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Homework Help: Clebcsh Gordan Coefficients - QM

  1. May 14, 2014 #1
    I'm looking at how a l j1,j2,jm> state can be operated on by the ladder operators, changing the value of m, and how this state can be expressed as a linear combination of l j1, j2, m1, m2> states.

    Where the following relationships must be obeyed (x2):

    l j1-j2 l ≤ j ≤ j1+j2

    So I'm looking at an example where we start with state j1=1/2 and j2=1/2.

    From the above relations I know :


    I am in state l j1,j1,j,m> = l 1/2,1/2,1,1>.

    I operate with L- to get:

    l 1/2, 1/2, 1,0>

    Now I want to express this in terms of l j1, j2, m1, m2> bases. So looking at the possible values of m1 and m2, I can see this can be produced by a linear combination of l j1, j2, m1, m2> = l 1/2,1/2,1/2,-1/2> , l 1/2,1/2,-1/2,1/2>

    So I need to figure out the coefficients.

    Now here is my question:

    I have the relation: J[itex]\pm[/itex] l j1,j2,j,m> = (j(j+1)-m(m[itex]\pm[/itex]1))[itex]^{\frac{1}{2}}[/itex] l j1,j2,j,m[itex]\pm[/itex]1>, *

    Which would give the coefficients as [itex]\sqrt{\frac{3}{4}}[/itex]

    So I dont understand where the coefficients come from : ([itex]\sqrt{\frac{1}{2}}[/itex] is the solution).

    I can see (i think ) that this makes sense from a normalization point of view. But then what happens to the coefficients attained from * ?

    Many thanks for any assistance, greatly appreciated !
  2. jcsd
  3. May 15, 2014 #2
    I'm afraid you are a little confused here. You are still in the [itex]|j_{1},j_{2}, j ,m \rangle[/itex] basis and not the [itex]|j_{1},j_{2}, m_{1} ,m_{2} \rangle[/itex] basis, so what you've got there is not the Clebsch-Gordan coefficient.

    The correct way to derive the Clebsch-Gordan coefficients is to recognise that the state [itex]|j_{1},j_{2}, j ,m \rangle = |\frac{1}{2},\frac{1}{2},1,1 \rangle [/itex] is equivalent to [itex]|j_{1},j_{2}, m_{1} ,m_{2} \rangle = |\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2} \rangle [/itex]. Now, you apply the lowering operator on this state. Give it a try, if you still need any help, I'll be happy to provide more details.
  4. May 15, 2014 #3
    Okay thanks.
    So do you consider the lowering operator acting on m1 and m2 individually, so effectively defining two different ladder operators - one for the '1 space' (L[itex]_{1}[/itex][itex]_{\pm}[/itex])and (L[itex]_{2}[/itex][itex]_{\pm}[/itex]) for the '2 space'

    If we then operate on l 1/2, 1/2, 1/2, 1/2 > in turn with (L[itex]_{1}[/itex][itex]_{\pm}[/itex])and (L[itex]_{2}[/itex][itex]_{\pm}[/itex])

    we get L[itex]_{1}[/itex][itex]_{-}[/itex] = l 1/2, 1/2, -1/2, 1/2 >

    and L[itex]_{2}[/itex][itex]_{-}[/itex]= l 1/2, 1/2, 1/2, -1/2 >

    So this gives a coefficient of 1 for both
    Would this be along the right lines?
    Last edited: May 15, 2014
  5. May 15, 2014 #4
    Yup! The trick is to consider the total lowering operator, which is just the sum of the lowering operators on each subspace. So we have:
    [tex]\ell^{-} \left|\frac{1}{2}, \frac{1}{2}, 1, 1 \right\rangle = \ell^{-}_{1} \left|\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2} \right\rangle + \ell^{-}_{2} \left|\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2} \right\rangle[/tex]
    where the LHS is in the [itex]\left|j_{1}, j_{2}, j, m \right\rangle[/itex] basis and the RHS in the [itex]\left|j_{1}, j_{2}, m_{1}, m_{2} \right\rangle[/itex] basis.
  6. May 15, 2014 #5
    Ok thanks. and the coefficients are 1, so does (1/2)^1/2 come from normalization?
  7. May 16, 2014 #6
    No, there is no need for an additional step. The factor that you are looking for comes out naturally from the expression that i wrote. If you compute it explicitly, it yields
    [tex]\sqrt{2}\left|\frac{1}{2}, \frac{1}{2}, 1, 0 \right\rangle = \left|\frac{1}{2}, \frac{1}{2}, -\frac{1}{2}, \frac{1}{2} \right\rangle + \left|\frac{1}{2}, \frac{1}{2}, \frac{1}{2}, -\frac{1}{2} \right\rangle[/tex]
    Move the factor of [itex]\sqrt{2}[/itex] to the RHS and it yields you exactly what you want.
  8. May 17, 2014 #7
    ah. thanks.
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