QM: Clarifying Pauli Exclusion Principle, State of System

[Maybe_Memorie]

Just want to clarify my understanding of these things.

Fermions obey the Pauli Exclusion Principal, meaning meaning only one particle can occupy a single state. This means the wave function is anti-symmetric under particle exchange. That's the part that isn't making much sense. Is it because if we have two particles designated by quantum numbers n, l, m₁ and n, l, m₂ and we swap the particles the only way for the states to be different is if m₁ = -m₂?Also, when we speak of the state of a system...
Consider the basic problem of a spin 1/2 particle in a constant magnetic field (0, 0, B).
H = γS_z , γ = geB/2mc, or something like that..

We write H|E> = E|E>
H²|E> = E²|E>
From this we determine the eigenvalues E₁ and E₂, and the state at some time t is given by

|ψ> = C₊e^aE₁t|E₁> + C_-e^aE₂t|E₂>What does this actually mean? Is it the energy at some later time t? The spin?

 

[vanhees71]

The logics concerning Fermions is somewhat the other way. The point is that identical particles are indistinguishable, which means that under exchange of any two particles in an N-particle state the wave function must not change (except through a phase factor). As it turns out by some complicated arguments, if space has 3 (or more) dimensions there can only be two realizations of this principle, namely either the wave function is totally symmetric under interchange of particles or totally antisymmetric. The corresponding particles are called bosons or fermions, respectively.

For the fermions with antisymmetric wave functions, it's clear that the interchange of two particles that are in the same one-particle state doesn't change the wave function at all, but according to the rule for fermions it must get an additional minus sign. That means the wave function \psi=-\psi, but this can only be true if \psi=0. Thus, there doesn't exist a state, where two fermions occupy the same one-particle state.

 

[Maybe_Memorie]

The logics concerning Fermions is somewhat the other way. The point is that identical particles are indistinguishable, which means that under exchange of any two particles in an N-particle state the wave function must not change (except through a phase factor). As it turns out by some complicated arguments, if space has 3 (or more) dimensions there can only be two realizations of this principle, namely either the wave function is totally symmetric under interchange of particles or totally antisymmetric. The corresponding particles are called bosons or fermions, respectively.

For the fermions with antisymmetric wave functions, it's clear that the interchange of two particles that are in the same one-particle state doesn't change the wave function at all, but according to the rule for fermions it must get an additional minus sign. That means the wave function \psi=-\psi, but this can only be true if \psi=0. Thus, there doesn't exist a state, where two fermions occupy the same one-particle state.

That explanation was perfect, thanks!