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I came across the term "diagonal to the computational basis" in my reading recently. I'm not entirely sure what this term means. I know that a diagonal matrix is one with only non-zero elements on the diagonal and I know that the computational basis is $\alpha \left| 0 \right> + \beta \left| 1 \right>$ but I'm not sure how these terms relate to one another nor how a particular gate could be said to be diagonal to the computational basis.

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Matrix just encodes linear operation that transforms basis vectors to some other vectors. For example, matrix $M$ can transform vector $|0\rangle$ to vector $m_{11}|0\rangle + m_{12}|1\rangle$ and vector $|1\rangle$ to $m_{21}|0\rangle + m_{22}|1\rangle$. In this case, this matrix is written as $\left(\begin{matrix} m_{11} & m_{21} \\ m_{12} & m_{22} \end{matrix}\right)$. And vectors $|0\rangle$, $|1\rangle$ are encoded as $\left(\begin{matrix} 1 \\ 0 \end{matrix}\right), \left(\begin{matrix} 0 \\ 1 \end{matrix}\right)$ respectively.

In a different basis the same linear operation has different matrix encoding, diagonality is not preserved under basis change.

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If the computational basis is the vectors $\{|0\rangle,|1\rangle\}$, then this means that $\rho$ is a diagonal matrix when written in this basis. In other words, $\rho=p|0\rangle\langle 0|+(1-p)|1\rangle\langle 1|$ for some real number $0<p<1$.

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    $\begingroup$ Original question had a gate, not density operator. So not rho with p and 1-p. Replace p's with phases. $\endgroup$ – AHusain Mar 20 at 14:08

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