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I was told that the concept of identical particles is because Hamiltonian is unchanged when permutating two particles, but if you imagine a classic mechanics system of N particles, it is also true that the Hamiltonian is unchanged when permutating arbitrary two particles, so it makes me wonder why there is no concept of identical particles in the classic mechanics.
My questions are:

  1. Does the concept of identical concepts require that the permutation between two particles should be feasible physically? For example, I could image two electrons transit into each other's position simultaneously, but it couldn't be feasible for classic particles.
  2. For two ions in the Lattice of a solid crystal, are they identical particles?
  3. If I have a bunch of electrons, when two electrons transit into each other's position, how long will the transition last? And during this transition, are they still identical particles?
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Classical physics has the concept of identical particles. What it doesn't have is the concept of indistinguishable particles.

Informally, the difference is as follows: Man can not tell identical particles apart. When it comes to indistinguishable particles, God can't tell them apart.

Suppose you have a proton (No. $1$), and I have proton (No. $2$). We bring them together and add an electron to make $^1H_2^+$. Then we break it and each take our protons home.

Classically there are two outcomes: We have our original protons, or we swapped. Only God knows.

Quantum mechanically, they're entangled. I have $|p_1p_2\rangle$, and you have $e^{i\phi}|p_2p_1\rangle$. Nature cannot tell them apart.

It's worse if you have a proton and I have a neutron. We make deuterium; hit it with a 2.2 MeV gamma ray and take our particles home. Surely your proton is the one you came with, as well as my neutron.

Well, no. The isospin wavefunction of $D$ is:

$$|I=0,I_3=0\rangle=\frac 1{\sqrt 2}(|pn\rangle-|np\rangle$$

So if you grab particle 1, it 1/2 p and 1/2 n, while mine's 1/2 n half p. (In theory, neither of us know until one of us looks...and then we a combination of Schrödinger's Cat and the EPR paradox....let's not go there).

Seriously, though, the proton you walk away with is not the proton you walked in with, thanks to pions being isovector, and:

$$ p+n\rightarrow (n+\pi^+)+(p+\pi^-)\rightarrow n+p$$

As long as the particles can be entangled, you need to worry about their indistinguishable nature. If we instead brought electrons together, you with a beam electron, and with a stationary target electron, and conduct a Møller scattering experiment:

enter image description here

Surely the scattered electron belongs to the beam?Alas, no again:

enter image description here

Crossing symmetry ensures the target and beam electrons are entangled, even when they are far, far, apart. Too far to ever be in physical contact/or exchanged again (or not for a long while).

A question is: at what scale does this break down? I believe Young's double slit experiment has been conducted with $C_{60}$ Buckey balls...some would consider them, at 0.7nm, as macroscopic objects.

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  • $\begingroup$ They're up to 2000-atom molecules for the double slit experiment now (I guess it's actually more complicated than the simple double slit, but still matter-wave interferometry): nature.com/articles/s41567-019-0663-9 $\endgroup$
    – llama
    Jun 28, 2021 at 16:48
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Your question 2. raises a profound issue.

The atoms in a crystal can be identified by where they are, so they are usually counted as distinguishable when doing statistical mechanics such as computing the vibrational contribution to the crystal's specific heat.

However, there is nothing in principle to stop them from quantum tunneling and exchanging places. This has been speculated as leading to a "supersolid", but some experimental evidence has turned out illusory, so it is still speculative.

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    $\begingroup$ any link to the related experiments? $\endgroup$
    – Jack
    Jun 27, 2021 at 17:03
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    $\begingroup$ The wikioedia article on supersolids has some links: en.wikipedia.org/wiki/Supersolid $\endgroup$
    – mike stone
    Jun 27, 2021 at 17:28
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Two or more particles can still be identical in classical mechanics. For example, we can look at the Hamiltonian, like you suggest. But the concept is not discussed in classical mechanics because there is no motivation for indistinguishable states.

The concept of identical particles appears in quantum theories because of the more important concept of indistinguishability of states; it turns out quantum theory works much better when configurations that are results of identical particle index permutation are pronounced the same configuration and thus the psi functions are limited to those that obey this permutation symmetry. For example, electrons in single helium atom are identical, and the allowed psi functions are such that from studying any such psi function, the two electrons are indistinguishable.

  1. Does the concept of identical concepts require that the permutation between two particles should be feasible physically? For example, I could image two electrons transit into each other's position simultaneously, but it couldn't be feasible for classic particles.

No. Permutation is exchange of indices on some theoretical quantities, such as coordinates and momenta, not a real process.

  1. For two ions in the Lattice of a solid crystal, are they identical particles?

It depends on the model. If the model says the ions are identical particles, then they are identical particles. This is not necessary; for example, one ion can be silicon, the other phosphorus, and then they are not identical. Or one ion can be silicon isotope 1, the other silicon isotope 2. Then they are not identical. Whatever the case, two ions in lattice are never indistinguishable, because they can be distinguished by their different positions in the lattice.

  1. If I have a bunch of electrons, when two electrons transit into each other's position, how long will the transition last? And during this period, are they still identical particles?

Electrons have no variable degrees of freedom, they all have the same mass and charge and thus are always identical particles, it does not matter what happens to them.

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I would say that classical mechanics has the concept, at least theoretically, of indistinguishable particles. Indistinguishability is just kind of trivial.

Consider the Lagrangian formulation of quantum mechanics in terms of two indistinguishable fermions. In Lagrangian mechanics we consider all possible paths, given the initial and final positions of the fermions, and weight them in our path integral by $\exp(i S/\hbar)$, where $S$ is the action of the path. In order to account for fermi statistics and indistinguishability, we should subtract off the paths where the particles are permuted. In the end, we say that the particles have followed all possible paths, and we add them up weighted by phase factors. The fact that we subtracted some of them instead of adding is just another part of the phase factor.

Now what about indistinguishable classical particles obeying deterministic paths. We could calculate the paths that minimize the classical actions subject to the two permutations of the boundary conditions. We could then say that depending on whether the particles exchanged places or not, they must have followed one of these paths or the other. But we couldn't assign probabilities to either of these scenarios without additional information, such as the probability distributions of the initial velocities.

To extend this notion of classical mechanics with uncertainty, we could also incorporate classical uncertainty, such as a Gaussian white noise. In this way we could say that even though the particles followed exactly one path, we could calculate the probability of a particular path and therefore its weight in the ensemble. See, for example, Kleinert's massive text on path integrals, which discusses classical uncertainty at some length.

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  • $\begingroup$ I would argue that the path integral of two fermions is something inherently quantum, not classical. Indistinguishable implies 1-dimensional rep of the permutation group, and is baked in the Fermi statistics, which is quantum. $\endgroup$ Jun 28, 2021 at 0:45
  • $\begingroup$ I would certainly agree that "the path integral of two fermions" is inherently quantum mechanical (though I have used it to calculate observables in purely classical systems, the math is more broadly applicable) but I don't see how the concept of indistinguishability, much less the concept of identicalness that the question mentions, implies a one-dimensional rep of the permutation group. It certainly doesn't imply Fermi statistics, if that's what you're suggesting. $\endgroup$
    – David
    Jun 28, 2021 at 2:49
  • $\begingroup$ It certainly does apply as fermion states must transform by the 1-dimensional alternating irrep of $S_n$. $\endgroup$ Jun 28, 2021 at 3:11

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