Entanglement: some (at least one) axes of the separable complex Hilbert space of the system of both will have two different coefficients and therefore probabilities to be avoided the contradictory according to which of both is measured. If both are measured by a relevant apparatus, a phenomenon of entanglement will be registered. 

Vasil,

the question is different. The two atoms are identical fermions, and in the overlap region they belong to the same cell of the phase-space for the x direction. This is forbidden. So, the two fermions should not reach the overlap region. But how can this be done? No field is present to repel them.

What comes to my mind is that they can behave as if they collide. However, for colliding, they still have to come into contact. Obviously, inside a gas of fermions, collisions happen. But, I believe that in such a gas each fermion occupies a different cell in the phase space, unless the gas is very dilluted and the fermions practically don't meet one another.

Two comments (or questions): 

1) What does it mean that the fermions are "polarized"? 

2) When applying the Pauli Exclusión Principle (PEP), one should not confuse phase space with  real space.

Sofia, anyway I believe that the question is the same. to which I have tried to answer. The difference is that I uses the separable complex Hilbert space rather than phase space. My answer can be translated in terms of phase space. Its sell at issue has to be able to obtain two different values, which seems to be contradictory and should be "forbidden" as you write. In fact, one can measure the one value, the other value, and one third value corresponding to the phase space of the entangled system. The contradiction does not appear because any of the three disjunctive values needs three disjunctive apparatuses to be measured. Even one tries to combine cleverly some of those three apparatuses in a single new apparatus, another entangled system will be obtained and its corresponding value will be what is measured.

One can uses the Heisenberg uncertainty to visualize the situation. To collide, both fermions occupy one and the same space position. Since the phase space cell cannot be smaller than the Planck constant,  the difference between their momenta is arbitrary, so three (and even arbitrarily many) different values can be reconciled.  

Raul,

An example of polarized fermion is a 3He atom with spin along z direction.

The exclusion principle says that two identical fermions can't occupy the same cell in the phase-space.

Vasil,

the question is how do the two fermions avoid getting into the overlapping region where both obey ΔPΔr = ħ.

OK. Thanks, Sophia.

Sofia, they do not avoid that overlapping because the value of the entangled system of both is what is measured in the case of both overlapping and corresponding apparatus for measuring the correlations and thus the value of entangled system. The prohibition of overlapping was suspended after the concept of entanglement had become commonly accepted.

That prohibition was due to the alleged contradiction which is the real value in the case of overlapping as if "by itself", i.e. independently of the measuring set.  If the set is that measuring only the first electron, its value will be what is measured independently of existing or not a second electron and entanglement with it. Analogically as to the second electron.

If one would like to visualize by Pauli's principle, the pair of two fermions in one and the same state is not a new fermion and admits measuring as a single entangled system. The spin of entangled system is a problem. I personally accept the opinion that it is neither a boson nor a fermion. and its spin is not (a few) halves of the Planck constant, but an arbitrary real number eventually in an interval. This corresponds to the conjecture that entanglement conserves energy-momenta as general relativity rather than energy and momenta separately. 

Pauli's principle is not refuted, but only generalized by complementing with the case of measuring the entangled system relevantly,  Indeed if one uses apparatuses measuring each of electrons from their entangled system separately, the experimenter would obtain two different values and thus corroborate Pauli's principle with good conscience. However, if other (or the same) researcher tests experimentally the entangled system, a single value different from the former two ones would be obtained with not less clear conscience.

Summarizing, our apparatuses avoid the contradiction being disjunctive to each other, but they or the electrons do not avoid the overlapping if the apparatus is chosen to measure the (degree of) overlapping.   

Furthermore, Sofia, the description of what happens [according to the text of your question] might be isomorphically interpreted in terms of space-time (if that is a hidden sense of your question), which is at least almost everywhere smooth and the concept of reference frame makes sense also almost everywhere:

Three different descriptions will appear. They might be reconciled to each other only if the two fermions are situated alternatively in the two disjunctive cones of Minkovski space admitting a domain of overlapping of both cones, which obeys general relativity and therefore constitutes a pseudo-Riemannian space. Then the conflicting three descriptions might be situated consistently each in a corresponding domain. After that. you will need a certain theory of quantum gravity (including GR among those) to describe what happens in a smooth manifold and both tangent and co-tangent spaces in a neighborhood of the cross point of  both trajectories each situated in one of the tangent and co-tangent spaces. .      

Sofia, in order to answer your question, it should be formulated physically, i.e. in  terms of the outcome of a certain experiment. For example: two beams of He³ atoms cross each other, what are the scattering events that can be detected? (The diameter, velocity, and density of the beams should be specified). The general form in which your question was formulated is rather vague.

The Pauli exclusion principle forbids two identical fermions to be in the same state/position.

In order to solve the question, it is necessary to explain the Pauli principle.

I never saw a useful explanation of the Pauli principle.

Entanglement does not help much because elementary fermions are clearly point-like objects. So, what is meant by the crossing of the paths of the two objects?

Vasil and Michael,

If one places a detector in the overlapping region, under certain conditions one won't detect two fermions. let's take a numerical example.

Let our fermions be 3He, whose mass μ = 5×10-24g. Let also the x-component of the group velocity of each fermion be 4cm/s. Then the difference in the x-component of the linear momentum of the two fermions is ΔPx = 2×5×10-24×4 = 4×10-23gcm/s.

The constraint that two fermions can't occupy the same cell in the phase-space imposes ΔPx Δx ≥ ħ. Therefore, the fermions can't get closer than 0.25×10-4cm. If they approach this distance, they can't cross this interval, they have to change direction and fly away - scattering.

Take also in consideration that the Gaussian form of the wave-packet also imposes δPx σ = ħ, where δPx is the spread of the x-linear momentum in the wave-packet, and σ is the wave-packet width. In short, the two fermions can't get closer than the biggest one between 0.25×10-4cm and σ. Usually, σ is the bigger one.

