Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer

I don't think there are six boson operator terms.

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.

share|improve this answer
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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.