Do bosons contradict basic probability laws?

In summary, the conversation discusses the concept of mutually exclusive events and how they relate to probability theory. It is explained that for distinguishable particles, the probabilities of different distributions are mutually exclusive and add up to the total probability. However, for bosons which are indistinguishable, the probabilities of different distributions do not add up in the same way. This is because the probability distribution depends on the state in which the particles are prepared, and different states can have different probabilities for the same event. The conversation also touches on the idea that what can be measured in an experiment can affect the probabilities observed.
  • #71
PeterDonis said:
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.
 
Last edited:
Physics news on Phys.org
  • #72
Vanadium 50 said:
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
 
Last edited:
  • #73
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.
 
Last edited by a moderator:
  • #74
Vanadium 50 said:
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.
 
  • Skeptical
Likes PeroK
  • #75
Philip Koeck said:
Then I'm not clear about what you mean by "identical" particles.
One shouldn't say "identical particles" when one means "indistinguishable particles"...
 
  • #76
vanhees71 said:
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.
 
  • #77
Well they can't be tracked in QM so...
 
  • #78
Philip Koeck said:
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.
 
  • Like
Likes mattt
  • #79
Philip Koeck said:
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.
 
  • Like
Likes mattt
  • #80
vanhees71 said:
One shouldn't say "identical particles" when one means "indistinguishable particles"...
vanhees71 said:
"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?
 
  • #81
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.
 
  • Like
  • Informative
Likes hutchphd and Dr.AbeNikIanEdL
  • #82
vanhees71 said:
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"!
 
  • #83
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.
 
  • #84
PeroK said:
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!
 
  • Like
Likes PeroK
  • #85
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.
 
  • Like
Likes PeroK
  • #86
vanhees71 said:
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?
 
  • #87
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]
 
  • Like
Likes DrClaude, mattt, vanhees71 and 1 other person
  • #88
Imo Shankar explained it pretty well. At least I feel like I understood the difference.
 
  • #89
AndreasC said:
Imo Shankar explained it pretty well. At least I feel like I understood the difference.
What difference?
 
  • Like
Likes vanhees71
  • #90
Philip Koeck said:
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
 
  • Like
Likes PeterDonis, Lord Jestocost and Philip Koeck
  • #91
Philip Koeck said:
What difference?
Between identical particles and indistinguishable particles.
 
  • #92
There is no difference. Both expressions are used synonymously in the literature.
 
  • #93
vanhees71 said:
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.
 
  • Like
Likes vanhees71
  • #94
AndreasC said:
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.
 
  • #95
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.
 
  • Like
Likes Lord Jestocost, PeroK and Dr.AbeNikIanEdL
  • #96
Philip Koeck said:
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.
 
  • Like
Likes Philip Koeck
  • #97
"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
 
Last edited:
  • Like
Likes Philip Koeck and PeroK
  • #98
Dr.AbeNikIanEdL said:
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.
 
  • #99
DrClaude said:
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?
 
  • #100
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.
 
  • #101
vanhees71 said:
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?
 
  • #102
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?
 
  • Like
Likes PeroK
  • #103
vanhees71 said:
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.
 
  • Like
Likes Stephen Tashi
  • #104
Philip Koeck said:
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.
 
  • #105
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.
 
  • Like
Likes Philip Koeck

Similar threads

Replies
12
Views
1K
Replies
3
Views
1K
Replies
0
Views
782
Replies
7
Views
2K
Replies
5
Views
1K
Replies
17
Views
3K
Replies
21
Views
3K
Back
Top