The Hamiltonian of the Heisenberg model in a spin coherent state Asume that the Hamiltonian of Heisenberg model is $H = \sum_{ij}\frac{1}{2}J_{ij}\hat{S_i} dot \hat{S_j}$
I want to get the exception value of the Hamiltonian of such Heisenberg model in a spin coherent state and the correct answer is:

I want to use the statement that $| \hat {\Omega }\rangle $ is an eigenstate of the spin component in the $\hat {\Omega } $ direction, so I decompose the spins in the $\hat {\Omega } $ direction first but I don't know how to deal with two other transverse directions and get the final result.
 A: Sorry if I am assuming too much or to little. As I understand it, your difficulty is in showing that $\langle\mathbf{\Omega}|\hat{\mathbf{S}}_i|\mathbf{\Omega}\rangle_i = S\mathbf{\Omega}_i$, correct?
The statement that $|\Omega\rangle$ is an eigenstate of the spin component along $\mathbf{\Omega}$ can be expressed as $\mathbf{\Omega} \cdot \hat{\mathbf{S}} |\Omega\rangle = S|\Omega\rangle$. If we then act from the left with $\langle\Omega|$ and use the fact that the states are normalized, we have $\mathbf{\Omega} \cdot \langle\Omega|\hat{\mathbf{S}} |\Omega\rangle = S$. We further know that $\langle\Omega|\hat{\mathbf{S}} |\Omega\rangle$ is (pseudo-)vector of length $S$. The only way for both these properties to be true is if $\langle\Omega|\hat{\mathbf{S}} |\Omega\rangle = S\mathbf{\Omega}$ since for $\langle\Omega|\hat{\mathbf{S}} |\Omega\rangle$ pointing in any other direction the dot product would have magnitude less than $S$.
We may then use this expression to for each site to arrive at your final result.
