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Why spin 1/2 nucleus has no quadrupole?

  1. Jan 23, 2009 #1
    Hi:
    I am confused that we talk in general about only nuclear spin larger than 1/2 which has electric quadrupole. For spin half, there is no quadrupole because of its spherical shape. Anyone can comment on why they are sphere? what is physical significance behind? I read some research articles. They mentioned the recent research shows proton shape maybe is sphere, peanut or bagel shape. How can we conclude spin 1/2 nucleus is sphere?
    thanks

    xfshi
     
  2. jcsd
  3. Jan 23, 2009 #2

    malawi_glenn

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    general talk? eh? Source please
     
  4. Jan 23, 2009 #3
    Please see Slitcher "Principles of magnetic resonance". on the chapter which talks about quadrupole, in the first or second page of that chapter( I don't remember clearly), he mentioned that only nucleus with spin angular moment with more than 1/2 has electric quadrupole. that is source why I question this problem. why don't spin half nucleus has electric quadrupole?
    thanks
     
  5. Jan 23, 2009 #4

    malawi_glenn

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    oh, but that is just basic quantum mechanics, just sandwich the quadropole moment operator between general spherical harmonics with equal L, only the one with L = 0 will contribute. (if i remember this correctly)
     
  6. Jan 27, 2009 #5

    clem

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    The electric quadrupole operator is a Legendre polynomial P_2. The matrix element
    <1/2|P_2|1/2> vanishes for any spin 1/2 state. This says nothing about the shape of the nucleus, just that this matrix element vanishes
     
  7. Jan 27, 2009 #6
    Thanks everyone's answer. Hi clem: Could you explain more detail? As I know, commutation [S_z z]=0, we can only derive <1/2,m'|z|1/2,m>=0 unless m'=m. For individual spin 1/2 nucleus, there is no orbital angular momentum(L=0). Intrinsic parity of spin half particle is +1.
    If we want to get <1/2,1/2|Q|1/2.1/2>=0, how do you derive <1/2,1/2|z|1/2,1/2>=0?
    thanks
    xf
     
  8. Jan 27, 2009 #7
    Hmm, does it have spherical shape? It does not make sense to me. Proton is uud, right? but charge of u=+2/3e and therefore two u-quarks cannot be at the same place because of Coulomb repulsion. I modeled this once assuming a linear or quadratic quark-quark attractive force to find the equilibrium structure, and i found that it take the shape of a line/rod "u-d-u". I guess that the quadropole term will be very small anyway. You may excuse me here because of my very small knowledge on quark-quark interaction.
     
  9. Jan 27, 2009 #8
    This must have a very restricted validity, assuming some old-fashioned shell-filling model? Assume three electrons in a weak parabolic potential. Then the structure would be a Wigner crystal, i.e., three sharp Gaussian like packages arranged in a triangle for Stot=1/2 (You can carry out this with Hartree-Fock. Shell filling approach using hydrogen orbitals would fail to describe a structure like this). Clearly the quadropole moment is not zero in this case.
     
  10. Jan 27, 2009 #9
    Sorry, I checked my claim (I did a fast estimation many years ago) with the line/rod structure u-d-u and its wrong! (sorry for the mess). It is most likely that the orbitals are bound together in the way that each quark has a spherical symmetric shape, like spherical Gaussians for example.

    The pure Coulomb problem collapses the quarks (u-d-u) to a point, but quantum kinetic energy broadens it to typical hydrogen-like structure. Adding the quark-quark attraction U=k*r, with k chosen so that the size of the wave functions are roughly a=1e-15m, gives a positive total binding energy of about 200MeV, which is of same order as mc^2 (Uquark, Ucoul and Ekin are roughly equal at the energy minimum). Has to be considered relativistic I think...
     
  11. Jan 28, 2009 #10
    thanks for your helpful answer. Now I focus on the whole nucleus. That means adding all the quark orbital angular momentum and spin angular momentum, nucleus spin are reached. If nuclear spin is half. Now I don't know how to prove <1/2.1/2|Q|1/2,1/2>=0.
     
  12. Jan 28, 2009 #11
    Do you think my understanding is correct? In general, we say spin angular momentum refers to total angular momentum(orbital angular momentum plus spin angular momentum). When we sandwich the quadrupole operator, wavefunction is the direct product of spatial wavefunction and spinor wavefunction. <J,J_z|Q|J,J_z>. By parity property, we know, x,y z is odd parity, the first J must be different from the second J. So that is reason why matrix element is zero.
    thanks
     
  13. Jan 28, 2009 #12

    clem

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    Q is a parity even operator. The Mel you gave is non zero for all J>1/2.
     
  14. Jan 28, 2009 #13

    clem

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    J=1/2 can come from L=0 or L=1. <0|Q|0> is easily seen to be zero.
    For L=1, s=1/2 form the eigenstate |1/2,1/2>. Then <1/2,1/2|P_2|1/2,1/2>=0.
     
  15. Jan 28, 2009 #14
    Thank you clem: let me repeat your solution.
    Because J=1/2, J=L+S, L can be either 0 or 1. How about S? For example, If L=0 and S=1/2, J could be 1/2. if L=1, S=3/2, we still can acquire 1/2 from the rules J~|L+S| to |L-S|. Ok now I assume S=1/2.
    (1) If L=0, S=1/2, J=1/2 and quadrupole operator is even parity, wave function |L,m> also is even parity due to Y_L^m=(-1)^L. <L1,m1|Q|L2,m2> where L1=0,L2=0,m1=0,m2=0. From selection rules, m1=m2 are satisfied, L1 is same to L2. How do you conclude <0,0|Q|0,0>=0? (where m1 and m2 are magnetic quantum number)

    (2) If L=1, S=1/2, we still get J=1/2. <L1,m1|Q|L2,m2> L1=1, L2=1, m1 can be 0, +/-1.
    m2 can be 0,+/-1. <1,1|Q|1,1>=<1,0|Q|1,0>=<1,-1|Q|1,-1>=0. Notice Q is only function of coordinate. Why are they equal to zero? Sorry I cannot follow your idea.
     
  16. Jan 29, 2009 #15

    clem

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    It's hard for me to do the details here. Look at a book on angular momentum.
    I gave one example, but the rule is more general. The three angular momenta in the matrix element <J_1|J_2|J_3> must obey a triangle rule that no J is greater than the sum of the other two.
     
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