1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Spin of single particle state of free Dirac Field

  1. May 6, 2013 #1
    1. The problem statement, all variables and given/known data
    Show that the state [itex] d^{\dagger}_{\alpha}(0)\mid 0\rangle [/itex] describes a postrion at rest by showing that it is an eigenstate of the operators [itex] P^{\mu}, Q, J^z [/itex].


    2. Relevant equations
    The fourier expansion of [itex] \psi, \psi^{\dagger}[/itex]:
    [tex]
    \psi = \int \frac{d^3k}{(2\pi)^3} \frac{m}{k_0} \sum_{s}\left(b_{s}(k)u_{s}(k)e^{-ikx} + d_s^{\dagger}(k)v_s(k)e^{ikx}\right)
    [/tex]
    [tex]
    \psi^{\dagger} = \int \frac{d^3k}{(2\pi)^3} \frac{m}{k_0} \sum_{s}\left(b_{s}^{\dagger}(k)u_{s}^{\dagger}(k)e^{ikx} + d_s(k)v_s^{\dagger}(k)e^{-ikx}\right)
    [/tex]

    3. The attempt at a solution
    I have shown without porblem that the state is an eignestate of [itex] P^{\mu} [/itex] and [itex] Q [/itex], however I am stuck at showing the same for the momentum.

    I know how to get, based on the conserved noether current, that the operator [itex] J^z [/itex] for the Dirac Lagrangian can be written as:

    [tex]
    J^z = \int d^3x \psi^{\dagger}(x\partial_y - y\partial_x + \frac{1}{2}\Sigma^3)\psi
    [/tex]

    However, if I expand [itex] \psi, \psi^{\dagger} [/itex] like I did for the momentum and charge, I get nowere since the way [itex] \Sigma^3 [/itex] acts on [itex] u_s(k), v_s(k) [/itex] is not simple.

    I have searching in books and have found that Peskin (page 61) provides some arguments to perform de calculation. He argues that the orbital angular moementum for the particle at rest is zero, and the only term that is left is the spin. Then he argues that since [itex] J^z \mid 0 \rangle [/itex] must be zero, then [itex] J^z d^{\dagger}_{\alpha}(0) \mid 0 \rangle = [J^z, d^{\dagger}_{\alpha}(0)] \mid 0 \rangle [/itex]. Then, he continues, the only nonzero term is the commutator [itex] [d(p) d^{\dagger}(p), d^{\dagger}(0)] [/itex]. However, I do not understand how he gets to this last step. Why doesn't [itex] \Sigma^3 [/itex] appear between the creator and destructor operators in this commutator ? As far as I understand the spin operator shouldn't commute with the creator/annihilator operators.

    Any help in understanding Peskin's arguments or any other way to achieve the same result is greatly appreciated.

    Thanks,

    Fedcer
     
    Last edited: May 6, 2013
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted



Similar Discussions: Spin of single particle state of free Dirac Field
Loading...