Commutation Relationships and Operator Functions

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SUMMARY

The discussion centers on the commutation relationships between operators in quantum mechanics, specifically examining whether the commutation of the Hamiltonian H with Lz implies commutation with L^2. It is established that since [H, Lz] = 0, by symmetry, [H, Lx^2] and [H, Ly^2] also equal zero, leading to the conclusion that [H, L^2] = 0. Furthermore, if the Taylor expansion of F(A) converges, F(A) can be treated as a polynomial in A, ensuring its commutation with H.

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  • Understanding of quantum mechanics and operator algebra
  • Familiarity with Hamiltonians and angular momentum operators
  • Knowledge of commutation relations and their implications
  • Basic concepts of Taylor series and polynomial functions
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Quantum physicists, students of quantum mechanics, and researchers focusing on operator theory and commutation relationships in quantum systems.

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There are 2 operators such that [A,B] = 0. Does [F(A),B]=0 ?

Specifically, let's say we had the Hamiltonian of a 3-D oscillator H and L^2. We know that L^2 = Lx^2+Ly^2+Lz^2, and it is known that [H,Lz] = 0. Can we say that since H and Lz commute, H and Lz^2 also commute, by symmetry H and Lx^2,Ly^2 commute also and therefore H and L^2 commute?
 
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If the Taylor expansion of F(A) converges, then you can essentially assume that it is a polynomial in A, so it will commute. Your argument about H and L^2 sounds right.
 

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