this same exp(ipx) must be carried over into problems which involve interactions instead of simply suggesting that one has .5 probability to reflect and .5 to refract because the two processes are independent and hence random. They are random, but with conserved momentum and hence an exp(ipx) scheme is required, we argue.
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).
We have argued in previous notes that the free particle wavefunction exp(-iEt+ipx) is a complex probability used to describe interactions. As a result, quantum calculations are statistical ones and only provide part of all possible information. If an interaction only involves space, one may use exp(ipx), as in a 2-slit calculation which uses an OR probability approach to deal with probabilistic interactions with both slits (if they are about hbar/p apart) by adding exp(ip dot r1) to exp(i p dot r2), where r1 and r2 are vectors from the center of each slit to the same point on a screen far away. The result (exp(ip dot r1) + exp(i p dot r2) * (exp(-ip dot r1) + exp(-i p dot r2)) gives an ensemble result- it is a statistical calculation. A single photon or particle interferes and only goes to one point and one does not know what that point is (except that it does not go to minima points). Only by considering an ensemble can one give meaning to the idea that there is a certain probability for the single photon to move to one peak versus another. In short, one performs a statistical calculation and needs to interpret it as such, we argue. For a single particle bound state, one again has an ensemble average because an exp(ipx) interacts with V(x) in a statistical manner. One creates an ensemble average to find average kinetic energy: {1/2m Sum over p a(p) pp/2m exp(ipx) } / { Sum over p a(p) exp(ipx)}. Again, this is a statistical result. For the overall bound state, exp(-i E average t+ i P x) where P=0 deals with ensemble averages and must be considered in a statistical sense. Also, this calculation involves the particle moving in and out of the bound region and so seems to be a clear large N ensemble calculation.Furthermore, there is information lost in this calculation which one should be aware of. For high energy levels, the particle should be moving first forwards, then backwards, and this is not shown. That does not mean that it does not occur. A statistical calculation does not need to show all information and one sometimes has to obtain this extra information outside of the statistical calculation. The reason we belabour the notion of a statistical calculation is that we believe that it also applies to the resonance decay factor exp(- gt). In textbooks (1), one often sees exp(-iEt) with E-> E + i g and exp(-iEt) -> exp(-iEt -gt). exp(-iEt) is a quantum factor which applies to a single particle and so we argue that it is a little confusing to make sense of exp(-gt) in a single particle context. In particular, one may also obtain exp(-gt) from dN/dt = -g N(t), which is a statistical equation applied to radioactive decay (among other things), i.e. a large N ensemble calculation which holds for a low rate of decay. Thus exp(-gt) from E+ig should be understood in terms of an ensemble calculation, we argue, which does not seem to be stressed (we suggest). In particular, if one has some photons uniformly dispersed in an n1=1 index of refraction medium (1-dimension) region of length L surrounded by n2>>n1, with a very small probability for refraction, exp(-gt) would need to be understood as applying to an ensemble because exp(-gt) is continuous and photons refract (leave the resonance area) in a discrete manner. Thus, exp(-gt) is a statistical solution which removes quite a bit of information. In particular, if the probability to refract is very small, dN/dt= -1/ delta t1 = N(t) 2 Probability(refract). The solution to this is not exp(-gt). Rather at t=0, N(0) = N. Then one may find a delta t for which, on average (ensemble average) one photon has been lost. Then N(delta t) = N(0) -1 and one may find the delta t2 etc. This approach uses the quantum calculated Probability(refraction), but not exp(-gt) even though this example should be a resonance. As a result, we suggest that quantum calculations are statistical in nature and as such leave out information as they describe an ensemble. In some cases, as in the reflection-refraction case, one may obtain a more detailed description of what happens than given by the resonance factor exp(-gt) which one sees in textbooks and in the Gamow factor calculation etc.
In Part 1, we argued that the Heisenberg uncertainty principle (standard deviation p) * (standard deviation x) >= hbar/2 only applies when a measurement (i.e. interaction) has occurred. Otherwise, the notion of standard deviation does not exist. Thus, if one has a Gaussian wavepacket, it is the measuring device which has created it and constrained is standard deviations of momentum and position. A free particle, which has not been measured, is described by exp(-iEt+ipx), as argued in Part 1, and not by a wavepacket. Here we ask: How does one interpret a wavepacket calculation? In (1), we argued that exp(-iEt) is a free particle energy conserving probability and exp(ipx), a momentum conserving one. Then, a sum of exp(ipx - iEt) represents an OR situation in probability theory. There is superposition and so one obtains a weighting of possible outcomes which is the probabilistic interpretation of the scenario. If one considers two slit interference, the viewpoint is the same. exp(ipx) is a probability which allows for momentum hits of p in a range of dx=hbar/p outside the center-of-mass, which follows x=vt. This built-in stochasticity of impulse delivery leads to the notion that probabilistically, a particle may interact with both slits if their separation is about hbar/p. One writes an OR probability statement: W(x)= exp(i p dot r1) + exp(i p dot r2), where r1 is the position vector from the center of slit 1 to a point far away on a screen, and r2, a vector from the center of slit 2 to the same point. The result, W*(x)W(x) is a probabilistic expression which describes the possible outcomes probabilistically. (1). Seen in this way, the Gaussian wavepacket should be interpreted in a similar manner and not as a physical description of the particle which moves freely. A single bound state in a potential V(x) is localized by a potential and one may consider only OR cases of exp(ipx), i.e W(x) = Sum over p a(p)exp(ipx). In the measurement case, there is no quasi-permanent localization and one uses a linear superposition of exp(-iEt+ipx)s. Nevertheless for a Gaussian wavepacket, two probability distributions describing possible outcomes are created. One has W*(x)W(x)= exp(- (x-u)(x-u) / (2 sigma(t) sigma(t))) where sigma is the x standard deviation. In addition, W(x) = Sum over p a(p) exp(ipx), leading to a momentum outcome distribution of a*(p)a(p). As written, sigma(t) for x which means that the Gaussian is expanding. We called this unphysical if one considers the wavefunction as representing a description of a single physical entity (2) at times beyond t=0. Like in the 2-slit case, we argue that it represents an outcome distribution. Thus, one would not consider any more time values than t=0, at which time the measurement is completed. An exp(-iEt+ipx), describing a free particle, is measured by a specific device which creates the Gaussian form and gives it a specific standard deviation in space and momentum. The measurement is then done and W*(x,t=0)W(x, t=0) represents the possible outcome positions of the particle at t=0 (i.e. probabilities for x portions of a phase shift). The momentum distribution yields possible p outcome values. As soon as the measurement is finished, the particle moves as exp(-iEt+ipx), with p being probabilistically linked to a*(p)a(p) (a Gaussian as well), where W(x,t) = Sum over p a(p)exp(-iEt+ipx).If one has a second measuring device which creates a tiny momentum standard deviation, one would be able to carry out experiments which would determine roughly the standard deviation of p created by the first measuring device. Furthermore, one would be able to measure phase shifts as well by comparing a free particle measured by a device with a tiny p standard deviation and a large x one with free particles which were not measured.