Past exam of the quantum physics course of the University of Oxford / 2016 / 10 / a / Solution

With $\mathbf{B}=B_0\hat{\mathbf{z}}$ and $S_z=(\hslash /2){\sigma }_z$,ThusFor $B_0>0$, spin down is the lower energy eigenstate. The negative electron magnetic moment is already accounted for in the positive coefficient multiplying $\mathbf{S}\cdot \mathbf{B}$ in the given Hamiltonian.