Exercise 6.2

Consider the operator \(U = 2|\psi\rangle\langle\psi| - I\) where \(|\psi\rangle\) is the equal superposition state

(1)\[ |\psi\rangle = \frac{1}{N^{1/2}}\sum_{x=0}^{N-1}|x\rangle \]

where \(\{|x\rangle\}\) is computational basis set. The operator \(U\) can be obtained by

(2)\[\begin{split} \begin{align} 2|\psi\rangle\langle\psi| - I &= \frac{2}{N}\left(\sum_{x=0}^{N-1}|x\rangle \right)\left(\sum_{x'=0}^{N-1}\langle x'|\right) - I \\ &= \frac{2}{N}\sum_{x, x'=0}^{N-1}|x\rangle \langle x'| - \sum_{x=0}^{N-1}|x\rangle \langle x| \end{align} \end{split}\]

Applying operator \(U\) to an arbitrary state \(\sum_k\alpha_k |k\rangle\) gives

(3)\[\begin{split} \begin{split} (2|\psi\rangle\langle\psi| - I)\sum_k\alpha_k |k\rangle =& \frac{2}{N}\left(\sum_{x, x'=0}^{N-1}|x\rangle \langle x'|\right)\left(\sum_{k=0}^{N-1}\alpha_k |k\rangle \right) \\ &- \left(\sum_{x=0}^{N-1}|x\rangle \langle x|\right)\left(\sum_{k=0}^{N-1}\alpha_k |k\rangle \right) \\ =& \frac{2}{N}\sum_{x, x',k=0}^{N-1}\alpha_k|x\rangle \langle x'|k\rangle -\sum_{x,k=0}^{N-1} \alpha_k |x\rangle \langle x|k\rangle \\ =&\frac{2}{N}\sum_{x, k=0}^{N-1}\alpha_k|x\rangle \langle k|k\rangle - \sum_{k=0}^{N-1}\alpha_k|k\rangle \langle k|k\rangle \\ =&\sum_{x=0}^{N-1}\left(\sum_{k=0}^{N-1}\frac{2}{N}\alpha_k\right)|x\rangle - \sum_{k=0}^{N-1}\alpha_k|k\rangle \\ =&\sum_{x=0}^{N-1}2\langle \alpha\rangle|x\rangle - \sum_{k=0}^{N-1}\alpha_k|k\rangle \end{split} \end{split}\]

Note that change the index would not change the state, so we could re-write eq. (3) as

(4)\[ (2|\psi\rangle\langle\psi| - I)\sum_k\alpha_k |k\rangle = \sum_{k=0}^{N-1}2\langle \alpha\rangle|k\rangle - \sum_{k=0}^{N-1}\alpha_k|k\rangle = \sum_{k=0}^{N-1}\left[2\langle \alpha\rangle-\alpha_k\right]|k\rangle \]