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I am wondering if we fix the locations of indistinguishable particles, will the identical particles become distinguishable? Say, put each one of the indistinguishable particles into a small box.

When I study Statistical Mechanics, I am confused by why earlier we treat spins as distinguishable particles. The only way it seems to make sense to me is that the spins are in a state of solid - their locations are fixed (or I was just shown some kind of a toy model?). But then, I am even more confused - as I recall from quantum mechanics, particles in crystal are also identical? Also fixing locations doesn't seem to change any of the quantum numbers to me. So when the indistinguishable particle distinguishable? Will fixing their locations makes them distinguishable?

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  • $\begingroup$ When you talk about "spins" are you referring to the Ising model and such? $\endgroup$ – valerio Dec 30 '17 at 18:39
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    $\begingroup$ "Also fixing locations doesn't seem to change any of the quantum numbers to me." Ah, but it essentially is. $\endgroup$ – DanielSank Dec 31 '17 at 1:37
  • $\begingroup$ @DanielSank You mean the energy eigenvalues become discrete? Would you mind elaborating a bit, please? $\endgroup$ – Shing Jan 10 '18 at 16:45
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In principle, it is true that you cannot distinguish between two identical particles. However, if the overlap between the wavefunctions of the two particles is close to $0$, you can often treat the particles as distinguishable. There are mainly two cases in which this happens:

1. Particles separated by a "high enough" potential barrier

If two identical particles are separated by a "high enough" potential barrier (for example a "small box", but also a potential well corresponding to a lattice site), the overlap between the respective wavefunctions is very small (in the ideal case of an infinite potential barrier, the overlap is rigorously $0$). This means that we will always be able to tell with a high degree of accuracy which particle is where, i.e. we can treat them with an high degree of accuracy as distinguishable.

See also this answer by Arnold Neumaier.

2. "Far away" particles

If two identical particles are "far away" from each other, we can treat them as distinguishable.

Take for example two far away electrons: since they are indistinguishable fermions, their wavefunciton is

$$\Psi(\mathbf r_1, \mathbf r_2)=\frac 1 {\sqrt 2} [\psi_1(\mathbf r_1)\psi_2(\mathbf r_2)-\psi_1(\mathbf r_2)\psi_2(\mathbf r_1)]$$

Now consider the expected value of some observable $\mathcal O$:

$$\langle \mathcal O \rangle = \int \int d\mathbf{r}_1 d \mathbf{r}_2 [\Psi^*(\mathbf r_1, \mathbf r_2) \ \mathcal O \ \Psi(\mathbf r_1, \mathbf r_2)]=\\ \int \int d\mathbf{r}_1 d \mathbf{r}_2 [\psi_1^*(\mathbf r_1)\psi_2^*(\mathbf r_2) \ \mathcal O \ \psi_1(\mathbf r_1)\psi_2(\mathbf r_2)] - \int \int d\mathbf{r}_1 d \mathbf{r}_2 [\psi_1^*(\mathbf r_1)\psi_2^*(\mathbf r_2) \ \mathcal O \ \psi_1(\mathbf r_2)\psi_2(\mathbf r_1)]$$

Let's assume that $\psi_1$ and $\psi_2$ are sufficiently localized, i.e. that $\psi_1$ is appreciably different from $0$ only in a domain $D_1 \in \mathbb R^3$ and that $\psi_2$ is appreciably different from only $0$ in a domain $D_2 \in \mathbb R^3$, and that $D_1$ and $D_2$ don't overlap. We will then have for the second term

$$\int \int d\mathbf{r}_1 d \mathbf{r}_2 [\psi_1^*(\mathbf r_1)\psi_2^*(\mathbf r_2) \ \mathcal O \ \psi_1(\mathbf r_2)\psi_2(\mathbf r_1)] \\ \approx \int_{D_1} \int_{D_2} d\mathbf{r}_1 d \mathbf{r}_2 [\psi_1^*(\mathbf r_1)\psi_2^*(\mathbf r_2) \ \mathcal O \ \psi_1(\mathbf r_2)\psi_2(\mathbf r_1)] \approx 0$$

since $\psi_1 \approx 0$ in $D_2$ and $\psi_2 \approx 0$ in $D_1$. It follows that

$$\langle \mathcal O \rangle \approx \int \int d\mathbf{r}_1 d \mathbf{r}_2 [\psi_1^*(\mathbf r_1)\psi_2^*(\mathbf r_2) \ \mathcal O \ \psi_1(\mathbf r_1)\psi_2(\mathbf r_2)]$$

