A Do bosons contradict basic probability laws?

[PeroK]

In other words, the change from a possible two to a possible three states must be a real physical change. Not just an arbitrary decision to categorise the results of where you find particles when you look for them.

PS to see the problem more clearly, imagine you gather the data first. Then you analyse it. You have only one set of data and it cannot meet all your equally likely criteria.

If, however, you physically subdivide the box, then you can only obtain data for that one configuration. And, there is no need for that configuration to be compatible in with a different physical configuration.

 

[Philip Koeck]

Let's take a close look at the derivation on Wikipedia. They just give an expression for the number of actually different distributions of n particles among g states on the same energy level. Clearly particle 1 in state 1 and 2 in 2 is the same as 1 in 2 and 2 in 1, since they are bosons. They do this for each energy level.
Then they just look for the distribution of particles among energy levels with the largest total number of distributions among states with fixed total energy and particle number. This is reported as the most likely distribution, without considering that different distributions of particles among states might have different probabilities. Isn't that right?

 

[AndreasC]

To understand a bit better what I believe Pero is saying, think of fermions. Let's say we have two identical electrons in a box. Let's call the state of them being in the box A. So they're both in state A. Except that's impossible. Now let's say we call the left part of the box A and the right B. Well, suddenly they can both be in the box by one being in state A and the other in B. What's happening? Well, neither A or B are real states. Instead, an infinite number of states can correspond to the particle being found in the box or in the left department or in the right department. There has to be some extra physical change to really separate the system into 2 distinguishable states, and the whole discussion is a bit confusing because we're jumping back and forth from a continuous particle in a box example to distinguishable states to thermodynamic equilibrium etc.

 

[Dr.AbeNikIanEdL]

That

the whole discussion is a bit confusing because we're jumping back and forth from a continuous particle in a box example to distinguishable states to thermodynamic equilibrium etc.

@Philip Koeck , you seem to just take isolated blocks from one of those scenarios and assume that it is true in a completely different one, just because one can broadly headline both with "has something to do with QM". What is the scenario we are actually supposed to discuss?

Particles in a box, prepared in a state that yet has to be specified?
If yes, please specify that state.

Particles in a system with n discrete states, distributed according to a distribution that yet has to be specified?
If yes, please specify that distribution.

Particles in a system with n discrete states, distributed according to thermal equilibrium?
If yes, please state that.

 

[Philip Koeck]

That

@Philip Koeck , you seem to just take isolated blocks from one of those scenarios and assume that it is true in a completely different one, just because one can broadly headline both with "has something to do with QM". What is the scenario we are actually supposed to discuss?

Particles in a box, prepared in a state that yet has to be specified?
If yes, please specify that state.

Particles in a system with n discrete states, distributed according to a distribution that yet has to be specified?
If yes, please specify that distribution.

Particles in a system with n discrete states, distributed according to thermal equilibrium?
If yes, please state that.

I'll go for the last one. I think that fits the discussion of BE-statistics best.

 

[Dr.AbeNikIanEdL]

In that case the last few answers apply. What you actually want to have is something like a system with 2 discrete states, you then make an actual physical change to that system so that it has 4 discrete states afterwards. You then wait until thermal equilibrium is reached again. After this, probabilities have changed, which should be expected since you changed the system, and the 2 states you started with are not occupied in the same way as before.

 

[Philip Koeck]

PS to see the problem more clearly, imagine you gather the data first. Then you analyse it. You have only one set of data and it cannot meet all your equally likely criteria.

If, however, you physically subdivide the box, then you can only obtain data for that one configuration. And, there is no need for that configuration to be compatible in with a different physical configuration.

Let's turn the problem around:
I have 2 bosons and 4 states with the same energy. According to the assumptions implicit in the derivation of BE-statistics the 10 possible distributions all have the same probability of 1/10.
If I carry out a series of measurements I'll find that both bosons are in the states 1 or 2 with a probability of 3/10 (or 3/10ths of the time).
Now I do a second set of measurements with a less accurate instrument. I cannot say whether a particle is in state 1 or 2, and likewise I cannot say whether a particle is in state 3 or 4. Effectively the instrument only distinguishes between 2 states.
Should I now get the probability 1/3 for finding both bosons in states 1 or 2 (which I can't tell apart) or should I get 3/10.
Notice that I haven't changed the physics of the system. I've only changed the accuracy of my measurement.

