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I'm reading the Wikipedia article about Fermionic field and have some troubles to understand the meaning following phrase:

We impose an anticommutator relation (as opposed to a commutation relation as we do for the bosonic field) in order to make the operators compatible with Fermi–Dirac statistics.

I understand that the (anti)commutator relations for operatorys are posed as algebraic conditions in the operator algebra with "composition" as binary relation there. So up to now that's a purely "algebraic data".

But what does it mean that the operators should be compatible with Fermi–Dirac statistics? Compatible in which sense? I not understand in what is the concrete connection between the obtained the operators obtained from second quantization procedure and the Fermi–Dirac statistic, in mathematical terms a distribution. How these concepts fit together?

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My understanding is that the anticommutator relationship shows that the many-particle wave function is asymmetric relative to particle exchange, which is related to the Pauli exclusion principle and also half-integer spin, which, in turn, defines Fermi-Dirac statistic, where the maximal value can't exceed one, unlike in the case of the Bose-Einstein statistics.

The spin statistic theorem talks about this, you can find it here.

To answer your question, they are compatible in the sense that two particles can't occupy the same quantum state.

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  • $\begingroup$ I'm not completely sure if I understand you first statement correctly. Say we start with a many-particle wave function and want to turn it into an operator via second quantization. At this stage we a priori "don't know yet" from naive point of view, if we should impose commutator or anticommutator rule. Should your statement be read as that if we would be "somehow" able to figure out, that the only possible relation we can impose for the quantized field must be anticommutative, $\endgroup$
    – user267839
    Commented Jul 10, 2023 at 15:53
  • $\begingroup$ then we would be able to deduce that the wave function we initially started with should have been asymmetric wrt particle exchange? Do I understand your statement correctly? $\endgroup$
    – user267839
    Commented Jul 10, 2023 at 15:53
  • $\begingroup$ If many many-particle wave function of indistinguishable particles is given (you say you start with it), it possesses certain symmetry relative to particle exchange in it. This will affect the commutation rules for the second-quantization operators. $\endgroup$
    – freude
    Commented Jul 11, 2023 at 2:29
  • $\begingroup$ In other words, to build a many-body theory of particles you need to know what those particles are in terms of spin and mass and how they interact. $\endgroup$
    – freude
    Commented Jul 11, 2023 at 2:31
  • $\begingroup$ You may ask why particles with half-spin have to have asymmetrical wave function relative to particle exchange, but this is another question. It has something to do with the phase of the wave function and rotations I believe. What we have established here is the commutator rules are governed by the symmetry of the wave function relative to particle exchange and also this symmetry defines what statistics should be. $\endgroup$
    – freude
    Commented Jul 11, 2023 at 2:33

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