# Why are identical particles indistinguishable?

1. Nov 8, 2011

### nonequilibrium

Why are identical particles indistinguishable? (Note that I only take the usual Hilbert space formalism as a given)

I've heard the very vague argument of "two waves overlapping and you can't say which is which" but of course physicists should be able to do better than that.

So does it follow from the formalism? Or is it a separate postulate based on observation?

2. Nov 8, 2011

### Ken G

I would put it the other way around-- since no observation can distinguish them, they are indistinguishable (you release two identical particles, you detect two identical particles, how do you know which is which?). Normally, indistinguishability doesn't matter, because the particles don't interfere with each other's actions, so they may as well be distinguishable (by their actions) for all the difference it would make. But, when they overlap with each other, their indistinguishability has dramatic and observable consequences. Examples are the periodic table (elements with more electrons have a different structure because the additional electrons cannot be in the same states as the previous ones, the Pauli exclusion principle is a ramification of indistinguishability), and white dwarf stars (again Pauli exclusion). An example of the indistinguishability of bosons (which have no Pauli exclusion) is radio interfererometry, where radio photons can interfere with each other because they are indistinguishable so must be ruled by their joint wave function, not their individual wave functions.

3. Nov 8, 2011

### csmcmillion

"Feynman, I know why all electrons have the same charge and the same mass" "Why?"

"Because, they are all the same electron!"

4. Nov 9, 2011

### nonequilibrium

Thanks for the reply, but I think my original question was misunderstood: my question was whether it's a consequence of the quantum formalism, or if it's a separate postulate (based on experiment).

5. Nov 9, 2011

### Ken G

Both, it is a separate postulate that is an inherent part of quantum formalism. There wouldn't be any other way to do quantum formalism-- to pretend that indistinguishable particles had their own individual wave functions could not be consistent with quantum formalism, because the wave function represents all the information we can know about the system, not more information than we can know about the system.

6. Nov 9, 2011

### nonequilibrium

That seems like a statement that should be proven, no? Can I find a proof somewhere?

Maybe I can also state my problem with what you say: isn't it the case that by symmetrizing the wave function we make identical particles indistinguishable? Because if we choose to not only use symmetrical wave functions (or at least the resulting probability functions), then they are distinguishable. It might be that then we're not describing reality (in the sense that our predictions would contradict experiments), but it doesn't seem like it would make the theory itself inconsistent.

7. Nov 9, 2011

### tommyli

Imagine that you have two spinless patricles and you want to write the Schrodinger equation. The Hilbert space consists of product wavefunctions
$$\Psi(x_1,y_1) = \Psi_1(x_1)\Psi_2(x_2)$$
as well as all linear combinations of such functions. Formally you can say that the basis of the Hilbert space is
$$|a\rangle | b\rangle = \delta(x_1 - a) \delta(x_2 - b)$$
Now suppose that the particles are non-interacting. Then you can always write the wavefunction as a product of single-particle wavefunctions. In this sense they are completely distinguishable. However as soon as you introduce any kind of interaction, it is impossible to write the wavefunction as a product of two single-particle wavefunctions: in the region where the interaction is strong the two particles are fully entangled. If you start with two identical particles which do not interact, but then come together and interact, in the region where the interaction takes place you can no longer assign a separate identity to the particles, you have a wavefunction which corresponds to a coherent superposition of products of single-particle states.

8. Nov 9, 2011

### Ken G

Indeed, I would add to that that even if the particles are non-interacting, treating them as having separate wavefunctions is just a kind of convenience-- similar to choosing an overall phase factor to the wavefunction. In practice wavefunctions we actually use are going to have an overall phase factor, but it's not really "part of the wavefunction" in quantum formalism, any more than the wave functions being distinct is. The formalism must reduce the wave function to the information that is actually extractable, or else the theory is not parsimonious.

9. Nov 9, 2011

### nonequilibrium

@ tommyli: Aha! Interesting :)

Then I don't understand why my professor tried to argue indistinguishability by using non-interacting identical particles (without stating it as a separate postulate based on experiment, but rather by some vague description using the statements "the waves overlap and we can't say which is which").

Thank you for that remark.

