Distinguishing identical particles

Question

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 the 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?

Answer 1

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.

Answer 2

(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://arnold-neumaier.at/physfaq/physics-faq.html

Answer 3

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?

Answer 4

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.

Answer 5

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.

Answer 6

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.

This motivation for wavefunction (anti)symmetrization seems wrong to me.

Whether two configurations are truly the same or are merely indistinguishable by the instruments available to us is a philosophical question in classical physics. But it's experimentally testable in quantum mechanics, because in the former case you get probabilities of the form $|α+β|^2$ and in the latter case you get probabilities of the form $|α|^2+|β|^2$.

Shankar argues that the classical billiard balls can be distinguished by their histories and so their state space must be $\{(x,x')\mathop|x,x'\in\mathbb \ldots\}$, while the quantum particles can't be and so their state space must be $\{\{x,x'\}\mathop|x,x'\in\ldots\}$. Both of those are wrong. In the classical case it doesn't matter which state space you use (if your instruments can't tell the balls apart) because you get the same $p+q$ predictions either way. In the quantum case, only one of the state spaces is correct and you must use that one. If $\{\{x,x'\}\mathop|x,x'\in\ldots\}$ is the correct state space, it follows that the particles have no distinguishing properties that were missed by your instruments, but Shankar is trying to argue in the opposite direction, which doesn't work.

You can't measure the positions of quantum particles continuously, but you can measure them frequently enough that they can't sneakily swap places between measurements even at the speed of light. If you see two well separated tracks in a cloud chamber, it's unambiguous which particle went where. That makes no difference to the state space; you must still symmetrize it, or antisymmetrize it, or do neither, depending on the particle species, or else you're using the wrong one.