# Composing three ket vectors

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!

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Avodyne
Yes, you can combine 3 kets together; just do it in stages, which appears to be what you did.

But there are two different |1,0> states in the product of the 3 kets that are orthogonal to each other, so you shouldn't add those two coefficients.

Hmm ok that's interesting...what makes the two |1,0> kets orthogonal to each other? Don't they both represent the same three-particle state?

Avodyne
No, they don't. You're combing three j=1 states. Let's allow all m values, as counting the states is helpful. Each of the three states has m = -1, 0, or +1, so there are 3*3*3=27 possible states.

First, let's combine two of the j=1 states. The allowed values of j of the combined state are j=0,1,2. Now we have to combine each of these with the third j=1 state. Combing 0 and 1 yields 1. Combining 1 and 1 yields 0,1,2. Combining 2 and 1 yields 1,2,3. So we end up with one j=3 multiplet, two j=2 multiplets, three j=1 multiplets, and one j=0 multiplet. The number of states in a j multiplet is 2j+1. So the total number of states is 1*(2*3+1)+2*(2*2+1)+3*(2*1+1)+1*(2*0+1) = 27, which is correct. So we have to include 3 different j=1 multiplets to get the right state counting. The states in each of these multiplets consist of different, orthogonal linear combinations of your original m1 m2 m3 states.

Okay, thanks, this is really helpful. One last question: say you started in a |1,0> isospin state, and this decayed into the three particle state above. How would we know which final-state |1,0> ket the initial state will couple with under the strong interaction? Could it couple with all three? Or would it just couple with the 0+1=1 state, for instance, since that state didn't really get shifted into an orthogonal orientation?

Also, I find that the set of final-state kets you get, and their coefficients, depends on the order in which you compose them, as if the algebra is non-associative. So, for instance, if I compose the set

$$|{1,-1}\rangle|{1,0}\rangle|{1,0}\rangle$$

and I start by composing the first two kets, I get a completely different result than if I start by composing the final two kets. How can I tell which is correct? Does it depend on the situation somehow?

Thanks!

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Avodyne
say you started in a |1,0> isospin state, and this decayed into the three particle state above. How would we know which final-state |1,0> ket the initial state will couple with under the strong interaction?
It would depend on the form of the interaction among the particles. In general, any isospin-allowed interactions are present.

Also, I find that the set of final-state kets you get, and their coefficients, depends on the order in which you compose them
This is because in the two cases you are looking at different linear combinations of the three different j=1 multiplets and two different j=2 multiplets. If you sum the squares of the coefficients of the different j=1 states in the two cases, the results should match. Ditto for the j=2 states. The j=3 state is unique and so the coefficient should match here as well.

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If you sum the squares of the coefficients of the different j=1 states in the two cases, the results should match. Ditto for the j=2 states. The j=3 state is unique and so the coefficient should match here as well.
I don't find this to be true. For instance, if I combine the first two kets first, I get

$$\left(|1,-1\rangle|1,0\rangle\right)|1,0\rangle=\sqrt{\frac4{15}}|3,-1\rangle-\sqrt{\frac1{12}}|2,-1\rangle-\frac12|2,-1\rangle-\sqrt{\frac3{20}}|1,-1\rangle+\frac12|1,-1\rangle,$$ whereas if I combine the last two kets first, I get

$$|1,-1\rangle\left(|1,0\rangle|1,0\rangle\right)=\sqrt{\frac4{15}}|3,-1\rangle+\sqrt{\frac1{3}}|2,-1\rangle-\sqrt{\frac16}|2,-1\rangle+\sqrt{\frac2{30}}|1,-1\rangle+\sqrt{\frac16}|1,-1\rangle.$$

These two results do give the same squared coefficient for the |3,-1> term, but for the j=1 and j=2 terms the sums of the squared coefficients are not the same. Either I'm not understanding something or this seems non-associative.

Actually, never mind, I found an error in my work. They are actually

$$\left(|1,-1\rangle|1,0\rangle\right)|1,0\rangle=\sqrt{\frac4{15}}|3,-1\rangle-\sqrt{\frac1{12}}|2,-1\rangle-\frac12|2,-1\rangle-\sqrt{\frac3{20}}|1,-1\rangle+\frac12|1,-1\rangle,$$ whereas if I combine the last two kets first, I get

$$|1,-1\rangle\left(|1,0\rangle|1,0\rangle\right)=\sqrt{\frac4{15}}|3,-1\rangle+\sqrt{\frac1{3}}|2,-1\rangle+\sqrt{\frac2{30}}|1,-1\rangle-\sqrt{\frac13}|1,-1\rangle.$$