What do you meab by "overlapping region"? How are you going to control the fact that two aroms will come at contact at a certain time? Explain your proposed experiment. Supposing that I have two jars with He³ atoms; what should I do to check your ideas?

"The constraint that two fermions can't occupy the same cell in the phase-space imposes ΔPx Δx ≥ ħ." implies "under certain conditions one won't detect two fermions".

However, I would offer another experiment: Two detectors counting the coincidences for both fermions to have anyway passed the overlapping region simultaneously and to have reached the detectors simultaneously . The number of those coincidences should be zero if the premise holds, but  non-zero far over the statistic deviation according to me. The latter would be an experimental confirmation of entanglement as well as a refutation of the premise.

The short answer is that there is a, long range, interaction, between the two fermions, that describes the constraint  that their  two-body wavefunction must be antisymmetric upon exchange of their spatial labels-since their spin labels are symmetric-no matter what their separation.That's what the constraint means.  This then leads to the corresponding modification of the interference pattern, that's been measured. 

Corresponding statements apply to two bosons-since the constraint that their wavefunction be symmetric, also, means the existence of a long range interaction, that, also, has consequences for and can be detected by interference patterns. 

In both cases such patterns have been studied recently in experiments on condensates. E.g. http://users.physics.harvard.edu/~greiner/publications/trieste%20greiner%20fermions.pdf

My dear Stam,

Many thanks for the reference. However, my identical fermions are also identically polarized, no chance of Cooper pairs. Of course, in this case the wave-function of the two fermions is antisymmetrical. But what I need to know is how the two fermions avoid getting close to one another. Do they scatter on one another, thereby changind direction and flying away from one another?

You say: "This then leads to the corresponding modification of the interference pattern, that's been measured."

Can you be more specific? I looked in the material you sent me. But it insists on Cooper pairs, which is not my case.

Let me tell you an akin situation: in the Hanbury-Brown and Twiss experiment with fermions (see picture), if the phase condition is fulfilled,

(1) ϕA1 + ϕB2 = ϕA2 + ϕB1,

no two fermions can be detected at the same detector. Here ϕA1 is the phase acquired by traveling from the source A to the detector D1, and analogously the other phases. On the other side, if one would place the detector D1 under the blue circles, one yes could catch two fermions because the wave-packets of the two fermions simply don't overlap anymore, they are distinguishable in space-coordinate. And though, I saw nowhere any scattering mentioned.

With kind regards,

Sofia

The short answer is that these diagrams assume the particles are distinguishable, whereas identical fermions are not. The only consistent description is in terms of two-particle wavefunctions:one only can say that one is describing the evolution of a two-particle state. The Hanbury-Brown-Twiss experiment has been done recently, cf. another discussion thread, where papers were given. 

Stam,

I don't understand what you say. There is some misunderstanding here.

Let me remind that the fermions are identically polarized. Then, if the distance Δr between the blue circles is sufficiently big and so the difference ΔP = |pA1 - pB1|, s.t. Δr ΔP > ħ, the two fermions are distinguishable. If one places a detector beneath each blue circle, the detectors won't care of any entanglement, each of them would record the fermion on the respective path. 

At the detector D1 the situation is different. If the detector has sufficiently small volume s.t. in each direction Δr ΔP ≤ ħ, the fermions are indistinguishable. Only in this case the HBT effect works.

I am also sending you a personal message - please be kind and read it.

Best regards,

Sofia

That they're identically polarized, in fact, doesn't matter. What does matter is, whether they're identical, or not, which *means*, whether they have the same mass and the same charge. If they are identical, then the two-particle wavefunction must have certain properties, under exchange of the labels of the two particles, depending whether they'e in the singlet or triplet state-but, once more, what matters is that these exchange properties are independent of distance.This repulsive interaction implies that the paths don't cross. 

If the two fermions are not identical, i.e. have different masses or different charges, then there isn't any constraint for the labels of the two-particle wavefunction under exchange and there isn't such a long range force between them. However, in this case, too, in fact,  the uncertainty principle enters the picture and implies that it isn't possible to assert that the two particles are found in the same place at the same time. 

Yes, Stam, they are identical, the same type of particle and identically prepared. As you see in the figures, whatever may differ is the direction of their linear momentum.

The point-like elementary particles follow a stochastic hopping path and their average positions follow the paths described by the graph above. There is no crossing in the normal sense.

We do not need interaction to stop two particles from being observed in the same position: interference is enough. When a single particle is diffracted on two slits, there are regions on the screen where the particle cannot be observed, because the wave function vanishes there. I do not believe there is more to the ``repulsion'' between the two identical fermions than just this.

That the two-particle wavefunction does not factorize as a product of one-particle wavefunctions  over the particles defines what it means that they interact. This is what gives meaning to the statement that the 1-particle wavefunctions interfere. 

I  formulated the problem as two particles  approaching each other and have an interaction between them. For example I took two electrons both of them longitudinally polarized. Then I calculated the cross section.  They approach each other they scatter and continue their paths.  The cross section  is finite and well defined.

  How is the exclusion principle satisfied? The particles interact all the time even when they do not overlap and thus the final result is an effect over long times when the eigenfunctions do nor need to be antisymmetric because they are at different point of space.

I do not think in makes sense to consider no interacting particles.

It becomes semantics: one can consider non-interacting particles, but impose the appropriate constraint on the many-body eigenfunctions, or one can solve the constraint and write the interaction, that will, automatically, select the appropriate combinations of eigenfunctions. In non-relativistic quantum mechanics it does make sense to consider non-interacting particles, since the spin-statistics theorem isn't a theorem, but a postulate in that case, http://ejde.math.txstate.edu/conf-proc/04/w1/wightman.pdf

@ Stam: Agreed, except, from a purely semantic perspective, I would not use the word ``interact'' for something that depends merely on the nature of the particles' state and not on possible forces existing between them. Rather, it is the latter I would prefer to call interactions.

