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Hamiltonian in Landau gauge

  1. Dec 18, 2015 #1
    1. The problem statement, all variables and given/known data
    Define n=(x + iy)/(2)½L and ñ=(x - iy)/(2)½L.
    Also, ∂n = L(∂x - i ∂y)/(2)½ and ∂ñ = L(∂x + i ∂y)/(2)½.
    with ∂n=∂/∂n, ∂x=∂/∂x, ∂y=∂/∂y, and L being the magnetic length.
    a=(1/2)ñ+∂n and a=(1/2)n -∂ñ
    a and a are the lowering and raising operators of quantum mechanics.

    Show that H=ħωc(aa + ½)

    2. Relevant equations
    L=ħc/eB, ωc=eB/mc (cyclotron frequency), e for the charge of the electron
    H = Px2/2m + ( Py2 + eBx/c )2/2m

    3. The attempt at a solution
    I have tried to find x,y,∂x,∂y in terms of n,ñ,∂n,∂ñ. But I ended up getting only some if the right terms to come out but not all, is my first step wrong? Any suggestions?
     
  2. jcsd
  3. Dec 19, 2015 #2

    blue_leaf77

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    Should the exponent "2" of ##P_y## be there?
     
  4. Dec 19, 2015 #3
    Sorry, it was a typo. Do you have any suggestions?
     
  5. Dec 19, 2015 #4

    blue_leaf77

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    You should post your initial attempt before we can discuss further. In particular, how the old variables look like in terms of the new ones?
     
  6. Dec 19, 2015 #5
    This is what I've done so far. My problem is that everything is there except for the ½. I wrote ∂ for ∂n and ∂(bar) for ∂ñ.
     

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  7. Dec 20, 2015 #6

    blue_leaf77

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    According to this link https://en.wikipedia.org/wiki/Landau_quantization, the Gauge you should be using is the symmetric gauge and hence the original Hamiltonian should be different than that you are using. For instance, in Landau gauge, the operator ##{y}## is not present.
     
  8. Dec 20, 2015 #7
    Why can't I show it using the Landau gauge? The choice is just for simplification of computation right?
     
  9. Dec 20, 2015 #8

    blue_leaf77

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    ##x## and ##y## appear symmetrically in the gauge transformation, but they do not in the original Hamiltonian.
     
  10. Dec 20, 2015 #9
    Oh I see that, then I'll try it again using the symmetric gauge. Thanks!
     
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