Energy Eigenstates of an $\ell=1$ System in a B-field

Recall that the Hamiltonian for a magnetic moment in an external B-field is

$$\bgroup\color{black}H={\mu_B B\over\hbar} L_z.\egroup$$

As usual, we find the eigenstates (eigenvectors) and eigenvalues of a system by solving the time-independent Schroedinger equation $H\psi=E\psi$. We see that everything in the Hamiltonian above is a (scalar) constant except the operator $L_z$, so that

$$\bgroup\color{black}H\psi={\mu_B B\over\hbar} L_z \psi= constant * (L_z \psi).\egroup$$

Now if $\psi_m$ is an eigenstate of $L_z$, then $L_z\psi_m=m\hbar\psi_m$, thus

$$\bgroup\color{black}H\psi_m={\mu_B B\over\hbar} * (m\hbar\psi_m)=(m\mu_B B)\psi_m\egroup$$

Hence the normalized eigenstates must be just those of the operator $L_z$ itself, i.e., for the three values of $m$ (eigenvalues of $L_z$), we have

$$\bgroup\color{black}\psi_{m=+1}=\left(\matrix{1\cr 0\cr 0}\ri... ...ht)\qquad \psi_{m=-1}=\left(\matrix{0\cr 0\cr 1}\right).\egroup$$

and the energy eigenvalues are just the values that $E=m\mu_B B$ takes on for the three values of m i.e.,

$$\bgroup\color{black}E_{m=+1}=+\mu_B B\qquad E_{m=0}=0\qquad E_{m=-1}=-\mu_B B.\egroup$$