1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Clebsch Gordan Coefficients

  1. Jan 2, 2009 #1
    Problem

    Express [tex]\bold{S}^2[/tex] in the [tex]|j_1 j_2 ; m_1 m_2\rangle[/tex] and [tex]|j_1 j_2; s m\rangle[/tex] bases and find the unitary matrix, [tex]U[/tex], which takes the [tex]|j_1 j_2 ; m_1 m_2\rangle[/tex] basis to the [tex]|j_1 j_2; s m\rangle[/tex] basis, for spin [tex]1/2[/tex] (s corresponds to the total angular momentum while m corresponds to the total z angular momentum). This is basically "adding" together two angular momenta of spin 1/2.

    My attempt at a Solution

    Well, we know that [tex]\bold{S}^2 = \bold{S_1}^2+\bold{S_2}^2 + 2 S_{1z} S_{2z} + S_{1+}S_{2-} + S_{1-}S_{2+}[/tex]. Now, using this and the fact that [tex]j_1 = j_2 = 1/2[/tex], I got that

    [tex]\bold{S^2} = \hbar^2\begin{bmatrix} 2 & 0 & 0 & 0\\ 0 & 1 & 0 & 1\\ 0 & 1 & 0 & 1\\ 0 & 0 & 2 & 0\end{bmatrix}[/tex]​

    in the [tex]|j_1 j_2 ; m_1 m_2\rangle[/tex] basis. The above columns are, from left to right, |+ +>, |+ ->, |- +>, and |- ->, respectively. Now, for the other basis, I got

    [tex]\bold{S^2} = \hbar^2\begin{bmatrix} 2 & 0 & 0 & 0\\ 0 & 2 & 0 & 0\\ 0 & 0 & 2 & 0\\ 0 & 0 & 0 & 0\end{bmatrix}[/tex]​

    to be the representation; the columns above represent |1 1>, |1 0>, |1 -1>, and |0 0>, respectively. First of all, are these correct? If they are, how can there be a unitary transformation from one matrix to another? Two of the first matrices' columns are equal, but the corresponding columns in the second matrix are not equal. How can we resolve this?

    Thanks very much for the help.
     
    Last edited: Jan 2, 2009
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you help with the solution or looking for help too?