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Question about angular momentum

  1. Apr 10, 2007 #1
    1. The problem statement, all variables and given/known data
    The deuteron consists of a proton and a neutron in a state of spin 1 and total angular momentum J = 1. What are the possible angular momentum states for this system.

    Another question: Consider a wave function with azimuthal dependence
    [tex] \psi(r,\theta,\phi) \propto \cos^2 \phi [/tex]
    What are the possible outcomes of a measurement of the z component of the
    orbital angular momentum and what are the probabilities of these outcomes?

    2. Relevant equations
    J = L + S

    3. The attempt at a solution

    since J =1 and S = 1, and thus -1 and 0. (this becuase suppose a particle had spin 3/2 then [itex] j=l\pm \frac{1}{2}[/itex] and [itex] j = l \pm \frac{3}{2} [/itex]

    so then the possible values for l are 2,0,1??

    seemingly trivial but i dont quite udnerstand this completely yet soo...

    For the second question - it appears that [itex] m_{l}=1 [/itex]

    and thus we can calcculate the probability of this measurement using
    [tex] c_{n}=\frac{1}{\sqrt{2\pi}} \int_{0}^{2\pi} e^{-i\phi} \cos^2\phi d\phi[/tex]
    For ml=1, then
    [itex] \left<L_{z}\right>=\hbar [/itex]

    Is that right??

    Thanks for the help!! it is greatly appreciated!!
     
    Last edited: Apr 11, 2007
  2. jcsd
  3. Apr 10, 2007 #2

    Meir Achuz

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    "so then the possible values for l are 2,0,1??
    seemingly trivial but i dont quite udnerstand this completely yet soo.."

    L could be either a mix of 0 and 2 (the actual case) or L=1.
    Even L and odd L cannot mix in the same state because parity is a good quantum number for nuclear binding.
    L=1 is ruled out if Isospin is introduced.
     
  4. Apr 10, 2007 #3

    Meir Achuz

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    cos^2\phi=[1+cos(2\phi)]/2.
    This mean m_L= a combiination of +2, -2, 0.
    For the probabilties, do your integrals using the appropriate Y_LM.
    The \Psi in your question should not depoend on phi.
     
  5. Apr 11, 2007 #4
    how are we 'mixing' even and odd L if we took L = 0, and L =1 and L = 2 separately??

    i dont quite see how parity is involved here
    also i edited the second question
     
  6. Apr 11, 2007 #5

    cgw

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    I think your reasoning is a little off.
    j = l+s or l-s

    If the particle had a spin of 3/2 then j= l +or- 3/2. There is no 1/2 spin.

    Right?
     
  7. Apr 12, 2007 #6

    Meir Achuz

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    Parity implies that the ground state can be a miixture of L=0 and L=2
    (the actual case), but if L=1, no even L could be mixed in.
     
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