This means that the wavefunction

$$\tilde \Psi(\mathbf r_1, \mathbf r_2) = \psi_1(\mathbf r_1)\psi_2(\mathbf r_2)$$

which is a wavefunction for distinguishable particles, would have given us approximately the same result! Therefore, we can consider far away particles to be distinguishable.

See also this answer by tparker.

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  • $\begingroup$ @Shing Well there shouldn't be a factor $1/2$, because the wavefunction wouldn't be normalized. $\endgroup$ – valerio Feb 2 at 18:27
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Identical particles are indistinguishable. This is true in the spin model as well. In that model, we only consider the states of the system in which there are particles only at the lattice sites (exactly one at each site). For simplicity consider a two-site model with spin. If the particles were distinguishable, there would be 8 states: first, pick the positions of the two particles (first one in the first site, the second one in the second, or the other way around), then pick the spin for each (up or down). But we ignore which particle is at which site, we only ask whether the one at the first site has spin up or down and the same for the other. That is 4 states.

When you restrict the position of your particles, you can talk about the particle at each site without compromising indistinguishability. Because we don't care which one is at that site, we only care about (for example) the spin of the one at that site.

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So when the indistinguishable particle distinguishable? Will fixing their locations makes them distinguishable?

Particles are distinguishable when they are in different eigenstates: an excited hydrogen atom that has absorbed a photon can be distinguished from a hydrogen atom with the electron in the ground state.

Also spin can distinguish a particle from another . A pi0 decaying into two gammas: if one gamma has spin+1 the other must have spin -1 and thus is distinguishable from the first one.

Locations are a continuum. In a lattice again energy (excited or at ground level) and spin may distinguish particles, depending on the experiment.

Particles can be distinguished if they have different eigenvalues for some observables, otherwise are indistinguishable.

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Since all fermionic creation operators anticommute, the wavefunction of any two fermions must always be antisymmetrized. This holds even if the fermions have different spins, different colors, or even if they're completely different particles, such as a down quark and a tau neutrino.

However, in many cases this can be neglected. The general rule is that a fermion can be excluded from the antisymmetrization if it has a property that no other fermion shares.

For example, consider an atom where all the electrons are fixed to have spin up, except for one which is fixed to have spin down. Then the Pauli exclusion principle doesn't place any constraints on the spatial state of the spin down electron, so nothing goes wrong if we treat that electron as distinguishable from the rest; you don't get any illegal configurations. Conceptually, the particle is distinguishable by its spin. (This has nothing to do with whether a specific experimental setup can do the distinguishing. This is a mathematical fact that was true even back in the stone age.)

Similarly, the quarks in a meson can be distinguished because only one is an antiquark, but the quarks in a baryon cannot, see here. In the case of spins whose positions are fixed on a lattice, each spin has a unique position, so they're all distinguishable by their positions.

A subtler version of this trick is pulled for the classical ideal gas. The wavefunction of each particle is of the form $\psi_{\text{spatial}} \psi_{\text{spin}}$. We assume that each particle has a distinct spatial state; once that's accounted for, the particles are distinguishable by virtue of their positions, for the purposes of assigning spins. But of course this is false: two particles can have the same spatial state and opposite spins. The approximation made for the classical ideal gas is that the gas is sparse enough that these configuration are negligible. If they aren't, you have to use the full Fermi-Dirac statistics.

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protected by Qmechanic Dec 31 '17 at 14:05

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