2
$\begingroup$

I am asked the following question: A kaon $K^0$ decays into two pions $\pi^0$, being this two pions particles with spin zero. Using the conservation of angular momentum and the fact that the two pions are identical particles, analyze which restrictions are imposed over the spin of the $K^0$.

This question is in an exam of an introductory course on Quantum Mechanics, so the answer should be simple and not involve anything about quarks. Here is what I did:

First of all I wrote the most general initial and final state as

enter image description here

Knowing that the pions are particles of spin zero, because the question specifies this, we have that $s_{\pi^0} \equiv 0$ so the final state can be reduced to

enter image description here

Now I consider the center of mass system. In this system we should have all the orbital angular momentums to be zero because 1) the kaon is at rest and 2) the pions are emitted in radial directions (I assume this). Thus the initial and final states become

enter image description here

Finally I apply the conservation of angular momentum. The final angular momentum is zero and so it must be the initial total angular momentum, thus the spin of the kaon must be zero.

I have doubts about this answer because I have never used the fact that the pions are identical particles... Do you think that what I did is correct?

Improve this question
$\endgroup$
0

3 Answers 3

Reset to default
4
$\begingroup$

Because the pions are spinless they must be in a state whose orbital angular momentum is equal to the spin of the kaon.

Because they are identical bosons, they must be in a state which is even if their labels are interchanged.

This means they must be in a state where L is even. Hence the spin of the kaon must be 0,2,4... And not 1,3,5...

Note. The $\rho^0$, which has a spin of 1, decays to $\pi^+ \pi^-$ but not $\pi^0 \pi^0$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ The orbital angular momentum you are assigning to the pions has to do with the orbit of the quarks inside each pion? Otherwise I cannot see how two non interacting particles that fly apart radially can have angular momentum. $\endgroup$
    – user171780
    Jul 30, 2018 at 11:44
  • $\begingroup$ Nothing to do with the quarks. The pions do not fly apart radially. They have equal and opposite momentum but there is a (too small to be directly measurable) offset between their trajectories. $\endgroup$
    – RogerJBarlow
    Jul 30, 2018 at 16:59
0
$\begingroup$

The only logical seems to me to be that the angular momentum goes into the movement of the quarks the pions consist of.

Improve this answer
$\endgroup$
0
$\begingroup$

A system of two pions can have angular momentum. Without measuring this no conclusion can be drawn for the kaon spin.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ How can two non interacting particles moving radially have angular momentum other than spin? $\endgroup$
    – user171780
    Jul 29, 2018 at 17:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.