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If $\{|u_1\rangle, ..., |u_n\rangle \}$ are an orthonormal basis for $\mathbb{C}_n$, then

$$ \sum_{j=1}^{n} |u_j\rangle\langle u_j| = I_n$$

I can see that this is true in the standard computational basis, but I'm having trouble seeing it intuitively when generalized to any basis, nor can I prove it. Can anyone help?

Brian Carmicle
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2 Answers2

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Let $A$ be the operator give by your sum.

  • Can you see that $A | u_i \rangle = | u_i \rangle$ for each $i$?
  • Can you see why the above observation implies that $A$ is the identity?
Mike F
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Given any orthonormal basis {$|i\rangle$}$_{i = 1,2,..,n}$ of $\Bbb{C}^n$, we can express any vector $|\psi\rangle$ as, $$|\psi\rangle = a_1|1\rangle + a_2|2\rangle + ... +a_n|n\rangle= \sum_{i=1}^n a_i|i\rangle$$ Now, let's define your sum as $\mathbf{A} = \sum_{j=1}^n |j\rangle\langle j|$. Then we want to find $\mathbf{A}|\psi\rangle$. $$\mathbf{A} |\psi\rangle = \left( \sum_{j=1}^n |j\rangle\langle j| \right) \left( \sum_{i=1}^n a_i|i\rangle\right) = \sum_{j=1}^n \sum_{i=1}^n a_i |j\rangle\langle j|i\rangle$$ $$=\sum_{j=1}^n \sum_{i=1}^n a_i |j\rangle \delta_{ij} = \sum_{i=1}^n a_i |i\rangle = |\psi\rangle$$ Hence, as $\mathbf{A}|\psi\rangle = |\psi\rangle,\space \forall \space |\psi\rangle \in \Bbb{C}^n,$ Then, $\mathbf{A} = \mathbf I_n$

Benjamin Carroll
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