Feelingfine Messages 2 Reaction score 1 Thread starter Jun 25, 2020 #1 I know that in the case of central potential V(r) the hamiltonian of the system always commutes with l^2 operator. But what happends in this case?
I know that in the case of central potential V(r) the hamiltonian of the system always commutes with l^2 operator. But what happends in this case?
A. Neumaier Science Advisor Insights Author Messages 8,731 Reaction score 4,848 Jun 25, 2020 #2 The Hamiltonian ##H=L_z## commutes with ##L^2## but is not rotation invariant. Adding a multiple of ##H=L_z## to a rotation invariant ##H## gives other counterexamples.
The Hamiltonian ##H=L_z## commutes with ##L^2## but is not rotation invariant. Adding a multiple of ##H=L_z## to a rotation invariant ##H## gives other counterexamples.
hilbert2 Science Advisor Insights Author Messages 1,612 Reaction score 611 Jun 25, 2020 #3 It's shown here that ##\hat{L}^2## and ##\hat{x}^2## commute https://physics.stackexchange.com/questions/93533/commutator-of-l2-and-x2-p2 So put a particle in three dimensions to a field described by a harmonic oscillator potential, but only in the x-direction ##V(x,y,z) = ax^2## That's not rotation invariant, but still commutes with the ##\hat{L}^2## (and hence the whole ##\hat{H}## commutes with ##\hat{L}^2##).
It's shown here that ##\hat{L}^2## and ##\hat{x}^2## commute https://physics.stackexchange.com/questions/93533/commutator-of-l2-and-x2-p2 So put a particle in three dimensions to a field described by a harmonic oscillator potential, but only in the x-direction ##V(x,y,z) = ax^2## That's not rotation invariant, but still commutes with the ##\hat{L}^2## (and hence the whole ##\hat{H}## commutes with ##\hat{L}^2##).