# The Pauli exclusion principle and the Pfaffian

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?

• " spinless fermion " ???? fermions by definition have spin 1/2 those are the ones obeying pauli exclusion. spinless=boson , no pauli exclusion . – anna v Dec 16 '18 at 4:18
• 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. – MannyC Dec 16 '18 at 4:27
• Maybe some additional information about Pfaffians would be welcomed. – ZeroTheHero Dec 16 '18 at 4:43
• 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 – wwwjjj Dec 16 '18 at 4:54
• 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. – wwwjjj Dec 16 '18 at 4:59

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}