Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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. – Lupercus Sep 16 '12 at 12:30
up vote 5 down vote accepted

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 ofte studies effective theories where there are additional means of distinguishing particles. Two important examples:

  1. 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.

  2. 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.

share|cite|improve this answer
"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... – aquirdturtle Jan 26 at 9:41
@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. – Arnold Neumaier Jan 26 at 15:51
Ah, yes, I misread that statement; I was thinking of the electrons. – aquirdturtle Jan 27 at 6:43

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.

share|cite|improve this answer

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

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..

share|cite|improve this answer

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.

share|cite|improve this answer
what you meant by molecules can't be labeled? Is it gaseous molecules – Eka Dec 11 '13 at 6:11

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.

share|cite|improve this answer

protected by Qmechanic Apr 2 '14 at 13:29

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.