i.e. $L = n$. For a cyclotron orbit of radius $r$, it is easy to convince oneself that $$L = \frac{e B}{2} r^2 = n \ \Rightarrow \ r = \sqrt{\frac{2 n}{e B}}.$$ Consequently, in the semi-classical treatment the allowed radii of cyclotron orbits are now quantized. Using eq.(\[Hamiltonian\]) one finds for the energy $$E = H = \frac{1}{2} M \omega^2 r^2 = n \omega.$$ As is well-known, up to a constant $\frac{\omega}{2}$, the semi-classically quantized energy values are those of a harmonic oscillator with the cyclotron frequency $\omega$.

Quantum Mechanical Treatment
----------------------------

Finally, we consider the problem fully quantum mechanically. The Schrödinger equation then takes the form $$- \frac{1}{2 M} \left[\partial_x^2 + 
\left(\partial_y + i e B x\right)^2\right] \Psi(\vec x) =
E \Psi(\vec x).$$ We now make the factorization ansatz $$\label{factorization}
\Psi(\vec x) = \psi(x) \exp(i p_y y),$$ and we obtain $$\left[- \frac{{\partial}_x^2}{2 M} +
\frac{1}{2} M \omega^2 \left(x + \frac{p_y}{M \omega}\right)^2\right] 
\psi(x) = E \psi(x).$$ Indeed, this is the Schrödinger equation of a shifted harmonic oscillator. Hence, the quantum mechanical energy spectrum takes the form $$E = \omega \left(n + \frac{1}{2}\right).$$ Interestingly, the energy of the charged particle is completely independent of the transverse momentum $p_y$. As a 