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It says explicitly on Wikipedia here and is implied in the lecture notes I am following (undergraduate level) that 'if a subset of M particles are indistinguishable from each other, then the M! possible permutations or possible exchanges of these particles will be counted as part of a single microstate'.

If the particles are indistinguishable by all relevant physical characteristics, but do actually have distinct, individual existences, then wouldn't allowing permutations of them within a microstate potentially cause that microstate to be more likely than others (contradicting the fundamental assumption of statistical mechanics)?

For example, there would be many more ways microstate 111 could occur than 101, as

101 includes

  • 1 0 1
  • 1 0 1

and 111 includes

  • 1 1 1
  • 1 1 1
  • 1 1 1
  • 1 1 1
  • 1 1 1
  • 1 1 1

I have a suspicion the answer may be along the following lines, but am not confident enough in my deductive skills to agree with myself and move on:

The order of the components here is a notational tool used to represent identity. The identity of the components is not distinct from their written order, which is what I assumed when I tried to denote identity with extra formatting. In reality, the identities of the components is what matters, not the order in which they are arranged (unless their spatial position was an important factor considered specifically). A system of spin-up and spin-down electrons, once selected by probability, can be moved about to any physical arrangement in space and be the same microstate because the identities of each electron are constant, and it was this that the probability of being 1 or 0 concerned. That is, it is each individual electron that has a probability of being spin-up or spin-down, not each individual electron for each position in the arrangement.

Does this make sense? Is this the case?


I have just come across this question in the sidebar. The answer uses the reasoning 'particles are indistinguishable, therefore their permutations give the same microstate'. I don't understand how being indistinguishable leads to loss of individual identity such that ignoring all the probabilities that each re-arranged configuration has of occurring is justified.

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  • $\begingroup$ Does order matter or does it not matter for these indistinguishable particles? Because you are not being consistent. If order matters, then those are 8 different microstates that are all equally likely. If order doesn't matter, then those are 2 different and 6 different ways of saying the same two things, respectively. In fact, if order doesn't matter, then why did you leave out the 110, 110, 011, and 011 microstates? Either way - however you define "microstate" - all microstates are equally likely. You just need to be consistent. $\endgroup$ – Geoffrey Jul 27 '17 at 9:26
  • $\begingroup$ @Geoffrey I think my issue was both this lack of certainty in whether order mattered in the examples I was reading, and the assumption that order did not correspond to spatial position and therein would not make two elements represented by the same symbol but at different positions in the list distinguishable. $\endgroup$ – perilousGourd Aug 7 '17 at 9:09
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The reason for this is that our macroscopical world, and the model of "particles" is just an approximation that arrises from quantum mechanics. What we in a classical non quantum mechanically treated way would describe as a system consisting of 3 electrons, that we can asign 3 positions and 3 spin states to, in quantum mechanics this is just a wild mixture of wave functions (to be precise, a symmetrised or antisymmetrised tensor product of single-electron states).

While we can asign an identity to each of the 3 electrons (as you indicated in your example, encoded by different styles), the same is not true for quantum mechanics. If we do give electrons names: "Electron1", "Electron2" and "Electron3", we could think of a classical configuration with those 3 electrons being at distinct places. Such a configuration would be forbidden in quantum mechanics: In quantum mechanics, the system will always be in a superposition of all the permutations possible, in the exact way that this quantum mechanical state will not change if I swap the identity of two particles. This means, that quantum mechanically, the different permutations you gave in your question, really just are one state. And as this one state, those 8 permutations do contribute to the number of microstates just one time.

To make it plain and simple, and to answer to your question: " I don't understand how being indistinguishable leads to loss of individual identity":

The 8 ways you wrote down your permutations allready make the 3 particles distinguishable. That's why a quantum mechanical system can't be in any of those states you have shown, but instead it will always be in a certain unique superposition of all of them, which does then count just as one microstate (because it is just one state).

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  • $\begingroup$ Awesome! This is starting to make a good deal of sense, though I have some trouble visualizing the specifics. For a quantum system of spin up and spin down electrons where spatial position mattered, would you have one probability density distribution over space for all spin ups and a second distribution for all spin downs? (If this is the case, wouldn't there be some chance that the electrons, upon collapsing the system, were not in the specified order? Also, how do you encode within the probability density the number of particles of that type that it can collapse to?) $\endgroup$ – perilousGourd Aug 7 '17 at 9:22
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    $\begingroup$ In the Formalism, you have a probability distribution over the space of all microstates. Imagine one quantum state $\Psi(x_1, x_2 .. x_n) = \Psi_1(s_1)\Psi_2(x_2) \Psi_3(x_3)... \vec{s}_1 \otimes \vec{s}_2 \otimes \vec{s}_3 \otimes .... $ (A Tensor product of single particle electron states). Obviously, particles are distinguishable inside this state (changing two indices would change the function). That's why your microstate is a superposition of all those states with permutations possible. $\endgroup$ – Quantumwhisp Aug 10 '17 at 12:22

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