I Counting Total Spin for N Two-Level Systems (TLS)

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The discussion focuses on counting total spin states for N two-level systems (TLS), particularly examining the implications of fixed r values on state counts. For r = N/2, the system has N + 1 states, while for r = N/2 - 1, it results in (N - 1)^2 states, leading to increased complexity as r decreases further. Participants express confusion regarding the validity of r = N/2 - 1 due to individual spins being half and the unclear multiplicity of states across irreducible representations (irreps) of SU(2). The correction from 2m + 1 to 2r + 1 is noted, emphasizing the importance of accurate terminology in the context of spin counting. Overall, the conversation highlights the intricacies of state counting in composite quantum systems.
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TL;DR
I am having considerable trouble understanding the addition of spins. The context in which I am studying this is the Tavis-Cummings Model.
Statement:
"Assume that r and m mean total spin and projection of spin along z, respectively. For N-TLS the total spin (r) can assume N+1 to 1/2 or 0 spin depending on N being even or odd. For a fixed r the value of m varies from +j to -j in integer steps. R is the operator whose eigen-values are r. The basis choice is |r,m>."
Now then, if I intend to make a matrix pretaining to single transitions of the composite system, i align the states with fixed r. For fixed r I have 2m+1 states. When r=N/2 my states are N+1 as simple substitution verifies. However, when r=N/2 -1 the number of states are (N-1)^2. It gets weirder for N/2 -2

Questions:
1) Why is r =N/2 -1 valid as individual spin is half not 1.
2) The counting on main diagonals are pretty confusing. For each irrep of SU(2) there's 2m+1 states and that's fine. But the multiplicity of each state is entirely vague to me.

I have linked the original paper and the figure 1 is where the counting is shown.

1730022772893.png
 

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Correction: its 2r+1 instead of 2m+1
 
For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

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