Identical Particles: Symmetry & Antisymmetry of Wave Function

[Careful]

Uh

The idea is simply that we have "two electrons, one in the state "left to me and flying off to the left" and one in the state "right to me and flying off to the right".

The problem is that the way we write the (non-relativistic) wavefunction, we have to assign certain degrees of freedom as "belonging to electron 1", and other degrees of freedom as "belonging to electron 2".
But clearly, the physical description:

(A) we have "two electrons, one in the state "left to me and flying off to the left" and one in the state "right to me and flying off to the right"

and

(B) we have "two electrons, one in the state "right to me and flying off to the right" and one in the state "left to me and flying off to the left"

are equivalent physical descriptions ; in other words, they should not correspond to *different* physical states.

Sure, but this is entirely similar in classical mechanics, and there I will call the one flying to the left number one, and the one flying to the right number two. After this, I am never going to say that number one is flying to the right. Here, by definition both particles are distinguishable, since the labelling depends upon their physical properties.

Now, suppose I would color two particles in my box red and green then I have 50 percent chance (given random dynamics within the box) that the green (red) one is flying to the right (left) and 50 percent chance that it flies to the left (right) (in case the red one would always fly to the left for some reason then the following reasoning does not apply). So in this case, I have a statistical mixture which is very different from a superposition. So, now you are left with a dilemma, either you assume your wavefunction to be a description of an ensemble of particles giving rise to independent emissions and then you might proceed as usual (but then the wavefunction is extremely nonlocal in a *spatio-temporal* sense). On the other side, on the individual level, it clearly makes no sense and you are left with the conclusion I draw. In any case, the reason for the single particles building up the correlations of the symmetric/antisymmetric state has to be very different from the reasoning employed previously.

But the way we write the wavefunction, there are a priori two different wavefunctions corresponding to (A) and to (B), simply because when making a cartesian product of hilbert spaces (for "electron 1" and for "electron 2") there's a difference between H1 x H2 and H2 x H1. So we have a whole lot of states in H1 x H2 which are different from those in H2 x H1, and that shouldn't be.

Why not ?! Imagine the situation of two different nuclei (call them 1 and 2) and two different electrons red and green bound to 1 and 2 respectively. Then, clearly particles red and green are distinguishable since they are bound to different nuclei and swapping them requires work to overcome the potential barriers of both nuclei. In order to get different statistics here, I would have to imagine that although the nuclei are fixed, the electrons were somehow placed there randomly in a previous moment in time or that in a statistical ensemble of two nuclei boxes I have freedom in identifying the nuclei.

So we have to work "modulo the flip" of H1 and H2, because in H1 x H2, there are different states which correspond to identical physical situations (A) and (B).
Doing this "modulo" thing would give us a lousy space, so another way to do the "modulo" thing is to find in each equivalence class a "preferred representative". This is the state that is "symmetrical" in the equivalence class. It can be the symmetrical, or the anti-symmetrical one.

Like I explained before, this depends upon your notion of distinguishability as well as your notion of what quantum mechanics is supposed to mean. It is not that simple (and was actually much debated upon) as you try to present it.

The whole disagreement boils down to (a) when to work with statistical ensembles (b) when to work with single wave equations (and this is not silly as epicurus tries to tell us) and (c) how to label particles. Classically, I could do the same : suppose I would be studying wave dynamics in which I had two wave types red and green which emerge through one of two opposite holes of the source respectively. Then I could have emissions RG and GR each with chance 1/2. Now, nobody in his right mind would say that actually RG - GR occurs with probability one, but that our detectors only allow for R or G to be observed (in a consistent way). Now in case something strange happens, then a classical physicist would say that something else occurs which is not properly taken into account yet. This is all I wanted to say.

I do not understand, as epicurus suggests, why these considerations can be confusing (actually it is rather worrying that some do not make them).

Cheers,

Careful

 

[masudr]

As far as I know, two electrons bound to different nuclei do not satisfy the principle: even if they have the same , their total state vector is specified by the position of the center of mass of each atom as well: and these will be different.

This is a blatant misunderstanding of QM and is one usually performed by chemists. We use the single particle orbitals to approximate the full many-particle Hilbert space. They are not in the same state.

Yes that is exactly the point I was making: that two electrons belonging to the different atoms are not in the same state. Where is the misunderstanding?

 

[Careful]

Yes that is exactly the point I was making: that two electrons belonging to the different atoms are not in the same state. Where is the misunderstanding?

You said in post 23 that a two electron wave of electrons in different orbitals does not antisymmetrize while in the textbook treatment this is assumed to be the case (see the Hartree Fock approximation). As I said in my previous post, the entire subtlety is in one's interpretation of distinguishability and QM. Similarly, you agreed that two electrons around different nuclei spaced a micron apart do not obey spin statistics, while many people believe spin statistics also to apply in this case.

Careful

 

[vanesch]

Sure, but this is entirely similar in classical mechanics, and there I will call the one flying to the left number one, and the one flying to the right number two. After this, I am never going to say that number one is flying to the right. Here, by definition both particles are distinguishable, since the labelling depends upon their physical properties.

The notion of "distinguishable" or not is indeed a physical notion which is largely independent of quantum theory per se. As you do somewhere else in your post, you "color" the particles to keep them distinguishable in principle. Of course, if in quantum mechanics, I have an internal degree of freedom which colors them, then the entanglement with this degree of freedom will ALSO make them behave as if they are distinguishable. So these tricks are not allowed (no coloring, painting, numbering them).
So what's left ? What's left is that we use the dynamical state itself to distinguish them.

Note that if two electrons are (postulated to be) physically indistinguishable (no colors, paint, numbers... on them), then in classical physics too we've made an "error" in setting up the configuration (or phase) space, because the point corresponding to "electron 1 to the left and electron 2 to the right" (where the 1 and 2 simply come from the order in which we label the coordinates in phase space) is a different point from the point in classical phase space where "electron 1 to the right and electron 2 to the right". So our "phase space" contains different points which correspond to identical physical situations, which shouldn't be the case. To be entirely correct we should have a phase space "modulo physical equivalence" also in classical phase space.