 

[Dr.AbeNikIanEdL]

If you don't change the system, the probability for a given state to find both bosons in state 1 or 2 does not change. Whether you actually resolve between state 1 and 2 does not matter for this.

 

[PeroK]

Let's turn the problem around:
I have 2 bosons and 4 states with the same energy. According to the assumptions implicit in the derivation of BE-statistics the 10 possible distributions all have the same probability of 1/10.
If I carry out a series of measurements I'll find that both bosons are in the states 1 or 2 with a probability of 3/10 (or 3/10ths of the time).
Now I do a second set of measurements with a less accurate instrument. I cannot say whether a particle is in state 1 or 2, and likewise I cannot say whether a particle is in state 3 or 4. Effectively the instrument only distinguishes between 2 states.
Should I now get the probability 1/3 for finding both bosons in states 1 or 2 (which I can't tell apart) or should I get 3/10.
Notice that I haven't changed the physics of the system. I've only changed the accuracy of my measurement.

This is going off at a tangent now. To take an example. Suppose we have 9 states that all have similar energies and one outlier. Let's say with equal probability. An accurate measurement might identify all ten states and the outlier would be found 1/10.

If we now have less sensitive equipment that cannot distinguish the nine states, then we still get the outlier only 1/10 and the amalgamation of the other states 9/10.

 

[Philip Koeck]

I think I'm satisfied with the take home message of this discussion: States are real things and not a matter of how one groups them together or how accurately one can measure. What led to the original problem in my first post is that Kazu and I grouped states together and treated these groups as if they were physical states.
Thank you all very much for input.

 

[bbbl67]

Summary:: The probabilities of mutually exclusive events are additive. For bosons this does not seem to be the case. How can we explain this?
...
How can we understand this?
Does it mean that different distributions of bosons in compartments are not mutually exclusive?
That would mean that there is some sort of overlap between the event of 2 bosons being in the upper compartment and the event of 1 being in the upper and 1 in the lower, somehow both distributions can exist at the same time.

I won't go into the probability math of this, I rather look at it physically rather than mathematically. And the way I picture bosons is that they are not mutually exclusive, several such particles can occupy the same space at the same time as each other. For example, the most common example of a boson is the photon. Within a given space and time, you can fit a nearly infinite number of photons in at the same time, and they don't interfere with each other. But with a fermion, that's not the case, there only one particle of a certain type can occupy the same time and space, due to the Pauli exclusion principle. The Pauli exclusion principle doesn't apply to the bosons, only to the fermions.

However, the amount of bosons you can fit into a certain space and time is not infinite (just nearly so, a very high number). The limit seems to be the Planck density, that is if you pack in the Planck energy's worth of bosons into a Planck volume of space, then the boson capacity is reached and you can't go any further, and a black hole is produced instead.

 

[Philip Koeck]

I won't go into the probability math of this, I rather look at it physically rather than mathematically. And the way I picture bosons is that they are not mutually exclusive, several such particles can occupy the same space at the same time as each other. For example, the most common example of a boson is the photon. Within a given space and time, you can fit a nearly infinite number of photons in at the same time, and they don't interfere with each other. But with a fermion, that's not the case, there only one particle of a certain type can occupy the same time and space, due to the Pauli exclusion principle. The Pauli exclusion principle doesn't apply to the bosons, only to the fermions.

However, the amount of bosons you can fit into a certain space and time is not infinite (just nearly so, a very high number). The limit seems to be the Planck density, that is if you pack in the Planck energy's worth of bosons into a Planck volume of space, then the boson capacity is reached and you can't go any further, and a black hole is produced instead.

I meant that the events are mutually exclusive. For example one event could be that there is 1 boson in state A and 1 boson in state B. Another event could be that there are 2 bosons in state A.
If these two events can't happen at the same time then they are mutually exclusive in the sense of probability theory, if I understand correctly.
I'm not completely convinced that this has to be true for bosons, but I don't want to make them weirder than they are.