My point is that anything that leads to the multiparticle wavefunction not being factorizable over the particles into  a single product of one-particle wavefunctions describes an interaction, because it leads to scattering-free particles don't scatter.  That's all. 

In the cross point region the two Fermions quantum states are interference. In the cross point, where both particle have identical position, there is a non-constructed interference. 

Yes, Stam,

I see your point and it is smart. Scattering requires a scattering potential.

Though, I see a problem with identical fermions (and identically polarized). I attach again the picture for clarity. Let's imagine them comming close to one another, so close that considering also their linear momenta, they would occupy the same cell in the phase space. Are they still free particles?

The separating distance imposed by their refusal to occupy the same cell, would NEVER be crossed, it is a forbidden region - the orange circle. The two fermions would never get into it. So, in continuation of their movement, they have to return to the half-space where from they came. That means, if they come from the left (right) of the z-axis they return to the left (right). This is scattering, isn't it?

I have no objection to your saying that they become entangled. But I place a question mark on the statement that we can't know whether the fermion moving toward the bottom-left of the picture, came or not from the top-left of the picture. I think that yes we know, because the fermion didn't cross the orange circle.

By the way, my calculi show that the diameter of the forbidden region may be a couple of orders of magnitude bigger than the atom radius.

So, how do you see this problem?

Best regards again,

Sofia

I would think that the two particles cross without interaction since they are not perfectly identical. Everything is identical except the perpendicular momentum. Thus the two particles do not have the same quantum numbers and so the exclusion principle does not apply. The two particles should cross as if the other is not there, however since they are indistinguishable you will not be able to tell if they swapped momentum or not (in the single particle picture).

No, Edward!

Unfortunately, the things are more subtle. When we have to do with continuous variables, as linear momentum and position, we have to say what we mean by "perfectly identical" values. If two tables are each one 1m long, are they of identical length? They may be so if the resolution is 1cm, but if we can measure them with a resolution of 1micron, we may find them very different in length.

Now, I will jump to the concept of identical particles. Two particles of the same type are defined as identical if all the quantum number characterizing their states, are equal, i.e. spin, energy, position, linear momentum, etc. But with position and linear momentum we have the problem mentioned above. So, we use the concept of phase-space cell. In the 6-dimentional space of position and linear momentum, we define cells, each one of 6D volume equal to ħ3. If two particles, A and B, have all the discrete quantum numbers equal, and also satisfy |(rA - rB)(PA - PB) ≤ ħ3, they are considered identical.

Thus, even if PA ≠ PB in the transversal plane, there exists a distance rA - rB in the transversal plane, for which the two particles would belong to the same phase-space cell.

@F. Leyvraz,

You say:

==> When a single particle is diffracted on two slits, there are regions on the screen where the particle cannot be observed, because the wave function vanishes there. I do not believe there is more to the ``repulsion'' between the two identical fermions than just this.

No! Something is wrong! Single-particle interference you get also with bosons. Here we speak of two-particle interference, and it goes differently for identical fermions than for identical bosons. Only the fermions have the property of keeping one another at a distance.

I won't participate in a discussion whether the term "interaction" is appropriate, but it seems to me obvious that fermions "smell" one another from a distance. The distance that they keep from one another may be by orders of magnitude bigger than the atom radius.

This effect is explained as the behavior of the fermion at a rotation around itself. But, practically, it indeed looks like a repulsive interaction. We all know the phrase "Pauli repulsion".

Good point, but I don't see how they can fit into the box.

In 2D we'd have (using h for hbar)

(dx,dy) * (dpx,dpy) < h2 for the particles to be identical

Let us say they are colliding in the y so dy=2Py which is of course non-zero since they are heading towards each other.

We thus get (after taking the dot product)

dy = (h2 - dxdpx)/(2 Py)

The smallest dy can be is 0 when dxdpx = h2 which is too small (too little uncertainty for a particle in 1D). Thus, I don't see how they can get into the same phase-space box.

What I meant was simply that, according to the rules of quantum mechanics, you must add the amplitudes of two fermions 1 and 2 as a_1(x) a_2(y) - a_1(y) a_2(x) and then take the squared modulus. So when x->y this goes to zero, at least if the momenta are similar. So there is nothing there apart from an amplitude vanishing. No force, no potential. In particular, when you say the two fermions bounce off and do not cross the ``red region'', this is misleading: for this sentence to make sense, you should be able to identify the fermions, which precisely is impossible. I agree that you will not observe either particle in the red region, but you cannot make a difference between the particles bouncing off and the particles going through the forbidden region, since the final state is exactly the same.

@F. Leyvraz

You say: "when x->y this goes to zero.  . . .  So there is nothing there apart from an amplitude vanishing".

Which "this" goes to zero? The wave-function goes to zero? No, the wave-function doesn't go to zero. The wave-function represents two particles carying energy, the particles don't disappear neither the energy they cary.

Neither x - y goes to zero. To the contrary, the minimal distance |x - y| at which the fermions can approach one another, may be a couple of orders of magnitude bigger than the atom radius.

Next:

==> "when you say the two fermions bounce off and do not cross the ``red region'', this is misleading: for this sentence to make sense, you should be able to identify the fermions, which precisely is impossible."

It is precisely possible to identify the fermions as long as their wave-packets are enough far from the orange region, and also, they are distinguishable by definition as belonging to different cells in the phase-space. Put a detector on each path. Also, use the continuity principle. 

Edward,

I don't know what you talk about. There is no dimension y in the picture. There are x and z. As to dy = 2Py, distance cannot be equal with linear momentum.

Ariadne, good day,

I appologize for telling you that this is not a useful answer. Nobody reads an article just because somebody indicated it. People are busy. You have to explain the main idea in the article and by what it is connected to the question.