However, this doesn't affect, because of its structure, the results of dynamics in classical physics. This didn't have to be the case, but it is. In other words, the dynamics of classical physics allows you to be sloppy with the phase space (that is, having different points which correspond to the same physical situation) and nevertheless the results always come out all right. So never anybody bothered about this error in setting up a phase space of identical particles, because it doesn't affect the results.

In quantum theory, there's no difference in approach: to each dimension in hilbert space must correspond a different configuration, and we have exactly the same error as in classical phase space when we compose the hilbert space of two identical particles by their tensor product.
We should have enumerated the dimensions of hilbert space by different configurations, and we didn't: we have introduced an unwanted degeneracy by having different dimensions which correspond to the same physical situation.
And, rats, quantum mechanics doesn't have the same forgiving property that classical physics had: it doesn't produce the same results with and without the error (while classical physics did).

We can of course look for the "origin" of the classical forgivingness: it is probably due to the fact that, apart from some pathological cases, the hamiltonian flow in phase space will never "confuse" the trajectory of the (r1,p1,r2,p2) point with the (r2,p2,r1,p1) point - there are no collisions between trajectories in classical phase space. So, dynamically, you can use one of both, and never bother about the other one, and this will not bring you dynamical problems: whether you've explicitly "cut away" the other trajectory of classical phase space or not, it doesn't affect the dynamics of the other one. But this is not true in quantum theory, where the dynamics of the state |r1,r2> is potentially influenced by the existence or not of the |r2,r1> dimension.

So, the point I wanted to make, is this:
if we take the physical assumption that the state (r1,r2) is physically identical (is the same physical state) as (r2,r1), then, keeping the two different points in state space (whether it is classical phase space, or quantum hilbert space) is an error. But in the classical case, the error has no consequences, while in the quantum case, it does.

 

[vanesch]

Similarly, you agreed that two electrons around different nuclei spaced a micron apart do not obey spin statistics, while many people believe spin statistics also to apply in this case.

Well, of course they obey spin statistics. That is:

We shouldn't write the state of the two electrons as:
|atom1,orb1>|atom2,orb2> but rather:
|atom1,orb1>|atom2,orb2> - |atom2,orb2>|atom1,orb1>

However, nobody does this, because in this case, even quantum mechanics will (in most cases) forgive you the error by considering the two different hilbert dimensions ||atom1,orb1>|atom2,orb2> and |atom2,orb2>|atom1,orb1>, and most results you could potentially be interested in will come out all right (just as in classical mechanics), no matter what "representative" you pick from the "physical equivalence" class.

So the poor guy/gal who applies his principles entirely correctly will hit himself simply with more involved algebra, to find the same results (in most cases). That's why people don't bother symmetrising between electrons in different atoms... unless they are solid-state physicists, where it does make a difference again, when the atoms get close together!

The rule when to be allowed to make the error (and hence simplify the algebra), is this:

when, during the entire dynamics, there's never a serious non-orthogonality of the evolutions of the "erroneous" different representatives of the equivalence class, then you'll be allowed to the error (and you don't have to symmetrise).

So as long as "atom1" stays away from "atom2", then the states will remain essentially orthogonal, and you can do your thing with or without symmetrising, as you like.

This is also why in classical mechanics, you can (apart from very pathological cases) always get away with "making the error".

 

[Careful]

The notion of "distinguishable" or not is indeed a physical notion which is largely independent of quantum theory per se. As you do somewhere else in your post, you "color" the particles to keep them distinguishable in principle.

The coloring is random, I am just labelling the particles 1,2,3, ... Once I have chosen such labelling, there is no reason why the state should be invariant under a reassignment (as the physical properties belonging to different labels might be different).

Of course, if in quantum mechanics, I have an internal degree of freedom which colors them, then the entanglement with this degree of freedom will ALSO make them behave as if they are distinguishable. So these tricks are not allowed (no coloring, painting, numbering them).
So what's left ? What's left is that we use the dynamical state itself to
distinguish them.

Now, I do not follow you, the (anti) symmetrized state does not leave any room for distinguishing the particles. Moreover, I disagree that I would not be allowed to color them : in classical physics I can simply follow the tracks of all my particles, therefore the latter are distinguished by the different tracks they trace out (in a bubble chamber). As I said in the beginning, if I label my tracks and declare that particle k is on track k, there is no room for (anti) symmetrization.

Note that if two electrons are (postulated to be) physically indistinguishable (no colors, paint, numbers... on them), then in classical physics too we've made an "error" in setting up the configuration (or phase) space, because the point corresponding to "electron 1 to the left and electron 2 to the right" (where the 1 and 2 simply come from the order in which we label the coordinates in phase space) is a different point from the point in classical phase space where "electron 1 to the right and electron 2 to the right".

No, you keep on missing that by labelling my particles by having the physical property ``1 = flying to the left'' and ``2 = flying to the right'', I get rid of the undesired states in another way. In this case no ambiguity arises and actually, this is also how we know we have two particles and not just one. In other words the product of Hilbert spaces is an *ordered* product, we take away all states in H_1 corresponding to right flyers as well as all states in H_2 corresponding to left flyers (this is just a particular representation of your equivalence classes).

So our "phase space" contains different points which correspond to identical physical situations, which shouldn't be the case. To be entirely correct we should have a phase space "modulo physical equivalence" also in classical phase space.

