# Can I express the Hamiltonian in terms of $L_z$ operator only, not $L_x$ and $L_y$? Is it generally true, that $-\omega L_z = H$?

I encountered the relation in the Solution of Problem 5.1 in the book by Kyriakos Tamvakis titled "Problems and solutions in quantum mechanics":

$$\frac{i}{h}[H,\textbf{L}]=-\frac{i \omega}{h}[L_z, \textbf{L}]$$

From which I conclude that $$-\omega L_z = H$$. Dimension-wise, it makes sense. Though we is it that we chose the $$L_z$$ operator to express the Hamiltonian with, and excluded $$L_x$$ and $$L_y$$?

Is it just the peculiarity of the problem and solution I am looking at, or this is something more general? If it is more general, how can we derive it?

• Thx, updated it, if still doesn't, you can choose one from here: libgen.io/… – zabop Mar 22 at 16:16
• Did you read the statement of the problem? The hamiltonian is a rotationally invariant piece, which drops off inside the commutator, and this extra piece, which he keeps. – Cosmas Zachos Mar 22 at 16:48