 

[Philip Koeck]

Looking at what you've done, you got the same answer for distinguishable and indistinguishable fermions. That can't be correct, because identical fermions (or to be precise, fermions with a symmetric spin state, hence anti-symmetric spatial wave-function) have a repulsive exchange force, so would tend to be found further apart.

That's an interesting detail. I was also surprised.
In my derivation I don't use the concept of space. I just have two groups of states that I call "volumes".
Both distinguishable and indistinguishable fermions block the states they are in, so maybe one shouldn't expect any difference in their behaviour.
I'm not even sure whether distinguishable fermions correspond to anything real anyway.

 

[PeterDonis]

I'm not even sure whether distinguishable fermions correspond to anything real anyway.

Of course they do; consider, for example, an electron and a proton.

 

[PeterDonis]

Both distinguishable and indistinguishable fermions block the states they are in

If by "block" you are referring to the Pauli exclusion principle, that only applies to indistinguishable fermions. An electron and a proton would not "block" each other from being in the same state.

 

[Philip Koeck]

If by "block" you are referring to the Pauli exclusion principle, that only applies to indistinguishable fermions. An electron and a proton would not "block" each other from being in the same state.

In my derivations I always talk about identical particles even if they are distinguishable.

 

[PeterDonis]

In my derivations I always talk about identical particles even if they are distinguishable.

Then I'm not clear about what you mean by "identical" particles.

 

[Philip Koeck]

Then I'm not clear about what you mean by "identical" particles.

Identical simply means the same in all properties. Two large objects (maybe atom clusters) could be identical, but it's still possible to keep track of them and see which particle is where or in which state. So they are distinguishable.
Indistinguishable means that it's principally impossible to tell which of two particles is in state A and which is in state B. There is no measurement that would give you that information.

 

[PeterDonis]

Identical simply means the same in all properties.

It can't literally be "all properties" unless the state of the particle is not one of its properties. But perhaps you mean just all invariant properties like charge or magnitude of spin.

 

[Vanadium 50]

Identical simply means the same in all properties. Two large objects (maybe atom clusters) could be identical, but it's still possible to keep track of them and see which particle is where or in which state. So they are distinguishable.

No, that's not true. The wavefunction is such that you can't tell which is which. Whatever this derivation of yours is, it's not quantum mechanics.

It's evident you don't know QM. That's fine. None of us know everything about everything. But maybe before coming here and telling su it's wrong it would be a good idea to learn what it says? It is probably also not a good idea to call this an A-level thread if you don't have a graduate-level background in QM.

 

[Philip Koeck]

It can't literally be "all properties" unless the state of the particle is not one of its properties. But perhaps you mean just all invariant properties like charge or magnitude of spin.

Yes, that's what I meant. One could think of identical atom-clusters having different kinetic energies, for example.

 

[Philip Koeck]

No, that's not true. The wavefunction is such that you can't tell which is which. Whatever this derivation of yours is, it's not quantum mechanics.

It's evident you don't know QM. That's fine. None of us know everything about everything. But maybe before coming here and telling su it's wrong it would be a good idea to learn what it says? It is probably also not a good idea to call this an A-level thread if you don't have a graduate-level background in QM.

I didn't actually intend my question to be about QM. I was thinking in terms of statistical physics, both classical and quantum.
I'm also not saying that anybody is wrong. On the contrary, I'm interested in finding a possible contradiction between the concept of distinguishable particles and QM. Does QM actually say that identical, distinguishable particles can't exist?
The ideas I discuss are not my own either. Many authors have discussed the statistical physics of distinguishable particles. A starting point could be Robert Swendsen. Here's one of his papers: https://www.researchgate.net/publication/325700825_Probability_Entropy_and_Gibbs_Paradoxes.

