Proving Fermion Occupation with Commutation Relations

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The discussion focuses on proving that the number operator for fermions yields eigenvalues of 0 and 1, while for bosons, it can yield any non-negative integer. The participant initially struggles with the concept, mistakenly thinking that the number operator could yield higher values for fermions. They clarify that the proof pertains specifically to the occupation of a single state, where the application of anticommutation relations leads to the conclusion that the eigenvalue must be either 0 or 1. The conversation highlights the importance of understanding the distinction between the total number of particles and the occupancy of a single quantum state. Ultimately, the participant acknowledges their misunderstanding and gains clarity on the topic.
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Homework Statement

prove , using appropriate commutation relations that the number operator yields the values 0 and 1 for fermions , and any non - negative values for bosons.



Homework Equations

the commutation relations for bosons and fermions.



The Attempt at a Solution

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