We are talking about spinless fermion many-body wavefunctions.

The determinant is a very nice structure for the Pauli exclusion principle, this is because when two single-particle states are the same, the many-body wave function will become zero automatically. We start from a completely orthogonal set of single-particle state to construct Slater determinant.

However, I am not very clear about Pfaffian. Because we use the two-particle wave function as the building block. And we only use one building block.

For example:

Suppose we have the two-particles wavefunction $ g(x_1,x_2)=e^{ik_0 x_1} e^{-ik_0 x_2}-e^{ik_0 x_2} e^{-ik_0 x_1} = \sin(k_0 (x_1 -x_2)) $

to save typing, let set $k_0=1$ and $g_{12}=g(x_1,x_2)$

$g_{12}=\sin(x_1-x_2)$ means we have two particles occupying $k_0=\pm 1$ states.

Now, let's use Pfaffian to construct a 4-body wavefunction:


Or the shorter notation:

$\Psi_4:=g_{12}g_{34}-g_{13}g_{24}+ g_{14}g_{23}:= \textbf{Pf}[g_{ij}]$

My question, we have only two single-particle states $k_0=\pm 1$, but there are 4 particles. Is it contradict with the Pauli exclusion principle?

  • $\begingroup$ " spinless fermion " ???? fermions by definition have spin 1/2 those are the ones obeying pauli exclusion. spinless=boson , no pauli exclusion . $\endgroup$
    – anna v
    Dec 16, 2018 at 4:18
  • 1
    $\begingroup$ No that's ok, the spin-statistic theorem applies to theories with Lorentz invariance. In a generic quantum many body hamiltonian we don't even need to have the "spin" as a quantum number. By fermion he just means that those are anticommuting degrees of freedom. $\endgroup$
    – MannyC
    Dec 16, 2018 at 4:27
  • $\begingroup$ Maybe some additional information about Pfaffians would be welcomed. $\endgroup$ Dec 16, 2018 at 4:43
  • $\begingroup$ Pfaffian is similar to the determinant, they are both summations over permutations. But Pfaffian also takes care of the "pairing" structure. They both serve as a tool to anti-symmetrize many-body wave functions. en.wikipedia.org/wiki/Pfaffian $\endgroup$
    – Jian
    Dec 16, 2018 at 4:54
  • $\begingroup$ To simplify the problem, I use "spinless fermion". My question remains the same if you add the spin, $ \{k_0=\pm 1 \} \otimes \{ s=\uparrow ,\downarrow \}$ then looking at eight-bodies wavefunction $\Psi_8=\textbf{Pf}[g_{ij}]$ , 4 states but 8 particles. Of course, you can always add more particles. $\endgroup$
    – Jian
    Dec 16, 2018 at 4:59

1 Answer 1


It is not a contradiction because the Pfaffian you computed vanishes. Let me call $x_{ij} = x_i-x_j$. For $k \neq i$ or $j$ we have $$ \sin(x_{ij}) = \sin(x_{ik}+x_{kj}) = \sin(x_{ik})\cos(x_{kj}) + \cos(x_{ik})\sin(x_{kj}) $$ Building on OP's notation let me further define $h_{ij} = \cos(x_{ij})$. The Pfaffian reads $$ \begin{align} \Psi_4 &= g_{34}(g_{13}h_{23} - h_{13}g_{23}) - g_{13}(g_{23}h_{34}+h_{23}g_{34})+g_{14}g_{23} \\&= g_{34}(g_{13}h_{23} - g_{13}h_{23} - h_{13}g_{23}) + g_{23}(g_{14}-g_{13}h_{34})\\&= g_{23}(g_{14}-g_{13}h_{34}-h_{13}g_{34}) = \\&= g_{23}(g_{14}-g_{14}) = 0\,. \end{align} $$


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