More papers:
J R Ray 1984 Eur. J. Phys. 5 219

https://www.researchgate.net/profile/Robert_Swendsen?_sg%5B0%5D=x7WFLGQBnj_yLtNDHRaXQQYKCOqu316_LjBDrGl0PYITlXC8S3W-sbNGMA3CMW29v465TG0.6B40YFlyYiUotq2XMAlsuT2t5O1YXy4ASAboH31H85_EEmGgRalec7vc7Z6XbpHqg91V9SgX9J6jZ04oJQZO0g&_sg%5B1%5D=GcrHY0XV7OCP7C2n8xnBhPit4dE6SX4S7Kp6drZPaKSxojJJyxa8Ykc0mPz6OEfphiG2shk.pKdWZxrCY6Yp_lBNbUiG465NexdKEmT7Uotg8mKNW8Buw6FSAA6liFd3R7iLYckyxafPW4x-GHIF4jjRLKV3Ng
The ambiguity of "Distinguishability" in statistical mechanics
June 2015, American Journal of Physics 83(6):545-554
DOI: https://www.researchgate.net/deref/http%3A%2F%2Fdx.doi.org%2F10.1119%2F1.4906793?_sg%5B0%5D=kun7bmb5aWZ3M59C3kpOVpn7ATEF9fIia4kxGevGSN3SonF0SSd-2uI2kOSwRwc7IBCG7AriQK4CQtIsKRbA-XnZYw.OFTUyFQabl3fbuend8y9vkstQk_teOAk6MTuUHWw3hCEEQWAjhz-IAn_zFqKkafoG2OPXNqpQE_A5Hf1De65aw

 

[Vanadium 50]

Sure, you can imagine a world with classical indistinguishable particles, as turn-of-the-last-physicists did when wrestling with the Gibbs paradox. But that's not the world in which we live. The world in which we live is fundamentally quantum mechanical.

 

[Philip Koeck]

I have no idea. You're the one who says this isn't QM but should be in the QM section.

I never said the original question had nothing to do with QM. I said it was intended as a question about statistical physics (both classical and quantum) rather than QM. I'm very open towards whatever QM has to say about it. For example if QM can show that identical, distinguishable particles can't exist then a lot of the discussion around the statistical physics of distinguishable particles might be completely meaningless at least within physics.

 

[vanhees71]

Then I'm not clear about what you mean by "identical" particles.

One shouldn't say "identical particles" when one means "indistinguishable particles"...

 

[Philip Koeck]

One shouldn't say "identical particles" when one means "indistinguishable particles"...

I agree. I would say identical particles are not necessarily indistinguishable. If they can be tracked they are distinguishable.

 

[AndreasC]

Well they can't be tracked in QM so...

 

[Nugatory]

I would say identical particles are not necessarily indistinguishable. If they can be tracked they are distinguishable.

As far as the math of quantum mechanics (quantum field theory if you want to do it properly) is concerned, if they can be tracked they aren't identical.

 

[vanhees71]

I agree. I would say identical particles are not necessarily indistinguishable. If they can be tracked they are distinguishable.

"Identical particles" is just a often used misnomer. The expression is used synonymously with "indistinguishable particle", and indistinguishable particles are particles with all intrinsic properties (i.e., the quantum numbers needed to label an asymptotic free one-particle state for vanishing momentum, i.e., the particle at rest). These are within the Standard Model: spin, electric charge, color charge, flavor, and weak hypercharge.

 

[Dr.AbeNikIanEdL]

One shouldn't say "identical particles" when one means "indistinguishable particles"...

"Identical particles" is just a often used misnomer. The expression is used synonymously with "indistinguishable particle", and indistinguishable particles are particles with all intrinsic properties (i.e., the quantum numbers needed to label an asymptotic free one-particle state for vanishing momentum, i.e., the particle at rest). These are within the Standard Model: spin, electric charge, color charge, flavor, and weak hypercharge.

Now you got me confused tbh. Are you arguing that one should just never use the term “identical particles”? Or do you actually have some situation in mind where “identical particles” would not be used synonymously with “indistinguishable particles”, and want to reserve “identical” for this?

 

[vanhees71]

As I said, I never use the expression "identical particles". I only wanted to say that you find this expression in many textbooks and papers meaning "indistinguishable particles".

