Statement about Forbidden Alpha decay transitions.

[Silversonic]

Hi, I'm confused about a statement about the change in final/initial states of the daughter/parent atom in an alpha decay. It is the following;

"The spin between the parent ($I_i$) and daughter ($I_f$) can change by $lh$ (h being h-bar, l is the orbital angular quantum number of the alpha particle), where;

$\vec{I_i} = \vec{I_f} + \vec{l}$

and the parity changes by $(-1)^l$ "

I'm confused because if, for example, we take the initial state of the parent to be $0^+$, then there are the following cases;

$\vec{I_i} = \vec{I_f} + \vec{l}$ means

$|\vec{I_i}| = |\vec{I_f} + \vec{l}|$

Coupling angular momentum together would surely mean that the total orbital quantum number would be of multiple values;

$L = I_f + l, I_f + l - 1, I_f + l - 2, ... I_f - l$

So if $I_i = 0^+$ we could have multiple $I_f$ for a given $l$. For example $l = 1$, then

$L = I_f + 1, I_f , I_f - 1 = I_i = 0$

Meaning $I_f$ could take on values $0$ or $1$. My notes seem to suggest only the $I_f = 1$ state is possible. Am I looking at this in completely the wrong way? I think I don't fully understand what it means by "can change by $lh$", what is the signficance of the "can" change?

I looked on the internet and in my textbook, not much to a detail that I can understand. Beta decay forbidden decays seemed to be the closest I could find which might explain it but I don't know how applicable it is to this situation.

 

[mfb]

$I_i=0$ gives $|\vec{I_f}+\vec{l}|=0$, which has the solution $\vec{I_f}=-\vec{l}$ only.

 

[Silversonic]

$I_i=0$ gives $|\vec{I_f}+\vec{l}|=0$, which has the solution $\vec{I_f}=-\vec{l}$ only.

Ah yes clearly. I should've realized that. Here's a better example which might help me understand it for general $l$ and $I_f$.

Say the initial state was $I_i = 1^+$. We must have the coupled orbital angular momentum quantum number $L$ equal to $1$. Then say the alpha particle had $l = 2$

The coupled orbital angular momentum quantum number of

$\vec{I_i} = \vec{I_f} + \vec{l}$

Would be

$L = I_f + l, |I_f + l - 1|, |I_f + l - 2|, ... |I_f - l|$

So if we had $I_f = 3$

$L = 5,4,3,2,1$

Corresponding to a possibility to have $L = 1$

If we had $I_f = 2$

$L = 4,3,2,1,0$

Corresponding to another possibility to have $L = 1$

And lastly; If we had $I_f = 1$

$L = 3,2,1$

Corresponding to another possibility to have $L = 1$So in a transition from a state $1^+$ with the emission of an alpha with $l = 2$, it's possible to have 3 final states for the daughter ($3^-, 2^-, 1^-$)? Is this correct?

 

[mfb]

Looks correct.

$|l-I_i| \leq I_f \leq |I+I_i|$