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.

  • 2
    $\begingroup$ Pauli exclusion principle -> antisymmetry of fermionic wavefunction -> spin statistics theorem in quantum field theory $\endgroup$
    – user26143
    Sep 20, 2013 at 3:56
  • $\begingroup$ 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 -> ? $\endgroup$
    – Hasan
    Sep 20, 2013 at 4:01
  • 2
    $\begingroup$ 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. $\endgroup$
    – Andrew
    Sep 20, 2013 at 4:05
  • $\begingroup$ "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. $\endgroup$
    – user26143
    Sep 20, 2013 at 5:02
  • 3
    $\begingroup$ 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. $\endgroup$
    – Trimok
    Sep 20, 2013 at 8:40

1 Answer 1


To perform canonical quantization of a fermion field, we write the field in creation and annihilation operators. Writing the Hamiltonian in those creation and annihilation operators, we find that the energy of a field is unbounded from below (it can be as negative as you like).

This would be a disaster; a field could forever decay to lower energy states by e.g. the emission of a photon. That is not what we see.

If, however, we insist that the field obeys an anti-commutation rule (the Pauli exclusion principle), the energy is bounded from below (cannot be as small as you like). The situation is saved.

To summarise: physically, fermions must obey the Pauli exclusion principle, because if they did not, they could forever decay to lower energy states. For detail and the mathematics, see any introductory book on quantum field theory.

  • $\begingroup$ Is there a proof for this? Can you point me to a source where this is shown explicitly? $\endgroup$
    – SuperCiocia
    Apr 13, 2015 at 13:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.