No, we shouldn't for exactly the reasons I told you before. Actually, there is another reason why we should not (anti) symmetrize blindly : it resides in the ergodic hypothesis which says that if I follow my *labelled* system long enough, its time averaged properties correspond to the phase space averages. There is no labelling invariance in following my particles 1,2,3,... so there is no a priori reason why this should be the case for the phase space treatment. Now, if my N particles are residing in a box, then after sufficiently long times almost all symmetric partners of every configuration have more or less occurred ; but that is a dynamical result which does not occur when the particles are truly free !

This didn't have to be the case, but it is. In other words, the dynamics of classical physics allows you to be sloppy with the phase space (that is, having different points which correspond to the same physical situation) and nevertheless the results always come out all right. So never anybody bothered about this error in setting up a phase space of identical particles, because it doesn't affect the results.

Euh There is a vast difference in the statistics by appling the symmetrization trick at the level of the partition sum ; as far as I know this has nothing to do with the distinction quantum - classical.

We can of course look for the "origin" of the classical forgivingness: it is probably due to the fact that, apart from some pathological cases, the hamiltonian flow in phase space will never "confuse" the trajectory of the (r1,p1,r2,p2) point with the (r2,p2,r1,p1) point - there are no collisions between trajectories in classical phase space. So, dynamically, you can use one of both, and never bother about the other one, and this will not bring you dynamical problems: whether you've explicitly "cut away" the other trajectory of classical phase space or not, it doesn't affect the dynamics of the other one.

This is an elaborated way of simply saying that I can track my particles (as I said previously).

But this is not true in quantum theory, where the dynamics of the state |r1,r2> is potentially influenced by the existence or not of the |r2,r1> dimension.

Well, that depends upon your interpretation of QM, this has nothing to do with our previous arguments of labelling invariance, since I could also say : ``well I have a particle which is more or less in r_1 and one more or less in r_2; let's follow these tracks through a bubble chamber.´´ Then of course the whole discussion starts whether an extremely light fog ``measures'' the particle or not, but I am not going to enter this one. Actually, I can put it still another way, the whole discussion goes around wheter in a two particle situation we *can* measure particle one and particle two (based upon abstract labelling). According to me this is not the case, we measure a right flyer or a left flyer depending on where our detector is standing (and then label the particles accordingly) and that is very different from what you say.

 

[Careful]

Well, of course they obey spin statistics. That is:

We shouldn't write the state of the two electrons as:
|atom1,orb1>|atom2,orb2> but rather:
|atom1,orb1>|atom2,orb2> - |atom2,orb2>|atom1,orb1>

However, nobody does this, because in this case, even quantum mechanics will (in most cases) forgive you the error by considering the two different hilbert dimensions ||atom1,orb1>|atom2,orb2> and |atom2,orb2>|atom1,orb1>, and most results you could potentially be interested in will come out all right (just as in classical mechanics), no matter what "representative" you pick from the "physical equivalence" class.

Well, you still did not provide any convincing argument why this should be so, but I agree that the difference between physical outcomes provided by both states is small. And of course I also am aware that the anti symmetrization term improves results in atomic physics, but that is not the topic of this discussion. We are trying to deduce wether this antisymmetrization term has something to do with ``indistinguishability'' of particles and that is where we disagree.

Careful

 

[vanesch]

The coloring is random, I am just labelling the particles 1,2,3, ... Once I have chosen such labelling, there is no reason why the state should be invariant under a reassignment (as the physical properties belonging to different labels might be different).

There is the *hypothesis* of identical particles, which means exactly that the physical state is the same upon "exchange" of them. Even the notion of "exchange" of identical particles is, by hypothesis, an impossibility, and is only made possible by the erroneous construction of the space of all possible physical states, in which two different points have been constructed for the same state.

You could compare the situation with (electronic) bank accounts. If Joe has 10 dollars, and Jack has 15 dollars in his account, then you could construct something like:
the first dollar is in Joe's account
the second dollar is in Jack's account
the third dollar is in Joe's account
...
the twentyfifth dollar is in Jack's account

and then we have the law that "dollars are identical particles" and that it doesn't make any difference whether the first dollar is in Joe's account and the second dollar is in Jack's account, or vice versa.
We say that the only genuine different "bank state" is: Joe has 10 dollars, Jack has 15 dollars.
Now, (classical mechanics) all kind of accounting can be done by making the superfluous assumption that dollar 1 was in Joe's account, and dollar 2 was in Jack's account and so on, and when a silly accountant uses this scheme, he will nevertheless arrive at correct books.
But in "quantum accounting" it makes a difference (think of path integrals, where you go through all POSSIBLE DIFFERENT physical states) whether there are several states (Joe = 10 dollars and Jack = 15 dollars) or whether there is only one.

If you take it that there "is no reason why the state should be invariant under their "exchange", that means that you consider that the two points hence constructed DO correspond to two different physical states, and hence you don't consider that the particles are "identical".
In other words, you consider that it makes somehow a difference whether the first dollar is in Joe's or Jack's account.
That's your good right, but then the discussion is moot, because then we are not discussing the consequences of identical particles.
This IS for instance entirely justified if the bank accounts are not electronic, but contain genuine bank notes of 1 dollar each.

Now, I do not follow you, the (anti) symmetrized state does not leave any room for distinguishing the particles. Moreover, I disagree that I would not be allowed to color them : in classical physics I can simply follow the tracks of all my particles, therefore the latter are distinguished by the different tracks they trace out (in a bubble chamber). As I said in the beginning, if I label my tracks and declare that particle k is on track k, there is no room for (anti) symmetrization.

You cannot "label the different particles" with their dynamical states, if the entire object of the discussion is to know whether two dynamical states are in fact, the same physical situation!
By stating that you can "follow the tracks" you are simply stating that classical dynamics will give you the same outcome, whether or not you consider ALSO the exchanged dynamical state as a member of the phase space or not. That's what I said: in classical physics, it often doesn't make a difference whether or not you have superfluous phase space states or not: the results come out the same.

No, you keep on missing that by labelling my particles by having the physical property ``1 = flying to the left'' and ``2 = flying to the right'', I get rid of the undesired states in another way.

