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States with minimum energy for electrons in mag field with nonzero Lz

  1. Sep 8, 2014 #1
    1. The problem statement, all variables and given/known data

    For an electron in a uniform magnetic field, say [itex]B\hat{z}[/itex] with no angular momentum, the Hamiltonian can be expressed as [itex]\hat{H}=\frac{1}{2m}\Big(\hat{p}_x^2+\frac{mω^2}{2}\hat{x}^2\Big)+\frac{1}{2m}\Big(\hat{p}_y^2+\frac{mω^2}{2}\hat{y}^2\Big)[/itex]

    Which is equivalent to two separate harmonic oscillators.

    Now the ground state is [itex]Ψ=Ne^{\frac{mω}{\hbar}r^2}[/itex]

    where [itex]r=\sqrt{x^2+y^2}[/itex] and [itex]ω=ω_B/2=eB/2mc[/itex]

    which yields energy equal to [itex]\hbarω[/itex]

    Now there are other states with nonzero angular momentum [itex]L_z[/itex] which yield the same energy. Those states are [itex]Ψ_n=Nr^ne^{inφ}e^{\frac{mω}{\hbar}r^2}[/itex]

    (n is not the quantum number)

    The question is to prove that [itex]\hat{H}Ψ_n=\frac{\hbarω_B}{2}Ψ_n[/itex]

    since [itex]Ψ_n[/itex] is also an eigenstate of the Hamiltonian.

    My question is, how do we prove that? I tried using the Hamiltonian with polar coordinates but I can't seem to get to the result.

    Is there any other way (using the creation/annihilation operators)?
     
  2. jcsd
  3. Sep 15, 2014 #2
    I'm sorry you are not finding help at the moment. Is there any additional information you can share with us?
     
  4. Sep 16, 2014 #3

    nrqed

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    Did you try simply staying in cartesian coordinates and applying the hamiltonian?
     
  5. Sep 16, 2014 #4
    Yes, it's still a mess.

    By the way, the correct ground state is [itex]Ψ=Ne^{-\frac{mω}{ℏ}r^2}[/itex] (forgot a minus). I can't see the edit button any more...
     
  6. Sep 16, 2014 #5

    nrqed

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    Sorry, I did not notice your ## \phi## dependence on the one you must check.
    Then you must use polar coordinates (you are working in two dimensions, right?).
    You need ## \nabla^2 ## in polar coordinates. Once you reexpress the hamltonian in polar coordinates, it should be easy to check.
     
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