Proving Fermion Occupation with Commutation Relations

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SUMMARY

The discussion focuses on proving that the number operator yields eigenvalues of 0 and 1 for fermions, while allowing non-negative values for bosons. Participants clarify the application of commutation relations, specifically noting that the number operator for fermions must consider anticommutation relations due to the Pauli exclusion principle. The confusion arises from the interpretation of the number of fermions in a single state, leading to the realization that only one fermion can occupy a given quantum state.

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  • Understanding of quantum mechanics principles, particularly fermions and bosons.
  • Familiarity with commutation and anticommutation relations.
  • Knowledge of the number operator in quantum field theory.
  • Basic concepts of eigenvalues and eigenstates in quantum systems.
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  • Study the implications of the Pauli exclusion principle on fermionic systems.
  • Explore the mathematical formulation of anticommutation relations in quantum mechanics.
  • Learn about the number operator's role in quantum field theory for both fermions and bosons.
  • Investigate examples of fermionic systems and their occupation numbers in quantum states.
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Students and researchers in quantum mechanics, particularly those studying quantum field theory, as well as physicists interested in the behavior of fermions and bosons in various quantum states.

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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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