Well, I could say that I prefer the state "1 = flying to the right" and "2 = flying to the left", and eliminate YOUR preferred state.

Now, as in classical physics, this is only a discrete symmetry, I could do all calculations with YOUR preferred state, or with mine. And, surprise, all of our physical conclusions would be the same. This is what I said: it doesn't matter, in classical physics, which one one takes. This didn't have to be the case: one could think of, say, a different way of defining inertial frames for states of your kind and states of my kind, so that Newton's law would apply differently to your states as to mine. As such, we would have different Hamiltonian flows in the slices of phase space corresponding to your states, and to mine. That's not inconceivable. But this is not the case. Things do come out the same. As such, in classical physics, it doesn't matter. Whether we keep only your states, or only mine, or whether we keep both, the results come out the same.

In this case no ambiguity arises and actually, this is also how we know we have two particles and not just one. In other words the product of Hilbert spaces is an *ordered* product, we take away all states in H_1 corresponding to right flyers as well as all states in H_2 corresponding to left flyers (this is just a particular representation of your equivalence classes).

You cannot do that, because in H1, there ARE also states "going to the left" AND states "going to the right". So you'd need to "cut away" part of the H1 space, and part of the H2 space, in order to avoid two different states of H1xH2 which correspond to the same physical reality.
That would be a possibility - in fact, that IS a possibility ! But we would now have to introduce quite a strange dynamics (unitary flow on H1_cut x H2_cut) in order to avoid that the state vector "leaks out of" H1_cut x H2_cut. I would think that that is entirely possible, by introducing a "re-ordering" operator and other stuff !
But it turns out that it is way simpler to make *another* selection in H1 x H2 which is not "H1_cut" x "H2_cut" but which has a cut in H1 x H2 (namely the symmetric, or anti-symmetric states). This space "doesn't leak out" under the normal unitary dynamics. We do not need to introduce "reordering during the dynamical evolution" that way.

No, we shouldn't for exactly the reasons I told you before. Actually, there is another reason why we should not (anti) symmetrize blindly : it resides in the ergodic hypothesis which says that if I follow my *labelled* system long enough, its time averaged properties correspond to the phase space averages. There is no labelling invariance in following my particles 1,2,3,... so there is no a priori reason why this should be the case for the phase space treatment.

You CAN identify the phase space points A = (1 left, 2 right) and B =(1 right, 2 left). When you start with A and evolve into B, you might just as well introduce, in your dynamics, an automatic jump from B to A, and you would STILL obtain the correct physical predictions.

Now, if my N particles are residing in a box, then after sufficiently long times almost all symmetric partners of every configuration have more or less occurred ; but that is a dynamical result which does not occur when the particles are truly free !

A free system is not ergodic !

Euh There is a vast difference in the statistics by appling the symmetrization trick at the level of the partition sum ; as far as I know this has nothing to do with the distinction quantum - classical.

Indeed, that's what I also say: the hypothesis of identical particles (that is: the assumption that physical states in which we "interchange" identical particles are the SAME SINGLE PHYSICAL STATE) has nothing to do with quantum mechanics per se, and is as applicable to classical physics as it is to quantum mechanics. Only, in classical physics, for the dynamics it doesn't make any difference whether we take this into account or not, while the dynamics of quantum mechanics does come out differently).

 

[Careful]

There is the *hypothesis* of identical particles, which means exactly that the physical state is the same upon "exchange" of them.

Nope, that is exactly what I disagree with, particles are distinguished by their dynamical content relative to an observer (even if they have the same quantum numbers such as mass, spin etc.).

Even the notion of "exchange" of identical particles is, by hypothesis, an impossibility, and is only made possible by the erroneous construction of the space of all possible physical states, in which two different points have been constructed for the same state.

Quite true, but you have to cut out states in the right way.

You could compare the situation with (electronic) bank accounts. If Joe has 10 dollars, and Jack has 15 dollars in his account, then you could construct something like:
the first dollar is in Joe's account
the second dollar is in Jack's account
the third dollar is in Joe's account
...
the twentyfifth dollar is in Jack's account

and then we have the law that "dollars are identical particles" and that it doesn't make any difference whether the first dollar is in Joe's account and the second dollar is in Jack's account, or vice versa.

What you say does not really make sense : classically, I would have partition a OR partition b OR partition c
which would amount to an impure density matrix description quantum mechanically ! I would not say that a AND b AND c occur at the same time.

Now, (classical mechanics) all kind of accounting can be done by making the superfluous assumption that dollar 1 was in Joe's account, and dollar 2 was in Jack's account and so on, and when a silly accountant uses this scheme, he will nevertheless arrive at correct books.

Huh ?? That is not a silly assumption, if you want to do physics then you have to give entities a name: you are employing a description where a coin which belongs to Joe could equally well belong to Jack (while they are physically in different accounts !).

But in "quantum accounting" it makes a difference (think of path integrals, where you go through all POSSIBLE DIFFERENT physical states) whether there are several states (Joe = 10 dollars and Jack = 15 dollars) or whether there is only one.

I said that myself at least three posts ago (and also in this post). It is because you do not properly distinguish AND from OR (statistical density matrix description versus single wave).

If you take it that there "is no reason why the state should be invariant under their "exchange", that means that you consider that the two points hence constructed DO correspond to two different physical states, and hence you don't consider that the particles are "identical".

But they are different through the context they are embedded in! Money in Jack's account does not belong to Joe. A right flyer is not a left flyer.

In other words, you consider that it makes somehow a difference whether the first dollar is in Joe's or Jack's account.
That's your good right, but then the discussion is moot, because then we are not discussing the consequences of identical particles.

Sure the money is ``identical'' (same weight, color, shape...) but they can also be distinguished.

This IS for instance entirely justified if the bank accounts are not electronic, but contain genuine bank notes of 1 dollar each.

