quantum mechanics - Magnetic susceptibility for spin-1 particle - magnetic moment and field interaction

A particle with spin 1 described in eigenstates of $S^2,S_z ^2: |-1\rangle,|0\rangle,|1\rangle$ is applied a constant homogeneous magnetic field $\bar B=B_0\hat u$. The magnetic moment is given by $\bar M=\gamma\bar S$, and the interaction is given by $W=\bar M\cdot \bar B$, and let $\omega_0 = \gamma B_0$. Write the matrix elements of $W$ in the above basis.

I figured I need to use the fact that before applying the external field, the problem is spherically symmetric. Then, I can define $\hat u = \hat z$ such that $W = \omega_0 S_z$, but I'm not sure of this reasoning.

If so, then $$\langle n|W|m\rangle = \omega_0\langle n|S_z|m\rangle = \omega_0\hbar m\delta_{nm}$$ Is this correct? Also, if upon choosing in advance a coordinate system with a predefined $\hat z$ that is not necessarily equal to $\hat u$, is there an elegant approach in general? Thanks

Edit: another thing that troubles me is that if the above is correct, which I believe is not, then I get that the first order perturbation of the wave function is zero... I'm not sure what exactly I'm misinterpreting.

Answer

Everything in the OP is correct. As the system has no preferred direction, we may convene that the magnetic field is along the $z$ axis; or rather, define the $z$ axis to be along the direction of $B$, whatever this direction is. Note that this doesn't affect the physics, because the eigenvalues of $\boldsymbol n\cdot\boldsymbol S$ are independent of $\boldsymbol n$: they are $0$ and $\pm1$. Choosing $\boldsymbol n=\hat{\boldsymbol z}$ is just a matter of convention.

Moreover, your calculation of $\langle n|W|m\rangle$ is correct, and your realisation that the first order correction of the wave-function vanishes is correct as well. The reason is that the unperturbed eigenstates are also eigenstates of the perturbation; or, in other words, the new Hamiltonian is diagonal in the old basis.