I'm trying to derive the Landau level by applying semiclassical dynamics and the time-dependent Schrodinger equation. From that, I success to derive $E = \hbar\omega_c n$, but I fail to derive the form $$E = \hbar\omega_c(n+1/2).$$
Firstly, I used the semiclassical dynamics of the Lorentz force: $$dk_{\parallel}/dt = -ev_{\perp}B/\hbar.$$
Secondly, I used the time-dependent Schrodinger equation: $$i\hbar\frac{d}{dt}|\Psi \rangle = \hat{H}|\Psi \rangle.$$
For $|\Psi \rangle$ having the eigenvalue $E$, $$|\Psi(t) \rangle = \exp(-\int^t_0i\frac{E}{\hbar}dt)|\Psi(0)\rangle.$$
I changed the integral variable from $t$ to $k$ by the semiclassical dynamics of the Lorentz force.
$$|\Psi(k)\rangle = \exp(-\int^k_{k_0}i\frac{E}{\hbar}\frac{dt}{dk}dk_{\parallel})|\Psi(0)\rangle = \exp(\int^k_{k_0}i\frac{E}{\hbar}\frac{\hbar}{ev_{\perp}B}dk_{\parallel})|\Psi(0)\rangle.$$
I think if I conduct the integral over a closed path, the single-valueness of the wave function will force the relation below
$$\oint \frac{E}{\hbar}\frac{\hbar}{ev_{\perp}B}dk_{\parallel} = 2\pi n.$$
For the calculation, I used $v_{\perp}=\frac{1}{\hbar}\frac{dE}{dk_{\perp}}$, $\oint dk_{\perp}d{k_{\parallel}} = dA$, and $m_c=\frac{\hbar^2}{2\pi}\frac{dA}{dE}$. After the calculation, I get $E=\hbar\omega_c n$, where $\omega_c = eB/m_c$.
I wonder why the above method gives a wrong answer without 1/2. If possible, I hope to know the right method similar to the above method, which consider a path in $k$-space.