Nah, the bits and bites are simply stored in different files.

You cannot "label the different particles" with their dynamical states, if the entire object of the discussion is to know whether two dynamical states are in fact, the same physical situation!

You do not seem to realize that the dynamical states of the particles are the only way for us to know about their existence. Your separation of kinematics and dynamics is contradictory to how we do physics.

By stating that you can "follow the tracks" you are simply stating that classical dynamics will give you the same outcome, whether or not you consider ALSO the exchanged dynamical state as a member of the phase space or not.

But here again, you are saying you allow particle 1 to go left AND right, so your particle one becomes entirely a nonlocal entity; next you claim to be able to measure this particle 1. Pure nonsense, sure classically, I don't care wether I call the left mover one or the right mover two, but at least I am going to say that 1 is my left OR right mover. That is the point !

That's what I said: in classical physics, it often doesn't make a difference whether or not you have superfluous phase space states or not: the results come out the same.

Because a classical physicist underderstands the difference between AND and OR.

Well, I could say that I prefer the state "1 = flying to the right" and "2 = flying to the left", and eliminate YOUR preferred state.

Sure, but that is an OR and not an AND; we are in perfect agreement. The rest of your message turns around the same point, so I am not going to comment it again. Now, you seem to have the idea that I cannot make the same mistake classically, sure I can and I already told you how to do it ! I previously gave the example of two waves R(ed) and G(reen) which can emerge either through the left or right hole of my source (of course G and R cannot go through the same hole). Now instead of GR OR RG, I can say that 1/sqrt{2}(RG - GR) occurs each time as a superpostion (so RG AND - GR) but that my detectors can only measure R or G in such a way that RR and GG do not occur (so here I put action at a distance in by hand). Of course I can extend this measurement principle to consistent superpositions of R and G and as such violate Bell inequalities to my liking.

Cheers,

Careful

 

[vanesch]

Nope, that is exactly what I disagree with, particles are distinguished by their dynamical content relative to an observer (even if they have the same quantum numbers such as mass, spin etc.).

So you agree with me that the physical state "particle 1 is flying to the left side and particle 2 is flying to the right side" IS THE SAME PHYSICAL STATE as "particle 2 is flying to the left side and particle 1 is flying to the right" I take it ?

In other words, that labelling them "1" and "2" is an artifact of language and that we actually want to say:
"there is ONE particle flying to the left, and ONE particle to the right", and that the previous two descriptions are in fact the same one as this one ?

What you say does not really make sense : classically, I would have partition a OR partition b OR partition c
which would amount to an impure density matrix description quantum mechanically ! I would not say that a AND b AND c occur at the same time.

Ah, but that's because in the "anti-symmetrical state" you still attach too much importance to the labels 1 and 2.
The anti-symmetrical state
|1a>|2b> - |1b>|2a>

is not to be understood as:
we have a superposition of the state "particle 1 in state a and particle 2 in state 2" with the state "particle 1 in state b and particle 2 in state a".

No, the anti-symmetrical state is simply the state:
"there is ONE particle in state a and ONE particle in state b".

The fact that it is written as a superposition in the Hilbert space H1 x H2 is an artifact because of the way we do tensor products.
The Hilbert space H1 x H2 contains an entire subspace of different points which could correspond to the single state "there is one particle in state a and one particle in state b", namely all states of the form:
U |1a> |2b> + V |1b> |2a>

and we have to pick out ONE representative ray to do so. Now, it would be nice if the way of picking this representative ray were invariant under the dynamics we'd obtain formally if we just did as if particles were distinguishable (the normal hamiltonian as H1 + H2 + H12 and so on).
Well, one possible solution is the totally anti-symmetrical ray.
Another possibility is the totally symmetrical ray.
So this antisymmetrical "superposition" is not a genuine superposition, but just the representative state we picked from the entire set of possibilities that corresponds to "one particle in state a and another particle in state b". Turns out that the way we pick our representative state has an influence on the dynamics in some cases. So we could handle this in two ways: change the dynamics, or try to pick the "right" representative so that we can use the usual dynamical rules.

I said that myself at least three posts ago (and also in this post). It is because you do not properly distinguish AND from OR (statistical density matrix description versus single wave).

No, AND or OR considerations would indicate that there are potentially different physical situations |1a>|2b> and |1b>|2a>, and that we have (as usual in QM) a "superposition" of these different physical states in a particular case. But this is not the case: |1a>|2b> is NOT a physical state (it has not been chosen as the representative of the space) for instance.

Sure the money is ``identical'' (same weight, color, shape...) but they can also be distinguished.

Not in an electronic bank account, where there's just the number "10" associated with Joe.

Imagine that Jack gives 3 dollars to Joe. The state now changes to "13" for Joe and "12" for Jack. We started from the state (10,15) and we are now in the state (13,12). It is a total illusion to consider this as different possible transitions of "the first and the third and the 12th" dollar from Jack went into Joe's account and so on.

You do not seem to realize that the dynamical states of the particles are the only way for us to know about their existence. Your separation of kinematics and dynamics is contradictory to how we do physics.

I understand that. I'm trying to tell you that describing the situation:
"one particle goes to the left and one goes to the right" by

two different descriptions:
"particle 1 goes to the left and particle 2 goes to the right"
and
"particle 2 goes to the left and particle 1 goes to the right"

is an artifact of the way we set up the description.
There's no distinction to be made between these last two descriptions, so we should have only ONE single "point" in our state space corresponding to this situation.

But here again, you are saying you allow particle 1 to go left AND right, so your particle one becomes entirely a nonlocal entity; next you claim to be able to measure this particle 1. Pure nonsense, sure classically, I don't care wether I call the left mover one or the right mover two, but at least I am going to say that 1 is my left OR right mover. That is the point !

