Hamiltonian Commutation Question

[Diracobama2181]

TL;DR Summary

Why certain operators commute when they commute with the Hamiltonian.

Why is it the case that when some operators commute with the Hamiltonian (let's say A and ), it implies A and B commute, but even when each angular momentum component commutes with the Hamiltonian, it does not imply each the angular momentum components commute with each other?

 

[PeterDonis]

when some operators commute with the Hamiltonian (let's say A and ), it implies A and B commute

Why? Where are you getting this from?

 

[Diracobama2181]

Why? Where are you getting this from?

If two separate operators commute with the hamiltonian, then it implies they share a common set of eigenvectors. Using this reasoning, it can be shown that A and B commute. However, this proof fails for L_i. I just want to know why it fails. I believe it has to do with degeneracy.

 

[PeroK]

If two separate operators commute with the hamiltonian, then it implies they share a common set of eigenvectors. Using this reasoning, it can be shown that A and B commute. However, this proof fails for L_i. I just want to know why it fails. I believe it has to do with degeneracy.

This does not hold in general if the Hamiltonian is degenerate.

For example, all operators commute with the identity operator and all Hermitian operators share an eigenbasis with the identity operator. But, this does not imply they share an eigenbasis with each other.

Note that in the theory of spin 1/2 particles, for example, ##L^2## is represented by a multiple of the identity operator.

 

[hilbert2]

If the Hamiltonian of the system is something like

##H=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}##

and ##A## and ##B## are non-commuting 2x2 matrices, then ##A## and ##B## both commute with ##H## (as any matrix commutes with a matrix that is equivalent to scalar multiplication) but ##AB\neq BA## and there is no common eigenbasis of ##A## and ##B##.

 

[PeterDonis]

This does not hold in general if the Hamiltonian is degenerate.

And, furthermore, if there exist multiple non-commuting operators that all commute with the Hamiltonian, the Hamiltonian must be degenerate. So any Hamiltonian that commutes with all three of the angular momentum operators (or even any two of them) must be degenerate.

 

[vanhees71]

Take as an example the free particle with the Hamiltonian
$$\hat{H}=\frac{\hat{\vec{p}}^2}{2m}.$$
Obviously ##\hat{H}## commutes with all three momentum components ##\hat{\vec{p}}##. Since also the momentum components commute you have a complete set of energy eigenvectors given by the momentum-eigenvectors ##|\vec{p} \rangle##.

Since ##\hat{H}## is a scalar with respect to rotations, ##\hat{H}## also commutes with the three orbital-angular-momentum operators ##\hat{\vec{L}}=\hat{\vec{x}} \times \hat{\vec{p}}##, but the ##\hat{\vec{L}}## do not commute among themselves. That's a counterexample to your claim in #1. Of course there's a complete set of orthonormal eigenvectors of ##\hat{H}##, ##\hat{\vec{L}}^2##, and ##\hat{L}_3##. These are of course no eigenvectors of ##\hat{\vec{p}}## (but only eigenvectors of ##\hat{\vec{p}}^2=2m \hat{H}##).