@ Ken G: hm, but does tommyli's post support your point that QM would be inconsistent if two identical non-interacting particles were viewed as distinguishable? It seems you didn't really reply to my last post, or am I mistaken? (I'm assuming your "indeed" refers to tommyli, not to me)

10. Nov 9, 2011

### Ken G

I was saying that even when they are non-interacting they are not distinguishable, because the separable wave function does not distinguish the particles (it only distinguishes the two observable locations, x1 and x2, never the particles-- separating the wave functions is a convenience we can get away with, it isn't the correct wave function if the first function is treated as the wave function of the first particle). Also, I don't know what you mean by "making QM inconsistent." If all particles were somehow distinguishable, there would still be nothing inconsistent about quantum mechanics, it would just be the quantum mechanics of distinguishable particles (like in a Hydrogen atom with a proton and an electron, or positronium with an electron and a positron). However, nature also gives us indistinguishable particles, so we need the quantum mechanics of indistinguishable particles. Classically everything is distinguishable (no two snowflakes are alike, etc.), but a theory of elementary particles must confront their indistinguishability. Quantum mechanics doesn't make them indistiguishable, it responds to their indistinguishability. So QM wouldn't be inconsistent if it didn't do that, it would just be wrong.
The antisymmetry of the wave function under swapping two particles is not what makes the particles indistinguishable, they are already indistinguishable as soon as we write a joint wave function like psi(x1,x2). The antisymmetry is another postulate as well.

11. Nov 9, 2011

### tommyli

Imagine you have a collision between two identical spinless particles. Initially they are non-interacting, and their momenta are different. By measurement of their momentum it is possible to distinguish the two particles. Note that in the asymptotic state the wavefunctions are not entangled, they are two separate wavepackets localized in different regions of space. After the collision, you perform a measurement of momenta of the two particles, but how do you know which one is which? If the wavefunctions are always separable, it would be possible, but in the interacting region, their wavefunctions are completely entangled since there is no 'trajectory' to follow it is impossible match the final particle to the initial particle.

Let me ask you a question. Imagine you have a collision between two particles with opposite spin. If the interaction conserves spin, then it IS possible to distinguish the two particles by measurement of the spin.. you simply collapse the wavefunction to a product of two spin states and since spin remains constant you can match the final particle to the initial particle. In this sense you have a collision between two NON-IDENTICAL species of particle (spin-up/spin-down). This the same whether the quantum number is spin, weak isospin, electric charge, hypercharge, etc, etc. As long as the interaction does not couple the additional degree of freedom, particles with different quantum numbers are distinguishable. E.g. an electron can become a neutrino by emitting a W boson, but if the charge of the electron cannot change in the experiment then electron and neutrino are distinguishable. Now imagine you have a hypothetical collision between and electron and neutrino where (with some small probability) the electron can flip its weak isospin and become a neutrino and the neutrino can flip its weak isospin to become an electron via exchange of a W- boson. Does the electron in the final state correspond to the electron in the initial state?

Or imagine you have a collision between two ELECTRONS of opposite spin, and the scattering vertex includes the hyperfine interaction, so that the spins of both electrons can flip simultaneously with nonzero probability. Can you match the final electron to the initial electr on simply by measuring the spin and momentum?

12. Nov 13, 2011

### Ken G

No, it is never possible to distinguish the particles, they are not distinguishable. However, you can distinguish the momenta of the particles.

What this all boils down to is that when particles don't interact in any way, and have wavefunctions that don't interfere or overlap, then we can get away with imagining that the particles are distinguishable, even though they never are. It doesn't matter, in that case, if they are distinguishable or not, so we replace the indistinguishable particles with distinguishable ones and carry on with our analysis, simply because it simplifies the analysis, and we know we'll get the right answer-- to within whatever small error we make in assuming the particles don't overlap or interact.

The wavefunctions are always entangled, there is never two separate wavefunctions for two indistinguishable particles. However, we always treat it as though there were, because it simplifies our lives, and we know we can get away with it. This practice does cause a lot of confusion though, when we forget to mention we are doing it.

What you really mean here is that if they collide, we might no longer be able to get away with imagining they are indistinguishable. But even if they don't collide, they are still never distinguishable. It would be pure philosophy to assert that they "keep their same momentum", formal quantum mechanics does not make any such claim expressly because the particles are not distinguishable.
Again that is just a picture that you are choosing to adopt because in many situations it will not lead you to a wrong answer, and is a whole lot simpler. But it isn't a true picture in the axioms of quantum mechanics-- the particles are not distinguishable, and so you can only make statements about the spins you will observe, you can never make statements about which particle you will observe because you cannot empirically demonstrate which particle you observed to be able to check your statement. Formal quantum mechanics is careful not to assert any truths that cannot be checked experimentally, such as which particle had which spin, and even though this often doesn't matter, when it does, it is a good thing that quantum mechanics is built like that. You only measure spins, not "which particle" had it, and formal quantum mechanics is built to respect that fact.

