Rotational invariance and operator-squares My mind is drawing a blank right now.  In systems with spin and orbital angular momentum, I know that rotational invariance implies that $[H, \mathbf{J}]=0$ where $\mathbf J=\mathbf L+\mathbf S$.  But I don't see why rotationally invariant systems should necessarily commute with the square-magnitude operators:
$$[H,\mathbf{L}^2]=0\qquad\text{and}\qquad[H,\mathbf{S}^2]=0\,.$$
But the funny thing is, I am unable to come up with model Hamiltonians that are rotationally invariant, but do not commute with $\mathbf{L}^2$ and $\mathbf{S}^2$.  So my question is: is there an easy algebraic reason for this commutation relation? or are there systems that do not conserve these operators?  In the latter case, I need help coming up with examples.
 A: (I throughout assume $\hbar =1$.) Define, where $X_i$ is the position operator along ${\bf e}_i$ (the same result will arise replacing it for the momentum operator everywhere in that follows) $$H:= {\bf X}\cdot {\bf S}\:.$$
$$e^{i\theta {\bf n}\cdot {\bf J}}H e^{-i\theta {\bf n}\cdot {\bf J}}=
e^{i\theta {\bf n}\cdot {\bf L}}e^{i\theta {\bf n}\cdot {\bf S}}{\bf X}\cdot {\bf S} e^{-i\theta {\bf n}\cdot {\bf L}}e^{-i\theta {\bf n}\cdot {\bf S}} =
e^{i\theta {\bf n}\cdot {\bf L}}{\bf X} e^{-i\theta {\bf n}\cdot {\bf L}}\cdot  e^{i\theta {\bf n}\cdot {\bf S}}{\bf S} e^{-i\theta {\bf n}\cdot {\bf S}}=
R{\bf X} \cdot R {\bf S} = {\bf X} \cdot{\bf S}\:,$$
where $R\in SO(3)$ is the rotation of the angle $\theta$ around ${\bf n}$. Thus
$$e^{i\theta {\bf n}\cdot {\bf J}}H e^{-i\theta {\bf n}\cdot {\bf J}}=H$$
and consequently, taking the derivatives for $\theta=0$ we get:
$$[H, J_k]=0\qquad k=1,2,3\:.$$
Finally (I am not sure on coefficients please check)
$$[H, L^2] = \sum^3_{j,k=1}[X_j,L_k]L_k S_j + \sum^3_{j,k=1}L_k[X_j,L_k]S_j=
i\sum^3_{j,k=1}\epsilon_{jkr} S_j\left(X_rL_k + L_kX_r \right)
= i\sum^3_{j,k=1}\epsilon_{jkr} S_j\left(2X_rL_k + [L_k, X_r] \right)
= 2i\sum^3_{j,k=1}\epsilon_{jkr} S_j X_rL_k - ii \sum^3_{j,k,r=1}\epsilon_{jkr}  \epsilon_{rks}S_jX_s =
2i \sum^3_{j,k=1}\epsilon_{jkr} S_j X_rL_k + \sum^3_{j,k,r=1} (\delta_{jr}\delta_{rs} - \delta_{js}\delta_{rr})S_jX_s \neq 0\:. \quad (1)$$
Indeed the former sum is:
$$2iS_1\otimes (X_2L_3-X_3L_2) + 2i S_2\otimes(...) + 2i S_3\otimes(...)$$
so it cannot vanish. The latter sum in (1) produces $-2 \sum_{k=1}^3 S_k \otimes X_k$ and e.g., $cX_1 \neq X_2L_3-X_3L_2$ for any number $c$ (I put that $c$ just because I am not sure on coefficients I found).
A: $\newcommand{\tot}{\mathbf{J}}$$\newcommand{\orb}{\mathbf{L}}$$\newcommand{\spin}{\mathbf{S}}$$\mathbf{J}$ is the total angular momentum operator, $\mathbf{L}$ is the orbital angular momentum operator, and $\mathbf{S}$ is the spin angular momentum operator. 
Now we are assuming that $[H,\tot ]$ = 0; i.e., that the total angular momentum is conserved. But we are trying to see if there is a way that transition can occur which do not preserve, say, $\orb$. Well, if $\orb$ changes and $\tot$ stays the same, then it must be the case that $\spin$ changes. In other words angular momentum is transferred from orbital to spin.