People indicate articles only for further details, but they have to give an explanation, as I said.

Best regards!

Luiz,

A point-like object has little opportunity to store properties in its internals. Only the dynamic geometric data that characterize the location of the point as function of progression are the natural properties of the point-like object. This location is defined relative to its environment. A static point-like object has only one location as its property. A point-like object can only achieve extra properties through its dynamic behavior. For example it can perform a hopping dance under the control of a dedicated mechanism. The hopping dance corresponds to a hopping path and a swarm of landing locations. The swarm can be characterized by a location density distribution. If sufficient landing locations are involved, then the location density distribution can approach a continuous function, but this only happens if the controlling mechanism ensures sufficient coherence. If the location density distribution is a continuous function, then it can be interpreted as a function that specifies the probability of detecting the point-like object at the location, which corresponds to the value of the parameter of the function. If that function has a Fourier transform, then the swarm owns a displacement generator and as a consequence at first approximation the swarm can be considered as to move as one unit. That unit moves along its own path. It also means that the location density distribution can be considered as a wave package. It means that the swarm can generate interference patterns. With other words under special conditions it can behave like a wave. Moving wave packages usually disperse. This is not the case with this wave package because it is continuously regenerated.

The hopping path can be used to construct a path integral. This can be done by letting each hop be accompanied with a jump up and down from the configuration space into the Fourier space where the displacement is represented by  multiplication factor, which is close to unity. A sequence of such multiplications can be converted into a summation of the corresponding logarithms. The total sum will represent the path "integral".

Luiz and Hans

If I may, Sofia's question related to fermions identically polarized, like two electrons with parallel spin alignment. Your point like argument for bosons makes me recall that if you assume point-like behavior to elementary fermions, this does not necessarily imply point-like in the mathematical sense as you seem to assume Hans.

The reason why the notion of point-like behavior came to be associated to electrons, for example, which caused them to be considered elementary, is due to the fact that the more energetic non-destructive mutual collision encounters between 2 electrons were during experiments, the closer they came to each other's "point-like center" before rebounding, even during head-on collisions, without having reached any unbreachable limit at some distance from their center. This seems to imply some "elasticity" of the material that they are made of, but not necessarily point behavior in the mathematical sense.

Recall that even the trajectory of the moon about the earth is calculated as if the masse of both bodies was mathematically concentrated in their centers.

Now, the Gaussian wave packet representation that Sofia specified implies mediation of the interaction by means of virtual photons. Feynman defined virtual photons as combining the Coulomb force with the amount of energy induced by the force at the distance considered between the particles, which allows calculating this combination ob both force-and-energy at any given distance during motion of the particles by means of the Lagrangian.

This is the reason why there is a general impression that no force is in continuously progressive action between the particles. It is not from the virtual photons mediation perspective if considered to represent the real physical interaction.

However, if you go for the point-like behavior perspective at the physical level as you are considering, then the wave packet perspective becomes only an approximate mathematical representation, and you end up with point-like behaving particles following precise trajectories being submitted to a continuously present and continuously progressive application of the Coulomb force at the submicroscopic level, continuously acting as a function of the inverse square of the distance between the particles.

The force can now be seen as separated from the energy that it induces. In the case of two identical fermions (same sign charges and parallel spin), the force progressively induces more and more energy vectorially directed away from each other as the two fermions close in on each other until they energy is sufficient for them to overcome the translational energy that was causing them to close in, and they will be repelled.

In my view, this is a behavior that could possibly explain how and why two identical fermions cannot occupy the same cell in the phase space.

``Which "this" goes to zero? The wave-function goes to zero?'' Yes! That is, of course, the wave function evaluated at (x, y). In formulae

psi(x, y)->0 as x->y.

at least if the momenta are sufficiently close.

Of course, the particles do not disappear. Their probability of appearing close to each other vanishes, just as, in a double slit experiment, the probability of the single particle appearing in a region of negative interference vanishes.

``It is precisely possible to identify the fermions as long as their wave-packets are enough far from the orange region, and also, they are distinguishable by definition as belonging to different cells in the phase-space. Put a detector on each path.''

The crucial point about the answer to your question is that this is incorrect. The entire issue is that fermions are in principle indistinguishable. If they were not, the entire issue would not arise. Of course, if fermions are distant enough, the antisymmetrised function yields the same results as the factorised function. But in the region where the fermions are not observed, this is not the case. And I stand by my claim that it is impossible meaningfully to decide whether both fermions went through the red region or whether both bounced off. Using a detector does not help: no detector can tell which fermion is which.

Dear all,

point like objects are only simplistic representations of something which  cannot be better described. They are  confortable means to avoid further complications in describing phenomena, of which though, sooner or later, we are going to pay the price in term of the accuracies of the models used for describing such phenomena. A point like mass cannot exist, not even as a singularity which in principle should have a finite volume, the point like is forbidden by the Uncertainty priniciple  and it is forbidden by the localization of finite energy whose density would diverge in an infinitesimal volume.

Dear Luiz,

By the way , the trajectories on the Feynman Path integral do not have any physical interpretation , even for spin trajectories in two dimensions

yes they do not have a phisical representation in term of being directly measured in the same way the complex wave function does not have one. They cannot have, they are a step below the physical reality which we can have perception of, since they are candidates selected by the Hamilton least action principle to achieve what we can have a direct experience of.  

F. Leyvraz,

You can't introduce a definition of distinguishability, of your own. The standard definition is:

Two particles are indistinguishable when ALL the quantum numbers are equal.

As I told you already, for continuous variables we use the concept of phase-space cell.

Two distant particles, e.g. one flying over the Earth and one over Mars, won't jump from one planet to the other instantaneously. So, we can say which is which. The position quantum number differs. A stronger argument on this, is provided by the continuity equation.