In quantum mechanics the particles are really indistinguishable. It's a feature you cannot intuitively understand, because it's something we are not used to in our experience with macroscopic objects which obey to a very good approximation classical laws. Macroscopic objects can be individually followed. You just mark, e.g., a ball somehow and then you can distinguish it from other similar balls by this mark. Formally you can follow its trajectory from its initial position at some time $t_0$ and identify this individual object at any later time $t$.

This you cannot in general anymore for an individual particle in a many-body system. The many-body quantum state of indistinguishable quantities must be either symmetric or antisymmetric under exchange of two particles, describing either bosons or fermions (where the bosons have necessarily integer and fermions necessarily half-integer spin).

Take two indistinguishable particles in non-relativistic quantum mechanics. Then you can describe a pure quantum state with a two-particle wave function $\Psi(t,\vec{x}_1,\sigma_1, \vec{x}_2,\sigma_2)$, where $(\vec{x}_j,\sigma_j)$ are positions and spin-z components ($\sigma_j \in \{\pm s,\pm (s-1),\ldots \}$). The physical meaning is, according to Born's rule, given by the two-body probability distribution
$$w(t,\vec{x}_1,\sigma_1,\vec{x}_2,\sigma_2)=|\Psi(t,\vec{x}_1,\sigma_1;\vec{x}_2,\sigma_2)|^2,$$
where
$$\mathrm{d}^3 x_1 \mathrm{d}^3 x_2 w (t,\vec{x}_1,\sigma_1,\vec{x}_2,\sigma_2)$$
is the probability to find one particle with spin component $\sigma_1$ within a volume elment $\mathrm{d}^3 x_1$ around the position $\vec{x}_1$ and one particle with spin component $\sigma_2$ within a volume element $\mathrm{d}^3 x_2$ around the position $\vec{x}_2$.
You can only say that much about indistinguishable particles: It doesn't make sense to say you find a specific particle around $\vec{x}_1$ and another specfic particle at $\vec{x}_2$.

Indeed from $\Psi(t,\vec{x}_2,\sigma_2;\vec{x}_1,\sigma_1)=\pm \Psi(t,\vec{x}_1,\sigma_1;\vec{x}_2,\sigma_2)$ (upper sign bosons, lower sign fermions) you get
$$w(t,\vec{x}_2,\sigma_2;\vec{x}_1,\sigma_1)=w(t,\vec{x}_1,\sigma_1;\vec{x}_2,\sigma_2),$$
i.e., it's not observable which individual particle is which.

 

[PeroK]

As I said, I never use the expression "identical particles". I only wanted to say that you find this expression in many textbooks and papers meaning "indistinguishable particles".

You'll be pleased to learn that Chapter 5 of Griffiths' QM Book is entitled "Identical Particles"!

 

[vanhees71]

Well, I'm not in favor of this textbook anyway,... As I said, it's unfortunately common jargon in the physics literature. You cannot help it. One only has to carefully explain the meaning to the students.

 

[DrClaude]

You'll be pleased to learn that Chapter 5 of Griffiths' QM Book is entitled "Identical Particles"!

So does Sakurai.

I think that "indistinguishable" comes from classical statistical physics, and is independent from identical particles. It just turns out that QM showed us that identical particles are fundamentally indistinguishable!

 

[vanhees71]

Yes, I know. Even Weinberg uses "identical particles". I don't say that any textbook that uses this phrase is bad. Then there'd be almost no textbook left, I guess.

 

[Philip Koeck]

In quantum mechanics the particles are really indistinguishable. It's a feature you cannot intuitively understand, because it's something we are not used to in our experience with macroscopic objects which obey to a very good approximation classical laws. Macroscopic objects can be individually followed. You just mark, e.g., a ball somehow and then you can distinguish it from other similar balls by this mark. Formally you can follow its trajectory from its initial position at some time $t_0$ and identify this individual object at any later time $t$.

This marking, does it have to be an actual change to the objects or could it just be in an image of the objects? For example if one could follow the trajectories of large molecules or clusters (for example in a solution) using some high speed, high resolution, imaging technique and then just mark the objects in the computer and track them, would that make normally indistinguishable particles distinguishable?

