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I Quantum numbers of Landau levels

  1. Nov 13, 2016 #1
    I have been reading about Landau levels for a two-dimensional system of charged particles in a perpendicular magnetic field and I have trouble understanding why there is degeneracy in the system. Let me provide some background to my problem.



    In the presence of a magnetic field, the momentum of a charge particle changes from ##p_{i}## to ##\pi_{i}\equiv p_{i}+eA_{i}##, where ##e## is the charge of the particle.

    It is also possible to define another kind of momentum ##\tilde{\pi}_{i}=p_{i}-eA_{i}##.


    For a two-dimensional system with a magnetic field pointing in the ##z##-direction, we can work in the symmetric gauge

    $${\bf{A}}=\left(-\frac{yB}{2},\frac{xB}{2},0\right),$$

    and define two sets of creation and annihilation operators ##a, a^{\dagger}## and ##b, b^{\dagger}## such that a generic state of the system is given by

    $$|n,m\rangle=\frac{a^{\dagger n}b^{\dagger}n}{\sqrt{n!m!}}|0,0\rangle,$$

    where ##|0,0\rangle## is the unique ground state annihilated by both ##a## and ##b##.



    Now, ##|0,0\rangle## is the unique ground state annihilated by both ##a## and ##b## since ##\pi_{i}## and ##\tilde{\pi}_{i}## commute in the complex plane defined by ##(z=x-iy,\bar{z}=x+iy)## and commuting observables have a common basis of eigenfunctions.

    Does this not mean that the quantum numbers labelled by ##n## and ##m## give the same wavefunctions?

    Why does the energy of the state depend only on ##n##, but not on ##m##?
     
  2. jcsd
  3. Nov 14, 2016 #2

    vanhees71

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    It's much easier to solve the energy-eigen problem in the gauge, where ##\vec{A}## depends only on one spatial component, e.g.,
    $$\vec{A}=-By \vec{e}_x,$$
    because of the higher symmetry of the corresponding Hamiltonian. In this gauge the Hamiltonian reads
    $$\hat{H}=\frac{1}{2m} [(-\mathrm{i} \partial_x+q By)^2-\partial_y^2-\partial_z^2],$$
    i.e., you can seek for common eigenvectors of ##\hat{p}_x##, ##\hat{p}_z## and ##\hat{H}##. This leads to the ansatz
    $$u_{E,p_x,p_z}(\vec{x})=C \exp(\mathrm{i} p_x x+\mathrm{i} p_y y) Y(y),$$
    leading to an effective one-dimennsional harmonic-oscillator energy-eigenvalue problem for ##Y(y)##.

    Of course, the energy eigenvalues are gauge invariant, and the wave function transforms with ##\psi'(t,\vec{x})=\exp[\pm \mathrm{i} q\chi(\vec{x})] \psi(t,\vec{x})## while ##\vec{A}'=\vec{A} + \vec{\nabla} \chi##. I'm a bit unsure about the sign in the phase factor. Just figure it out by direct calculation. :wink:
     
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