We need a good test system where we can add a transition to convert orbital angular momentum to spin angular momentum. Let's look at the hamiltonian for the hydrogen atom. Let's not introduce any coupling (spin orbit, hyperfine, zeeman, ...) so that the hamiltonian eigenvalues in the usual $|n l m s_z \rangle$ depend only on the principal quantum  number $n$. Now let's restrict our attention to the $n=2$ subspace. We will introduce our transition in this subspace.
In fact we will restrict ourselves to the $j=1/2$ subspace of the $n=2$ subspace. To see why, notice that the total angular momentum quantum number can take on one of two values in this subspace: $1/2$ and $3/2$. The only way the system can be in the $j=3/2$ state is if $l=1$, and if $j$ is conserved, then when $j=3/2$ initially, $j$ must still be $3/2$ after the transition, so $l$ must be $1$ after the transition and $l$ will be conserved.
Now for the case $j=1/2$. For symmetry reasons, it should be sufficient to consider just the states where $j=1/2$ and $m_j = 1/2$. There are two ways this can happen: $l=1$ and $l=0$. If $l=0$, then $s_z$ must be $1/2$. Let's call this state $|n=2,l=0,j=1/2,m_j=1/2 \rangle$. This will be the final state of the transition. If $l=1$, then we know that when we go to the coupled basis, we can get a $j=1/2$, $m_j=1/2$ state, which will be a linear combination of $|l=1,m=0,s_z=1/2 \rangle$ and $|l=1,m=1,s_z=-1/2 \rangle$. Let's call the state given by this linear combination $|n=2,l=1,j=1/2,m_j=1/2 \rangle$.
Now we are ready to come up with our hamiltonian. Let the unperturbed hydrogen hamiltonian be $H_0$, and now let's define an interaction hamiltonian $$H_{int} = |n=2,l=0,j=1/2,m_j=1/2 \rangle\langle n=2,l=1,j=1/2,m_j=1/2 | + h.c.$$. Then the hamiltonian $H_0 + H_{int}$ (or I guess just $H_{int}$ by itself) preserves total angular momentum, but not spin or oribital angular momentum by itself.     
Edit
I did not satisfactorily show that $[H_{int}, \tot]=0$. In fact this does not hold. I must modify my hamiltonian by adding a new term for $m_j= -1/2$. Then I have $$H_{int} = |n=2,l=0,j=1/2,m_j=1/2 \rangle \langle n=2,l=1,j=1/2,m_j=1/2 | + |n=2,l=0,j=1/2,m_j=-1/2 \rangle \langle n=2,l=1,j=1/2,m_j=-1/2 | + h.c.$$. I must show that this commutes with $J_z$, $J_+$, and $J_-$ but not $L^2$. (It is sufficient to check $J_+$ and $J_-$, because linear combinations of those give $J_x$ and $J_y$.)
For $J_z$ just notice the $m_j$ quantum number is preserved by this hamiltonian.
For $J_+$, notice that if $m_j=1/2$ then both $H_{int} J_+$ and  $ J_+ H_{int}$ annihilate the state. If $m_j=-1/2$ Then $J_+$ has the effect of increasing $m_j$ by one and $H_{int}$ has the effect of flipping $l$ between 0 and 1. Notice that $H_{int}$ flips $l$ the same way regardless of the value of $m_j$, so the final state has $l$ flipped and $m_j=1/2$ no matter the order of operations.
Seeing that $J_-$ commutes is exactly analogous logic except $m_j$ is initially $-1/2$ and you are increasing instead of decreasing.
To see that $L^2$ is not conserved, simply see that the value of $l$ is switched between $0$ and $1$.
A: Commutation with each of the $\mathbf{J}$ components implies commutation with any linear combination of the functions thereof, in particular with $J_x^2+J_y^2+J_z^2$.
An example on non-rotationaly invariant $H$ that that does not conserve total $\mathbf{S}$ would be Ising-like spin: $H=- K S_z^2$. You can add a 4th order term to limit $\langle S^2 \rangle$, if you like.  
