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I Nuclear vibrations

  1. Aug 10, 2016 #1
    In a passage of our book (Krane page 141) they add two quadrupole phonons to a ##0^+## state. So as I understand it these phonon can be written in the form ##Y_{\lambda \mu}## with ##\lambda=2##. It makes sense that this corresponds to two units of angular momenta. Then they talk about the possible ##\mu## values for these phonons and get the list below. But I don't understand how they get the list below. I can understand how ##+\mu = \pm 4## forces ##l=4## but not the rest. Why isn't ##l=1## or ##l=3## permitted?

    ##l=4 \; \; \; \mu = +4, +3 ,+2 ,+1 ,0 ,-1,-2,-3,-4##
    ##l=2 \; \; \; \mu = +2 ,+1 ,0 ,-1,-2##
    ##l=0 \; \; \; \mu = +0##

    They also say that if we instead add ##3## quadrupole the possible states are
    ##0^+, 2^+, 3^+, 4^+,6^+##.
    But how is ##3^+## possible? Shouldn't the parity be ##(-1)^l##?
    Even parity makes some sense with the total wave function of the phonons must be symmetric since they have integer spins and ##0^+## being symmetric the combination should be symmetric but I don't get why this contradicts the rule above.
     
  2. jcsd
  3. Aug 10, 2016 #2
    i was looking up and found that the allowed states are due to angular momentum correlations....
    <Angular-momentum selection rules allow for the values of λ = 0,1,2,3,4.
    However, it turns out that not all of these values are possible

    the wave functions for odd values of λ vanish: such states do not exist !
    The two-phonon states are thus restricted to angular momenta 0, 2, and 4,
    forming the two-phonon triplet.

    This effect is an example of the interplay of angular-momentum coupling and
    symmetrization (or, for fermions, antisymmetrization).>

    To get a detail treatment of modes of vibrations (both surface vibrations and spherical vibrations)
    one can look up the following....

    <http://th.physik.uni-frankfurt.de/~svogel/lecture_ws_2011_12/slides_bratkovskaya_3.pdf> [Broken]
     
    Last edited by a moderator: May 8, 2017
  4. Aug 11, 2016 #3
    That would explain it. I've read through the pdf too. I haven't read about Clebsch-Gordan coefficients before so I guess the explanation is a bit beyond me. But at least now I know a reason for why we only have even values of angular momenta. Kind of makes me even more curious about the three quadrupole phonon thought with their non even state.

    I now remembered the parity rules explaining the ##3^+## state as well. Parity is ##\pi = \Pi_i \pi_i## so since the quadrupole phonon is even so is the total parity.
     
  5. Aug 11, 2016 #4
    I found that the following pdf explain it quite nicely in a simpler way
    https://www.icts.res.in/media/uploads/Talk/Slides/Regan-lecture2.pdf
    So for example for two phonons we only had a single ##\mu = 3## state that is already accounted in ##l=4## so we got nothing over for ##l=3##. You can see the same reasoning applied to three phonons in the pdf. So just accounting for all the ##\mu## states we arrive at the right conclusion.
     
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