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--- science:phd-notes:2025-02-17-obtd [2025/02/18 08:24] – Continue thought process jon-dokuwiki
+++ science:phd-notes:2025-02-17-obtd [2025/03/31 14:56] (current) – Elaborate on decays and excites jon-dokuwiki
@@ Line 1: / Line 1: @@
-===== Another possible bug in the OBTD calculations =====
+===== Another possible bug in the OBTD calculations: Why are excitations stronger than decays? =====
+==== Making sure that initial and final states are ordered correctly ====
 Feast your eyes on the below figure:
@@ Line 38: / Line 40: @@
 so if we take $B(EM) \rightarrow$ as the //decay// probability then it seems that the $0^-$ state is the initial state of the decay, while the $1^-$ state is the initial state of the excitation. This seems contrary to the header of the table...
-Now consider that for this transition, the energy of the final state is actually higher than the energy of the initial state, aka. $E_x$ is negative. A decay where $E_f > E_i$ is no decay at all, but rather an excitation, and I would dare say that $B(EM) \rightarrow$ in the table consequently is the excitation probability. To make sense of it all, I will for transitions of negative $E_x$ flip the roles of initial and final, and decay and excite, and say that the initial state is $0^-$ and the final state is $1^-$
+Now consider that for this transition, the energy of the final state is actually higher than the energy of the initial state, aka. $E_x$ is negative. A decay where $E_f > E_i$ is no decay at all, but rather an excitation, and I would dare say that $B(EM) \rightarrow$ in the table consequently is the excitation probability. To make sense of it all, I will for transitions of negative $E_x$ flip the roles of initial and final, and decay and excite.
+While conceptually simple, it proved to be a bit of a riddle to solve this programmatically because I want it to fit with my existing dictionary structure. I think it is solved now, and as any programmer should do, I implemented [[https://github.com/GaffaSnobb/kshell-utilities/blob/00eb1aa3762fb1685454e939c64fe7cbce67ad47/kshell_utilities/test_loaders.py#L124|tests]] which I do believe covers all of the possible cases. The tests are [[https://github.com/GaffaSnobb/kshell-utilities/blob/00eb1aa3762fb1685454e939c64fe7cbce67ad47/kshell_utilities/kshell_utilities.py#L365|called]] every time $\verb|_read_obtd|$ is called which is practically speaking every time $\verb|loadtxt|$ is called.
+I have now double and triple checked that the OBTDs are organised so that $E_i > E_f$ and just to be sure I have put checks into the code that generates the OBTD grids that $E_i > E_f$ is indeed true, and something has happened to the values, however not what I was aiming for in this text... (protons left, neutrons right)
+{{ :science:phd-notes:proton_neutron_grid.png |}}
+==== Wth is going on ====
+So after all that making sure that the levels are correct, I still get that the excitation "strength" is larger than the decay. So let us now elaborate on why I'm calling it "strength" and not strength. In the plots we're just looking at the OBTDs which are not the complete expression for the (reduced) transition strength. So maybe we are led astray by that fact? Let's see what happens if we plot not only the OBTDs but the OBTDs multiplied by the single-particle transition matrix elements (see the definitions [[science:phd-notes:2025-03-03-lee|here]]). We're still not looking at the actual reduced strength as we would need to sum up over all the orbitals / single-particle contributions to get the actual strength and then we would no longer be able to make the grids.
+{{ :science:phd-notes:obtd-times-ls-protons-and-neutrons-yes-abs.png |}}
+Hmm, it seems that the excitation "strength" is still not afraid of being larger than the decay "strength"...  What if we sum up without taking the abs of each term?
+{{ :science:phd-notes:obtd-times-ls-protons-and-neutrons-no-abs.png |}}
+Nope, does not seem to solve anything. Let's think about the original assumption, namely that the decay strength should be larger than the excitation strength. That is an alright assumption, but remember that we are not actually looking at true strengths here, even if we multiply the OBTDs by the L, S terms. So maybe the assumption is bad for this situation?