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I've been going through Shankar's Principles of Quantum Mechanics. In the section of the system of identical particles, he uses an example of billiards to illustrate the difference between identical particles in classical vs. quantum mechanics.

He argues that in classical mechanics, we can track the history of a particle (a billiard ball) to distinguish it from another particle with no intrinsic differences. In quantum mechanics, however, he argues since continual observation is not possible, we can't use the same method to distinguish identical particles.

A potential counter-example I thought was...suppose we have two non-interacting particles in a same square well. And at the end of some measurement, we find that one particle 1 is in a stationary state $\psi_1$, and particle 2 in $\psi_2$. We measure the system again after some time t, then we know whichever particle that's in $\psi_1$ must be particle 1 from the previous measurement. And the same goes for particle 2. Thus we can distinguish the two "identical" particles.

What conceptual mistakes am I making here?

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Excellent question :-) –  David Z Oct 19 '12 at 0:08
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5 Answers 5

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Let me try to understand what you are proposing: If there are two different stationary states, then they must have different energies, unless they are degenerate states. You stated that the two identical particles are non-interacting, thus they cannot exchange energy. This would mean that each particle would be stuck in their stationary states and we'll be able to distinguish them apart.

So I see that there's no other energy involved, both particles have different energies and therefore will be in different stationary states. I think the problem is that you state they are non-interacting. The two particles are distinguishable by construction.

At first I thought you were talking about position measurements, if so read the following:

You can see the problem clearly when their probability density overlaps in the same region of space. Notice that if you draw out the stationary state of both particles (no matter which states you choose), you have to overlap them because both of these are in the same well! There will be overlap for certain places. So, if you find out that one particle's position is where the overlap is, you won't know if that came from the first stationary state or the second.

Try this example: Take an O2 molecule. How can you tell which electron belongs to which nuclei?

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Thanks, the oxygen analogy is really helpful. –  Asy Oct 19 '12 at 0:48
    
Your first two paragraphs do indeed describe my example. So you're saying in that case, the two particles are indeed distinguishable by construction, but my assumption of their non-interacting is problematic? –  Asy Oct 19 '12 at 1:45
    
My best educated guess is that yes, it is problematic that you say they are non-interacting. After taking any two identical particles, I can't think of why they wouldn't interact with each other. I guess two photons of different energy wouldn't, but they can't be in a stationary state, they move. Also, learning about stationary states is just the preliminaries of QM. You have to add on the time portion. SO even if the math does says something weird, it maybe because the math isn't accurately describing physical reality. Remember, math only describes reality, it does not dictate it. –  QEntanglement Oct 19 '12 at 6:45
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The problem is within the description of the system after your first measurement. You say particle A is in state $\psi_{1}$ and particle B is in state $\psi_{2}$. This state could be written down as

$\psi_{1}(x_A)\psi_{2}(x_B)$

However, given that the two particles are identical, this is not a valid quantum state. The reason this state is not valid is not straightforward, but it boils down to the fact that you would get different predictions for some observables depending on the labels you chose for the particles. Since the particles are identical this would not be physically possible. The only valid quantum states are of the form

$\psi_{1}(x_A)\psi_{2}(x_B) \pm \psi_{2}(x_A)\psi_{1}(x_B)$

The problem with your reasoning is that you assumed that your first measurement allowed you to distinguish the particles and assign a label to them. That is in fact not possible, and there lies the loophole in your argument.

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Nicely done, Pedro. I can't believe how much nonsense other people are posting in response to this very straightforward question. –  Marty Green Oct 19 '12 at 15:27
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I'd say that between measurements, the system goes back to an "unobserved" state where the two particles are not associated with certain states. When you do your second measurement, the dice are rolled again. So you can't say that a particle you find in state $\psi_i$ is particle $i$ from the previous measurement -- this description is only valid for the "snapshot" you are currently looking at.

Also, wave functions describing particles in a quantum-mechanical system aren't really that picky about telling them apart either. If you are dealing with fermions, then the overall wave function will just change by a factor of $-1$ if you exchange any two particles; in the case of bosons, the wave function doesn't even change at all.

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If $\psi_i$ is a stationary state, that means after some time, it will remain in that state. So if two particles start in two different stationary states, then we'll be able to track them by that. I guess the question is then, are two particles still identical if they are in two different stationary states? –  Asy Oct 19 '12 at 0:37
    
The more I think about it, the more I get confused. Yes, I see your point. I think in this case, the problem lies in the definition of "observing" a system, and whether stationarity and time-independence of states extends beyond a measurement. In terms of a famous, over-used analogy: If Schrödinger's cat was dead when we opened the box the first time -- may it be alive the second time? –  Antimon Oct 19 '12 at 1:20
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(In)distinguishability is always relative to a chosen level of description.

Once you label particles by the eigenstate they occupy they become distinguishable. This means that for their description you changed the assumptions about the accessible state space, restricting it to eigenstates (by not allowing external forces that would perturb the Hamiltonian).

This happens generically for electrons in quantum chemistry, where they are classified as to which shell they belong to, based on spectral information in the Hartree-Fock approximation. These labelled electrons behave like distinguishable particles, as the effective state space has been reduced.

See also the entry ''Indistinguishable particles and entanglement'' in Chapter B3: Basics on quantum fields of my theoretical physics FAQ at http://www.mat.univie.ac.at/~neum/physfaq/physics-faq.html

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So does that by "indistinguishable", people usually mean indistinguishable in space? –  Asy Oct 19 '12 at 21:39
    
No. Particles which are distinguished by any property are already distinguishable. Space is one possibility; direction, mass, charge, etc. are other possibilities. For example the two H atoms in H_2O are indistinguishable, whereas the H's are distinguishable from the O. –  Arnold Neumaier Oct 21 '12 at 8:00
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I can see one problem here. How do you tell that one particle is in one particular state? The usual process requires perturbing the particle and from that determining what state it was in.

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I thought that by perturbing a particle, we collapse the wavefunction and thus measure what state the particle is in right AFTER the measurement. –  Asy Oct 19 '12 at 4:20
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