Operator that commutes with the Hamiltonian

[Riemann9471]

Homework Statement

In the case of the quantum harmonic oscillator in 3D , does the z-component of the angular momentum of a particle commute with the Hamiltonian? Does the fundamental state has a well defined value of L_z (variance = 0) ? If you said no , why? If you said yes , what is the value?

Relevant Equations

L_z = x p_y - y p_x (position and impulsion in x and y direction)

Homework Statement: In the case of the quantum harmonic oscillator in 3D , does the z-component of the angular momentum of a particle commute with the Hamiltonian? Does the fundamental state has a well defined value of L_z (variance = 0) ? If you said no , why? If you said yes , what is the value?
Homework Equations: L_z = x p_y - y p_x (position and impulsion in x and y direction)

So I was able to prove that the z component of angular momentum operator L_z = (x p_y - y p_x) (They are all operators ) commute with Hamiltonian, but now I struggle to explain why it implie that the variance of the operator L_z is zero. I could compute it and prove that , in the case of the harmonic oscillator in 3D , the variance is 0 by computing the expectation value of L_z and (L_z)^2 , but that would be a lot of work and I know that there's a simple explanation to why the variance of an operator that commute with the Hamiltonian is 0 , but I don't see it.

 

[mitochan]

Hi.
The question should say "Say there exists an eigenstate of energy or Hamiltonian. CAN this eigenstate of enerygy be an eigenstate of L_z?" It CAN be but not always is. The eigenstate of energy whose L_z has not zero variance takes place.

For an example, for free particle H=p^2/2m, obviously [H,p]=0.
|p>,|-p> and a|p>+b|-p> are all energy eigenstates. The last one is not an eigenstate of momentum.

 

[DrClaude]

The question relates to common eigenstates of commuting observables.

You don't need to actually calculate the variance to figure out the result. There are symmetry considerations you can use.

 

[PeroK]

Homework Statement: In the case of the quantum harmonic oscillator in 3D , does the z-component of the angular momentum of a particle commute with the Hamiltonian? Does the fundamental state has a well defined value of L_z (variance = 0) ? If you said no , why? If you said yes , what is the value?
Homework Equations: L_z = x p_y - y p_x (position and impulsion in x and y direction)

Homework Statement: In the case of the quantum harmonic oscillator in 3D , does the z-component of the angular momentum of a particle commute with the Hamiltonian? Does the fundamental state has a well defined value of L_z (variance = 0) ? If you said no , why? If you said yes , what is the value?
Homework Equations: L_z = x p_y - y p_x (position and impulsion in x and y direction)

So I was able to prove that the z component of angular momentum operator L_z = (x p_y - y p_x) (They are all operators ) commute with Hamiltonian, but now I struggle to explain why it implie that the variance of the operator L_z is zero. I could compute it and prove that , in the case of the harmonic oscillator in 3D , the variance is 0 by computing the expectation value of L_z and (L_z)^2 , but that would be a lot of work and I know that there's a simple explanation to why the variance of an operator that commute with the Hamiltonian is 0 , but I don't see it.

If two operators commute, then they have a common eigenbasis. I.e. you can find a basis of eigenvectors of both operators. Now, unless you have degeneracy of the spectrum, this means that an eigenvector of one is an eigenvector of the other.

In this case, unless you have degeneracy of the "fundamental" state, this eigenstate of the Hamiltonian must be an eigenstate of any operator that commutes with it.

For an example of the degenerate case, consider the angular monentum operators $L^2$ and $L_z$. The spectrum of $L^2$ is degenerate. I.e. the eigenspace associated with certain eigenvalues is multi-dimensional. The $L_z$ operator, although it commutes with $L^2$, may have several different eigenvalues on this eigenspace of $L^2$.