Undergrad Double Check Normalization Condition

Click For Summary
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
Messages
130
Reaction score
20
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.
 
Physics news on Phys.org
Ok I took another crack at the problem and this is indeed the correct normalization condition.
 
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 0 ·
Replies
0
Views
1K
Replies
3
Views
1K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
6
Views
1K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K