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I read in an online portal about $^{112}$Sn nucleus making a transition from $0_{g.s}^{+} \rightarrow 2_{1}^{+}$ state. Also, some higher excited states were named as $0_{2}^{+}$, $3_{1}^{-}$, etc. From shell model, it is known that the nucleus has a ground state $0^+$ and first excited state $2^+$. What is the significance of the subscripts $g.s.$,$1$ etc. used in the nomenclature?

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  • $\begingroup$ The g.s is ground state. $\endgroup$
    – Jon Custer
    Commented Sep 24, 2015 at 12:45
  • $\begingroup$ It's usually another quantum number. In the light nuclei you make a point to distinguish states by isospin, but tin is heavy enough that isospin shouldn't be a good symmetry. If you can give us the reference you're reading we can help you puzzle it out. $\endgroup$
    – rob
    Commented Sep 24, 2015 at 13:24
  • $\begingroup$ I understand $g.s.$ is the ground state, but what is the need of writing that if $0^+$ is the known ground state. What does the state $0_2^+$ signify? Soes the subscript indicate the $l$ value corresponding to $J=0$? $\endgroup$
    – Ana
    Commented Sep 25, 2015 at 10:20

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As an example, the subscripts 1,2 and 3 after a $0^+$ would mean the first, second and the third $0^+$ state, respectively in increasing order of excitation energy. The states containing subscript 1 is usually known by the name 'yrast' state whereas others are known as 'yrare' states.

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    $\begingroup$ Thanks for your answer! Could you provide a link to a paper which uses and explains this notation? $\endgroup$
    – rob
    Commented Jun 29, 2019 at 21:32
  • $\begingroup$ To be honest, this notation is quite trivial in itself. Having said that, you can get a very good idea about it by understanding how the various $0^+$ states were formed in the first place. For that, I suggest you to go through the classic example of formation of $0^+$ states in $^{18}O$, with $^{16}O$ as the core and two nucleons outside it in the $sd$-shell. It is a very easy problem to solve. I recommend the book by Brussaard and Glaudemans. Please fell free if you need any further explanations. $\endgroup$ Commented Jun 29, 2019 at 23:32

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