# How to understand the elementary excitation of spin density wave?

From Ch.4 of Interacting electrons and quantum magnetism by Auerbach, the elementary excitation of spin density wave can expressed as: $$\alpha^\dagger_{k+}=\cos\theta_kc^\dagger_{k\uparrow}+\sin\theta_kc^\dagger_{k+q\downarrow}$$ $$\alpha^\dagger_{k-}=-\sin\theta_kc^\dagger_{k\uparrow}+\cos\theta_kc^\dagger_{k+q\downarrow}$$

But I cannot understand the motivation of such transformation and cannot relate this excitation operator with its boson operator: $$\rho_{sq\alpha}=\sum_ka^\dagger_{s(k+q)\alpha}a_{sk\alpha},\alpha=\uparrow,\downarrow$$.

What's more, I know the state after above transformation has a "wave" behavior in x-y plane, i.e. $$\langle S^+_i\rangle=m_qe^{-iqx_i}$$: But the picture of spin wave is also similar: So, I am confused that what the difference between this two picture?

That said, Auerbach introduces that transformation without much ado, so you'll have to look elsewhere in the literature for a motivation. Personally, I think it's described quite clearly in chapter 2 of the book "Quantum Theory of the Electron Liquid" by Giuliani and Vignale. The basic idea is to to do a Hartree-Fock decoupling of the Hubbard model (or a more general theory), and consider a simplified, non-interacting mean-field problem. The resulting Hamiltonian can be rewritten $$\hat{H}_{HF} = \sum_\vec{k} \epsilon_{\vec{k}+\vec{q},\downarrow} c^\dagger_{\vec{k}+\vec{q},\downarrow}c_{\vec{k}+\vec{q},\downarrow} + \sum_\vec{k} \epsilon_{\vec{k},\uparrow} c^\dagger_{\vec{k},\uparrow}c_{\vec{k},\uparrow} + \sum_\vec{k} g_\vec{k} \left[ c^\dagger_{\vec{k}+\vec{q},\downarrow}c_{\vec{k},\uparrow} + c^\dagger_{\vec{k},\uparrow}c_{\vec{k}+\vec{q},\downarrow}\right]$$ The transformation is then introduced to diagonalize this Hamiltonian.