I Confused about a matrix element

[Malamala]

Hello! My questions is in the context of a matrix element in a diatomic molecule. I will rephrase it as well as I can to remove any non needed complexity. I denote the sperical harmonic $Y_m^l = |m,l>$. I have the operators $n_{0} = Y_1^0$ and $n_{\pm 1} = Y_1^{\pm 1}$. I also define the operator $\mathbf{N}$ which is the raising/lowering operators for $|m,l>$. For example, $N_+|m,l>\propto |m,l+1>$ (the prefactors don't matter for my question). Now, I build the operator $H = i(n_+N_- + n_-N_+)$. If I calculate the matrix element $<0,0|n_+N_- + n_-N_+|1,0> = <0,0|n-|1,1> + <0,0|n+|1,-1> = 2<0,0|n+|1,-1>$, which is some non-zero value. However, if I calculate $<1,0|n_+N_- + n_-N_+|0,0>$ I get zero, simply because $N_{\pm 1} = |0,0> = 0$. What am I doing wrong? This is a Hermitian operator so the 2 matrix elements should be the same. What am I missing?

 

[div_grad]

This is a Hermitian operator so the 2 matrix elements should be the same. What am I missing?

Your operator $H=i(n_+ N_- + n_- N_+)$ is not Hermitian. This is because the ''raising/lowering'' operators $N_\pm$ act on the spherical harmonics, and since your $n_0$ and $n_\pm$ are spherical harmonics themselves, the operators $N_\pm$ do not commute with $n_0$ and $n_\pm$ . Compute $H^\dagger$ to see this.

More importantly, you did not specify what you mean by $l$ and $m$ in your question. However, the standard convention found throughout the literature is that, rather, $l$ denotes the angular momentum quantum number while $m$ is the angular momentum projection quantum number (the projection onto some "Z-axis"). The spherical harmonics, which I will denote as $Y_{l,m} \equiv |l,m\rangle$, are the eigenfunctions of the angular momentum operators: $\mathbf{L}^2$ and $L_z$. Now, both of these operators commute with the Hamiltonian of, e.g., the hydrogen atom. But if you are interested in diatomic molecules:

My questions is in the context of a matrix element in a diatomic molecule.

then you must remember that $\mathbf{L}^2$ do not commute with the molecular Hamiltonian (only the operator $L_z$ does). So for a diatomic molecule, you do not even have a well-defined notion of a quantum number $l$ $-$ assuming, of course, that you are referring in your question to the orbital anguar momentum of the molecular electrons. If this is the case, then the "raising/lowering" operators do not act in a way that you assumed in your question (the result is not a proportionality, which you denoted by "$\propto$").