I Double Check Normalization Condition

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The discussion focuses on the normalization condition for the state defined as ##\ket{\Psi} = \sum_{1 \leq n_{1} \leq n_{2} \leq N} a(n_{1},n_{2})\ket{n_{1},n_{2}}##, where the coefficients ##|a(n_{1},n_{2})|## are proportional to ##\cosh[(x-1/2)N\ln N]##. It is claimed that coefficients with ##n_{2}-n_{1} > 1## approach zero as ##N\rightarrow\infty##, raising concerns about normalization. The proposed normalization condition is $$C^{2}\frac{1}{4}\sum_{n_{1},n_{2}}^{N} |a(n_{1},n_{2})|^2 = 1$$, with the factor of ##1/4## accounting for the ordering in the sum. After further analysis, it is confirmed that this normalization condition is indeed correct. The discussion highlights the importance of proper normalization to avoid divergence in the coefficients.
thatboi
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Consider the state ##\ket{\Psi} = \sum_{1 \leq n_{1} \leq n_{2} \leq N} a(n_{1},n_{2})\ket{n_{1},n_{2}}## and suppose $$|a(n_{1},n_{2})| \propto \cosh[(x-1/2)N\ln N]$$ where ##0<x=(n_{1}-n_{2})/N<1##. The claim is that all ##a(n_{1},n_{2})## with ##n_{2}-n_{1} > 1## go to ##0## as ##N\rightarrow\infty##. Clearly we need some kind of normalization constant, otherwise the cosh function should just blow up. So is the right normalization condition then $$C^{2}\frac{1}{4}\sum_{n_{1},n_{2}}^{N} |a(n_{1},n_{2})|^2 = 1$$ where ##C## is our normalization constant (I introduced the ##1/4## because I removed the ordering in the sum)? Because I tried doing the calculation and making the plot but I still cannot see this exponential decay.
 
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Ok I took another crack at the problem and this is indeed the correct normalization condition.
 
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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