# Decay of spin-1 particle into two spin-0 particles

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.

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:

$$l = 1$$

and

$$l_z = +1$$