However, if you disagree with the standard definition, I see no point in arguing.

Best regards!

The wave function is a high level concept that was deduced from reinterpretation of formulas that existed in classical physics. This reinterpretation is called quantization and is not based on a proper foundation. It is a methodology that appears to work. Many methodologies that became in common use are generated in this way and one of them is the path integral.  The problem with proper foundations of a model of physical reality is that its structure and the corresponding phenomena are not accessible for observation. Thus in no way this foundation can be verified by experiments. On the other hand the foundation must be rather simple and must be easily comprehensible by skilled scientists. The trick is to find a foundation that can only be extended in a stringent restricted way such that the model developer cannot be accused from pursuing his fantasies. The nice circumstance is that such foundation exists and is already known for a long period. In 1936 the orthomodular lattice was discovered by Birkhoff and von Neumann and this relational structure extends immediately to a separable Hilbert space. The set of closed subspaces of a Hilbert space has exactly the structure of this orthomodular lattice. The Hilbert space introduces number systems into the model, but it can only cope with division rings and the most elaborate of the division rings is the number system of the quaternions. The orthomodular lattice is an atomic lattice. These atoms are elements that cannot be split in lower level elements. In the Hilbert space these atoms are represented by closed subspaces that are spanned by a single Hilbert vector. This vector is eigenvector of a special operator that attaches a quaternion as the eigenvalue to the considered Hilbert vector. The quaternion can be interpreted as a combination of a scalar progression value and a three dimensional vector that represents a location. Thus in this interpretation the atom of the orthomodular lattice is represented by a dynamic geometric datum at a single progression instant. At a different progression instant the atom is represented by a different Hilbert vector and a different location. Thus the atom is represented by a point-like object that is hopping around in the eigenspace of a normal operator that resides in a separable Hilbert space. The landing locations of these hops form a swarm and together with the hopping path this swarm represents the atom of the orthomodular lattice. The model can be further extended by adding a mechanism that controls the coherence of the hopping behavior of the atom. For example it may ensure that the generated swarm can be described by a location density distribution which conforms to a continuous function. This can go so far that this function has a Fourier transform. In this case the swarm (and thus the atom), owns a displacement generator. On its turn this means that at first approximation the swarm (and thus the atom) can be considered to move as one unit, which follows its own path. This story takes the mystery away that is always attached to the notion of the wave function. The mystery is now shifted to the mechanism that controls the hopping behavior of the point-like object that represents the corresponding atom of the orthomodular lattice.

Hans,

==> The problem with proper foundations of a model of physical reality is that its structure and the corresponding phenomena are not accessible for observation.

Something is wrong. How the phenomena are not accessible to observation? We do experiments.

You speak a lot about "hopping". Please be clear, what is "hopping"? You also say "the landing locations of these hops . . .". What you mean? Between two "hops" the atom doesn't "land"? Does it disappear from the ordinary space? Nothing disappears!!!!

But, in general, what are you trying to say, practically? You gave us a high theoy but what is, practically, the answer to my question? Do my fermions cross paths, or do they run away from one another (scattering)?

Hopping easily enters a mathematical test model that starts with a foundation in the form of an orthomodular lattice. That lattice is atomic and these elements stand for elementary objects. The orthomodular lattice extends in a straightforward way into a separable Hilbert space and Hilbert spaces can only cope with number systems that are division rings. The set of closed subspaces of this Hilbert space has exactly the relational structure of an orthomodular lattice. Thus the atoms of the lattice are represented by one dimensional subspaces. These are spanned by a single Hilbert vector. That Hilbert vector can be eigenvector of a special normal operator that attaches a quaternion as eigenvalue. The real part of this eigenvalue can be interpreted as a progression value and the imaginary part can be interpreted as a spatial location. Thus, at this stage the test model represents the elementary object at a single progression instant. At another progression instant another Hilbert vector represents that elementary object and this new progression value corresponds with a new spatial location. In this way the test model represents the atom of the  orthomodular lattice as an elementary object that hops through the eigenspace of a normal operator that resides in a quaternionic separable Hilbert space. After a period the landing locations of the hops have formed a swarm and the hopping path as well as the location swarm represent the elementary module and the atom of the orthomodular lattice. If the swarm is coherent enough, then it can be characterized by a location density distribution. 

I do not disagree with the standard definition: rather I think you do. Two electrons are always identical. So are two protons. The quantum numbers in that standard definition are the ones pertaining to the type of particle. Momentum, or position do not count.

Thus, when studying a metal, we in principle antisymmetrise over all electrons, even though they may be very distant in phase space (an electron having momentum close to the lowest possible momentum is very different from one having Fermi momentum).

Yes, I agree that *in practice* we do not need to antisymmetrise over particle that are exceedingly distant, because the same result is obtained whether we do or we don't. But the correct thing is always to antisymmetrise, for any two Fermions. And, I insist, no detector will ever be able to answer the question whether it was electron 1 or 2 that went through it. So in your setup, bouncing and going through are equivalent.

A similar example was given by Feynman: if 2 Fermions are scattered off each other, they cannot, in the lab frame, fly off at 90 degrees. That is because the process in which Fermion 1 goes to the right and Fermion 2 to the left is indistinguishable from the opposite process and interferes destructively with it.

F. Leyvraz,

I am sorry, metals are not my domain of competence. I can't discuss a domain in which I am no specialist.

About Feynman's example, I have to think. You posed me an challenging problem.

Now, what I know is that in the HBT experiment, the astronomers do estimations of phase-space cell to decide if two photons are identical or not, because photons from different parts of the disk of a star come with some difference in linear momenta. As to fermions, if the wave-packets are wide enough s.t. in the crossing region there is no problem of belonging to the same phase-space cell, then the particles are considered as completely distinguishable and pass through that region unaware of one another. No anti-symmetrical wave-function is written in the crossing region in that case, I know that from experimenters.

