Before decay, the nucleus is in state with total angular momentum J and symmetry axis quantization M :

\Phi_{JM}

Say, the emitted radiation (can be EM wave or particle ) carries angular momentum  l and axis quantization m, its wavefunction is:

\phi_{lm}

then the daughter nucleus has angular momentum j and m_j, the wave function is

\Psi_{j m_j}

their relation is:

\Phi_{JM} = \sum_{m, m_j}{\phi_{lm} \Psi_{j m_j} \left< l m j m_j | JM \right>}

where the $Latex \left< l m j m_j | JM \right> $ is Clebsch-Gordan coefficient.

The wave function of the emitted radiation from a central interaction takes the form:

\phi_{lm} = A_0 u_{nl}(r) Y_l^m(\theta,\phi)

The angular distribution is:

\int |\phi_{lm}|^2 dx^3 =A_0^2 \int |u_{nl}|^2 r^2 dr |Y_l^m|^2

for a fixed distance detector, the radial part is a constant. Moreover, not every spherical harmonic contribute the same weight, there is weighting factor due to Clebsch-Gordan coefficient.   Thus, the angular distribution is proportional to:

W(\theta) \propto \sum_{m_j=M-m}{|Y_l^m|^2 |\left<l m j m_j |JM\right>|^2 }

For example, JM=00, The possible (l, j) are (0,0), (1,1), (2,2) and so on, the m=-m_j . The C-G coefficient are,

\langle 0000 | 00 \rangle = 1

\langle lml-m | 00 \rangle = \frac{1}{\sqrt{2l+1}}

 

thus,

Y_0^0 = \frac{1}{4\pi}

\displaystyle \sum_{m}{\left|Y_l^m\right|^2  \left|\langle l m l -m |0 0 \rangle \right|^2 } =  \sum_m|Y_l^m|^2 \frac{1}{2l+1}= constant

Thus, the angular distribution is isotropic.

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