If we consider the decay of a spin-1 particle with spin projection $m_s=1$ into two (distinguishable) spin-0 particles, what are the possible values of the orbital angular momenta $l$ of the resultant particles?

Using the rules for addition of angular momentum, $m_s=m_1+m_2$ so for $(m_1, m_2)$ we have $(1,0)$ or $(0,1)$.

But the total angular momentum is initially $j=1$ so $|l_1-l_2|\leq j\leq l_1+l_2$. So naively, I'm thinking that $l_1$ and $l_2$ can be arbitrarily large.

I don't think this is right, but I can't figure out what I'm missing.


1 Answer 1


There is no $l_1$ and $l_2$, there is just $l$.

This problem has an $\vec r_1$ and $\vec r_2$, but you solve it in terms of:

$$\vec R =(m_1\vec r_1 + m_2\vec r_2)/(m_1 + m_2)$$

which is the center-of-mass coordinate, so set it to 0 and forget about it.

The other coordinate is:

$$ \vec r = \vec r_1 - \vec r_2 $$

and solve for that coordinate using the reduced mass:

$$ \mu = \frac 1 {\frac 1 {m_1} + \frac 1 {m_2}} $$

When is all said an done, you should find:

$$ \Psi(\vec r_1) = \psi(r)Y_1^1(\theta, \phi) $$

so that:

$$ l = 1 $$


$$ l_z = +1$$


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