# Why are particles in Quantum Mechanics indistinguishable?

I'm currently reading about tensor products in Quantum Mechanics and composite systems and I've read that in Quantum Mechanics particles are indistinguishable while in Classical Mechanics that's not the case. This leads to the requirement that the wave function be symmetric or antisymmetric and thus, to the classification of particles as fermions and bosons.

Now this might seem totally silly but why are particles in QM indistinguishable while in CM they are not?

In CM we describe possible configurations of a particle as points in $\mathbb{R}^3$ while in QM we describe possible states as vectors in a Hilbert space $\mathcal{H}$.

If we have two particles in CM we'll be able to say that the first is at $\mathbf{r}_1$ while the second is at $\mathbf{r}_2$. In QM we'll be able to say that the first has state $|\varphi\rangle$ and the second has state $|\psi\rangle$.

It seems we can distinguish them in QM using the states inasmuch as we can distinguish them in CM using the points.

So why are particles in QM indistinguishable? What am I missing here?

• Let $H$ be the state space for one particle. If the state space for a pair of particles were $H\otimes H$, then the particles could occupy state $|\phi>|\psi>$ and be distinguishable. But if the state space consists only of the symmetric (or antisymmetric) part of $H\otimes H$, then $|\phi>\psi>$ is not a state. So the question is: "Why is the state space for a pair one thing rather than another?". I'm pretty sure the only answer is that the state space is chosen to match observed behavior --- but will be interested if someone says it's derivable from something more fundamental. – WillO Mar 2 '16 at 4:00
• My impression would be that particles in CM aren't distinguishable; I'd take position and momenta to be an extrinsic property not an intrinsic one like mass or charge; another view, would be that CM doesn't really deal with particles but large objects; in which case they're distinguishable - but then the Sorites paradox then plays a part in determining what is accounted as identity. – Mozibur Ullah Mar 2 '16 at 4:13
• "In QM we'll be able to say that the first has state |φ⟩ and the second has state |ψ⟩" - for example, let the two states be position eigenstates so, initially, there are two particles, each with a definite position (so the momentum of each particle is maximally uncertain). Suppose the system evolves and then a detector registers the passing of a particle. Which of the particles was detected? – Alfred Centauri Mar 2 '16 at 4:30
• @AlfredCentauri: One can certainly imagine a setup in which it makes sense to ask which particle was detected. Let E be the eigenstate that was detected. Then we know that the pair is either in state $\phi'\otimes E$ or $E\otimes\psi'$ (where the prime indicates the result of the time evolution), but we don't know which. As far as I can see, that would be a perfectly sensible theory, but ultimately not consistent with observations. Am I missing something? Is there a reason this would not make sense? – WillO Mar 2 '16 at 5:22
• Definitely read this other Physics Stack Exchange question and the accepted answer. The title may not sound related but as you'll see from the answer, it is. – DanielSank Mar 2 '16 at 5:47

The statement arose to fit the experimental observations. The mathematical framework of quantum mechanics has to have the elementary particles indistinguishable because that is what has been observed, i.e. the standard model of particle physics that has indistinguishable elementary particles fits the data.

The reason within the mathematical model is that there does not exist an underlying complexity that could distinguish one electron from another by an individuated quantum number. As complexity increases, for example an excited Hydrogen atom and one in the ground state, quantum mechanical states can be distinguished, for the example because of the differences in energy at that point of time t, and individuation at specific coordinates appears.

When larger masses are involved the complexity increases exponentially and individuation of matter is evident. As far as the quantum mechanical framework goes, mass systems of high complexity become classical because the probability of finding two billiard balls with the exact wavefunction is infinitesimally small. Note that classical radiation still carries mathematically a simplicity, just frequency identifies it.

first of all, particles are the same because they can be treated as the excitations of the same underlying quantum field.

also, considering how we experimentally identify different species of particles, we need to measure the mass, charge, etc. of the particle. Now if we shoot two electrons, which are labelled $e_1$ and $e_2$ from two electron guns, against each other. Note that even though the two electrons have exactly the same mass and charge, they can in fact be distinguished initially because they come from two different sources, just like the different spins in a spin chain. However, when the scattering happens at some place and two electrons come out after the scattering, we will lose track of which one is 1 and which is 2. One simple way to think about this is by uncertainty principle. We know that the scattering happen at a certain region, then we know the $\Delta x$ and therefore will have some $\Delta p$. Since we have some uncertainty in the momentum, we can not say for sure in which way each of the electrons would go. As a result, the two particles with the same properties got completely mixed up.

"Indistinguishable" is the property that, given two particles, we have no way to figure out "which one" of them we detect when we measure a single particle.

The claim that in classical mechanics we can always say that the first particle is at $r_1$ and the second at $r_2$ is only true for distinguishable particles. For indistinguishable particles, you have to divide out the discrete $S_2$-permutation symmetry out of the phase space, which promptly becomes singular at the points where both particles are at the same place. There is slight confusion about whether to remove those points from the resulting space, see this and this question.

