electronquark
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- TL;DR
- Im not sure how one determines the angular distribution of the decay products in a particle decay
For decays of particles one can use angular momentum conservation to find the final states via Clebsch Gordan coefficients for example for the decay of a spin zero particle into spin 1/2. The Wigner d matrices are used to find the angular distributions. What I don’t get is given some final state how does one properly find the angular distribution I can’t seem to find many good sources on this. Does one just assume each spin product state is a product of spherical harmonics if not what is the correct way to find the total angular distribution. Is it a sum of squares is it weighted how does one determine this.
For example a spin zero particle decaying into 2 spin one particles we know the coupled spin state is given as via the Clebsch Gordan Coefficients
$$|0,0\rangle = \frac{1}{\sqrt{3}}(|-1,1\rangle+|1,-1\rangle-|0,0\rangle$$
Where the state on the left is denoted as
$$|j,m\rangle$$
And on the right as
$$|m_1,m_2\rangle$$
where j1 and j2 are 1
From this how does one determine the angular distribution does one replace m1 and m2 states with the joint spherical harmonics Y1m1Y1m2 if not what is the way to do this. What I want is the joint angular distributions or the reduced one body angular distribution.
$$p(\Omega_1,\Omega_2)$$
For a decay of a spin 1/2 particle into spin 0 and 1/2 as seen in Merzbacher ch. 17 this is straightforward and determined by the probability the decay product is aligned with the axis. But for more general spin like spin 1 it doesn’t seem clear since either rotated m=1 m=0 contribute how does one properly determine which one to use. Also how does one include orbital angular momentum in this and how does one define the angular distribution in this case is it the spatial distribution due to orbital l or due to how the spin is aligned which one. I just want to have the right understanding.
For example a spin zero particle decaying into 2 spin one particles we know the coupled spin state is given as via the Clebsch Gordan Coefficients
$$|0,0\rangle = \frac{1}{\sqrt{3}}(|-1,1\rangle+|1,-1\rangle-|0,0\rangle$$
Where the state on the left is denoted as
$$|j,m\rangle$$
And on the right as
$$|m_1,m_2\rangle$$
where j1 and j2 are 1
From this how does one determine the angular distribution does one replace m1 and m2 states with the joint spherical harmonics Y1m1Y1m2 if not what is the way to do this. What I want is the joint angular distributions or the reduced one body angular distribution.
$$p(\Omega_1,\Omega_2)$$
For a decay of a spin 1/2 particle into spin 0 and 1/2 as seen in Merzbacher ch. 17 this is straightforward and determined by the probability the decay product is aligned with the axis. But for more general spin like spin 1 it doesn’t seem clear since either rotated m=1 m=0 contribute how does one properly determine which one to use. Also how does one include orbital angular momentum in this and how does one define the angular distribution in this case is it the spatial distribution due to orbital l or due to how the spin is aligned which one. I just want to have the right understanding.