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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?

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  • $\begingroup$ Thx, updated it, if still doesn't, you can choose one from here: libgen.io/… $\endgroup$ – zabop Mar 22 at 16:16
  • $\begingroup$ 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. $\endgroup$ – Cosmas Zachos Mar 22 at 16:48
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The applied magnetic field points only in the z-direction. Therefore the energy of the electron only depends on the z-component of its angular momentum. Another way to look at it is that due to the symmetry of the system the energy cannot depend on the direction of spin in the x and y directions. There is nothing that makes any direction other than the z-direction "special"

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