# Physical implications behind the exchange antisymmetry condition of fermions

Explain the Physical implications behind the exchange antisymmetry condition of fermions. This condition forms the basis of the pauli principle but I can't find/understand what happens physically that requires then the presence of a minus sign upon particle interchange.

• Pauli exclusion principle -> antisymmetry of fermionic wavefunction -> spin statistics theorem in quantum field theory Commented Sep 20, 2013 at 3:56
• found this: physics.stackexchange.com/q/4049 seems like people prefer to skip this question. maybe it's not supposed to be asked this way. comment on the previous comment: spin stats theorem -> ? Commented Sep 20, 2013 at 4:01
• It's not at all obvious that you should get this minus sign for identical fermions: this is a very nontrivial result of quantum field theory that doesn't really have a simple explanation. (Essentially, if you try to write down a quantum field theory for identical fermions with a plus sign under exchange you find the energy is unbounded below so the system is unstable). The Pauli exclusion principle is very important though: matter is stable and doesn't collapse because the Pauli exclusion principle prevents fermions (like electrons and protons) from occupying the same state. Commented Sep 20, 2013 at 4:05
• "spin stats theorem -> ? " In QFT, the spin statistical theorem is based physical considerations, such as the energy spectrum bound from below (quantizing Dirac field) and causality (quantizing Klein-Gordon field). I don't know if there is any deeper origin. Commented Sep 20, 2013 at 5:02
• In quantum field theory, if you write the hamiltonian for a fermionic field (for instance a Dirac field), you will find something like $H = \sum_{k,s} k^o (b^+_{k,s}b_{k,s} - d_{k,s}d^+_{k,s})$ ($b$ concerns particles and $d$ concerns anti-particles). But this hamiltonian has to be bounded below, and you have to choose anti-commutation relations, to have $H = \sum_{k,s} k^o(b^+_{k,s}b_{k,s} + d^+_{k,s} d_{k,s})$, up to a (infinite) constant. Commented Sep 20, 2013 at 8:40