lueffy said:
Why fermion has anti-symmetric wave function? I've read that this is related to the matter of indistinguishability of particles, but if we can assign certain quantum numbers to an electron, we can say that they're distinguishable, can't we?
(Pardon me for jumping in the middle of this thread, but all i can say is that I'm interested in MiGUi's statement that fermion has an anti-symmetrical wave function, and i can't find out why...)
Cause fermions are 'defined' with the antisymmetry condition for its wavefunction and the bosons with the symmetric one.
After, the spin-statistics theorem proofs that particles with an integer spin have to obey Bose-Einstein's statistic and particles with semi-integer spin have to obey Fermi-Dirac's one.
A wavefunction is symmetric if changing the position of two particles, the wavefunction remains the same. And it is antisymmetric if doing that, the wavefunction is the same but with the sign changed. If you link this with the indistinguishability of particles, you have the Pauli's Exclusion Principle.
Mathematically, indistinguishability of particles means this:
\Psi(\xi_1, \ldots, \xi_i, \ldots, \xi_j, \ldots, \xi_n) = \Psi(\xi_1, \ldots, \xi_j, \ldots, \xi_i, \ldots, \xi_n)
And symmetric and antisymmetric condition is:
\Psi_{boson} (\xi_1, \ldots, \xi_i, \ldots, \xi_j, \ldots, \xi_n) = \Psi_{boson} (\xi_1, \ldots, \xi_j, \ldots, \xi_i, \ldots, \xi_n)
\Psi_{fermion} (\xi_1, \ldots, \xi_i, \ldots, \xi_j, \ldots, \xi_n) = - \Psi_{fermion} (\xi_1, \ldots, \xi_j, \ldots, \xi_i, \ldots, \xi_n)
If you link the first condition with the third condition, you see that the wavefunction is equal to 0. This means that particle i and particle j can't be in the same individual state.
MiGUi