 

[Lord Jestocost]

In Walter Greiner's "QUANTUM MECHANICS An Introduction" (Fourth Edition) one reads:

"One characteristic of quantum mechanics is the indistinguishability of identical particles in the subatomic region. We designate as identical particles those particles that have the same mass, charge, spin etc. and behave in the same manner under equal physical conditions. Therefore, in contrast with macroscopic objects, it is not possible to distinguish between particles like electrons (protons, pions, α particles) on the basis of their characteristics or their trajectory. The spreading of the wave packets that describe the particles leads to an overlapping of the probability densities in time (Fig. 15.1); thus we will not be able to establish later on whether particle no. 1 or no. 2 or another particle can be found at the point in space r. Because of the possible interaction (momentum exchange etc.), dynamical properties cannot be used to distinguish between them, either." [Italics in original, LJ]

 

[AndreasC]

Imo Shankar explained it pretty well. At least I feel like I understood the difference.

 

[Philip Koeck]

Imo Shankar explained it pretty well. At least I feel like I understood the difference.

What difference?

 

[DrClaude]

This marking, does it have to be an actual change to the objects or could it just be in an image of the objects? For example if one could follow the trajectories of large molecules or clusters (for example in a solution) using some high speed, high resolution, imaging technique and then just mark the objects in the computer and track them, would that make normally indistinguishable particles distinguishable?

You can of course follow individual particles and make a distinction between identical particles in different locations. For instance, when researchers trap an electron in a Penning trap and keep it there for days on end, they know it is always the same electron until it escapes the trap.

Likewise, if I follow a single carbon dioxide molecule that I expire, I can differentiate it from one you just breathed in.

It is only when these identical particles are part of the same system or can somehow interact that the indistinguishability plays a role.

I recommend reading Feynman on the subject: https://www.feynmanlectures.caltech.edu/III_04.html

 

[AndreasC]

What difference?

Between identical particles and indistinguishable particles.

 

[vanhees71]

There is no difference. Both expressions are used synonymously in the literature.

 

[AndreasC]

There is no difference. Both expressions are used synonymously in the literature.

In classical mechanics you can call two particles "identical" (in terms of intrinsic features) but they can still be distinguishable, unlike quantum mechanics where they are indeed used synonymously. Although as Dr Claude said, even in QM you can sometimes distinguish between them.

 

[Philip Koeck]

In classical mechanics you can call two particles "identical" (in terms of intrinsic features) but they can still be distinguishable, unlike quantum mechanics where they are indeed used synonymously. Although as Dr Claude said, even in QM you can sometimes distinguish between them.

I wonder if you've noticed what you have written. You say that in QM "identical" and "indistinguishable" are synonymous, and that identical particles can be distinguishable even in QM, according to what Dr Claude wrote.
Essential you've stated that in QM indistinguishable particles can be distinguishable.

 

[DrClaude]

Let me clarify some things before this turns into a Brian Cox moment.

Consider an experiment where two electrons are in different traps in different sections of a lab. Technically, the wave function describing the two electrons should have a definite symmetry under the exchange of the two electrons, meaning that both electrons have to be in a superposition of being in both traps, and we cannot distinguish one from the other, as they are fundamentally indistinguishable. However, just like Schrödinger's cat is never really in a superposition of alive and dead, in practice the two electrons are not in a superposition of being in both traps.

If we take the electrons and send them hurtling towards one another, we will still talk of the electron coming in from the left and the electron coming in from the right. After the collision, however, we cannot distinguish between the two electrons passing by each other or bumping and reversing direction.

Fundamentally, identical particles are indistinguishable. In practice, there are plenty of cases where we can distinguish them, if they are not part of what we would consider the same system. In other words, even if the electrons are in a superposition of being in both traps, if we measure electron L in the left trap and electron R in the right trap (at $t=0$), for well-separated traps the later ($t>0$) probability of now measuring electron R as being in the left trap and electron L in the right trap is so low that it can be neglected FAPP.

 

[Dr.AbeNikIanEdL]

You say that in QM "identical" and "indistinguishable" are synonymous

No, he said that the terms are used synonymously [by many sources]. As @vanhees71 pointed out, this is probably not a good idea in the first place. Obviously, in the special cases where particles with identical properties can approximately be distinguished one has to be particularly careful with one’s nomenclatur.

