Clarification neutral Kaon decay (Griffith, Elem. Particles)

[Dr.AbeNikIanEdL]

I was reading in Griffit's "Introduction to Elementary Particles" when I stumbled over the following Remark in chapter 4.4.3.1. CP eigenstates are constructed by

$|K_1> = \frac{1}{\sqrt{2}}(|K^0> - |\bar{K^0})$
$|K_2> = \frac{1}{\sqrt{2}}(|K^0> + |\bar{K^0})$

and, assuming CP conservation, the only possible decay modes should be

$K_1 \rightarrow 2\pi$
$K_2 \rightarrow 3\pi$

In the footnote at the end of the page (146) it is mentioned that "with the right combination of orbital angular momentum, it is possible to construct a $CP=+1$ state of the [$3\pi$] system, [...] this might allow $K_1$ to decay (rarely) into $3\pi$ [...]".
Now my question is, as Kaons as well as pions have spin 0, wouldn't angular momentum conservation forbid any orbital angular momentum in the final state? Do I miss something or is $K_1 \rightarrow 3\pi$ then rigorously forbidden (under the assumption of strikt CP conservation)?

 

[nrqed]

I was reading in Griffit's "Introduction to Elementary Particles" when I stumbled over the following Remark in chapter 4.4.3.1. CP eigenstates are constructed by

$|K_1> = \frac{1}{\sqrt{2}}(|K^0> - |\bar{K^0})$
$|K_2> = \frac{1}{\sqrt{2}}(|K^0> + |\bar{K^0})$

and, assuming CP conservation, the only possible decay modes should be

$K_1 \rightarrow 2\pi$
$K_2 \rightarrow 3\pi$

In the footnote at the end of the page (146) it is mentioned that "with the right combination of orbital angular momentum, it is possible to construct a $CP=+1$ state of the [$3\pi$] system, [...] this might allow $K_1$ to decay (rarely) into $3\pi$ [...]".
Now my question is, as Kaons as well as pions have spin 0, wouldn't angular momentum conservation forbid any orbital angular momentum in the final state? Do I miss something or is $K_1 \rightarrow 3\pi$ then rigorously forbidden (under the assumption of strikt CP conservation)?

Why do you think that orbital angular momentum would be forbidden in the final state? This is unrelated to the particles being spin 0 or not. You can in principle imagine a bound state of two spin 0 particles and this bound state could have excited states with nonzero angular momentum. Can you clarify your question?

 

[Dr.AbeNikIanEdL]

My reasoning was as follows:
Total angular momentum should be conserved in this decay, so $J = l + s$ should not change. The Kaon has $l = 0$ and $s = 0$, so $J = 0$. Therefore the final state must have $J = 0$ as well. Pions are spin 0 particles, so there are no spins to add up, $s = 0$ in the overall final state. So any nonzero angular momentum $l$ would make the total angular momentum $J$ being some nonzero value, violating total angular momentum conservation.
I am clear that three pions could form a state with nonzero angular momentum, I just don't understand how this state could be produced in a Kaon decay, sorry for not being clear about that.