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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?
Yes, a Hamiltonian with non-spherical potential can commute with l^2. This is because the commutation between the Hamiltonian and l^2 is determined by the symmetry of the potential. If the potential has spherical symmetry, then it will commute with l^2. However, if the potential does not have spherical symmetry, then it may not commute with l^2.
A non-spherical potential in a Hamiltonian can represent a more realistic and complex system. It allows for the consideration of different types of interactions and forces, such as anisotropic interactions or external fields. This can lead to a more accurate description of physical systems.
The shape of the potential directly affects the commutation with l^2. If the potential has spherical symmetry, then it will commute with l^2. However, if the potential does not have spherical symmetry, then it may not commute with l^2. This is because the commutation is determined by the symmetry of the potential.
One limitation of a non-spherical potential in a Hamiltonian is that it can be more mathematically complex and difficult to solve compared to a spherical potential. This can make it challenging to obtain analytical solutions and may require numerical methods. Additionally, the physical interpretation of the potential may also be more complex.
The commutation with l^2 affects the energy levels of a system by determining the allowed quantum numbers and corresponding energy levels. If the Hamiltonian commutes with l^2, then the energy levels will be quantized and the system will have discrete energy states. If the Hamiltonian does not commute with l^2, then the energy levels may not be quantized, leading to a continuous spectrum of energy states.