4
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ " 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 '18 at 4:18
  • $\begingroup$ 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 '18 at 4:27
  • $\begingroup$ Maybe some additional information about Pfaffians would be welcomed. $\endgroup$ – ZeroTheHero Dec 16 '18 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 $\endgroup$ – wwwjjj Dec 16 '18 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. $\endgroup$ – wwwjjj Dec 16 '18 at 4:59
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.