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I am trying to decompose the isospins of a three particle state using Clebsch-Gordan coefficients such as:

$|1,1\rangle \otimes |1/2,-1/2\rangle \otimes |1,0\rangle$

Decomposing the first two states gives:

$|1,1\rangle \otimes |1/2,-1/2\rangle = \sqrt{\frac{1}{3}}|3/2,1/2\rangle + \sqrt{\frac{2}{3}}|1/2,1/2\rangle$

And then these combined with the third state give:

$|3/2,1/2\rangle \otimes |1,0 \rangle = \sqrt{\frac{3}{5}}|5/2,1/2\rangle + \sqrt{\frac{1}{15}}|3/2,1/2\rangle - \sqrt{\frac{1}{3}}|1/2,1/2\rangle$

$|1/2,1/2\rangle \otimes |1,0 \rangle = \sqrt{\frac{2}{3}}|3/2,1/2\rangle + \sqrt{\frac{1}{3}}|1/2,1/2\rangle$

When I combine these all together I get:

$|1,1\rangle \otimes |1/2,-1/2\rangle \otimes |1,0\rangle = \sqrt{\frac{1}{5}} |5/2,1/2\rangle + \frac{10+\sqrt{5}}{15} |3/2,1/2\rangle + \frac{-1+\sqrt{2}}{3}|1/2,1/2\rangle$

Which has to be incorrect as this state is not normalised. Basically my question is, what am I doing wrong?

Edit: What I'm attempting to calculate is amplitudes for processes like $\Lambda p \to \Lambda p \pi^0$ using isospin states for all of the particles.

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  • $\begingroup$ One assumes you are using Racah's W-coeffs?. $\endgroup$ – Cosmas Zachos Jan 30 '19 at 19:00
  • $\begingroup$ I've been using Wolfram Alpha's CG coefficient calculator wolframalpha.com/input/?i=Clebsch-Gordan+calculator $\endgroup$ – CT1234 Jan 30 '19 at 19:41
  • $\begingroup$ You are right that you should get a normalized answer if you did it right, with the proper (sparse!) 18x18 orthogonal matrix which changes basis! (See, e.g. this answer for the addition of 3 spin 1/2s.) In your case, using dimensionalities, i.e. 2j+1 s, $3\otimes 3 \otimes 2= 6\oplus 4 \oplus 4 \oplus 2\oplus 2$, so there are two different spin 3/2s and two 1/2s ... And you have to 6-j distribute them suitably... That's what's wrong. $\endgroup$ – Cosmas Zachos Jan 30 '19 at 20:43
  • $\begingroup$ The issue here is that there are two copies of $s=3/2$ and $s=1/2$ that occur in the decomposition of $1\otimes 1/2 \otimes 1$ and you need to separate states in distinct copies of $s=3/2$ and $s=1/2$. As mentioned by @CosmasZachos this involves Racah technology. $\endgroup$ – ZeroTheHero Jan 30 '19 at 21:02
  • $\begingroup$ re: your latest edit... Isn't Λ an isosinglet? what's the fuss? $\endgroup$ – Cosmas Zachos Jan 30 '19 at 21:17
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Using software, I get:

for ket in Ket(1,1)*Ket(0.5,-0.5)*Ket(1,0): ...: print unicode(ket) ...:
...:

√(2∕9) |½, ½⟩

-⅓|½, ½⟩

⅔|1½, ½⟩

1∕25 |1½, ½⟩

√(⅕) |2½, ½⟩

whose coefficients add (in quadrature) to unity. So the question is, why can't you add the two $|\frac 1 2, \frac 1 2\rangle$ kets coherently (and likewise for the two $|\frac 3 2, \frac 1 2\rangle$ kets)?

See the comments--there are different mixed symmetry representations in the final answer with the same multiplicity.

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So the key point here is to realize that the coupling $$ 1\otimes \frac{1}{2}\otimes 1 $$ will contain some final $J$ values more than once. Indeed $$ 1\otimes \frac{1}{2}=\frac{3}{2}\oplus \frac{1}{2} \tag{1} $$ and coupling this to $1$ will produce, for instance, two types of states with final $J=\frac{1}{2}$, depending on the intermediate $J_{12}$ value. Thus, copy will come from the $J_{12}=\frac{3}{2}$ and the other will come from the $J_{12}=1$ states of (1).

To be systematic write $$ \vert 1,1\rangle \vert\textstyle\frac{1}{2},-\frac{1}{2}\rangle =\frac{1}{\sqrt{3}}\vert \frac{3}{2}\frac{1}{2}\rangle + \sqrt{\frac{2}{3}} \vert\frac{1}{2}\frac{1}{2}\rangle\, .\tag{2} $$ Coupling the $\frac{3}{2}$ state to $1$, the part proportional to final $J=\frac{1}{2}$ using the CG $C_{3/2,1/2;1,0}^{1/2,1/2}=-\frac{1}{\sqrt{3}}$ yields $\vert \frac{1}{2}\frac{1}{2};J_{12}=\frac{3}{2}\rangle$ with $$ \langle \textstyle\frac{1}{2}\frac{1}{2};J_{12}=\frac{3}{2}\vert 1,1;\frac{1}{2},\frac{1}{2} ; 1,0\rangle =-\frac{1}{3} $$ but going through the $J_{12}=\frac{1}{2}$ produces a different $J=\frac{1}{2}$ state with $$ \langle \textstyle\frac{1}{2}\frac{1}{2};J_{12}=\frac{1}{2}\vert 1,1; \frac{1}{2},\frac{1}{2} ; 1,0\rangle =+\frac{\sqrt{2}}{3} $$ You can actually check that $\vert\textstyle\frac{1}{2}\frac{1}{2};J_{12}=\frac{1}{2}\rangle$ is really different from $\vert\textstyle\frac{1}{2}\frac{1}{2};J_{12}=\frac{3}{2}\rangle$ by computing their explicitly expressions in terms of $j_1=1,j_2=\frac{1}{2}, j_3=1$ states; you will see that these are distinct linear combinations of the $j_1=1,j_2=\frac{1}{2}, j_3=1$ states.

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