# Why does Identity operator work in general in quantum mechanics?

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?

• Just reverse the step in your last equality, i.e. swap $\langle m|A|n\rangle$ with $|m\rangle$?! What is the problem? Jun 18 at 12:33
• @JasonFunderberker The problem was that I was unable to show $\sum_m\sum_n c_n\langle m|A|n \rangle|m\rangle=\sum_n c_nA|n\rangle$. I now understood that I had missed the point that the basis is complete. Though I am not able to understand what you intend to do by swapping. Jun 18 at 12:44
• I mean $\langle m|A|n\rangle |m\rangle = |m\rangle \langle m|A|n\rangle$, just what you've used in the very same equality...Then the desired result is easy to see, by summing over $m$. Jun 18 at 12:51
• I am trying to prove that summing over $m$ works like an identity operator in the case when it is operating on $A|\Psi\rangle$. For that I am trying to show that $\sum_m |m\rangle \langle m|A|\Psi\rangle=A|\Psi\rangle$. Jun 18 at 13:47
• I really don't understand, sorry. But as it seems, you got a good answer to your question. Jun 18 at 13:52

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.