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Welcome to The Tech Basement!
The Tech Basement is my personal knowledge base! It is organised as a wiki, but I am also using it like a blog. The subjects are programming, maths and physics, which are subjects highly related to my the PhD I'm taking in computational nuclear physics.
Check out the latest entries in The Tech Basement below!
Which values to choose for gl and gs?
The gyromagnetic ratios determine how strongly the nuclear magnetic dipole moment interacts with the electromagnetic field, which directly affects the M1 transition strength. They show up in the $M1$ operator as
$$ \hat{M1} = g_l \hat{L} + g_s \hat{S}. $$
Protons and neutrons have different gyromagnetic ratios. The orbital motion of a charged particle produces a magnetic moment. Since only protons have charge, they alone contribute to the orbital part of the nuclear magnetic moment. The magnetic moment of a moving charge is given by
$$ \mu = \frac{q}{2m}L $$
(continue with completing this text and look at what happens to the M1 GSF when the orbital g-factors change, calculate some magnetic moments for different spin g-factors and see what makes most sense!) TBC
Orbital contributions to the LEE
Let's get to the crux of the matter. We can analyse OBTDs all day, but what we're really wondering about is which orbitals are contributing to the enhancement in the low-energy region of the gamma strength function. Recall that in shell model calculations, the reduced transition strength is calculated by
$$ B(\sigma \lambda; \xi_i j_i \rightarrow \xi_f j_f) = \frac{1}{2 j_i + 1} \mid ( \xi_f j_f \mid \mid M_{\sigma \lambda} \mid \mid \xi_i j_i ) \mid^2, \qquad(0) $$
$$ \langle \Psi_f | \hat{O}_{\lambda \mu} | \Psi_i \rangle = \sum_{\alpha \beta} \langle \alpha | \hat{o}_{\lambda \mu} | \beta \rangle \langle \Psi_f | \hat{c}^\dagger_\alpha \hat{c}_\beta | \Psi_i \rangle. \qquad(1) $$
I so happen to be in possession of all the OBTDs and reduced matrix elements needed to re-calculate any transition of my desire. Now, what if I were to simply skip certain orbitals ($\alpha$ and $\beta$) in eq. (1)? I could for example decide to skip any mention of $0f7/2 \rightarrow 0f7/2$, or any other single-particle transitions for that matter. Re-calculating all of the transition strengths with the modified OBTD files consequently gives me the possibility to re-calculate the $M1$ GSF with the modified values. We can then directly see how the LEE is affected by the modification! That's what I call a home-run! schmack
[SOLVED] Possible bug in the OBTD calculations
So there is a possible bug that I am trying to wrap my head around. In the OBTD log files from KSHELL, the OBTDs are listed in blocks where each block represents an initial and a final state aka. one specific transition. For example
w.f. J1= 0/2( 3) J2= 2/2( 1) B(L;=>), B(L ;<=) 0.00622 0.00207 <||L||> 3 1 -0.13590 0.11892 i j OBTD <i||L||j> OBTD*<||> 1 1 0.00244 5.79655 0.01416 1 2 -0.00012 1.54919 -0.00018 1 9 0.00000 0.00000 0.00000 1 10 0.00000 0.00000 0.00000 ...
This block represents the OBTDs for the transition from the 3rd 0- state to the 1st 1- state (the parity is not listed in the block, but I know it is negative). Now, there is no guarantee that the 3rd 0- state has higher energy than the 1st 1- state. It might be so, but it is not guaranteed. If the initial state however has lower excitation energy than the final state then the OBTD indices in the blocks ($i, j$) have to be swapped in order to make it so that the final state is of lower energy.
So I made a short script which runs through all of the OBTD blocks and looks up the excitation energy of the initial and final state for the given block. The result was that 628337 of 1280000 (49.09 %) blocks had initial state excitation energy lower than the final state. Thats good, as expected, some of the final states are of lower energy and some are of higher. The puzzle however is that when I make a sub-set of transitions, selecting only the transitions which are present within the $E_\gamma = [0, 3]$ MeV region of the $M1$ GSF then only 3 of 230397 (0.00 %) transitions have initial energies lower than final. I don't understand where in the process of calculating the GSF that this kind of energy sorting happens.