To see this, just write the wave function of two electrons in a triplet state, versus two electrons in a singlet state. Are they not different wave functions, even if nothing is happening that can flip the spin of an electron?
This brings in a subtle but important issue, which is how we label particles (not how we distinguish them). For example, if we do a two-slit expriment on identically prepared laser photons, we get an interference pattern on the wall. Does this mean we cannot distinguish the photons? Of course we can distinguish the photons, we can distinguish them by where they hit the wall. So we could give all the photons that hit a certain point on the wall a label, after the fact, based on where they hit, and distinguish them that way. But the prediction of where they go requires they be yet undistinguished during the process, the distinguishing happened after the fact.

Now imagine placing polarizers in the slits, at right angles to each other. Now you do not get the interference pattern, and you can also distinguish the photons by their polarization. Does that mean the photons are suddenly not indistinguishable particles, because we can use our apparatus to distinguish them? No, we used our apparatus to distinguish them before, without the polarizers, but the distinguishing happened after the fact, just as with the polarizers. It is not fundamentally different to include polarizers (or an apparatus that can detect isospin, or hypercharge, etc.), those all constitute distinguishing the particles after the fact.

What makes this subtle is that in many situations, you get the same answer if you imagine the particles are distinguishable. When this is commonplace, like with isospin (and we say we can distinguish protons from neutrons, etc.), imagining the particles are distinguishable becomes so automatic that we really completely forget we are strictly applying the same after-the-fact distinguishing we would apply to laser photons in a two-slit apparatus. When the difference becomes moot, we always imagine we can distinguish the particles, even when we actually cannot (we can only distinguish the measurement outcomes, like isospin). This is an especially acute issue when the very label of the particle itself, like proton or neutron, is applied after the fact via that measurement outcome. Surely protons and neutrons are distinguishable particles, they have different labels! But they are not really distinguishable before the fact (during the process, like propagation in a beam), only after the fact of measuring their isospins. We just never need to confront this formal truth of quantum mechanics unless we have to worry about something like neutrons decaying into protons during our experiment! When we do have to worry about that is when we discover that neutrons and protons are not actually distinguishable-- because a neutron that decays into a proton can interfere with the protons already in some beam of both, in a way that requires we cannot distinguish which neutrons decayed into which protons. Thue the neutrons never had a separate identity, or they wouldn't be able to lose it.
Yet what about terms which involve loops with virtual particles? Can the electron not be a neutrino for a short while in that loop diagram? And while it is a neutrino, is it distinguishable from other neutrinos? The formal quantum mechanics says that electrons and neutrinos are not actually different things and are not distinguishable, but the outcomes of measurements are distinguishable, so electrons and neutrinos are distinguishable after the fact of a measurement, just like photons in a two-slit pattern. But the situations you are describing are situations where we can get away with ignoring these formal aspects, and instead treat the particles in the sensible way, imagining that the particles themselves are distinguishable, instead of just the outcomes of the measurements on them. We will always incur some kind of error when doing that, but oftentimes the error is so tiny that it is well worth the simplification of imagining the particles are distinguishable throughout the process we are modeling. It's not that they are in some sense "really" distinguishable, it is that we are choosing to treat them that way, invoking the standard tradeoffs we always invoke whenever making such choices.

Last edited: Nov 13, 2011
13. Nov 13, 2011

### tommyli

In my response i cared very little about formal field theory, i always think its better to start from a simple physical situation to elucidate a question like this. There is no need to go into formal details for this kind of question, when you think about what is observed in an actual experiment the answer becomes clear.

I do not understand your point about the impossibility of distinguishing particles. When you calculate a scattering process, you assume that particles are always fully entangled even in the asymptotic regime. This is a simplification to aid calculation. If particles are localized in different regions of space they are obviously distinguishable. Imagine I have two electrons on opposite sides of the universe. Strictly speaking formally they are both local excitations of a single field so you may say that they are indistinguishable in order to be consistent with commutation relations, but the physical picture and common sense tell you that they are obviously distinguishable. Otherwise you have to define what you mean by distinguishable particles.

Simple example, you have a hydrogen molecule, and the two electrons are fully entangled. Now pull the molecule apart so that the two protons are on opposite sides of the universe. What wavefunction do you write for the electrons? Do you write the wavefunction as a simple product of two Hydrogen atoms or must you explicitly account for electron exchange in the wavefunction?

Anyways we are now arguing about semantics.. but my point is that if I take your view to the extreme, there are many physical situations that make no sense; neutrons and protons are no longer distinguishable as you mentioned. To me this may be mathematically correct but very often we make simple approximations that are much more helpful to understand physics.

