The Pauli exclusion principle and the Pfaffian

Question

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:

$$\Psi(x_1,x_2,x_3,x_4)\\=\sin(x_1-x_2)\sin(x_3-x_4)-\sin(x_1-x_3)\sin(x_2-x_4)+\sin(x_1-x_4)\sin(x_2-x_3)$$

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?

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 $$ \begin{align} \Psi_4 &= g_{34}(g_{13}h_{23} - h_{13}g_{23}) - g_{13}(g_{23}h_{34}+h_{23}g_{34})+g_{14}g_{23} \\&= g_{34}(g_{13}h_{23} - g_{13}h_{23} - h_{13}g_{23}) + g_{23}(g_{14}-g_{13}h_{34})\\&= g_{23}(g_{14}-g_{13}h_{34}-h_{13}g_{34}) = \\&= g_{23}(g_{14}-g_{14}) = 0\,. \end{align} $$