l: 1..3
m: -3..3
s: 1..2
900 elements

baptiste
June 27, 2024
The convention is to use
so for a T-matrix with \(lmax=3\), we have
l: 1..3
m: -3..3
s: 1..2
900 elements

for each block we introduce a combined p-index such that,
\[
p(n,m) = n(n+1)+m
\] which here varies as p: 1..15, as each block is of dimension pmax = lmax(lmax+1)+lmax = 3(3+1)+3 = 15.
The whole T-matrix is indexed with a combined q-index such that,
\[ q(s,p) = (s-1)p_{max} + p \] which here gives us, q: 1..30. The total T-matrix dimension (along rows/columns) is thus \(q_{max}=2\times(lmax(lmax+1)+lmax)=30\).
In summary, for a given \(l_\text{max}\) the indices are given by:
Given \(q\) and \(l_\text{max}\) we can invert these indices as follows,
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 | 28 | 29 | 30 | |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| p | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
| l | 1 | 1 | 1 | 2 | 2 | 2 | 2 | 2 | 3 | 3 | 3 | 3 | 3 | 3 | 3 | 1 | 1 | 1 | 2 | 2 | 2 | 2 | 2 | 3 | 3 | 3 | 3 | 3 | 3 | 3 |
| m | -1 | 0 | 1 | -2 | -1 | 0 | 1 | 2 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | -1 | 0 | 1 | -2 | -1 | 0 | 1 | 2 | -3 | -2 | -1 | 0 | 1 | 2 | 3 |
| s | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 2 | 2 | 2 | 2 | 2 | 2 | 2 | 2 | 2 | 2 | 2 | 2 | 2 | 2 | 2 |