I'm simply saying that the element of H1 x H2 which describes the physical state "one particle in state a and one particle in state b" is represented by |a>|b> - |b>|a>. This is simply its representation in Hilbert space.

And that the elements |a> |b> or |b>|a> of H1 x H2 have been eliminated from the hilbert space of allowed states (and hence do not correspond to anything physical).

We could (I presume) have kept |a>|b> as the representative. We only have then to add some extra fiddling to the unitary evolution in order not for it to "leak out" of the allowed states. I haven't done this explicitly but I'd think that this is possible (but very messy).
The anti-symmetrical state seems to work well with the "usual" dynamics we would have on H1 x H2. So that's the most elegant "representative" of the family of possible representations: we can keep the hamiltonian unitary flow most closely to what we would expect.

 

[Careful]

So you agree with me that the physical state "particle 1 is flying to the left side and particle 2 is flying to the right side" IS THE SAME PHYSICAL STATE as "particle 2 is flying to the left side and particle 1 is flying to the right" I take it ?

Euh ?? Sorry Patrick you are wrong here, you are precisely saying that your state is the sum of a universe in which particle 1 has a, particle 2 has b AND a universe in which particle 1 has b and particle 2 a. That is how you get your correlations : take state a and b your two particle state is psi(x,y) = a(x)b(y) - a(y)b(x), x corresponds to particle 1, y to particle 2 (you can draw your own conclusions). Having a genuine OR would result in a density matrix description.

In other words, that labelling them "1" and "2" is an artifact of language and that we actually want to say:
"there is ONE particle flying to the left, and ONE particle to the right", and that the previous two descriptions are in fact the same one as this one ?

You still do not get it that my description is invariant under this too (although not manifestly). I am asking questions about the properties of the LEFT flying particle or RIGHT flying particle. It does not matter when I call these 1 or 2. You are asking questions about particle 1 or 2 (very different thing to do) and this one can be left or right flying depending in which universe you are.

Sorry but I do not really have time to get into the rest (since I have answered it already). Now, let me tell you that is easy to give a Hilbert space description of my classical R(ed) G(reen) example (and no Red and Green do not need to be distinguishable, they might just be different excitations of the same internal variable, just as an electron can have spin up and spin down in the z-direction or different frequency modes in a fluidum (remember photons of different frequency - in a box - do correspond to symmetric wave functions)). Now, in your reasoning (in which only Hilbert space arguments and symmetries of the Hamiltonian apply) I should end up with RG + GR or GR - RG which is clearly rubbish classically (there is nothing quantum mechanical about your argument - hbar does not enter anywhere) !

Careful

PS: The assignment of 10 and 15 coins respectively out of 25 is an OR operation and not an AND (partition 1 till 25!/(10!15!)). Of course, these partitions cannot be distinguished, but neither is the outcome of my questions dependent upon any suplementary labelling. Here the only physical question would indeed be ``how many coins do I have ?'' but if I would have labelled the coins then for sure the distribution of them would have given me certain numbers.

 

[vanesch]

Euh ?? Sorry Patrick you are wrong here, you are precisely saying that your state is the sum of a universe in which particle 1 has a, particle 2 has b AND a universe in which particle 1 has b and particle 2 a.

No, this is exactly what I'm trying to say that this state does not represent. The state |a>|b> - |b>|a> is NOT to be seen as a superposition of a state |a>|b> and ANOTHER state |b>|a> (even though it is written that way), for the simple reason that the state |a> |b> is postulated not to exist, just as well as the state |b>|a>.

In order for |u> = |psi1> + |psi2> to be a superposition of 2 states (2 universes, if you will), |psi1> and |psi2> need to be individually possible states. But we now say that |a> |b> is NOT a physical state, or is |b>|a>. Only, because we committed ourselves to use the space H1 x H2 (which was hence not a good idea, strictly speaking), we HAVE to write vectors in the basis |ei>|fj>. But we actually only want ONE vector that corresponds to the state "one particle in state a and one particle in state b". And we PICK, hear well, ONE SINGLE ray of the space H1 x H2 to REPRESENT this case, and that ray is given by |a>|b> - |b>|a>

But, it is not to be seen as a superposition of |a>|b> and |b>|a> because, although these vectors are part of H1 x H2, we decided that they do not correspond to any physical state.
In other words, just as we called the vector |x,y,z> in H_x x H_y x H_z, the vector corresponding to the physical state "my point is in (x,y,z), in the same way we now decide to call the vector |a>|b> - |b>|a>,
the vector corresponding to "one particle in state a and another in state b". It is not a superposition of two other states, it is just THE vector corresponding to the physical state "one in a and one in b".

We only write it down, when we consider the hilbert space of states, and we try to "map" the hilbert space of states onto H1 x H2 (which is, as I said, not strictly the correct hilbert space) as a "sum" of two vectors which do not correspond to any physical state, for convenience.

 

[vanesch]

PS: The assignment of 10 and 15 coins respectively out of 25 is an OR operation and not an AND (partition 1 till 25!/(10!15!)). Of course, these partitions cannot be distinguished, but neither is the outcome of my questions dependent upon any suplementary labelling. Here the only physical question would indeed be ``how many coins do I have ?'' but if I would have labelled the coins then for sure the distribution of them would have given me certain numbers.

Consider it this way: in electronic bank accounts, Joe and Jack have respectively 10 and 15 dollars in the bank. Now, imagine that Jack owes 3 dollars to Joe.

A classical bank would then "add 3 to Joe's account", "diminish Jack's account with 3 dollars + transmission costs, proportional to the amount due, (say 33% percent)".

So Joe ends up with 13 dollars, and Jack ends up with 15 - 3 - 1 dollar.
The 1 dollar transmission costs is because the transfer was about 3 dollars.

If you do it with individual coins which you can distinguish, you'd still arrive at this outcome. So it doesn't matter, because it is a classical bank.

