The Tech Basement

Feast your eyes on the below figure:

It might seem fine and dandy, but consider this: If you look at for example $0f7/2 \rightarrow 0f5/2$ with a value of 3.2 %, note that it is larger than the reverse $0f5/2 \rightarrow 0f7/2$ which is at 2.3 %. Since $0f7/2$ is at lower energy than $0f5/2$ and consequently that $0f5/2 \rightarrow 0f7/2$ should be more energetically favourable than the opposite, then how come the opposite has a larger percentage of the OBTDs? It might simply be that I incorrectly have swapped some indices, let's take a closer look at that!

Recall that kshell-utilities checks whether or not the final state has lower energy than the initial state and swaps them if that is not the case. However, the OBTDs are stored in separate files, also organised after angular momentum and parity, for example '…_j0n_…_j2n_…', and I have not implemented such a check and swap for the OBTD files! The reverse file '…_j2n_…_j0n_…' does not exist as KSHELL stores both 0n to 2n and 2n to 0n in the '…_j0n_…_j2n_…' file, and since all the entries in that file is organised with 0n to the left and 2n to the right, the entries representing 2n to 0n need to have their indices swapped! This is as I described, what is happening with the transitions after line 174 here, but I have to do the same for the OBTD files.