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Self-adjointness domain of P_r^2

  1. Dec 9, 2009 #1
    hi everybody, i'd like to discuss with you a problem occurred to me in the study of a central simmetry field in non rel. qm.
    at a certain point, i get the linear operator [tex] P_{r}^{2}=\frac{1}{r}\frac{\partial^{2}}{\partial r^{2}}r [/tex], which is the square of radial momentum, and i want to determine the domain in which it is self-adjoint, i.e. the maximal subset [tex] D(P_{r}^{2})\in L^{2}(\mathbb{R}_{+},r^{2}\ \mathrm{d}r) [/tex] such that [tex]D(P_r^2)=D(P_r^2\dagger)[/tex] and the two coincide in the domain above.
     
  2. jcsd
  3. Dec 9, 2009 #2

    dextercioby

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    Goerg Teschl (professor at TU in Vienna) wrote a book based on some lecture notes available somewhere on the internet for free. I think a link to these notes is right here on PF in the <Reference> session. You may search for the latest version of his notes. He has an extended coverage of the self-adjointess problem for different hamiltonians.

    I think the [itex] p_{r}^{2} [/itex] operator has the same domain of self-adjointness as the radial part of the H-atom operator. But you may check on that.
     
  4. Dec 9, 2009 #3

    Landau

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    Since it's Gerald Teschl, you may have trouble searching for Goerg Teschl, so here's the link ;)
     
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