Can a Hamiltonian with non-spherical potential commute with l^2?

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SUMMARY

The discussion centers on the commutation relations between Hamiltonians and angular momentum operators in quantum mechanics, specifically addressing non-spherical potentials. It is established that a Hamiltonian with a harmonic oscillator potential defined as V(x,y,z) = ax², which is not rotation invariant, still commutes with the angular momentum operator L². The example provided demonstrates that while the Hamiltonian H = L_z commutes with L², it does not maintain rotational symmetry, highlighting the nuanced relationship between potential forms and angular momentum in quantum systems.

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Feelingfine
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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?
 
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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.
 
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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##).
 
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