Pauli Exclusion Princple for a fermion and antifermion I understand that the Pauli Exclusion Principle applies only for identical particles, so that a fermion and an anti-fermion should be allowed to be in the same state.  However, when I look at the arguments for the principle from properties of commutators, it seems to me that it actually should prohibit a fermion and anti-fermion from being in the same state. My reasoning goes like this:
The Pauli Exclusion Principle for fermions can be justified as follows. Based on the fact that
$$\{{a^s_p}^{\dagger}, {a^r_q}^{\dagger}\}=0,$$
we have
$${a^s_p}^{\dagger}{a^s_p}^{\dagger}|0\rangle = -{a^s_p}^{\dagger}{a^s_p}^{\dagger}|0\rangle = 0,$$
so states where two fermions occupy the same state are just the zero vector, and we get the Exclusion Principle for two fermions.
But we also have
$$\{{a^s_p}^{\dagger}, {b^r_q}^{\dagger}\}=0,$$
so
$${a^s_p}^{\dagger}{b^s_p}^{\dagger}|0\rangle = -{a^s_p}^{\dagger}{b^s_p}^{\dagger}|0\rangle = 0$$
which by the same argument seems to imply that a fermion and an anti-fermion cannot be in the same state.
I'd appreciate if somebody could point out the flaw here.
 A: The last equation in the question is incorrect.  The creation operators for fermions and antifermions anticommute, which means that we have
$${a^s_p}^{\dagger}{b^s_p}^{\dagger}|0\rangle = -{b^s_p}^{\dagger}{a^s_p}^{\dagger}|0\rangle.$$
In fact, this relation is just a special case of what you would have with arbitrary creation operators.  A state with a single fermion and single antifermion is
$${a^s_p}^{\dagger}{b^{s’}_{p’}}^{\dagger}|0\rangle = -{b^{s’}_{p’}}^{\dagger}{a^s_p}^{\dagger}|0\rangle.$$
What you are missing is that a multi-fermion state vector is not just described by the occupation numbers of the fermion and anti-fermion modes; the order in which the occupied states are listed also affects the state vector.  The two kets, ${a^s_p}^{\dagger}{b^{s’}_{p’}}^{\dagger}|0\rangle$ and ${b^{s’}_{p’}}^{\dagger}{a^s_p}^{\dagger}|0\rangle$, represent the same ray in the Hilbert space, but they are not identical; they differ by a phase factor of $-1$.  The order of the operators tells you which mode excitation is created first, and switching the order of the creation operators changes the overall sign of the state.
The is no different from what you have when you create a state with two fermions in different modes,
$${a^s_p}^{\dagger}{a^{s’}_{p’}}^{\dagger}|0\rangle = -{a^{s’}_{p’}}^{\dagger}{a^s_p}^{\dagger}|0\rangle.$$
The state vector is nonzero unless the modes coincide (that is, ${a^s_p}^{\dagger}{a^{s’}_{p’}}^{\dagger}|0\rangle\neq 0$ unless $s=s’$ and $p=p’$).  But the order in which you apply the creation operators determines the sign.  This is, in fact, just what you need in order to have the two-particle wave function $\psi(x_{1},x_{2})$ satisfy $\psi(x_{1},x_{2})=-\psi(x_{2},x_{1})$.
