# Why are Maxwell Boltzmann particles distinguishable but bosons and fermions are not?

Maxwell-Boltzmann statistics is applicable for identical, but distinguishable particles.

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Why are particles which satisfy Maxwell Boltzmann statistics distinguishable while bosons and fermions are indistinguishable (in the Statistical Mechanics sense)?

Also, which particles classify as Maxwell Boltzmann particles? Would two hydrogen atoms be classified as distinguishable or indistinguishable. Hydrogen atoms can surely be of at least two types - ortho and para. (But @Arnold Neumaier says in his answer that two hydrogens atoms are indistinguishable.) I find this confusing.

• – user139621 Sep 19 '17 at 6:36

There are a few misconceptions here. First, the difference between Maxwell-Boltzmann statistics and Fermi/Bose statistics is not centered on distinguishability. Even in classical thermodynamics, one has to deal with questions of indistinguishable particles.

Instead, you should think of Maxwell-Boltzmann as the classical limit of quantum statistics, corresponding to high temperature and low density. Quantum effects become important when the average intermolecular distance is on the order of the de Broglie wavelength of the particles. Alternatively, one could say that classical statistics is valid as long as the probability that any given quantum state is occupied is much, much less than 1.

The question of indistinguishability can be approached classically or from a quantum mechanical viewpoint. In the quantum picture, particles are identical if they share precisely the same quantum numbers (labels, essentially), which are determined by the Hamiltonian of the system. These quantum numbers refer to intrinsic properties such as mass, spin, and electric charge, as well as extrinsic properties which correspond to values of observables which commute with the Hamiltonian such as total angular momentum. In solid state physics, particles which are localized to distinct lattice sites can also be distinguished by spatial location.

Classically, the question is a bit more vague. From the standpoint of the Hamiltonian physics, the state of the system is specified by the precise position and momentum of each constituent particle. As no two particles have the same position and momentum, all particles are distinguishable in the sense that swapping their labels results in a noticeably different state.

However, when we move to actual thermodynamics, we no longer define the state of a system as a singular point in $2N$-dimensional phase space. Instead, we deal with macrostates of the system, which are characterized by macroscopic quantities like energy and volume. If two particles exchange their position and momenta and the system remains in the same macrostate, the particles can be considered to be identical, and in fact need to be in order to resolve the Gibbs paradox.

Also, to answer your question - yes, two hydrogen atoms are indistinguishable if they have the same quantum numbers. Ortho/para-hydrogen refer to hydrogen molecules, not hydrogen atoms, but nonetheless they are also indistinguishable if they have the same quantum numbers.

• I suppose two or more molecules usually don't have the same quantum numbers ? (which we obtain by solving the Schrodinger's Equation) – user139621 Sep 19 '17 at 11:43
• @Blue As the complexity of the particles increases, you see more and more degeneracy - more states which the molecules could occupy without "overlapping". However, pack things in tightly enough and at sufficiently low temperatures and the quantum effects will take over, as is the case with superfluid liquid hydrogen. – J. Murray Sep 19 '17 at 12:07

Particles obeying M-B statistics are classical particles, whereas the bosons and fermions are quantum particles. Quantum particles have wave functions associated with them. (All particles do, even humans have wave functions, but they are undetectable) When the particles are close enough to each other so that their wave functions overlap, they loose classical nature and can no longer be explained by M-B statistics. You can treat atoms as classical particles and use M-B stat (e.g. in kinetic theory of gas). But this has limited applications i.e. cannot explain the full characteristics of the particles. To get realistic results you have to consider their quantum nature. Wien's theory and Rayleigh-Jeans theory could explain only part of the black body radiation (as they didn't assume the quantum nature of radiation), but Planck's theory could as he used quantum radiation.

• Thanks, but you didn't answer the second part of my question. Which statistics would you use for hydrogen atoms, and why? Do we consider QM effects for distinguishable H atoms? (Two H atoms can be distinguished on basis of whether they are para or ortho...) – user139621 Sep 19 '17 at 7:25
• If the H atoms are distinguishable then you don't need to consider QM effect as H atoms will be distinguishable only if they are far apart. But if they are in close proximity then QM effects has to taken into consideration. e.g. in Benzene ring H atoms are not indistinguishable and that is why consider the resonance effect. I think that ortho or para thing is actually about the relative position of H atoms rather than which atom is in which position. If you think of two identical tennis balls you can still distinguish them due to their size but H atoms are too small to be distinguished. – Samapan Bhadury Sep 19 '17 at 7:39
• Thanks. By the way I still have some confusion. Would two or more molecules of $H_2, N_2, O_2$ etc be distinguishable or indistinguishable ? – user139621 Sep 19 '17 at 7:51