So, one of my homework problems reads the following
Let $A$ and $B$ be commuting operators and $| \psi_i \rangle$ denote the eigenfunctions of $A$. Show that $\langle \psi_i |B| \psi_j \rangle=\delta_{ij}$
I have tried to approach the solution in the following way:
Since $A$ and $B$ commute, we have, for any vectors $| \psi_i \rangle$ and $| \psi_j \rangle$, $$\langle \psi_i |AB| \psi_j \rangle=\langle \psi_i |BA| \psi_j \rangle$$ Let $\lambda_i$ be the eigenvalues of $A$ for each vector $|\psi_i \rangle$. Therefore, solving the above equation, we get- $$\lambda_i^*\langle \psi_i |B| \psi_j \rangle=\lambda_j\langle \psi_i |B| \psi_j \rangle$$ This leaves us with- $$(\lambda_i^*-\lambda_j)\langle \psi_i |B| \psi_j \rangle=0$$ From here, I have certain questions on proceeding further-
- Is it to be assumed that $A$ is hermitian? If so, then since $\lambda_i^*=\lambda_i$, so for $i \neq j$, we have $\langle \psi_i|B|\psi_j \rangle=0$. But which theorem states that $A$ can,be considered Hermitian here.
- If $i=j$, how do I go further and prove $\langle \psi_i|B|\psi_j \rangle=1$