Visualising OBTDs
The one-body transition density (OBTD) is used to calculate the transition probability from an initial to a final state:
$$ \langle \Psi_f | \hat{O}_{\lambda \mu} | \Psi_i \rangle = \sum_{\alpha \beta} \langle \alpha | \hat{o}_{\lambda \mu} | \beta \rangle \langle \Psi_f | \hat{c}^\dagger_\alpha \hat{c}_\beta | \Psi_i \rangle \qquad(0) $$
with the OBTD defined as:
$$ \rho_{fi}(\alpha, \beta) = \langle \Psi_f | \hat{c}^\dagger_\alpha \hat{c}_\beta | \Psi_i \rangle. \qquad(1) $$
Assume $\alpha$ and $\beta$ ($i$ and $j$ in the list below) to be indices of the orbitals (not the $m$ substates) in the model space, which for sd-pf-sdg are 12 proton orbitals and 12 neutron orbitals. Let's select a subset of transitions so that we have transitions of gamma energies only within a certain gamma energy interval $[E_\text{min}, E_\text{max}]$. This might for example be the energy interval of the LEE. Let's also select transitions based on their multipolarities so that we look at only, for example, $M1$ transitions. Transitions of $0^- \rightarrow 1^-$ are within these requirements, and the raw OBTD data for one such transition is listed below. Note that the $M1$ operator has an $L$ and an $S$ term
$$ \hat{M1} = g_l \hat{L} + g_s \hat{S}.\qquad(2) $$
The file below lists some $L$ matrix element values for vanadium 50 and there is a corresponding file with the $S$ values.
w.f. J1= 0/2( 3) J2= 2/2( 1) B(L;=>), B(L ;<=) 0.00622 0.00207 <||L||> 3 1 -0.13590 0.11892 i j OBTD <i||L||j> OBTD*<||> 1 1 0.00244 5.79655 0.01416 1 2 -0.00012 1.54919 -0.00018 1 9 0.00000 0.00000 0.00000 1 10 0.00000 0.00000 0.00000 1 11 0.00000 0.00000 0.00000 ...
Click here to see the complete L list
Click here to see the complete S list
Since there are no transitions between proton and neutron orbitals (gamma decay cannot turn protons into neutrons), we can organise the OBTDs into two 12×12 grids, one grid for protons and one grid for neutrons, for example like this (protons):
Notice that the OBTD can be negative. If we don't care about the sign itself, we might take the absolute value of the OBTDs to get a feel of how large the overlap of the initial and final states are given the single-particle transition from $\alpha$ ($i$) to $\beta$ ($j$). One such 12×12 proton OBTD grid can be made for not just this one transition, but one can be made for each of the transitions we selected, which might be tens or hundreds of thousands of transitions. Let's then consolidate all of the thousands of 12×12 grids into a single 12×12 grid by adding the grids at each grid value. This results in some very large grid values whose magnitude doesn't really matter; what matters is how large the values are relative to each other. Let's then normalise by dividing each grid value by the sum of all the grid values and multiply by 100 to express them as a percentage of the sum of all the OBTDs:
Note that the percentages do not add to 100% because they are normalised to both proton and neutron values and the above plot shows only proton values. The sum of the proton and neutron grid values add up to 100%.
An important consideration to make is that the OBTD eq. (1) has no dependency on any transition operator. The OBTD simply tells us the overlap between the initial and final wavefunctions given a single-particle transition in the initial wavefunction. The OBTD does not care about electromagnetic transition rules, so if we want to consider OBTDs for a subset of transitions with an $M1$ characteristic we need to do some filtering.
Note that the $M1$ operator in eq. (2) consists of an $L$ and an $S$ term, the values for $L$ and $S$ for a single transition are given in the above lists. We see that in some cases a non-zero OBTD value has corresponding $L$ and $S$ values which are zero. In such a case, I think it would be wrong to include that OBTD value in the 12×12 grid since its contribution is zeroed by transition rules. Here I'm assuming that the angular momentum and parity transition rules manifest as the $L$ and $S$ terms being zero. If you look at eq. (0), where else could the transition rules come into play? There are only two terms.
Now what can we say about these two latest 12×12 grids… First, the single-particle transitions seem to respect the transition rules for $M1$ transitions. Recall that $M1$ transitions have no change in parity from the initial to the final state which means that the only allowed single-particle transitions are within each major shell. We can clearly see block structures within $sd$, $pf$, and $sdg$. Second, $M1$ transitions allow only $\Delta j = 1$, recall the rule
$$ \Delta j = | j_i - j_f |, ..., j_i + j_f $$
which means that, for example, a single-particle transition from $s1/2$ to $d5/2$ has a minimum $\Delta j$ of 2, and consequently that this single-particle transition is illegal for $M1$ transitions.
We can see that even before explicitly respecting the $M1$ transition rules by skipping OBTDs where $L$ and $S$ are zero, the OBTDs in the 12×12 grid already agree with $\Delta j = 1$ and $\Delta \pi = \text{no}$. This probably comes from the fact that I already made a selection of transitions which are $M1$ transitions, but just because I selected $M1$ transitions does not mean that all single-particle transitions within the $M1$ transitions (eq. (0)) are allowed.
A still open question to me is why the $1s1/2 \rightarrow 0d3/2$ OBTDs disappeared when
$$ \Delta j = 3/2 - 1/2, ..., 3/2 + 1/2 = 1, 2 $$
allows $M1$ transitions…
Discussion