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Calculating Clebsch Gordon Coefficients

  • Thread starter quasar_4
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  • #1
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Homework Statement



Find the Clebsch Gordon coefficients of [tex] 1/2 \otimes 1 = 3/2 \oplus 1/2 [/tex]

Homework Equations



We have some properties of the CG coefficients which might be useful:

1) they are nonzero only if j is between j1-j2 and j1+j2
2) m = m1+m2 for nonzero coefficients
3) they are real

The Attempt at a Solution



I am horribly confused. I know that the CG coefficients are given as the coefficients in the expansion

[tex] |j m, j_1 j_2 \rangle = \sum_{m_1} \sum_{m_2} | j_1 m_1, j_2 m_2 \rangle \langle j_1 m_1, j_2 m_2| jm, j_1 j_2\rangle [/tex]

or

[tex] \langle j_1 m_1, j_2 m_2 | j m \rangle [/tex]

and I know that the possible |j m j1 j2> states are for the product space with j1 = 1, j2 = 1/2:

[tex] | \frac{3}{2} \frac{3}{2} ,1 \frac{1}{2} \rangle, | \frac{3}{2} \frac{1}{2},1 \frac{1}{2} \rangle, | \frac{3}{2} \frac{-1}{2} ,1 \frac{1}{2} \rangle, | \frac{3}{2}\frac{-3}{2} ,1 \frac{1}{2} \rangle, | \frac{1}{2} \frac{1}{2} ,1 \frac{1}{2} \rangle, | \frac{1}{2} \frac{-1}{2} ,1 \frac{1}{2} \rangle [/tex]

but I don't understand what on earth to do or where to even start. Any help would be great.
 

Answers and Replies

  • #2
674
2
You usually just look them up in a table. I hope your assignment doesn't want you to actually derive them. Also, I am a bit confused on your notation in the first sentence.
 
  • #3
290
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Yes, I think the notation is what's confusing me. I computed some of the coefficients a couple of years ago in an undergraduate class, but I've never seen this notation. I guess what the author means is to find the coefficients for all the basis kets in the product space where j1 = 1/2, j2 =1. There are (2*j1+1)*(2j2+1)=6 such kets.

I actually have figured it out - but boy, I think Shankar has weird notation... :yuck:

All one needs do is to start with the maximal possible z component, and apply lowering operators for each given j value, along with applying orthogonality and normalization conditions. A bit tricky but doable...
 
  • #4
vela
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Start with the [itex]| 3/2\mbox{ }3/2 \rangle[/itex] state. You know that it has to correspond to [itex]|1/2\mbox{ }1/2; 1\mbox{ }1\rangle[/itex]. Then apply the lowering operator to get [itex]| 3/2\mbox{ }1/2 \rangle[/itex], and so on.


Edit: Oh sure, figure it out right before I post! ;)
 
  • #5
290
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Thanks anyway! It's good to know that my solution wasn't just pure nonsense :-)
 

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