Last edited: Nov 13, 2011
14. Nov 14, 2011

### Ken G

It is indeed a semantic issue, but they aren't-- not in quantum mechanics, nor in empirical physics either. If you think they are, then you must tell me an experiment that comes out A if the particles are distinguishable, and B if they are not. There is no such experiment, so in the situation you describe, we have no test of distinguishability, we have only the fact that the difference is moot. But here's the point: every time the prediction is A if they are distinguishable and B if they are not, then the experiment comes out B. Or, if A and B are the same, then we can say nothing on the issue.

This has been my point all along-- quantum mechanics says that particles that are indistinguishable must have joint wave functions, not individual wave functions, but when the overlap is negligible, the difference is moot. A moot difference is not the same thing as saying the theory changes into something different, it merely means we can get away with a simplification that is actually outside the theory but we know will give the same answer as the theory to a very high level of accuracy. Granted, we do that all the time, every time we talk about the wave functions of two distinguishable particles-- we always incur some kind of error when we do that, an error that QM can calculate, but when it is too small of an error to care about, we do it even though we know it's wrong. I'm not saying there's anything unusual about that, we do things we know are wrong in physics all the time, so you are certainly correct that those particles are going to be treated as distinguishable by any sane physicist. But they are still indistinguishable in formal quantum mechanics-- quantum mechanics does not distinguish particles, it distinguishes measurable outcomes.
I would simply say the theory says they are still quite indistinguishable, but there is just no need to either distinguish them or not distinguish them, so the issue is moot. If I claim that the two electrons swapped identity just before I did my measurement, no experiment can prove me wrong.

Now, it is certainly true that before quantum mechanics, mine would have been thought a silly position to take, and people would have invoked Occam's Razor to cut me down. But then came quantum mechanics, and it was discovered that whenever such exchange terms are not zero, they do indeed alter the outcome of the experiment. So we must choose to either claim we have a different theory for separated electrons as for overlapping ones, or we simply say we have the same theory but when the difference is moot we will enter into a standard kind of simplification that throws out the parts of the theory we don't care about at the moment. We always do that in physics, it's fine, but we don't need to claim the particles are actually distinguishable when the theory says they aren't and no experiment shows that they are. Instead, we say their indistinguishability simply doesn't matter in those situations, whereas they do matter in situations like the Bose-Einstein condensate in the OP.

I don't need that, I simply have a rule about joint wave functions, and a rule about when I can ignore joint wave functions and enter into the approximation of single-particle wave functions. That's just what quantum mechanics does. If I'm using a single-particle wave function, I'm ignoring the rest of the universe, whether distinguishable or indistinguishable-- I am choosing to treat that as a moot difference, but I should not claim this somehow proves distinguishability.
That's always up to the physicist. What accuracy are you looking for? These are standard issues in physics, but we don't need to imagine that some new theory is being invoked in which the electrons are distinguishable when separated but indistinguishable when overlapping. It's the same theory, and the same indistinguishability, it's just different choices about what approximations to invoke. In practice, we of course invoke the single particle wavefunction approximation, but that does not require we claim the electrons are distinguishable, it requires we claim we just don't care that they are indistinguishable because exchange terms wouldn't be important.
I'd say it's just quantum mechanics-- but we must also realize that all physics theories require approximations to be used in practice. So we have the formal theory, and we need to track when various aspects of it matter, and when we can get away with ignoring some aspects in favor of others, and invoke approximations like single-particle wavefunctions. But I'm also saying that a single-particle wavefunction is not a claim on distinguishability, because a single particle has a moot relationship with either distinguishability or indistinguishability-- it is the only particle in that treatment, and an only particle is neither distinguishable nor indistinguishable.

Last edited: Nov 14, 2011
15. Nov 14, 2011

### tommyli

Hi Ken,

It's an entirely semantic issue. On the level of formal field theory I absolutely agree with you, in a system of interacting particles there is no experiment that can distinguish particles. As long as you have one of those particles leaving the system and travelling to the other side of the universe, it its wavefunction will be prepared over multiple collisions and in the regions where it is not interacting with anything, its quantum state allows the particle to be identified and momentarily distinguished from other particles. In doing this we do not violate statistics, we are just attaching a label to the particle via its quantum numbers. This is very straightforward and its usually how all experiments are interpreted.

