PX262 - G1 - particle with angular momentum in magnetic field

considering the classical motion of an electron on a circular path with a radius, $r$ , and angular velocity, $ω$
the magnitude of the angular momentum will be:

l = m ω r^{2}

it direction will be perpendicular to the plane of the orbit
the circular movement will induce a current, $- \frac{e ω}{2 π}$ , when moving around a loop of area, $π r^{2}$
the magnetic moment will be equal to the product of the current and the loop area, its direction anti-parallel to $\vec{l}$ due to the negative sign: $\vec{μ} = - \frac{e}{2 m} \vec{l}$
using postulate 3, the magnetic moment operator: $\hat{\vec{μ}} = - \frac{e}{2 m} \hat{\vec{L}}$
the $z$ -component of the operator: ${\hat{μ}}_{z} = - \frac{e}{2 m} {\hat{L}}_{z}$
if a magnetic field, $\vec{B}$ , is applied to this system, the energy of the interaction will be:

Δ E = - \vec{μ} \cdot \vec{B}

in quantum mechanics, if the $B$ -field is aligned with the $z$ -axis: $\hat{H} = \frac{e B}{2 m} {\hat{L}}_{z}$
systems that are eigenstates of ${\hat{L}}_{z}$ , with eigenvalue $m ℏ$ , a measurement will yield: $Δ E = μ_{B} B m$
defining the bohr magneton

μ_{B} = \frac{e ℏ}{2 m}

if the state with $l = 1$ is put into the magnetic field, three transmission energies are expected in the spectra, for $m = - 1, 0, 1$