 

[Lord Jestocost]

"Equation (462) shows that the symmetry requirement on the total wavefunction of two identical bosons forces the particles to be, on average, closer together than two similar distinguishable particles. Conversely, Eq. (465) shows that the symmetry requirement on the total wavefunction of two identical fermions forces the particles to be, on average, further apart than two similar distinguishable particles. However, the strength of this effect depends on square of the magnitude of $\left< x \right>_{ab}$ , which measures the overlap between the wavefunctions $\psi(x,E_a)$ and $\psi(x,E_b)$. It is evident, then, that if these two wavefunctions do not overlap to any great extent then identical bosons or fermions will act very much like distinguishable particles."

From: Quantum Mechanics by Richard Fitzpatrick
http://farside.ph.utexas.edu/teaching/qmech/Quantum/node60.html

 

[vanhees71]

No, he said that the terms are used synonymously [by many sources]. As @vanhees71 pointed out, this is probably not a good idea in the first place. Obviously, in the special cases where particles with identical properties can approximately be distinguished one has to be particularly careful with one’s nomenclatur.

I have always given the definition: Two particles are called indistinguishable or identical if they have the same INTRINSIC properties. Intrinsic are all properties that are defined for a particle at rest, i.e., with vanishing momentum. So you have spin and various charge-like (electric charge, color, flavor) quantum numbers.

Of course you can always dinstinguish such particles by their momentum or position and their polarization/spin-$z$ component. For fermions the $N$-particle Hilbert space is spanned by the totally antisymmetrized product states ("Slater determinants").

In the above mentioned example of two electron, each trapped at different locations, you have a wave function like
$$\Psi(t,\vec{x}_1,\sigma_1;\vec{x}_2,\sigma_2)=\frac{1}{\sqrt{2}} [\psi_1(t,\vec{x}_1,\sigma_1) \psi_2(t,\vec{x}_2,\sigma_2) - \psi_2(t,\vec{x}_1,\sigma_1) \psi_1(t,\vec{x}_2,\sigma_2)].$$
If the single-particle wave functions have negligible spatial overlap you can distinguish the electrons by their position.

 

[Philip Koeck]

Fundamentally, identical particles are indistinguishable. In practice, there are plenty of cases where we can distinguish them, if they are not part of what we would consider the same system. In other words, even if the electrons are in a superposition of being in both traps, if we measure electron L in the left trap and electron R in the right trap (at $t=0$), for well-separated traps the later ($t>0$) probability of now measuring electron R as being in the left trap and electron L in the right trap is so low that it can be neglected FAPP.

I quote Dr. Claude's post above, but I'm also referring to the following posts with similar content.

I'm wondering whether it is possible to make identical particles distinguishable in such a way that they still form a system in the thermodynamic sense, specifically that they have a distribution of energies that's described by, e.g., the BE-distribution?

As a possible example I could imagine a gas of carbon-60 clusters. If one follows these clusters using some high-speed and high resolution imaging method one would be measuring the position of each cluster at regular time-points. If I understand correctly this would make the wave function of the system collapse every time an image is taken.

Would this make the clusters distinguishable?

What would one actually see in the video? Would the clusters appear to be following classical paths or would they change position randomly?

In case the observed paths appear like classical paths, would the kinetic energies follow the BE-distribution or the MB-distribution?

 

[vanhees71]

You get, of course the Bose-Einstein or Fermi-Dirac distribution from the formalism assuming thermal equilibrium (maximum Shannon-Jaynes-von Neumann entropy of the state) for bosons and fermions respectively, and indistinguishable particles are indistinguishable. The Hilbert space of the many-particle system has the completely symmetrized or antisymmetrized product states as a basis. Equivalently and much simpler in practice is to realize these spaces as the Fock space in a quantum-field theoretical framework.

I don't know which "classical paths" you are referring to. As in the operator formalism also in the path-integral formalism many-body systems are most efficiently described by the field-theoretical formalism, i.e., path integrals integrating over field configurations.