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Addition of Angular Momenta, problem with notation

  1. Dec 12, 2012 #1
    1. The problem statement, all variables and given/known data
    The electron in a hydrogen atom is described by the following superposition of two states:

    [itex]|\psi> = \frac{1}{\sqrt{2}}(|n=2,l=1,m=0,s_z= +1/2> + |n=2,l=0,m=0,s_z= +1/2>) [/itex]

    (b) Let J = L + S be the total angular momentum. Express state [itex]\psi[/itex] in basis [itex]|n, l, J, J_z>[/itex]
    Hint: use the Clebsch -Gordan coffecients from the table on pg. 188 of Griffiths.

    2. Relevant equations

    [itex] |s_1 m_1>|s_2 m_2> = \sum C^{s_1s_2s}_{m_1m_2m}|s m> [/itex]


    3. The attempt at a solution

    For the first state:

    [itex]|n=2,l=1,m=0,s_z= +1/2> ====> l=1, s = 1/2 [/itex]
    [itex]|1,0>|1/2,1/2> = \sqrt{\frac{2}{3}}|3/2,1/2> - \sqrt{\frac{1}{3}}|1/2,1/2> [/itex]

    For the second state:

    [itex]|n=2,l=0,m=0,s_z= +1/2> [/itex] ====> there's no orbital angular momentum

    My question is: I don't know if what I've done is correct and, if it is, I don't know how to transform to the notation he asks. For me, the "l" in the new notation doesn't make sense or have to be 0.
     
  2. jcsd
  3. Dec 16, 2012 #2

    vela

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    That's right.

    So what are j and jz equal to in this case?

    You have one basis with states of the form ##\lvert \ l\ m_l \rangle \lvert\ s\ m_s \rangle##. These states are eigenstates of L2, Lz, S2, and Sz. The other basis has states of the form ##\lvert \ l\ s\ j\ j_z\rangle##, which are eigenstates of L2, S2, J2, and Jz. These two bases are related through the Clebsch-Gordan coefficients.

    When you wrote
    $$\lvert 1,0\rangle \lvert 1/2,1/2\rangle = \sqrt{\frac{2}{3}} \lvert 3/2,1/2\rangle - \sqrt{\frac{1}{3}} \lvert1/2,1/2\rangle,$$ what it really means is
    $$\lvert 1,0\rangle \lvert 1/2,1/2\rangle = \sqrt{\frac{2}{3}} \lvert l=1, s=1/2, j=3/2, j_z=1/2\rangle - \sqrt{\frac{1}{3}} \lvert l=1, s=1/2, j=1/2, j_z=1/2\rangle,$$
     
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