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Let say we consider two distinguishable fermions(bi-fermions) in compact form. The case when both fermions are existing as free fermions, they will obey Pauli exclusion principle. In other case if pair of fermions are maximally entangled and making composite boson similar to pure boson. More than one composite boson can stay in same energy state. But what about the case when pair of fermions is making a composite boson but the properties of this composite boson are deviating from pure boson. And composite boson has less degree of compositeness. Will fermions obey pauli exclusion principle or not? Will more than one composite boson can stay in the same energy state or not in this case?

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  • $\begingroup$ Distinguishable particles are not subject to Pauli exclusion. $\endgroup$ – probably_someone Jan 23 at 17:49
  • $\begingroup$ In single composite boson we are taking distinguishable fermions. But if we consider N composite bosons they are identical. And fermions of two composite bosons will oppose each other because of pauli exclusion principle , in case when composite boson is deviating from pure boson @probably_someone $\endgroup$ – Tooba Jan 23 at 17:53
  • $\begingroup$ My comment was in relation to this: "The case when both fermions are existing as free fermions, they will obey Pauli exclusion principle." $\endgroup$ – probably_someone Jan 23 at 17:54
  • $\begingroup$ Ya, but what in case when we have N identical composite bosons but these composite bosons has less compositeness compare to pure bosons. @probably_someone $\endgroup$ – Tooba Jan 23 at 18:05
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    $\begingroup$ I encourage you to read the answer in the link I posted. $\endgroup$ – probably_someone Jan 23 at 18:31

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