I invite you to write the formula for the HBT effect, while considering that the wave-packets that pass through the crossing region produce an anti-symmetrical wave-function. How can one still obtain the HBT effect?

My best regards!

 Sofia

I didnt read none of the answers provided here. But it seems they are going the wrong way. Because it is very simple. The two electrons can not inresect their paths not because of Pauli, but because of Coulomb's law. I dont think their is Energy enough at any means to do this.

O sorry! I see you mean neutrinos and neutrons.

In fact the answer for atoms is clear! At close distance they can 'see' their charge structure and Coulomb law will come into play.

The same is true for neutrons = as they are composed by quarks caring electric charge and also color charge.

So from your question only the neutrinos remain. I am not a specialist on neutrinos. You know there are some things like oscillations etc.

I suspect that as every particle can for a brief moment  exist for example as e- and e+ which annihilate after (I am not precise but I think you see what I mean - Feynman diagrams) then they can see this charge structure before Pauli comes into play.

I think this resolved your question!

The uncertainity relations make the question about intersection senseless. Elementary particles have no paths.

Eugene

Just curious.

How can they accelerate protons at the CERN accelarators along very precise trajectories if they have no path?

The same for electrons and positrons at SLAC and all other high energy accelerators?

@ André: Eugene was referring to the uncertainty relations, which do limit the significance of the concept of path. At high energies, the path can be very accurate in position without violating the uncertainty relations. It is a quantitative issue, a question  of degree.

@F. Leyvraz,

You are right, but not clear. If I weren't acquainted with the topic, I wouldn't have understood anything from you said. What is so special about high energies that quantum objects behave as classical?

About path, if a wave-packet has a non-zero group velocity which keeps all the time greater in absolute value than the uncertainty in velocity, ∆v, then the wave-packet has a path. More exactly, one can see that the wave-packet moves in time. 

F.Leyvraz (sorry I don't know your first name)

To my knowledge, the wave function is totally unforgiving in this regard. Either the particle can be localized, then it is not moving, or else it is moving and is representable by the wave function only as a wave packet, which cannot possibly be conceived of as following the precise trajectories that are an experimental fact in high energy accelerators.

The conclusion seems simple to me, faced with experimental evidence. The wave function can only be a mathematical tool, that has its uses, of course, but an incomplete tool that by very nature is unable to represent the real trajectories that can verifiably be followed in an infinitesimally progressive manner by permanently localized charged particles in high energy accelerators.

In other words, there must be some real down to earth reason why wave mechanics cannot be used in dealing with real electromagnetic particles in high energy accelerators, which deals with "real" charged particles in motion following very precise trajectories.

Ref: Stanley Humphries, Jr.. Principles of Charged Particle Acceleration, John Wiley & Sons, 1986.

Francois is my name. If, on the other hand, you found my use of your first name offensive, I heartily apologise: this was not my intention.

The statement you make is, to my knowledge, misleading: a wave function determines a density in position space, given by |psi(x)|^2, which gives the probability of finding the particle at x. It also gives the probability of finding the particle with a momentum p, which is given by | hat psi(p) |^2, where hat psi(p) is the fourier transform of psi(x) evaluated at p.

Now, roughly speaking, the Fourier transform of a narrow function is a broad function, and viceversa. Or more exactly, the product of the ``breadth'' appropriately defined, of psi(x) with the breadth of hat psi (p) cannot be less than a fixed number, in that case hbar/2.

But there certainly do exist wave fiunctions which localise a particle quite well both in momentum and in position. In particular, there are wave functions for which the product of the breadths is exactly hbar/2.

For electrons, for example, you can localise the position to a Bohr radius and the momentum uncertainty will only correspond to a variation in kinetic energy of about 13 eV (one Rydberg) which is a negligible part of the momentum of a highly accelerated electron. But of course, no accelerator has localisation of its beam to Bohr radius precision (which is a fraction of an Angstrom). So there is no impediment to an appropriately accurate localisation of an electron in a beam.

So your question ``why wave mechanics cannot be used in dealing with real electromagnetic particles in high energy accelerators, which deals with "real" charged particles in motion following very precise trajectories.'' rests on a assumption that is not based in quantum mechanics as it is actually practised, but rather on an idealised view of it. You argue that it is not possible to specify both moentum and position. But this is only true if you insist on specifying both *to arbitrary accuracy*. If you ask concretely which combined accuracy can be tolerated, then you will find that accelerator beams do not, by far, contradict quantum mechanics. 

François

Contrary to latin countries in general (I think), in English speaking countries in general and in America in particular, use of first name for informal conversation between casual  acquaintances is quite common and not offensive at all, and is even considered friendly. So no apology required.

Regarding high energy accelerators, to my knowledge, they work with Maxwell, Lorentz, Coulomb and LC equations exclusively to control charged particles on totally precise trajectories, all involving continuous electric and magnetic force application. The ref I gave is one source describing what they use.

For example, I append below the equation being used to my knowledge to calculate the gyroradius of a permanently localized electron in a magnetic field.

To my knowledge also, they can control trajectory and scattering in the nanometer range. I am not actually looking down on QM and the wave function, which shows total precision in establishing stable states within energy gradients such as that corresponding to the coulomb force variation with proximity to nuclei in atoms.

So energy states are the domain of QM no questions asked. When motion is involved though, continuous electric and magnetic forces seem to be a better tool, which is why they seem be the prefered tool in high energy accelerators.

Even with the wave function, the ionization energy of the Bohr radius, or more precisely that of the hydrogen ground state can be established from QM at exactly 13.6 eV.

QM however, to my knowledge, and unless I missed out on someting, does not allow establishing the unreleasable amount of 27.2 eV of energy which is permently induced at the ground state mean orbital of the hydrogen atom.

