What's the proof that sum of all projection operators for orthonormal basis gives us identity operator?
Solution 1
Act with your sum of projection operators on an arbitrary state psi. Use completeness to expand psi into a sum of basis vectors. Use orthonormality to simplify the sum (with $\langle n m\rangle=\delta_{ij} $). Simplify. The sum you're left with is the original vector psi. By the arbitrariness of the initial vector, the operator must be the identity matrix.
Solution 2
Take the expansion of an arbitrary vector in an orthonormal basis like so $$ \alpha\rangle = \sum i\rangle\langle i \alpha\rangle=\left(\sum i\rangle\langle i \right)\alpha\rangle$$
But, this is true for any $\alpha\rangle$, so the sum over projections must be the identity operator.
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Qwedfsf
Updated on September 16, 2020Comments

Qwedfsf almost 2 years
\begin{align} \ I = \begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} \end{align}
\begin{align} \ {\sum_{i} i{\rangle}}{{\langle}i}=I \end{align}
$I$ is identity operator, while the $i{\rangle}$ is the orthonormal basis. How can I prove this sum?

ClassicStyle almost 7 yearsAh, the same answer. Sorry, shall I delete?

Admin almost 7 years@TylerHG I don't care either way. No biggie!