What are distinguishable and indistinguishable particles in statistical mechanics? While learning different distributions in statistical mechanics I came across this doubt; Maxwell-Boltzmann distribution is used for solving distinguishable particle and Fermi-Dirac, Bose-Einstein for indistinguishable particles. What is the significance of these two terms in these distributions?
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2$\begingroup$ Besides this problem, there are quite a few features of classical Statistical Mechanics which are reminiscent of QM, although they were developed one or two decades before QM. It would be very interesting to clarify once and for all how much actual QM is contained in these "classical" assumptions of Statistical Mechanics. Examples of these problematic classical features include all equations where $\hbar$ actually appears (e.g. the Sackur-Tetrode Eq., the integration measure in phase space ${\frac{1}{N!}\(\frac{dp\ dq}{2 \pi \hbar}\)}^N$) as well as the third principle of Thermodynamics. $\endgroup$– LupercusCommented Sep 16, 2012 at 12:30
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5 Answers
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On the deepest level, particles are indistinguishable if and only if they have the same quantum numbers (mass, spin, and charges).
However, in statistical mechanics one often studies effective theories where there are additional means of distinguishing particles. Two important examples:
In modeling molecular fluids, two atoms on the same molecule are distinguishable if and only if there is no molecular symmetry interchanging the two atoms, and two atoms in different molecules are distinguishable if and only if there is no congruent matching of the two molecules such that the two atoms correspond to each other.
In modeling the solid state, one typically assumes that the atoms are confined to lattice sites, and that each site is occupied at most once.. In this case, the position in the lattice is a distinguishable label, which makes all atoms distinguishable.
The computational relevance of the distinction is that permutations of (in)distinguishable particles (don't) count towards the weighting factor.
For an expanded discussion see my article at PhysicsForums.
answered Sep 16, 2012 at 10:43
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$\begingroup$ "the position in the lattice is a distinguishable label, which makes all atoms distinguishable." But in quantum mechanics the particles aren't localized to a single lattice site, and in most models I've seen the lattice is treated as periodic or infinite... $\endgroup$ Commented Jan 26, 2016 at 9:41
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$\begingroup$ @aquirdturtle: In solid state physics, the atoms are confined to a lattice site, apart from small vibrations. Thus they are distinguishable. The valence electrons can move freely and are indistinguishable. There may be defects that move, but this is a motion of distinguishable particles. In ferromagnetsim, the spins are localized at lattice sites. That the lattice is periodic or infinite has no bearing on distinguishability. $\endgroup$ Commented Jan 26, 2016 at 15:51
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$\begingroup$ Ah, yes, I misread that statement; I was thinking of the electrons. $\endgroup$ Commented Jan 27, 2016 at 6:43
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$\begingroup$ Nice answer. I still have some confusion. Would two molecules like say two $H_2$ molecules be distinguishable ? (I mean the molecule as a whole and not the individual particles) $\endgroup$– user139621Commented Sep 19, 2017 at 7:48
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1$\begingroup$ @Blue: Two hydrogen molecules as a whole, or two water molecules as a whole, are indistinguishable. But two DNA molecules are distinguishable since they are so large that they can be distinguished by location. $\endgroup$ Commented Sep 20, 2017 at 14:16
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Assume you have two particle A and B in states 1 and 2. If the two particle are distinguishable, then by exchanging the particles A and B, you will obtain a new state that will have the same properties as the old state i.e. you have degeneracy and you have to count both states when calculating the entropy for example. On the other hand, for indistinguishable particles, exchanging A and B is a transformation that does nothing and you have the same physical state. This means that for indistinguishable particles, particle labels are unphysical and they represent a redundancy in describing the physical state and that is why you would have to divide by some symmetry factor to get the proper counting of states.
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Suppose you have two distinct particles. If they are distinguishable (Like a helium-3 atom and a helium-4 atom), then you can switch their positions and the system changes. If they are indistinguishable (Like two protons), switching the two particles' positions makes no physical change because we do not know whether particles switched at all. I haven't studied advanced quantum mechanics, so I can't give a better explanation, but Wikipedia can http://en.wikipedia.org/wiki/Identical_particles#Distinguishing_between_particles.
The number of permutations of the distinguishable particles is n! more than that of indistinguishable ones, so quantities like entropy can change depending on whether we can distinguish particles in our system. All three distributions can be derived from the grand partition function, but the derivations for Bose-Einstein and Fermi-Dirac distributions uses indistinguishability..
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Since there is no way in which the molecules can be labeled, the particles are indistinguishable. On the other hand, if the assembly is a crystal, the molecules can be labeled in accord with the positions they occupy in the crystal lattice and can be considered distinguishable.
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$\begingroup$ what you meant by molecules can't be labeled? Is it gaseous molecules $\endgroup$– EkaCommented Dec 11, 2013 at 6:11
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In classical mechanics, one can keep track of individual particles. For example, if you have two particles in a system, you can color these particles red and blue and follow them. Notice this, for classical mechanics, dynamical variables are $(q_i,p_i)$ with $i=1,2$. There is nothing classical mechanics that prevents you from specifying more variables e.g $c_i\in \{r,b\}$.
In quantum mechanics, however, the maximum you can do is specify a complete set of commuting observables for each of the particles. There is also this problem of following the trajectory in quantum mechanics. There is no trajectory, owing to Heisenberg's Uncertainty. So, there will be regions in the system, where you will be blind and the two particles exchange their positions. If you look too carefully, the system would be disturbed, thus making your measurements useless.
P.S. I have channeled my answer from the first paragraph of J.J. Sakurai's Modern Quantum Mechanics. Look at the Chapter Identical Particles to get an answer with a picture.
answered Jun 22, 2015 at 15:37
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$\begingroup$ Can we keep track for particles also in thermodynamics? $\endgroup$ Commented May 17, 2020 at 19:39