André:

``Regarding high energy accelerators, to my knowledge, they work with Maxwell, Lorentz, Coulomb and LC equations exclusively to control charged particles on totally precise trajectories, all involving continuous electric and magnetic force application.''

What you are saying is that accelerators are designed by approximating the particles as classical. That is quite OK, because you do not need a high level of accuracy to keep a beam collimated. Not high is, of course, relative: it may well be fractions of a millimeter, or even possibly microns. But these distances are all very large compared to the scale at which quantum uncertainty effects become important. The breadth of a wave function can easily be very narrow on these scales, and yet the momentum be also perfectly well-defined, again at the required accuracy.

Once more, quantitative issues must be discussed. Such a question cannot be discussed in the abstract as a question involving only the basic principles of quantum mechanics.

François

Well, not really approximating the particles as classical, but as electromagnetic particles, that by nature have an electric aspect obeying the inverse square electric interaction law (Coulomb) with the electric aspect of other electromagnetic particles and a magnetic aspect obeying the inverse cube magnetic interaction law with the magnetic aspect of other electromagnetic particles. We all are well aware of the inverse square law for electric interaction, and for the magnetic inverse cube law I refer you to the paper I append below for recent experimental confirmation.

Could you give me a figure, even approximate of the actual scale at which the quantum uncertainty effect starts to become important in this context for a proton or an electron for example in motion at 80% of light speed on a precise trajectory in a high energy accelerators. I never could. how much in the nanoscale range for example.

http://www.nature.com/articles/nature13403.epdf?referrer_access_token=yoC6RXrPyxwvQviChYrG0tRgN0jAjWel9jnR3ZoTv0PdPJ4geER1fKVR1YXH8GThqECstdb6e48mZm0qQo2OMX_XYURkzBSUZCrxM8VipvnG8FofxB39P4lc-1UIKEO1

@André

All this talk about Coulomb forces is out of topic. Experiments relevant o my question are done with neutral atoms, e.g. 3He. Also, the uncertainties in distance are supposed to be much bigger than the distances where the atoms feel, each, the electron shell of the other. Such distances are of the order of the nanometer.

As to the order of magnitude of the distances of which I speak in my question, they are by far bigger. Experiments relevant to my question are done with very cold atoms - velocities of the order of a few centimeters/second.

For fermions, it is forbidden that two identical fermions occupy the same cell in the phase-space. The phase space has 6 dimensions, 3 for position and 3 for linear momentum. A cell in the phase-space has the volume

Δx ∙ Δy ∙ Δz ∙ Δpx ∙ Δpy ∙ Δpz = ħ3

where Δx, Δy, Δz, are uncertainties in the respective coordinates, and Δpx, Δpy, Δpz, are the uncertainties in the linear momentum projections.

With these data you can calculate by yourself the uncertainties in position. You can take the uncertainty in velocity also as 2-3 cm/second.

Sofia

Thanks for the call to order. I had completely forgotten about your question.

Also, I had been considering velocities in the millions m/s or more, so such low velocities had not crossed my mind. The equation you quote is neat. I can work with this.

Sofia, the Pauli principle is applied when the particles are in the same place with zero distance between them. This has nothing to do with Heisenberg. So then electromagnetic structure is surely to be felt.

Two electrons cannot be in the same state. If their path's cross, then by virtue of having paths they will be in momentum eigenstates not position eigenstates. There is no problem reconciling this with the Pauli rule.

If I look at this from your view-point as meaning that the wave packets "runaway" from each other then yes you can indeed think of this as a force. It has even been given the name "exchange force". Is it really a force? That's a matter of semantics. You could ask if gravity is really a force if it is simply an acceleration frame change. (Einstein's equivalence principle.)

Ilian,

There is no such thing as "particles in the same place". How to you measure the distance between particles? With the cm? With the micron? With the femtometer (10-15m)?

Next, why don't you read other posts? I advice you to read my answer to André. At low velocities - see that answer of mine, the Pauli exclusion is of longer range that the Coulomb forces. Also, nobody is so ignorant as to do such experiments as I speak of, with charged particles, only with neutral atoms, e.g. 3He.

But, again, one has to read previous posts. It's tiresome to repeat again and again the same explanations.

Dear users,

Please read previous posts. People raise all the time the same arguments already answered. 

Sofia I never heard of Pauli exclusion range. If the paricles have the same spin etc. the space part of wavefunction is antisymmentrized leading to that the particles can not occupy precisly the same point in space whatever this might mean (I think about center of mass). Pauli is not intererraction in order to have a range. Please let me know where you get this and what it is. I can not find answer in previous post but instead some considerations about Heisenberg. But this is another thing. Heisenberg UP is not in any means connected with Pauli.

Heisenberg's uncertainty principle prohibits trajectory description for particles. This is because, if  particles have well defined trajectories, then both position and velocity can be determined at the same instant (velocity can be obtained by differentiating the trajectory position vector with respect to time), contradicting del x. del p > h/4 pi.

Therefore, trajectory description for fermions are ruled out. But one can ask what is the probability amplitude for  two identical fermions with identical spin state to have the same position? The answer is zero, since the wave function is anti-symmetric.

Ilan,

I advised you to read my answer to André. I explained there that two identical fermions can't occupy the same cell in the phase-space. It's from this restriction that one can get the distance up to which two fermions can get close to one another. The issue is extremely simple.

As to the saying that quantum objects can't have trajectories, that is not relevant. Again, the wave-packets can have trajectories, and how close the fermions can get when the wave-packets overlap, is established by the phase-space cell interdiction as I explained to André.

I regret, but my time is precious, I wont' explain again and again for each one that can't invest two minutes and read previous explanations.

Sofia: "The issue is extremely simple....but my time is precious"

Indeed it is simple but you seem to persist in trying to make it more complicated. Your question has been adequately answered. (By myself, at least, I can't answer for others as I don't have time to read them all.) For someone who's time is so precious, it seems a little perverse to ignore the good answers you get but continue arguing with others..

