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Elicity of dirac eigenstates

  1. May 18, 2009 #1
    Hi people,

    I was asking myself... is it true that the elements of the base of solutions of the dirac equation usually used are eigenstates of elicity?

    Yesterday I tried the calculation following the notation of this site (it uses the dirac representation) and its set of solutions:

    http://quantummechanics.ucsd.edu/ph130a/130_notes/node493.html

    I came up with this elicity operator:

    [tex]\frac{\vec{p}}{|p|}\vec{\Sigma}=((\frac{\vec{p}}{|p|}\vec{\sigma},0),(0,\frac{\vec{p}}{|p|}\vec{\sigma}))[/tex]

    (the one on the right is a matrix expressed in pseudo-matlab language, that is ((line1),(line2),...)) , [tex]\Sigma[/tex] is the spin operator and [tex]\vec{\sigma}[/tex] is the vector of pauli matrices)

    where I think the operators I used commute and so I can forget about the order and in particular I can first act with the momentum operator and then with the sigma matrices

    The point is that if for example I take the dirac spinor with the second component of [tex]u[/tex] equal to zero (the first one in the end of the link I posted before) and I act with the operator above this component won't remain zero... and this condition is required in order this to vector to be an eigenstate...

    what I'm missing? maybe these vectors are not eigenvectors of elicity? (as I supposed to know by common knowledge?)... or I made some mistake?
     
    Last edited: May 18, 2009
  2. jcsd
  3. May 19, 2009 #2
    Are you French? Do you mean helicity?
     
  4. May 20, 2009 #3
    yes... sorry I mean helicity.... :blushing:....

    I'm not french but surely english is not my mother tongue!... I know maybe I wasn't clear with my question... in case ask me!
     
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