Now, a quantum bank would do it this way:
"take a dollar from Jack's account and put one to Joe's account, and subtract a certain amount of money (0.33 cents) for the effort",
do this in all possible pathways (without front and fro) which are compatible with Joe receiving 3 dollars in the end.

Now, if "dollars are indistinguishable", then the quantum bank does first:
Joe + 1, Jack - 1.33
Joe + 1, Jack - 1.33
Joe + 1, Jack - 1.33

and we end up, again, with Joe having 3 extra dollars, and Jack having paid 4 dollars.

But if we keep all the extra states, we can do it in miriads of ways:

Joe gets dollar number 5 and Jack looses dollar number 5, plus pays 0.33
Joe gets dollar number 8 and Jack looses dollar number 8 ...
Joe gives dollar 6 back to Jack, Joe looses this time 0.33 dollars
Joe gets dollar number 9 and Jack looses dollar number 9, plus pays 0.33

Now, if for each potential way there could be shuffled dollars between Jack and Joe so that in the end, Joe ends up with 3 dollars more, the bank gets 0.33 dollars, their services are way way more expensive !

So the very "existence" of these "duplications of states" by giving numbers to identical dollars has changed the "bank dynamics".

It is this "path integral over all possible ways" which makes that the introduction of different vectors corresponding to the same physical situation gives different results from the case where we only keep one vector in the quantum case, while this is not the case in classical mechanics (and where hence we don't care whether or not we keep a few duplicata).

 

[Careful]

No, this is exactly what I'm trying to say that this state does not represent. The state |a>|b> - |b>|a> is NOT to be seen as a superposition of a state |a>|b> and ANOTHER state |b>|a> (even though it is written that way), for the simple reason that the state |a> |b> is postulated not to exist, just as well as the state |b>|a>.

I am afraid we are getting mixed up in words. Anyway, if you calculate only the permutation invariant observables from your antisymmetric state, you get the same answers as I do (for states with disjoint supports). We only disagree about the states whose supports are not disjoint. Here again, I use that that states a and b are representing statistical averages over an infinite number of observations and that -at least according to the Bohmian interpretation - particles are following well defined tracks in space according to a generalized Newton law. Therefore, even if two particles are in the same state, they will be distinguishable in principle (they have different tracks, now of course you can debate whether I could really follow them without disturbing the system too much), therefore my previous reasoning would equally apply to those cases. Only when you take a strict non deterministic point of view, you can - in my opinion - avoid this.

But again, I don't know why you don't realize you are actually taking an OR for an AND, and the way you do it is by appealing to the measurement paradox. The fact that you don't have a theory for single events forces you to make this ``mistake'' ; classically, we call this rubbish because we know we can't replace the behavior of the single event by the one of the ensemble (which I pointed out to you in this Red Green example).

It is a terrible thing to let a weakness of the original theory propagate that far. You should be aware that the quantum spookyness is not a result of the theory, you have put it by hand in from the very beginning at the single particle level (without any convincing reason why this should be so). You do it again by postulating that |a>|b> and |b>|a> are not allowed states even if |a> and |b> have disjoint support (so actually you have made no derivation of anything at all, merely you used a sick notion of identical particles).

By the way, just for fun, let me remark that no two interacting particles with identical bare parameters are physically the same (at least not classically) due to (very tiny) mass and spin corrections coming from the radiation back reaction energy momentum tensor. The point here is that swapping the momenta and positions of the particles is not enough since the equations of motion are (at least) third order in time (for example the Lorentz Dirac equation).

For completeness, I should add that also in my way of thinking about QM, the product state receives corrections even for non interacting particles. But the latter has nothing to do with the particles being indistinguishable, but which a specific local mechanism at the level of single events. Hestenes has been working on that for a long while too.

Careful


PS: you don't have to explain all you did to me, I am perfectly aware of all this. I simply disagree for the reasons mentioned above; a particle is not in two places at once. End of story.

PPS: In your bank example, you again take the quantum result as a starting point.

 

[vanesch]

Here again, I use that that states a and b are representing statistical averages over an infinite number of observations

Well, that already messes up then

and that -at least according to the Bohmian interpretation - particles are following well defined tracks in space according to a generalized Newton law. Therefore, even if two particles are in the same state, they will be distinguishable in principle (they have different tracks, now of course you can debate whether I could really follow them without disturbing the system too much), therefore my previous reasoning would equally apply to those cases. Only when you take a strict non deterministic point of view, you can - in my opinion - avoid this.

Well, this is an interesting example, which in fact shows that there can even be *classical* models (Bohmian) in which it matters what description one takes. You are right of course that you could follow the track of each particle individually in Bohmian mechanics. But the physical state in Bohmian mechanics is NOT given by only the particle trajectories: the wavefunction (over configuration space) is just as much part of it. And that wave function will be the symmetrical or antisymmetrical one - AGAIN. In other words, when the particles will be close together, and the factors corresponding to them in the wavefunction overlap, the forces on the particles will be different whether the wavefunction is symmetrised, or anti-symmetrised. I assume that the rules for setting up the Hilbert space of wavefunctions in Bohmian mechanics is identical with those in ordinary QM (except that we're only supposed to work in the position basis). That means then that the |a>|b> and the |b>|a> wavefunctions are, again, forbidden, and only the wavefunction |a>|b>-|b>|a> is allowed for.

So, in Bohmian mechanics, there is some redundancy left, for the particles: you can indeed have two "physical states" (particle configuration + wavefunction) which correspond to the same situation: you can swap the particles in the particle configuration and you'll have a redundant state. However, there's only one wavefunction for the two.

The next remark is that Bohmian mechanics is not automatically empirically equivalent to QM. It is only so when you allow for the "quantum equilibrium condition". So this means that you have to suppose that you have not sufficient knowledge of the particle configuration in order for you to be able to distinguish between trajectories.

