In reading about angular momentum and spin, I came across a derivation showing that the Dirac Hamiltonian does not commute with orbital angular momentum, and hence L is not conserved. What is it about commutation relations which allow us to conclude whether the orbital angular momentum is conserved or not?
[Physics] Commutation between Dirac hamiltonian and angular momentum
angular momentumhamiltonianquantum mechanics
Related Solutions
Did you try the wikipedia article about angular momentum operators?
The arrows schematically represent the internal state of the particle (its spin state). The blue and red dots represent two particles at different locations. Or, if you prefer, it could be two parts of the wavefunction of a single particle, with blue meaning negative phase and red meaning positive phase.
When you exponentiate Jz (the total angular momentum operator), you get real-world rotation about the z axis -- part A in the figure. The whole system and everything in it is rotated.
When you exponentiate Lz (the orbital angular momentum operator), you get "spatial-only" rotation about the z axis -- part B in the figure. The positions of particles get rotated but their spin states stay exactly the same.
When you exponentiate Sz (the spin operator), you get "internal-only" rotation about the z axis -- part C in the figure. The internal state of the particle is rotated, but the particle itself stays in the same place.
By the way, you should avoid phrases like "ket changes direction" and "rotate the ket". You're thinking about rotation in real three-dimensional space, but the ket is not an object in real three-dimensional space. The phrase "rotate the ket" sort of works for a spin-1/2 particle at a single point, but anyway using that phrase is a very bad habit.
Let's calculate $[L_x L_y, L_z]$. I'm going to use the property $[AB,C]=A[B,C]+[A,C]B$.
If you apply the property to our case, you obtain $L_x[L_y,L_z]+[L_x,L_z]L_y$. Now you can substitute the value of the commutators and find the correct answer.
Note that the quantum commutation relations are pretty similar to vector product in cartesian coordinates. For example, $[L_x,L_y]=i\hbar L_z$ is analogous to $\hat{x} \wedge \hat{y} = \hat{z}$. I'm not saying that this is the same, but the math is really similar.
So if you look the commutators above, and the property that I've used, you find that you're doing something similar to $(\hat{x}+\hat{y})\wedge\hat{z} = \hat{x} \wedge \hat{z} + \hat{y}\wedge\hat{z} = -\hat{y} + \hat{x}$. So from here you infere that the answer should be something related with $L_x -L_y$. (D) is the only option which has this, so this must be the correct answer.
As a little disclaimer, I find this really strange. If I were you, I would use only the commutators and forget about that "classical tricks". They will often confuse you in QM.
I think this is more or less the reasoning, but maybe somebody has a more complete answer.
Best Answer
A simple answer is to look in the Heisenberg picture with the Heisenberg equation of motion. In this picture, operators evolve instead of states as in the Schrodinger picture. Here the evolution of the operator is chosen so that its expectation value has the same time evolution as in the Schrodinger picture. To do so its time evolution is governed by:
$ \frac{d}{dt}L=\frac{1}{i\hbar}[L,H]$
so the expectation of the time derivitive of L is non 0 if it's commutator with the Hamiltonian is non 0 $ <\frac{d}{dt}L>=\frac{1}{i\hbar}<[L,H]>$
Thus it's not something specific to commutation relations its specific to the commutation relation with the Hamiltonian. For this explanation to be really satisfying you should read about the Heisenberg Picture.
Another way to look at it is that since the two operators don't commute they can't be diagonalized at the same time. This means an angular momentum eigen state isn't a energy eigen state and therefore you have to decompose the angular momentum eigen state into energy eigen states to get its time evolution. Since this state will be a sum of energy eigen states time will evolve the system away from it and the expectation value of angular momentum will change.