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I read about Bose-Einstein condensate consist of bosonic atoms at incredible low temperature do not obey Pauli exclusion, I am wondering what happens if it is possible to create fermionic photon for example so it obeys Pauli exclusion principle?

Qmechanic
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user6760
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    Comments to the post (v2): Boson is a noun & bosonic is an adjective. Is that what you are asking? What is a fermionic photon? – Qmechanic Jun 10 '20 at 09:28
  • @Qmechanic: like Cooper pair of electrons and I made up the last one to see if the transformation is vice versa. – user6760 Jun 10 '20 at 09:43
  • Although bosonic is an adjective, it has two meanings: 1) something characteristic of s boson, and 2) something that behaves like a boson, but not necessarily a boson – Roger V. Jun 10 '20 at 10:11
  • @Vadim: thanks actually I want to ask about the properties like how come the electrons can behave like boson and ignore the exclusion prinicple especially at extremely low temperature? – user6760 Jun 10 '20 at 10:20
  • Related: https://physics.stackexchange.com/q/81414/2451 , https://physics.stackexchange.com/q/142061/2451 and links therein. – Qmechanic Jun 10 '20 at 10:30
  • @Vadim: I have the doubt that the set of objects that "behave like a boson, but are not necessarily a boson" is actually empty. – fra_pero Jun 10 '20 at 10:30
  • @fra_pero You've read my answer: many quasiparticles are claimed to undergo bose-einstein condensation, although one cannot rigorously call them "bosons": https://en.wikipedia.org/wiki/Bose%E2%80%93Einstein_condensation_of_quasiparticles – Roger V. Jun 10 '20 at 11:26

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Boson and bosonic are a noun and an adjective, such as fermion and fermionic.

To answer your question directly: photon is always a boson, since it has zero spin.

However one uses sometimes term bosonic particle to refer to particles that are really not bosons, but exhibit boson-like behavior. For example, excitons do not really have a well-defined spin, since they are composite exitations of many electrons and holes. Nevertheless, in some conditions that can be viewed as a bound state of a spin-1/2 electron and spin-1/2 hole, in which case they exhibit behavior typical of bosons, such as Bose-Einstein condensation.

Roger V.
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  • Why excitons do not have well-defined spins? Also quasi-particle excitations of Fermi liquids are "composite excitations" of many electrons, and they have perfectly well-defined spin. – fra_pero Jun 10 '20 at 09:43
  • You have a band with N electrons interacting via coulomb interaction. We move one electron to a conduction band (i.e. we change its orbital state) - what is the spin of the whole system? Is it 1/2? or N/2? It is by no means obvious. Hydrogen-like picture of an exciton is an approximation: https://books.google.fr/books/about/Theory_of_Excitons.html?id=GuQ9AQAAIAAJ&redir_esc=y – Roger V. Jun 10 '20 at 10:08
  • My understanding is that Fermi liquid is rather an exception in the sense that the properties of its excitations are so similar to those of the particles composing it. Yet, for the same particles in 1D we have Luttinger liquid with bosonic excitations. – Roger V. Jun 10 '20 at 10:09
  • Of course a fermionic system such as the Fermi liquid may and in general will have both fermionic and bosonic excitations. My concern is: I do not know if your statement "excitons do not really have a well-defined spin" is wrong. I just do not know. I have the doubt. – fra_pero Jun 10 '20 at 10:27
  • I think that the burden of proof should be in the other direction: it seems to me as a non trivial statement that excitations of a many-body system should have a well-defined spin. Particularly taking into account that such excitations are often only well-defined at low densities. – Roger V. Jun 10 '20 at 11:23