But again, I don't know why you don't realize you are actually taking an OR for an AND, and the way you do it is by appealing to the measurement paradox. The fact that you don't have a theory for single events forces you to make this ``mistake'' ; classically, we call this rubbish because we know we can't replace the behavior of the single event by the one of the ensemble (which I pointed out to you in this Red Green example).

Well, it is a matter of opinion to say whether or not we have a theory of single events or not. But this doesn't matter for this discussion. I'm simply explaining here where the symmetry or anti-symmetry of the quantum state for different particles comes from: it comes from the fact that we have to assign *one single hilbert space dimension* to every considered physical state (of the one-event case), and that, if we do H1 x H2, we have redundancy, because of the hypothesis of identical particles which tells us that "flipping identical particles" doesn't change the physical state. That a nice way to "cut away" the redundancy in H1 x H2 without having the unitary flow "leaking out", is to use symmetrical or anti-symmetrical states. That's all.

It is a terrible thing to let a weakness of the original theory propagate that far. You should be aware that the quantum spookyness is not a result of the theory, you have put it by hand in from the very beginning at the single particle level (without any convincing reason why this should be so). You do it again by postulating that |a>|b> and |b>|a> are not allowed states even if |a> and |b> have disjoint support (so actually you have made no derivation of anything at all, merely you used a sick notion of identical particles).

This is just opinion. You are entitled to your opinion of course, but that doesn't alter the way the quantum formalism is set up. You are talking about other theories that might eventually reproduce some aspects of quantum theory. But we're talking here about *quantum theory*, not about its potential replacements. In *quantum theory* there is only one Hilbert dimension for each different physical configuration ; you're not allowed to have redundancy, because this can alter the unitary evolution.

By the way, just for fun, let me remark that no two interacting particles with identical bare parameters are physically the same (at least not classically) due to (very tiny) mass and spin corrections coming from the radiation back reaction energy momentum tensor. The point here is that swapping the momenta and positions of the particles is not enough since the equations of motion are (at least) third order in time (for example the Lorentz Dirac equation).

Well, in as much as the coupling with the EM field would "distinguish" the particles, that's what in quantum theory is taken as entanglement with the EM field, which produces the necessary decoherence in order to make them "distinguishable" (in other words, to make them entangle with essentially orthogonal EM field states, which "tag" them).

For completeness, I should add that also in my way of thinking about QM, the product state receives corrections even for non interacting particles. But the latter has nothing to do with the particles being indistinguishable, but which a specific local mechanism at the level of single events. Hestenes has been working on that for a long while too.

Don't understand me wrong: it might be that there is a totally different mechanism "in reality" at work than the one by "identical particles". But that's not the point: *in quantum mechanics*, that's how things are done, and it is entirely logical and follows from the basic postulates of quantum theory (the setting up of the hilbert space).
So in a universe where quantum theory is true, this is how things are done.
As far as we know, this works well concerning the outcomes in our universe too.

So I don't see why you are fulminating against this rule in quantum theory. It seems that, each time a technicality of the formalism of quantum theory is discussed, you feel the need to point out how this could be also obtained in a totally different way, outside of quantum theory. Fine, but that doesn't change how things are done *within* quantum theory.

 

[Careful]

Don't understand me wrong: it might be that there is a totally different mechanism "in reality" at work than the one by "identical particles". But that's not the point: *in quantum mechanics*, that's how things are done, and it is entirely logical and follows from the basic postulates of quantum theory (the setting up of the hilbert space).
So in a universe where quantum theory is true, this is how things are done.
As far as we know, this works well concerning the outcomes in our universe too.

So I don't see why you are fulminating against this rule in quantum theory. It seems that, each time a technicality of the formalism of quantum theory is discussed, you feel the need to point out how this could be also obtained in a totally different way, outside of quantum theory. Fine, but that doesn't change how things are done *within* quantum theory.

Patrick, in contrast to you, I do not see any point in repeating the absurdities of a theory which should be replaced (with often contradictory arguments). For someone who wants to make sense out of QM, the notion of identical particles has to go out of the window; the task at hand is to find a physical mechanism behind the one particle wave. Moreover, what you call ``QM'' is some specific (albeit most popular) interpretation of it, for example the ``Barut'' school had quite similar points of view towards spin statistics as I have - so I regret you simply put everything under one group name ``mainstream QM'' (and that for an MWI er ). Moreover, you keep on repeating the same nonsense that it has everything to do with the Hilbert space structure, you can do it on the ordered product as well; there are also other possibilities (which I shall not discuss here).

Concerning your comment on my example with the backcoupling, although you are formally correct (though I would like to see your Hamiltonian which does it), it simply shows how far QM ``thinking'' has gone astray from classical thought (indeed the dynamical mass is due to dressing with off shell particles which are not measurable and hence not existing in the QM state description).

Your comment on my example in BM : well you know BM (in its popular form) hasn't got off the ground in any way in solving particle - wave duality ; hasn't it ?! De Broglie (actually the real inventor of all this) realized it very well (in contrast to more modern followers) and the Paris school has done some interesting work in that direction as far as I know. Wave particle duality is the key to all of this and a solution of the latter is required before real progress can be made. Actually, de Broglie made related comments on these issues as I did now in his book ``non-linear wave mechanics''; hmmm... de Broglie must not be a quantum physicst when I understand you or better, he is a crackpot since he denied the existence of entanglement without hesitation (even I don't do that).

Look Patrick, my point of view on all of this is, that QM has many side branches, some of which have different things to say about many particle systems. It is pretty frustrating that people who do not bother about facing the most interesting problems allow themselves to declare these problems as inexistent and next use this excuse to set up absurd notions such as ``identical particles''. In such circumstances, it is the duty of every sensible person to protest and indeed, the notion of identical particles is a pretty central one.

If the aim of the thread was to discuss the ``derivation'' of statistics from the assumption of identical particles, then we were done in 4 lines (high school math).

Careful