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I have the following problem: If $| I I_3 \rangle = \alpha |1/2,-1/2 \rangle |1,1\rangle + \beta |1/2,1/2 \rangle |1,0\rangle $, are the isospin and the third component of the isospin, calculate $\alpha$ and $\beta$ when $I=3/2$ and $I_3 = 1/2$.

The solution is $\alpha = \sqrt{1/3}$ and $\beta = \sqrt{2/3}$, but I don't know how to arrive to it because I don't understand the notation ($|a,b\rangle|c,d\rangle$). Can you please explain me this notation and the method to solve it? Thanks!

Qmechanic
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1 Answers1

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This is basically the problem of composition of angular momentum states since the isospin algebra is identical to the algebra of angular momentum. The notation $\vert a,b\rangle\vert c,d\rangle$ thus refers to:

  1. $\vert a,b\rangle$ for particle 1, with isospin $a$ and isospin projection $b$,
  2. $\vert c,d\rangle$ for particle 2, with isospin $c$ and isospin projection $d$,
  3. $\vert II_3\rangle$ for the combined system, with isospin $I$ and isospin projection $I_3$.

Thus the coefficients $\alpha$ and $\beta$ of your linear combinations are the Clebsch-Gordan coefficients $C^{II_3}_{ab;cd}$. They can be looked up in tables or obtained using various programs (v.g. Mathematica).

ZeroTheHero
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