Parity of [tex]^{9}_{5}B^{5}[/tex]

[tirrel]

Hy... I'm reviewing some nuclear physics but don't remember certain topics.

This exercise for example:

Knowing that ^{9}_{4}Be^{5} and ^{9}_{5}B^{4} have both as quantum numbers \frac{3}{2}^{-} and their properties are determined by the odd nucleon, show that the ground state of ^{9}_{5}B^{5} can be \frac{3}^{+}.

My thinkings, which are correct?:

First of all we see that ^{9}_{5}B^{5} has as outermost nucleons the odd nucleon of ^{9}_{4}Be^{5} and the odd nucleon of ^{9}_{5}B^{4}. So I=I_1+I_2, and thanks to the rules of q.m. I \in \{0,1,2,3\}. 3 is present and we are happy.
So for both ^{9}_{4}Be^{5} and ^{9}_{5}B^{4} we have for the odd nucleon I=\frac{3}{2}. but thanks to the rules of angular momentum I=L+S and so since I=\frac{1}{2}, I \in \{l+\frac{1}{2},l-\frac{1}{2}\} and this forces l \in \{1,2\} , Now the parity is (-1)^l, so we are forced to have l=1 for both particles. I don't know is this is useful but I think is correct.
Now maybe we can say that the state in ^{9}_{5}B^{5} will be a superposition of vectors of the type v \otimes w, where the vector v refers to one nucleon and the vector w to the other. The parity on each member acts on each subspace: \Pi v \otimes w= \Pi_1 v \otimes \Pi w_2=(-1)^2v \otimes w. and so the parity is positive. This reasoning seem to tell that to find the parity I simply have to multplicate the parities of the individual nucleons. is it right? sounds strange...

 

[tirrel]

ehm... the firs messag behaves strange and I'm not able to modify it... can you delete and leave only the second?:


Hy... I'm reviewing some nuclear physics but don't remember certain topics.

This exercise for example:

Knowing that ^{9}_{4}Be^{5} and ^{9}_{5}B^{4} have both as quantum numbers \frac{3}{2}^{-} and their properties are determined by the odd nucleon, show that the ground state of ^{9}_{5}B^{5} can be 3^{+}.

My thinkings, which are correct?:

First of all we see that ^{9}_{5}B^{5} has as outermost nucleons the odd nucleon of ^{9}_{4}Be^{5} and the odd nucleon of ^{9}_{5}B^{4}. So I=I_1+I_2, and thanks to the rules of q.m. I \in \{0,1,2,3\}. 3 is present and we are happy.
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For both ^{9}_{4}Be^{5} and ^{9}_{5}B^{4} we have for the odd nucleon I=\frac{3}{2}. but thanks to the rules of angular momentum I=L+S and so since S=\frac{1}{2}, I \in \{l+\frac{1}{2},l-\frac{1}{2}\} and this forces l \in \{1,2\} , Now the parity is (-1)^l, so we are forced to have l=1 for both particles. I don't know is this is useful but I think is correct.
\\
\\
\\
\\
Now maybe we can say that the state in ^{9}_{5}B^{5} will be a superposition of vectors of the type v \otimes w, where the vector v refers to one nucleon and the vector w to the other. The parity on each member acts on each subspace: \Pi v \otimes w= \Pi_1 v \otimes \Pi_2 w=(-1)^2v \otimes w. and so the parity is positive. This reasoning seem to tell that to find the parity I simply have to multplicate the parities of the individual nucleons. is it right? sounds strange...