A Do bosons contradict basic probability laws?

[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.

 

[Philip Koeck]

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.

I mean:
Would one see the clusters flying in straight lines between collisions as in classical mechanics?
Remember that we are filming the clusters in 3D, so we are carrying out position measurements at regular time points. Doesn't the wave function collapse whenever we measure the position?

 

[vanhees71]

This is explained rather by the famous paper by Mott on why we see straight-line tracks of $\alpha$ particles when looking at a lump of $\alpha$-radiating matter in a cloud chamber. It's the interaction between the particles and the detector (in that case consisting of the vapour molecules). There's no collapse needed at all!

N. Mott, The Wave Mechanics of alpha-Ray Tracks, Proceedings of the Royal Society of London. Series A 126 (1929) 79.

https://doi.org/10.1098/rspa.1929.0205

What has this to do with indistinguishable particles though?

 

[Philip Koeck]

This is explained rather by the famous paper by Mott on why we see straight-line tracks of $\alpha$ particles when looking at a lump of $\alpha$-radiating matter in a cloud chamber. It's the interaction between the particles and the detector (in that case consisting of the vapour molecules). There's no collapse needed at all!

N. Mott, The Wave Mechanics of alpha-Ray Tracks, Proceedings of the Royal Society of London. Series A 126 (1929) 79.

https://doi.org/10.1098/rspa.1929.0205

What has this to do with indistinguishable particles though?

Thanks for the link to Mott's paper. I have to have a closer look at it.

I thought measurements in general made the wave function collapse. Is that not an accepted interpretation of the measurement process?

If every particle or cluster in a chamber appears to be moving in a straight line wouldn't that make the particles/clusters distinguishable?
If I saw some almost continuous lines in a video recording or a cloud chamber I would draw the conclusion that all the position measurements along such a line are measurements of the same particle and, essentially, I've labelled the particles by observing them.

 

[PeroK]

Thanks for the link to Mott's paper. I have to have a closer look at it.

I thought measurements in general made the wave function collapse. Is that not an accepted interpretation of the measurement process?

If every particle or cluster in a chamber appears to be moving in a straight line wouldn't that make the particles/clusters distinguishable?
If I saw some almost continuous lines in a video recording or a cloud chamber I would draw the conclusion that all the position measurements along such a line are measurements of the same particle and, essentially, I've labelled the particles by observing them.

Each particle in a bubble chamber is essentially an isolated system. The particles do not interact with each other and are clearly not in thermal equilibrium.

Also, these lines are not continuous. At the atomic level they are formed by a sequence of discrete collisions. They only look continuous at a macroscopic scale.

 

[vanhees71]

The collapse is part of some flavors of the socalled Copenhagen interpretation of quantum mechanics. I personally think it's superfluous and misleading, contradicting fundamental principles of relativistic quantum field theory (microcausality). I follow the minimal statistical interpretation of QT, which is the essence of the theory needed to do science. Everything else is philosophy and thus a matter of opinion rather than part of physics as a natural science.

 

[Philip Koeck]

Each particle in a bubble chamber is essentially an isolated system. The particles do not interact with each other and are clearly not in thermal equilibrium.

Also, these lines are not continuous. At the atomic level they are formed by a sequence of discrete collisions. They only look continuous at a macroscopic scale.

Let's think of the gas of clusters which is monitored by 3D-video.
Even if there are no long range forces between the clusters they would still collide and form a system that tries to achieve thermal equilibrium, just like an ideal gas. The question is whether this equilibrium will be that of distinguishable or indistinguishable particles.

I realize that the lines won't be continuous even in a video, but they would suggest a straight classical path, from which I would conclude that it's always the same particle on a given path.

 

[PeroK]

Let's think of the gas of clusters which is monitored by 3D-video.

And if you had a 3D video of a hydrogen atom then you'd see precisely where the electron was at all times and could plot its classical orbit round the nucleus.

You can't just pretend that there is no such thing as QM. That if you have a 3D video camera that QM will just "go away".

 

[vanhees71]

Well, such a 3D cam itself cannot exist because of QT. One should also be aware that classical statistical physics is incomplete. Boltzmann and Gibbs had a lot of problems with particularly the distinguishability of classical particles (Gibbs's paradox, how to gauge phase-space cells, etc.), all of which are solved with many-body QT of indistinguishable particles, and that the distributions are rather Bose-Einstein and Fermi-Dirac distributions than Maxwell-Boltzmann distributions is an empirical fact, explaining a lot of other problems related to a classical statistical model.

 

[PeterDonis]

Let's think of the gas of clusters which is monitored by 3D-video.

Let's close this thread since it is now degenerating into personal speculation.