Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 4 / 34C / b

These potentials give the uniform magnetic field $\mathbf{B}=B\hat{\mathbf{z}}$ and zero electric field. With the signed frequency ${\omega }_c=qB/m$, the Lorentz force law becomesThusand explicitlyThe transverse trajectory is a circle about the fixed guiding centre $(x+\dot{y}/{\omega }_c,y-\dot{x}/{\omega }_c)$. The positive cyclotron frequency is

Let $\mathbf{\Pi }=\mathbf{p}-q\mathbf{A}$ be the kinetic momentum. The quantum Hamiltonian isAssume $qB>0$ and zero longitudinal momentum, as in the displayed states. Sincewe have $[a,a^{†}]=1$ andFor $w=x+iy$,which annihilates everywith $f$ holomorphic. Hence all these states haveThe Landau levels are $E_n=\hslash {\omega }_B(n+1/2)$ for $n=0,1,\dots$; the arbitrary holomorphic function exhibits the degeneracy of the lowest level. For $qB<0$, the normalizable lowest-level states are anti-holomorphic instead.

For $f(w)=w^M=r^M{e}^{iM\phi }$, the angular momentum operator $L_z=-i\hslash {\partial }_{\phi }$ givesThe probability density is proportional towhose maximum occurs atIn a large disk of area $A=\pi R^2$, the states that fit inside satisfy $M\lesssim qB{R}^2/(2\hslash )$. Thus the degeneracy of each Landau level is asymptoticallywhere ${\Phi }_B=BA$ is the magnetic flux. There is one state per magnetic flux quantum $h/q$, up to boundary corrections.