Timeline for Why magnon band is half filled in XY model?
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| when toggle format | what | by | license | comment | |
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| Sep 10, 2022 at 8:02 | comment | added | Paweł Korzeb | I see, thank you! | |
| Sep 9, 2022 at 23:33 | comment | added | Meng Cheng | Applying the creation operator $c_k^\dagger$ changes the total $S^z$ by $1$, so in this sense it is like a magnon. I don't think there is a simple picture for the ground state. In fact, the wavefunction you wrote down in terms of $c_k^\dagger$'s is the simplest way to think about the wavefunction. It is a lot more complicated than the paramagnetic state of the Ising model. | |
| Sep 9, 2022 at 20:25 | comment | added | Paweł Korzeb | Thanks for the answer, that helps a lot. Could you explain why one could think of the creation of a Jordan-Wigner fermion as the creation of a single-magnon? And what image should one have in mind when thinking of the ground state of that system? Is it somehow comparable with e.g. the paramagnetic state of the Ising model? | |
| Sep 9, 2022 at 19:47 | comment | added | Meng Cheng | $\omega_k$ is the energy of single-magnon excitation. Think of $c_k$ roughly as the annihilation operator of a magnon (not quite, since it is really a Jordan-Wigner fermion, but this does not affect what follows). The lowest energy state is obtained by filling all negative energy modes (those with $\omega_k>0$, if $J_\perp>0$), which are precisely half of the k points, thus half-filled. The magnon is gapless (expand $\cos k$ around $k=\pm \pi/2$), and that is kind of like the Goldstone mode/spin wave. More precisely it is a Luttinger liquid. | |
| Sep 9, 2022 at 19:34 | history | asked | Paweł Korzeb | CC BY-SA 4.0 |