# Why magnon band is half filled in XY model?

This is my first contact with that field so I hope that I describe my problem clearly.

In the notes I follow we consider a 1-dimensional XXZ problem with $$$$\hat{H}_{XXZ}=-J_{\perp}\sum_{j}\left(\hat{S}^{x}_{j+1}\hat{S}^{x}_{j}+\hat{S}^{y}_{j+1}\hat{S}^{y}_{j}\right)-J_{z}\sum_{j}\hat{S}^{z}_{j+1}\hat{S}^{z}_{j}$$$$ which by Jordan-Wigner transformation could be written in the form $$$$\hat{H}_{XXZ}=\sum_{k}\omega_{k}\hat{c}^{\dagger}_{k}\hat{c}_{k}-\frac{J_{z}}{L}\sum_{k,k',q}\cos(q)\hat{c}^{\dagger}_{k-q}\hat{c}^{\dagger}_{k'+q}\hat{c}_{k'}\hat{c}_{k}$$$$ where $$\hat{c}^{\dagger}_{k}$$ is the creation operator of Jordan-Wigner fermion, L the size of the system and $$\omega_{k}=J_{z}-J_{\perp}\cos(k)$$ describes the excitation spectrum.

To my understanding, when we assume that $$J_{z}=J_{\perp}>0$$ the model reduces to the Heisenberg model that obeys $$SO(3)$$ continuous symmetry which ground states have all the spins set in one direction. The model breaks that symmetry by choosing one of these directions and therefore the ground state is fully magnetised. Therefore the gapless, continuous excitation $$\omega_{k}=J(1-\cos(k))$$ describes the Goldstone modes which we interpret as magnetons. Rewriting the first summand of the Hamiltonian in terms of $$\hat{S}^{+}_{j}\hat{S}^{-}_{j+1}$$ we see that $$$$|k\rangle=\sum_{n=1}^N e^{ikn}|...,\uparrow,\downarrow_n,\uparrow,...\rangle,$$$$ is its eigenstate, whereas the second term in the Hamiltonian corresponds to the interaction between magnetons. Is that a correct viewing of the problem?

Now comes the question. In the notes I follow one now takes $$J_{z}=0$$ therefore the problem reduces to the XY model. Thus the excitation spectrum takes the form $$\omega_{k}=-J\cos(k)$$. Then one states that the magnon band is half-hilled i.e. the ground state of the system is $$$$|GS_{XY}\rangle=\prod_{|k|<\frac{\pi}{2}}\hat{c}^{\dagger}_{k}|0\rangle$$$$ and thus the average magnetisation is zero.

While the second step is easy, one calculates the expected value of $$\hat{S}^{z}_{j}$$ on the ground state which indeed is zero, I don't understand at all the first one. What does it even mean that the magnon band is half-field and what is the reason for that? XY model also has continuous symmetry, so I would anticipate that it will also have a Goldstone mode is it that one? Why? And what are possible excitations of that mode?

• $\omega_k$ 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. Commented Sep 9, 2022 at 19:47
• 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? Commented Sep 9, 2022 at 20:25
• 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. Commented Sep 9, 2022 at 23:33
• I see, thank you! Commented Sep 10, 2022 at 8:02