I was reading the following article Fermion FIelds and discovered the following passage not fully explained to me :

It is these anticommutation relations that imply Fermi–Dirac statistics for the field quanta. They also result in the Pauli exclusion principle: two fermionic particles cannot occupy the same state at the same time.

What is the proof that the anticommutation relations of the Fermion Field gives rise to the Pauli Exclusion Principle?

  • 1
    $\begingroup$ It's called the spin-statistic theorem. The Pauli principle simply arises because, if the wavefunction(al) is anti-symmetric under swap and the two fermions are in the same state, then swapping changes nothing and the wavefunction(al) must have been zero already. $\endgroup$
    – ACuriousMind
    Jul 24, 2014 at 16:52

1 Answer 1


Keeping it simple, let's asume that $\psi(a)$ creates a particle in the state $a$ (i.e., characterized by some collection of quantum numbers that we call $a$),

$$ \psi(a)|0\rangle=|a\rangle .$$

and $\psi(b)$ does the same for $b$. We can create a state with two particles:

$$ \psi(b)\psi(a)|0\rangle = \psi(b)|a\rangle = |a;b\rangle $$

$$ \psi(a)\psi(b)|0\rangle = \psi(a)|b\rangle = |b;a\rangle $$

Since $\psi$ is anti-commutative,

$$ \psi(a)\psi(b) + \psi(b)\psi(a) = 0 .$$


$$ |a;b\rangle = -|b;a\rangle, $$

that is, the state is antisymmetric under particle change, it has fermionic statistic.

In particular, if $b=a$,

$$ \psi(a)\psi(a) + \psi(a)\psi(a) = 2\psi(a)\psi(a) = 0$$


$$ |a;a\rangle=0 .$$

This is Pauli's Exclusion Principle.

  • $\begingroup$ Thanks for the answer. I am stuck with understanding how to compute observables from a wavefunction of two fermions using the expressions provided above. For example how do you compute energy, momentum, and position? Thanks. $\endgroup$ Jul 24, 2014 at 20:35
  • $\begingroup$ I think you should open another post for that question, because of you are not familiar with QFT the answer could be quite large. $\endgroup$
    – dpravos
    Jul 24, 2014 at 21:04

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.