How to unify these two distinct pictures of spin waves?

Question

For convenience let's focus on an isotropic ferromagnet, $H = -\sum \boldsymbol{S_i \cdot S_j}$.

On a classical level we are often given the picture of spin waves as slowly rotating spins, each sufficiently close to perfect alignment as to keep the energy cost really small. This is also the intuition given for Goldstone modes: one can use the continuous symmetry to start with a given spin direction and then apply bigger and bigger rotations as one goes further along.

However the quantum-mechanical picture seems completely different to me. There a spin wave with momentum $\boldsymbol k$ is created by applying the operator $\sum e^{-i \boldsymbol{k \cdot r}} \; S^+_\boldsymbol{r}$ on a ground state of all spins down. In other words it is a massive superposition where each state only has a single perturbed spin.

Self-contained, I understand the logic of each picture separately (the second having to do with thinking of how the Heisenberg Hamiltonian induces `spin hopping' etc) but I really can't see how the first picture arises from the latter, even in a large-$S$ limit. Can anyone clarify this for me?

Answer 1

Classical waves are condensations of wave quanta. A single magnon (a spin wave quantum) do not have a well-defined classical spin wave behavior. The classical picture arises only if we have many magnons condensed in a coherent state.

In the second quantized language, a single magnon of momentum $\boldsymbol{k}$ is created by the creation operator $$ a_\boldsymbol{k}^\dagger = \sum_{\boldsymbol{r}}e^{-\mathrm{i}\boldsymbol{k}\cdot\boldsymbol{r}}S_{\boldsymbol{r}}^+,$$ from the ferromagnetic ground state $|0\rangle = \prod_{\boldsymbol{r}}|\downarrow\rangle_\boldsymbol{r}$ , where $|\downarrow\rangle_\boldsymbol{r}$ represents the lowest-weight state (the state of the spin $S^z$ quantum number $m_z=-S$ ) at site $\boldsymbol{r}$. So the state $a_{\boldsymbol{k}}^\dagger|0\rangle$ is a single mangon state, which is far from the classical limit and do not have a classical spin wave picture. To see the classical spin wave, one need to create many magnons of the same momentum, and condense them into a coherent state (much like creating a laser light by condensing photons). The magnon coherent state is described by

$$|\boldsymbol{k}\rangle=e^{-A a_{\boldsymbol{k}}^\dagger}\;|0\rangle = \prod_{\boldsymbol{r}}e^{-A \exp(-\mathrm{i}\boldsymbol{k}\cdot\boldsymbol{r}) S_{\boldsymbol{r}}^+}\hspace{8pt}|\downarrow\rangle_\boldsymbol{r},$$

where a parameter $A$ controls the strength of the condensation (i.e. the amplitude of the spin wave). Now we can evaluate the expectation values of the spin operators on this coherent state (to the leading order of the large-$S$ expansion):

$$\begin{split}\langle\boldsymbol{k}|S_{\boldsymbol{r}}^+|\boldsymbol{k}\rangle &= \sqrt{2S}A e^{\mathrm{i}\boldsymbol{k}\cdot\boldsymbol{r}}+\mathcal{O}(1/S),\\ \langle\boldsymbol{k}|S_{\boldsymbol{r}}^z|\boldsymbol{k}\rangle &= -S+A^2+\mathcal{O}(1/S).\end{split}$$

The equations describe a classical configuration, where each spin is tilted a little away from the order direction ($S^z=-S$ ) and circulates (precesses) around the $S^z$ axis in a wavy manner in the space. This is indeed the classical spin wave configuration in a ferromagnet.