As the answers to those questions show, if we insist on geometric quantization giving the correct quantum state space without implementing fermionic/bosonic behaviour by hand, we have to already classically take indistinguishability serious.

The reason you rarely see classical indistinguishability discussed is that classical mechanics mostly deals with marcoscropic objects where distinguishing them is as easy as throwing a bit of paint at one. Furthermore, the ability to classically continuously measure the position of a particle might also be seen as an argument that classical particles can't be really indistinguishable - decide at one point which of them is which, then continuously track their positions.

In quantum mechanics, you typically lose both abilities. Particles don't move along a unique trajectory if not measured, you can't decide which is which and then just track them, and you certainly can't paint electrons. So saying that "A is in state 1 and B is in state 2" is impossible to distinguish from "B is in state 2 and A is in state 1" even in principle in quantum mechanics, while classical mechanics allows you such outs as painting or using the existence of a unique trajectory.

• You know, I really don't think it's so complicated. If I write a fourier series $f(x) = \sum_n c_n \text{mode}_n(x)$, then I don't try to give each unit of excitation, i.e. the "units of $c_n$", individual identity. Indistinguishable particles only sounds weird because "first quantised" notation is stupid in that it tries to name the units of excitation of a mode instead of naming the modes and treating units of excitation the same way we do classically. – DanielSank Mar 10 '16 at 23:56
• @DanielSank: If the question were about the Fock space and why it is symmetrized/anti-symmetrized or why states in a Fock space correspond to indistinguishable particles, I would completely agree. But quantum mechanics (and also the concept of the (in)distinguishable particle) is usually developed without the assurance that a field-theoretic model (and hence a Fock space) is behind it. You're right that the "deep" underlying reason is that "particles" come from fields, but I felt that was not the line of reasoning this question was looking for. – ACuriousMind Mar 11 '16 at 0:07
• You don't need fields, you just need "second quatization" (terrible term) and everything is hunky-dory. Frankly put, first quantization is not only stupid from the "deep" perspective, it's also totally unnatural if you first think about any classicial problem with a mode structure. The fact that we brainwash our students with first quantization, confuse the crap out of them with (anti)symmetrized wave functions, and then teach them how to actually think about Nature, is totally historical and has absolutely no reason to live on in school curricula. – DanielSank Mar 11 '16 at 4:01

The point is not on the mathematical formulation (you could easily make quantum systems to have distinguishable particles), but on the physical interpretation of quantum mechanics.

Let's think for a moment what we do operatively to distinguish between to classical particles. First of all, we need to observe the state of each one, and since the state of a classical particle is described by a point of the phase space, we can easily do that just measuring two observables: position and momentum. In addition, we know/postulate that two classical particles cannot occupy the same state (i.e. point of the phase space) at the same time, and each one has to occupy a state. Placing detectors (in principle) at each point of the phase space, we can then attach a label "$1$" to the particle in the state $(x_1,p_1)$ and a label "$2$" to the particle in state $(x_2,p_2)$, both detected at the initial time. Now that we have attached the labels, we can follow their trajectory through time. That trajectory is uniquely identified by the initial condition, and therefore we always keep the two particles distinct (even collisions would not cause problems, because even if the particles then are in the same position, they must have different momenta since they started from different initial conditions and so we can still distinguish them).

Now, are we able to do the same in quantum mechanics? Are we able to attach labels to two identical particles in different states in order to follow them? The answer is, in general, no (because of the nature of quantum measurements). The first difficulty is to measure the state of each quantum particle, since now it is something that we can't measure directly and in general performing measurements modifies the state of the system (that last property will become crucial later). Nevertheless, suppose that we were able to prepare two particles, each one in a defined stationary state $\varrho_1$ and $\varrho_2$. Suppose in addition that each particle has a non-zero probability of being found in every small region (complying with the indeterminacy principle) $\Omega$ of the phase space (there are always states like that in quantum mechanics). Now the crucial problem arises: how can we attach a label to each particle? Suppose that at the initial time we detect the particles in the regions $\Omega_1$ and $\Omega_2$. It may very well happen that $\Omega_1=\Omega_2$! Now, how could we attach a label to each particle? If $\Omega_1=\Omega_2$, it is clearly not possible; if the two regions are different, we could, but then the quantum measurement has changed the state, and forced the particles to be confined to the regions $\Omega_1$ and $\Omega_2$ respectively. So they can't interact anymore, and any eventual coherence between them is completely destroyed. So we managed to distinguish the particles, but at the cost of isolating them from each other, therefore modifying completely the system. That is not the type of distinguishability we sought. And we actually have no other way to distinguish particles, and not because we are not able, but because the quantum theory (of measurements) prevents it.

Therefore it is quite natural, due to this operative indistinguishability, to formulate a theory that postulates quantum identical particles to be indistinguishable. As always, the then experimental success of the predictions of such theory is the best confirmation of it; however in this case the idea is a priori justified (in my opinion) by the above.

## protected by Qmechanic♦Mar 2 '16 at 11:20

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