Register to reply 
Isospin decomposition of K>ππ decay 
Share this thread: 
#1
Nov1912, 12:26 PM

P: 305

I'm studying the decay K>ππ and I have some doubts on the isospin decomposition. We know that the state [itex](\pi\pi)[/itex] can have total isospin 0 or 2. Now, if we remember that in the isospin representation we have [itex]\pi^+\langle=1,1\langle[/itex], [itex]\pi^0\rangle=1,0\rangle[/itex] and [itex]\pi^\rangle=1,1\rangle[/itex], then using ClebschGordan coefficients we find:
\begin{eqnarray} \pi^+\pi^\rangle=\frac{1}{\sqrt{6}}2,0\rangle+\frac{1}{ \sqrt{2}}1,0\rangle+\frac{1}{\sqrt{3}}0,0\rangle \\ \pi^0\pi^0\rangle=\sqrt{\frac{2}{3}}2,0\rangle  \frac{1}{\sqrt{3}}0,0\rangle \\ \pi^+\pi^0\rangle=\frac{1}{\sqrt{2}}(2,1\rangle + 1,1\rangle) \end{eqnarray} Now, my textbook says that we can decompose the decay amplitudes as follow: \begin{eqnarray} A_{K^0\rightarrow \pi^+\pi^}=A_0e^{i\delta_0}+\frac{A_2}{\sqrt{2}}e^{i\delta_2} \\ A_{K^0\rightarrow \pi^0\pi^0}=A_0e^{i\delta_0}\sqrt{2}e^{i\delta_2} \\ A_{K^+\rightarrow \pi^+\pi^0}=\frac{3}{2}A_2e^{i\delta_2} \end{eqnarray} where A_{0} and A_{2} are the aplitude referred to the final state with I=0,2. The problem is: why the decay amplitudes don't present the same coefficient as in the ClebschGordan decomposition? 


#2
Nov1912, 01:47 PM

Sci Advisor
Thanks
P: 4,160

Instead of the second line, Wikipedia gives
1,0> ⊗ 1,0> = √(2/3)2,0>  √(1/3)0,0> Does that help? 


#3
Nov1912, 02:08 PM

P: 305

My fault. I wrote wrong in the post but I did the calculation with the correct formula. I will correct it right now. Still if you see the coefficients of the firt and second group of equation don't match.



#4
Nov1912, 03:27 PM

Sci Advisor
Thanks
P: 4,160

Isospin decomposition of K>ππ decay
Well, how about this, the ClebschGordan coefficients give
0> = √(1/3)(1,1> + 1,1>  0,0>) 2> = √(1/6)(1,1> + 1,1>) + √(2/3) 0,0> Let K^{0} decay into the state Φ ≡ B_{0}0> + B_{2}2> = √(1/3)(1,1> + 1,1>)(B_{0} + √(1/2)B_{2})  √(1/3)0,0>(B_{0}  √2B_{2}) Rescaling A_{0} = √(1/3)B_{0}, A_{2} =  √(1/3)B_{2} we get Φ = (1,1> + 1,1>)(A_{0} + √(1/2)A_{2}) + 0,0>(A_{0}  √2A_{2}), which reproduces the first two of the textbook equations. 


#5
Nov2012, 02:08 AM

P: 305

Actually, if I haven't done wrong calculation, I think there are some signs that doesn't match. However I think the situation is a little more complicated as I read here: http://web.mit.edu/woodson/Public/Duarte_isospin.pdf
The article talk about some fictitious particles (spurion) that must be introduced because the decay is a weak one, while the isospin is a good quantum number for strong interaction. I'll see Thank you very much 


Register to reply 
Related Discussions  
Why is a deuteron an antisymmetric singlet in  High Energy, Nuclear, Particle Physics  4  
Particle Spin, Isospin, weak Isospin  Quantum Physics  2  
Jordan Decomposition to Schur Decomposition  Calculus & Beyond Homework  1  
Explain me what the nuclear isospin is?  High Energy, Nuclear, Particle Physics  4  
What exactly is isospin?  High Energy, Nuclear, Particle Physics  4 