My issue is if you extend this to say that particles on opposite sides of the universe have a wavefunction which must be the antisymmetric product rather than a simple product of single particle wavefunctions. This is clearly wrong, for example you can have a complex system where the phase coherence length is small compared to the extent of the system. You have a system with many interacting identical particles and these particles are indistinguishable, however you cannot write the wavefunction as the antisymmetric product; this clearly gives the wrong answer. This is also very well known. If I have a hydrogen molecule and I stretch the molecule so that the protons move to opposite sides of the universe, the electrons have non-overlapping wavefunctions and the wavefunction is a product of single particle wavefunctions, not the antisymmetric product. There is zero probability of one electron being close to the proton on the opposite side of the universe. This is not because the electrons are truly distinguishable, I do not claim that they are, they are not "different" electrons but still local excitations of the same electron field, but all observation will point to the fact that your electron will have zero probability of switching identities with the electron on the opposite side of the universe -- they are completely disentangled. Otherwise you would observe a dramatically different spectrum. Again, this is not to say that the electrons are truly distinguishable in the way that you are saying, it is a property of the way that the joint wavefunction is prepared.

16. Nov 14, 2011

### Ken G

Yes, in practice neither of us would solve any problem differently. So the issue is just, how best to understand quantum mechanics-- a theory that changes its conception of indistinguishability to fit the circumstance, or a theory that takes a very particular and surprising stance on indistinguishability, which we often don't need to confront in practical applications like the one you are talking about? I'm saying it is the latter.
That doesn't make it "wrong", in fact it is still "right", but it is right in a way that is moot. That's something different from wrong.

No, it gives the right answer. And it gives the same answer as what you are talking about.
Why do you think that? If you write the (formally correct but practically unnecessary) antisymmetric wave function for that situation, it makes all the right predictions. They just happen to be the same as the predictions made by separable wavefunctions, which also make the same predictions as single-particle treatments. Indeed, if you really did that experiment, you would not use separable wavefunctions either, you would just use a single-particle wavefunction for the calculation of each half of the stretched atom, and ignore the other half of the atom completely. This is standard practice, we replace the actual universe with a pretend one that is highly simplified, and we make choices about what we think we can get away with. Distinguishability is just like that.

17. Nov 14, 2011

### nonequilibrium

Very interesting discussion! Thank you for this. I haven't had the time to read all of it yet, but I certainly intend to. For now just one remark (on a post by Ken)

Ah, perfect, that was the kind of answer I was looking for. What you say makes sense. I was confused because my professor tried to argue that identical particles are indistinguishable based on the concepts of psi functions. When I asked him whether it was a separate postulate (based on experiment) he seemed to imply that it was not, and repeated his hand-wavy argument about two wave functions overlapping and thus being indistinguishable. But as you state it, it indeed seems to be a separate postulate based on experiment, and that makes me happy again. Correct me if I misunderstood you.

18. Nov 14, 2011

### tommyli

Ok I understand your point. You are saying that if you have non-overlapping wavefunctions then there is effectively no difference between writing the separable wavefunction or the antisymmetric product, but to be consistent you must choose the antisymmetric product for the interacting and non-interacting scenario. You are absolutely right.

19. Nov 14, 2011

### Ken G

That's it exactly. I hadn't thought about the isospin issue before though-- that's an interesting wrinkle. I'd say it really stretches just what we mean by distinguishability of particles with different isospins. Most likely, even those particles are indistinguishable in some formal sense, but it becomes almost pedantically stubborn to not just let them be regarded as distinguishable by virtue of being different particles! But I think it could be said that the discovery of isospin essentially relegates all particles to being indistinguishable at some level. Maybe that's part of why Heisenberg once said, mysteriously, that we should stop thinking in terms of elementary particles and start thinking in terms of elementary symmetries. It's probably all a question of how abstractly mathematical one likes to take things, versus remaining grounded in some degree of physical intuition and everyday language.

20. Nov 14, 2011

### Ken G

Yeah, that does sound kind of backwards to me. Instead, we find that we have no way of distinguishing two electrons, so we incorporate that fact into the psi function, not the other way around. The wavefunction is supposed to encode what we can know about a system, but it is not supposed to make assertions about a system that no experiment can establish without help from our interpretation, like which electron is which.
I think the above discussion is very relevant-- there's an important difference between indistinguishability that you can test (when the wave functions overlap) and indistinguishability that is only a kind of logical extension of what you can test (when the wave functions don't overlap). When the issue is moot, you are probably free to adopt whatever interpretation works for you, I'm just saying that the most unified interpretation uses the same indistinguishability in all situations, independent of overlap, because it seems more natural to require that distinguishability be demonstrable rather than that indistinguishability be demonstrable. But the OP makes a good point that the tests that the particles are indistinguishable come when there is a lot of overlap, like in Bose-Einstein condensations.

21. Nov 14, 2011

### PatrickPowers

Thank you for your explanation. It is always good to hear from an expert.

22. Nov 14, 2011

### Ken G

Thank you, but we can all explain these things to each other, I don't count myself an expert in this company.