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?

  • $\begingroup$ Just reverse the step in your last equality, i.e. swap $\langle m|A|n\rangle$ with $|m\rangle$?! What is the problem? $\endgroup$ Commented Jun 18, 2022 at 12:33
  • $\begingroup$ @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. $\endgroup$
    – s0meb0dy
    Commented Jun 18, 2022 at 12:44
  • $\begingroup$ 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 $m$. $\endgroup$ Commented Jun 18, 2022 at 12:51
  • $\begingroup$ 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 $\sum_m |m\rangle \langle m|A|\Psi\rangle=A|\Psi\rangle$. $\endgroup$
    – s0meb0dy
    Commented Jun 18, 2022 at 13:47
  • $\begingroup$ I really don't understand, sorry. But as it seems, you got a good answer to your question. $\endgroup$ Commented Jun 18, 2022 at 13:52

1 Answer 1


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.


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