Selection Rules (Time Dependent Perturbation Theory)

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The selection rules for the matrix element of an operator X in time-dependent perturbation theory state that the matrix element is zero unless specific conditions on the quantum numbers are met. Specifically, the rules require that the change in orbital angular momentum (Δl) and magnetic quantum number (Δm) are constants that depend on the operator X. For the position operator, these constants are Δl = ±1 and Δm = ±1. Therefore, the assumption that the matrix element is zero unless l' = l ± 1 and m' = m ± 1 is accurate only for the position operator. Understanding these selection rules is crucial for analyzing transitions in quantum systems.
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Homework Statement
Suppose that the system is a single spinless particle of mass $$M$$ and charge $$e$$ in a central Coulomb potential and the quantum numbers of the initial state are labeled $$|n, l, m>$$ in the usual way. It is subject to perturbation with magnetic field $$ B(t) = B_0e ^{−λt}$$ pointing in the x direction, which adds a term $$eL_xB(t)/(2Mc)$$ to the Hamiltonian. Find the quantum number selection rules for allowed transitions to a different state.
Relevant Equations
$$[L_x,X]=0$$
I suppose my question is, since X commutes for H, does this mean that the selection rules are $$<n',l',m'|X|n,l,m>=0$$ unless $$l'=l\pm 1$$ and $$m'=m\pm 1$$, as specified in Shankar?
 
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No, that's not correct. The selection rules for the matrix element of any operator $X$ are given by $$\langle n',l',m'|X|n,l,m\rangle=0$$ unless $$l'-l=\Delta l$$ and $$m'-m=\Delta m$$ where $\Delta l$ and $\Delta m$ are constants depending on the particular operator $X$. In the case of the position operator, $\Delta l=\pm 1$ and $\Delta m=\pm 1$.
 

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