Does Hamiltonian and momentum operator commute?

The angular momentum of the particle is a constant of motion (proved later on in the slides) the eigen states of the energy operator would be the same as the eigen states for the angular momentum. Angular momentum operator L commutes with the total energy Hamiltonian operator (H).

Does P commute with Hamiltonian?

Typical examples of an Hamiltonian that commutes with P is the free particle, or more generally any admissible function of P alone.

Does energy and momentum operator commute?

The position and momentum operators do not commute in momentum space. It is easy to show that this result is equal to i. The product of the position‐momentum uncertainty is the same in momentum space as it is in coordinate space.

Does momentum commute with momentum squared?

however, the square of the angular momentum vector commutes with all the components.

Does position commute with angular momentum?

The operator nature of the components promise difficulty, because unlike their classical analogs which are scalars, the angular momentum operators do not commute. Example 9–1: Show the components of angular momentum in position space do not commute.

Do rotation operators commute?

Thus, the noncommuting nature of the angular momentum operators is a direct consequence of the fact that rotations do not commute.

Does an operator commute with itself?

However the components do not commute themselves. An additional property of commuters that commute is that both quantities can be measured simultaneously. The physical quantities corresponding to operators that commute can be measured simultaneously to any precision.

Does commute and momentum spin?

The total angular momentum of a particle is spin plus orbital angular momentum. The orbital component doesn’t commute with momentum, but the spin component does.

Do momentum components commute?

Angular momentum and linear momentum don’t commute because the angular momentum operator contains the position operator in its definition. The spin operator isn’t defined in terms of r x p or anything like that.