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Can you combine three ket vectors together? The usual Clebsch-Gordan tables allow you to compose two ket vectors, but what if you wanted to compose a state like

$$|1,-1\rangle|1,+1\rangle|1,0\rangle$$

and get the possibilities for total spin or isospin, etc.? Is this even valid? I tried going through it, and I arrived at

$$\sqrt{\frac3{30}}|3,0\rangle-\sqrt{\frac13}|2,0\rangle+\left(\sqrt{\frac13}-\sqrt{\frac{2}{30}}\right)|1,0\rangle+\sqrt{\frac16}|0,0\rangle.$$

But I don't see how this can be right, because it's no longer correctly normalized as best I can tell (although interestingly if I neglect the cross-term from squaring the coefficient of the |1,0> ket, I do get the correct normalization). Am I just missing something?

Thanks!

$$|1,-1\rangle|1,+1\rangle|1,0\rangle$$

and get the possibilities for total spin or isospin, etc.? Is this even valid? I tried going through it, and I arrived at

$$\sqrt{\frac3{30}}|3,0\rangle-\sqrt{\frac13}|2,0\rangle+\left(\sqrt{\frac13}-\sqrt{\frac{2}{30}}\right)|1,0\rangle+\sqrt{\frac16}|0,0\rangle.$$

But I don't see how this can be right, because it's no longer correctly normalized as best I can tell (although interestingly if I neglect the cross-term from squaring the coefficient of the |1,0> ket, I do get the correct normalization). Am I just missing something?

Thanks!

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