I am tring to find the eigenvectors of a two spin system, with $j_1=3/2$ and $j_2=1/2$. To start, $$m_1 =-3/2,-1/2,1/2,3/2$$ $$m_2=-1/2,1/2$$ For $j_1$, there are 4 possible states, and 2 possible states for $j_2$. $(2j+1)$. Now, I get confused. I am looking for $$|S,m\rangle=\sum_{m_1+m_2=m}CG|s_1m_1\rangle|s_2m_2\rangle$$ I presume that these are the eigenvectors for total angular momentum and its z component. From here, I am not sure what to do. I write out all of the states, but then I am unsure of how to find the CG coefficients (we are not given a table). Any help would be appreciated.

  • $\begingroup$ I sympathize, addition of angular momentum is my single least favorite thing in physics. I don't have time for a long answer now, but there is a table here: en.wikipedia.org/wiki/Table_of_Clebsch–Gordan_coefficients, and wikipedia discusses how to compute them in the recursion relation section: en.wikipedia.org/wiki/Clebsch–Gordan_coefficients. $\endgroup$ – Andrew Mar 19 '14 at 15:39

I can get you thinking along the right path, but I'll leave most of the doing for you. I suspect that for much of this to make sense, you'll need to actually do it.

The idea is to start at the "top" of the ladder:

$$|2, 2\rangle = |3/2, 3/2\rangle_1\otimes|1/2, 1/2\rangle_2 =|3/2, 3/2\rangle_1|1/2, 1/2\rangle_2.$$

With this statement (and choice of phase), you're saying that the combined system can have maximum z-angular momentum only if each of the comprising particles also have maximum z-angular momentum.

Since we're at the top of the ladder, you can get the other $S=2$ states by applying the lowering operator $J_- = J_{1-} + J_{2-}$ to both the LHS and RHS. Here, $J_{1-}$ only acts on the first particle, and $J_{2-}$ on the second. (A more careful physicist would write $J_-=J_{1-} \otimes 1 + 1\otimes J_{2-}$, but hey.) The operator $J_-$ just acts on the LHS.

Doing this again and again should get you all of the $S=2$ states. The CG coefficients are just the numbers that sit out front of the states. You get these from carefully using the lowering operator. (There are some shortcuts one can use, but this is the hard but thorough way that I learned.) Try this once to get one set of CG coefficients, and check your work with a table on the internet.

Those were the $S=2$ states. To get the $S=1$ states, you'll have to use some orthogonality and a bit of convention. Very briefly, note that $\langle 2, 1\,|\,1, 1\rangle=0$. Also note that the vectors $(a,b)^T$ and $(b,-a)^T$ are orthogonal.

  • $\begingroup$ how can you get a $|5/2,5/2\rangle $ state. Doesnt the system have to be in either S = 2,1,-1? $\endgroup$ – yankeefan11 Mar 19 '14 at 16:25
  • $\begingroup$ Yes. In my head $3/2+1/2 = 5/2$. I suppose that isn't true. $\endgroup$ – BMS Mar 19 '14 at 16:28
  • $\begingroup$ So am I correct in saying that the state has to be in either s=2,1 or -1? $\endgroup$ – yankeefan11 Mar 19 '14 at 16:30
  • $\begingroup$ Are you asking if you combine spin 3/2 and spin 1/2, what the allowed values for spin are? No, those aren't the correct ones. Your textbook should discuss the allowed values, and there will be plenty of resources online. It will be helpful to have this idea in place prior to attempting to work out the CG coefficients; otherwise one is bound to lose sight in the gritty details. $\endgroup$ – BMS Mar 19 '14 at 16:37
  • $\begingroup$ So to get S=1 states, do you have any hints with the orthoganlity? $\endgroup$ – yankeefan11 Mar 19 '14 at 17:30

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