Timeline for The Pauli exclusion principle and the Pfaffian
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Dec 16, 2018 at 6:00 | history | tweeted | twitter.com/StackPhysics/status/1074182346906263552 | ||
| Dec 16, 2018 at 5:13 | vote | accept | Jian | ||
| Dec 16, 2018 at 5:06 | answer | added | MannyC | timeline score: 3 | |
| Dec 16, 2018 at 5:04 | history | edited | Jian | CC BY-SA 4.0 |
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| Dec 16, 2018 at 4:59 | comment | added | Jian | 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. | |
| Dec 16, 2018 at 4:54 | comment | added | Jian | 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 | |
| Dec 16, 2018 at 4:43 | comment | added | ZeroTheHero | Maybe some additional information about Pfaffians would be welcomed. | |
| Dec 16, 2018 at 4:27 | comment | added | MannyC | 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. | |
| Dec 16, 2018 at 4:18 | comment | added | anna v | " spinless fermion " ???? fermions by definition have spin 1/2 those are the ones obeying pauli exclusion. spinless=boson , no pauli exclusion . | |
| Dec 16, 2018 at 0:40 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
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| Dec 16, 2018 at 0:23 | history | edited | Jian | CC BY-SA 4.0 |
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| Dec 16, 2018 at 0:16 | history | edited | Jian | CC BY-SA 4.0 |
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| Dec 16, 2018 at 0:07 | history | asked | Jian | CC BY-SA 4.0 |