# What are the eigenvalues of $L_+$ and $L_-$?

I'm studying angular momentum in quantum mechanics. My question involves the operators $L_+=L_x+iL_y$ and $L_-=L_x-iL_y$; in a problem I have a Hamiltonian, $H$, depending an $L_y$, $L^2$ and $L_z$. The solutions suggest to write $L_y$ as a combination of $L_+$ and $L_-$ and then, using the eigenvectors of $L_z$ and $L^2$, write the matrix associated with $H$, and then diagonalize the matrix. How is this possible? How can $L_y$ and $L_z$ be diagonalized in the same basis? Sorry for bad English.

• After you diagonalize the Hamiltonian the resulting eigenfunctions will no longer be eigenstates of both $L_y$ and $L_z$. – Lewis Miller Dec 28 '17 at 16:22
• Note that the ladder operators $L_{\pm}$ are not normal operators, and hence are not diagonalizable in an orthonormal basis (except for the singlet representation). – Qmechanic Dec 28 '17 at 16:28
• Perhaps you can post the question exactly and not "rephrase/interpret" it here? – DanielC Dec 28 '17 at 16:31

If you write the matrix representation of $L_y$ in a basis where $L_z$ is diagonal, you should get (assuming $\ell=1$) something like $$\hat L_z= \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \\ \end{array} \right)\, ,\qquad \hat L_y=\frac{1}{2i} \left( \begin{array}{ccc} 0 & 1 & 0 \\ -1 & 0 & 1 \\ 0 & -1 & 0 \\ \end{array} \right) \tag{1}$$ Of course $\hat L_y$ is not a diagonal matrix there is nothing to prevent you from writing the Hamiltonian and diagonalize the resulting matrix. The eigenstates will not be eigenstates of either $L_y$ or $L_x$ but that's not a big deal: they will be expressed as combinations of your basis eigenstates of $L_z$

However, since $a L_y+b L_z$ commutes with $L^2$, the latter will still be diagonal.