# $J_1$-$J_2$ Heisenberg antiferromagnet

In this paper, the authors solve for the excitation spectrum in a $J_1$-$J_2$ Heisenberg antiferromagnet using the modified spin-wave theory in the Dyson-Maleev representation. As an intermediate step, the authors computed the neighboring-spin correlators $\left<\vec{S}_i\cdot \vec{S}_j\right>$, as shown in Eq. (4).

Since the quantization axes of neighboring spins are separated by an arbitrary angle $\phi_{ij}$, it seems that the spin correlators will always contain terms of the sixth power in the boson operators. However, such terms do not generate the quadratic form of Eq. (4). What am I missing here?

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Suppose you rotate your quantization frame by a angle $\phi$ about the $z$-axis, the spin operators will become: $$S^x \rightarrow S'^x=\cos\phi S^x+\sin\phi S^y,\\ S^y \rightarrow S'^y=-\sin\phi S^x+\cos\phi S^y,\\ S^z \rightarrow S'^z=S^z.$$ The correlation function $\langle \vec S_i\cdot \vec S_j\rangle$ is in fact $\langle \vec S_i\cdot \vec S'_j\rangle$, thus we have $$\vec S_i\cdot \vec S'_j=\cos\phi(S_i^x S_j^x+S_i^y S_j^y)+\sin\phi(S_i^x S_j^y-S_i^y S_j^x)+S_i^z S_j^z\\ =\frac{1}{2}e^{i\phi}S_i^+ S_j^- + \frac{1}{2}e^{-i\phi}S_i^- S_j^+ +S_i^z S_j^z.$$ Because the formula has only $S^+S^-$ and $S^-S^+$ terms (no $S^+S^+$ term since $[S_i^z+S_j^z,\vec S\cdot \vec S']=0$). Substitute the Dyson-Maleev representation of spin, you find at most four boson operator terms.
I believe we do get an $S^+S^+$ term if we rotate the quantization frame about the $y$-axis instead. –  leongz Nov 24 '12 at 1:10
Yes, you are right. I've corrected the reason for the absence of $S^+ S^+$ term. If we rotate about the $z$-axis, $[S_i^z+S_j^z,\vec S_i\cdot\vec S_j]=0$, thus there is no $S^+ S^+$ term. If rotate about the $y$-axis, total $S^z$ is not conserved, we do get $S^+ S^+$ term. But in this paper, the Neel ordered spins lie in the $xy$ plane. That's why we get only fourth power terms in boson operators I suppose. –  Tengen Nov 24 '12 at 5:58
In the paper, the quantization axes of neighboring spins are separated by an angle $\phi$, so the rotation is not about the $z$-axis. Even though the axes may be in the $xy$ plane, we will still get an $S^+S^+$ term. –  leongz Dec 5 '12 at 0:20