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Reading CG tables

  1. Jan 24, 2015 #1
    1. The problem statement, all variables and given/known data
    Ok, this my not be appropriate topic to post my question but it is related to QM (moderators, feel free to move the thread if needed). I don't have any calculations nor do I have anything interesting to write. I am just trying to learn to read the Clebsch-Gordan coefficients from the tables - but I am having hard time understanding them...

    Here is the table http://pdg.lbl.gov/2002/clebrpp.pdf

    2. Relevant equations


    3. The attempt at a solution
    Now using this table above, we wrote $$|\frac 3 2 , \frac 1 2 >=\sqrt{\frac 1 3}|\uparrow >|1>+\sqrt{\frac 2 3}|\downarrow >|0>$$ and $$|\frac 1 2 , \frac 1 2 >=\sqrt{\frac 2 3}|\uparrow >|1>-\sqrt{\frac 1 3}|\downarrow >|0>$$ Is there anybody that has 5 minutes of free time and could give me an idiot proof explanation? How do these tables work?
     
  2. jcsd
  3. Jan 24, 2015 #2

    DrClaude

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    Staff: Mentor

    I'm guessing a bit, because you didn't give your notation. You seem to be adding angular momenta 1/2 and 1, so I guess that you are using the notation
    $$
    | J, M \rangle = \sum_{m_{1}, m_{2}} c_{m_{1}, m_{2}} |m_{1} \rangle |m_{2} \rangle
    $$
    with ##j_1 = 1/2##, ##j_2 = 1##.

    The CG coefficients as expressed in the table as ##\langle j_1, j_2, m_1, m_2, | j_1, j_2, J, M \rangle##, which in your case means that the coefficients ##c_{m_{1}, m_{2}}## are
    $$
    c_{m_{1}, m_{2}} =\langle 1/2, 1, m_1, m_2 | 1/2, 1, J, M \rangle
    $$

    For example, lets find the coefficient in front of ##|\uparrow \rangle |1\rangle## that appears in the equation for ##|3/2 , 1/2 \rangle##. It will be
    $$
    c_{1/2, 1} = \langle 1/2, 1, 1/2, 1, | 1/2, 1, 3/2, 1/2 \rangle
    $$
    As you are adding ##j_1 = 1/2## with ##j_2 = 1##, you look at the values for ##1 \times 1/2## (the order of ##j_1## and ##j_2## is not important) and look for ##J = 3/2, M = +1/2## in an upper-right box, then for ##m_1=+1, m_2=+1/2## in the corresponding left box, and you don't find it! That's because the coefficient is 0! The triangle rule ##M = m_1 + m_2## is not followed, therefore the coefficient is zero, so you have made a mistake in your expansion for ##|3/2 , 1/2 \rangle##.

    Now, can you try again?
     
  4. Jan 24, 2015 #3
    No, I can't. :D
    The example I wrote is the example my professor wrote. Let me give you a bit background information about the problem we were solving with my professor.
    We had a Hamiltonian ##H=\frac{p^2}{2m}+\lambda \delta(x)\vec{S_1}\cdot \vec{S_2}##. Than we said that ##S_1=1/2## and ##S_2=1##. Meaning that basis vectors for particle with ##S_1## are ##\left \{ |\uparrow>,|\downarrow> \right \}## while for particle with ##S_2## the basis vectors are ##\left \{ |1>,|0>,|-1> \right \}##.
    We continued by saying that total spin ##S## can be ##\frac 3 2## or ##\frac 1 2 ##. (I am still trying to figure it out where these two numbers came from, but it is not important for the CG coefficients).

    This leaves us with two possible groups of basis vectors. First one, we called it product basis, is ##\left \{ |\uparrow >|1>,|\uparrow >|0>,|\uparrow >|-1>,|\downarrow >|1>,|\downarrow >|0>,|\downarrow >|-1> \right \}## and the second one expressed with the total spin ##S## is ##\left \{ |3/2,3/2>, |3/2,1/2>, |3/2,-1/2>,|3/2,-3/2>, |1/2,1/2>, |1/2,-1/2> \right \}##

    Now the whole idea is that using CG coefficients we can transform from one basis to antoher. And we wrote that ##|\frac 3 2 , \frac 1 2 >=\sqrt{\frac 1 3}|\uparrow >|1>+\sqrt{\frac 2 3}|\downarrow >|0>##.

    BTW. Of course my professor explained this to us, but obviously I wasn't paying any attention - now this is the result.

    EDIT: Oh, ok. I can see I made I mistake. We wrote ##|\frac 3 2 , \frac 1 2 >=\sqrt{\frac 1 3}|\downarrow >|1>+\sqrt{\frac 2 3}|\uparrow >|0>## and NOT ##|\frac 3 2 , \frac 1 2 >=\sqrt{\frac 1 3}|\uparrow >|1>+\sqrt{\frac 2 3}|\downarrow >|0>##
     
  5. Jan 24, 2015 #4
    Oh, ok, I get it now. Thank you very much my friend. :)

    I think this comes from ##|S_1-S_2|\leq S \leq S_1+S_2##.
     
  6. Jan 24, 2015 #5

    DrClaude

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    Staff: Mentor

    Correct!
     
  7. Jan 24, 2015 #6
    hmmm... One stupid question... (still the same example ##|3/2,1/2>##)

    You said, that I should look for ##J=3/2## and ##M=1/2## in the upper right box. But there are two of them. Following the wrong one gives me ##|3/2,1/2>=\sqrt{\frac{2}{3}}|\downarrow>|0>+\sqrt{\frac{1}{3}}|\uparrow>|-1>##.

    Am.. I think the dilemma is obvious. Why is the first box ok and second one not?
     
  8. Jan 24, 2015 #7

    DrClaude

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    Staff: Mentor

    The second box is gives you ##|3/2,1/2\rangle## and the third ##|3/2,-1/2\rangle##. So the correct equation is
    $$
    |3/2,-1/2\rangle \sqrt{\frac{2}{3}}|\downarrow\rangle|0\rangle+\sqrt{\frac{1}{3}}|\uparrow\rangle|-1\rangle
    $$
     
  9. Jan 24, 2015 #8
    Aaaa, so those J and M are written vertically and not horizontally!

    Oh god... I can even see it written now in the top right corner http://pdg.lbl.gov/2002/clebrpp.pdf ..

    -.-

    Thank you!
     
  10. Jan 24, 2015 #9

    DrClaude

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    Staff: Mentor

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