@ André:

assume an electron goes at about lightspeed. So

del p * del x approx hbar

Take

del p/p * del x approx hbar/p

p is approx mc. So we have

del p/p * del x approx hbar/(mc) = 2* 10^{1-2} m

for the electron. So assume you want a relative accuracy on p of 10^{-6}

del x approx 2 * 10^{-6} m

So a micron precision in position (hard to get in practice: think of those beams making trajectories that are km across. My memory tells me that you get millimeter accuracy for the beam width) would be compatible with a 10^{-6} relative accuracy in momentum. Combining those two, even with a factor 10 on each uncertainty to be comfortably above the Uncertainty principle, still leaves you with something that qualifies as a trajectory, indeed an accurate trajectory.

On the other hand, in the microscopic world, things are different: an electron bound to a proton in a hydrogen atom,  has a quite uncertain momentum, and a quite well-defined position, both having an uncertainty of the order stated by Heisenberg (slightly larger). Such a particle, having both a large relative uncertainty in position and in momentum, cannot truly be said to be on anything like a trajectory.

typo in the previous post: it should read

2* 10^{-12} m

not 2* 10^{1-2} m

Sorry

F. Leyvraz understood me very well.

Coinsidenss is event. Yes or no. "Very high" (comparing to what?) precision is no matter here. Precision is probability of our observation.

But in common it is the philosophical question as a whole QM. If we have descrete space, then we have exact coinsidensses. If we have continuous space, than we have not coinsidensses at all. What kind of space have we?

I don't know the answer on this question.

Eugene,

Position and linear momentum are continuous variables in QM. I don't speak here of advanced theories that quantify the space-time, but of the standard QM.

What yes introduces some sort of quantification of position and momentum, is the division of the phase-space into cells of volume Δx ∙ Δy ∙ Δz ∙ Δpx ∙ Δpy ∙ Δpz = ħ3. For two identical fermions IT IS FORBIDDEN to be in the same cell.

But, I think that I already explained this cca. 10 times until now.

READ PRVIOUS ANSWERS!!! That economizes time and provides a lot of information about the discussed issue, from different points of view.

Path is the time also, Sofia. There is no uncertainity relation between time and energy. It is severe problem of QM, mentioned already by Schreodinger. See also Landau, Peierls and so on. Acording to Landau (volumes 3,4) event of scattering (more precisely consequense of events during scattering, what You, I think, mean speaking about the coinsidenss) can not be defined at all. Only initial and final states (S-matrix) can be defined.

"Hamiltonian is dead" - last words of Landau.

P.S. What You wrote about phase-space cells is non-relativistic. There is no time here, as well as in QM as a whole. See once more Schreodinger.

Regards,

Eugene.

François,

With "assume an electron goes at about lightspeed", no doubt you mean "goes at just below light speed".

A quite interesting representation. It never crossed my mind to even consider representing an electron trajectory from the statistical spread angle. I always thought of each electrons as having to be permanently localized since they can be forced to follow well defined trajectories in high energy accelerators, for which both position and velocity can theoretically be individually determined at the same instant at any point of the magnetically constrained trajectory (velocity being obtainable by differentiating the trajectory position vector with respect to time, as highlighted by Patrick).

Of course they constrain "beams" of particles, and this is where I can see some uncertainty in position of each electron in the beam in the manner you describe, even it this would not invalidate the possibility of each electron remaining permanently localized within the beam (from my perspective, of course).

I agree that things are different for an electron stabilized in least action  rest orbital state in a hydrogen atom. The one thing we know about this state is the amount of unreleasable energy that it is induced with by the Coulomb force. Translational momentum is definitely uncertain since local electromagnetic circumstances may completely immobilize it translationally. What cannot be stopped is its axial momentum, which is what averages out at the mean orbital radius, equivalent to the Bohr radius.

In an isolated hydrogen atom, if translational motion was theoretically not restricted I easily conceive of a statistical spread of possible positions about a spherically spread mean distance with mean rest orbital radius being representable by the wave function, even if the electron proper remains permanently localized, but with axial extent limited by the inertia of the electron rest mass.

I do not agree that black holes ( rather space-time singularities) do not exist but I do believe that they cannot be explored by using the notion of three spatial coordinates and one temporal coordinate in side the black hole. I have derived the  space time transformations at the horizon (boundary) of the black hole and demonstrated  that  on crossing the horizon the role of spatial and temporal   coordinates get interchanged and consequently the inside space of black hole is prescribed by three temporal coordinates and one spatial coordinate such that in this space all signals and particles travel with the speed faster than light speed. 

Concept of path is denied in quantum mechanic and the microscopic objects like fermions are described in terms of their quantum states. Two or more than two fermions can not occupy the same quantum state simultaneously and on any attempt to bring them in the same quantum state (quantum analogue of crossing the paths) they will repel each other.

The discussion is taking a somewhat  philosophical direction. Physics answers clear experimental questions: what will we see (measure) if we do this and that. If the question were formulated in such physical terms, there would be a clear answer dictated by quantum physics, including Pauli principle. Not much to discuss really.

Is not Einstein philosopher, Michel?

We are all philosophers, in a way. However simple physical questions (that have clear experimental answers) should be separated from philosophy (which has not)

Seemingly simple questions, Michel. Experiments prove nothing. Ask Einstien.

Can You separate language from philosophy, Michel?

Or philosophy from language?

the answer to initial question is just QED ! nothing more..relativistic quantum mechanics gives the explanation.

Relativistic quantum mechanics is nonsense, Stephen. Logically contradictory...

" It seems to me that the two wave-packets should "run away" from the region of overlapping "

No, they don't. They only have (and indeed are)to be in states Orthogonal. Then they can pass one through the other without occupied the same location (\delta).