# What's the proof that sum of all projection operators for orthonormal basis gives us identity operator?

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## 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, 2020

• 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 years
Ah, the same answer. Sorry, shall I delete?
• Admin almost 7 years
@TylerHG I don't care either way. No biggie!