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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.

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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.

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