Can we interpret the spin wave dispersion as the allowed (or required) energy at a given momentum? In condensed matter or solid state physics, we usually meet the dispersion which is a function $\omega(\vec{k})$, for example the spin wave of the antiferromagnetic Heisenberg model in the square lattice has the dispersion $\omega=\sqrt{1-\gamma(\vec{k})}$, $\gamma(\vec{k})=(\cos{k_x}+\cos{k_y})/2$, as in the figure below (PRB 72,014403). It tells us that the energy is different with different momentum.
Question: 1. how to understand the dispersion?
For example, $(\pi,\pi)$ is gapless, can we say that the wave (excitation) along this (diagonal) direction needs no energy (since it is compatible with this (neel) order pattern)? And although $\vec{k}_1=(\pi/2,\pi/2)$ is also along the same direction, but the wave length do not match the order pattern, for example $exp(i\vec{k}_{1}\cdot \vec{R}_{1})=-1$ at $\vec{R}_{1}=(1,1)$, so it needs finite energy. And also $k_{2}=(\pi,0)$, $exp(i\vec{k}_{2}\cdot\vec{R}_{2})=1$ at $\vec{R}_{2}=(0,1)$, so this momentum has finite energy. But then why $(0,0)$ is gapless, only because it is the long wave length?


*For $(\vec{k},0)$, can we say the wave propagates along x direction should cost $\omega(\vec{k})$ energy?

 A: No, these kinds of graphs are supposed to be understood differently, although I agree it is not the most transparent representation of the SW dispersion. First, note that the horizontal axis is in the k-space, where $\pi$ in fact means the right end of the Brillouin zone, i.e. the largest available k-vector in the periodic lattice. Then, you have anisotropy of the SW propagation, so their dispersion is different in different directions of $\vec{k}$.
Now, look at the inset with the arrows. Let's say we are interested in the dispersion of the SWs propagating in the "$x$-direction", that is, SWs with $\vec{k}=(k_x,0,0)$. In the momentum space, this corresponds to the direction from the point $(0,0)$ toward the point $(\pi,0)$. Now, in the main graph, you look at the region between these two points (first section from left to right). Within this region, $(0,0)$ corresponds to $k_x=0$, and $(\pi,0)$ to  $k_x=\pi$ (in the units of $a^{-1}$). You see that in this direction, the SW dispersion is quadratic at small $k$ and more flat at large $k$, just as can be expected.
Other regions of the main graph should be understood in the same way, and arrows in the inset indicate in which direction the SW propagation is considered. For example, it might be argued from the data that the SWs along the direction from  $(\pi/2,\pi/2)$ to $(\pi,\pi)$ (3rd section) has somewhat larger stiffness that the one in the 1st section (from $(0,0)$ to $(\pi,0)$). The flat part of the dispersion between  $(\pi/2,\pi/2)$ and $(\pi,0)$ thus means that in that direction, the stiffness is very small. I'm not 100% sure, but it looks like in that direction you have no AFM periodicity in the Heisenberg model, so the SWs are of a FM nature, and then obviously the energy is way smaller.
I agree it is confusing if you are not used to these kinds of plots. Yet, this is a very concise way of representing calculated dispersion, and quite commonly used by theoreticians, not just for SWs but other excitations too.
