# Fermion occupation (1 Viewer)

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#### commutator

1. The problem statement, all variables and given/known data prove , using appropriate commutation relations that the number operator yields the values 0 and 1 for fermions , and any non - negative values for bosons.

2. Relevant equations the commutation relations for bosons and fermions.

3. The attempt at a solutionthe boson case is solved.
the fermions-i can understand this thing intuitively ,because on applying the number operator on a state containing n fermions in a single state , and then using commutation relation i am getting (-n) as eigenvalue which is absurd. but i do not know how to give a more compact proof.

The problem is not precisely stated. The number operator for fermions can as well have eigenvalue 100 - can't you have 100 of electrons? Of course you can. Probably what you mean is the number of fermions in a given state. In that case you indeed use anticommutation relations, but you end up with the eigenstate being equal to its minus, not with the eigenvalue equal to its minus!

#### commutator

yes. i meant single state.
i have understood my mistake. thanks.

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