As much as I could understand from reading quantum mechanics-related texts, the identity operator works on a state vector as follows
\begin{align}
|\Psi\rangle&=\sum_{n} c_n |n \rangle \\
\sum_m|m\rangle\langle m|\Psi \rangle&=\sum_m\sum_nc_n\langle m|n\rangle |m\rangle\\
&=\sum_m\sum_nc_n\delta_{mn}|m\rangle\\
&=\sum_nc_n|n\rangle
\end{align}
Basically, the operation is dependent upon the fact that we write $|\Psi\rangle$ as $\sum_n c_n |n \rangle$.
Now suppose I do $A$ operation on $|\Psi\rangle$
\begin{align}
A|\Psi\rangle&=\sum_{n} c_n A|n \rangle \tag{1} \\
\sum_m|m\rangle\langle m|A|\Psi \rangle&=\sum_m\sum_nc_n\langle m|A|n\rangle |m\rangle
\end{align}
My question is how do we get back (1)? I understand that if $A|\Psi\rangle=a|\Psi\rangle$ I can get back (1). But is it necessary that each operator have an eigenvalue?
Best Answer
Since the set of states $\{\left | n \right>\}$ forms a complete basis, you can always write $A\left | n \right>=\sum_kc'_k\left | k \right>$ for some coefficients $c'_k$. If you plug this into your equation for $A\left | \Psi \right>$ you should be able to get the answer.
Alternatively, you cay simply say directly that $A\left | \Psi \right>=\left | \phi \right>$, where $\left | \phi \right>$ is some state and hence can be written as $\left | \phi \right>=\sum_j c''_j\left | j \right>$ for some coefficients $c''_j$ and the the result follows as in your first calculation.