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2

Here is a more general view of your question that might offer some clarity: An operator $A$ on a Hilbert space $\mathcal{H}$ is called normal if it commutes with its adjoint, $AA^\dagger = A^\dagger A$. Equivalently, any operator $A$ decomposes as $A = B + iC$, where $B$ and $C$ are self-adjoint, and $A$ is normal if and only if $B$ and $C$ commute. This ...


4

To fix what we are talking about, I think you mean $$ H = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \quad S = \begin{pmatrix} 1 & 0 \\ 0 & i \end{pmatrix} \quad T = \begin{pmatrix} 1 & 0 \\ 0 & \exp(i\pi/4) \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & \frac{1+i}{\sqrt{2}} \end{pmatrix}.$$ If these are ...


2

I did it like this (no promises that it's minimal!): Basically, I realised that by swapping the second and last rows and columns of your target matrix (achieved using the first and last controlled-nots in the circuit), you get something this is very nearly of the form $I\otimes R_x(\phi)$, except that one of the $2\times 2$ blocks would need to have the ...


2

In the linked post, $|\psi_{mn}\rangle$ is defined as $$ |\psi_{mn}\rangle = a_{00}|0_m 0_n\rangle + a_{10}|1_m 0_n\rangle + a_{01}|0_m 1_n\rangle + a_{11}|1_m 1_n\rangle, $$ and the action of a CNOT operating between $m$-th and $n$-th qubit is written as $$ \mathrm{CNOT}^{(m,n)}|\psi_{mn}\rangle = a_{00}|0_m 0_n\rangle + a_{10}|1_m 0_n\rangle + a_{11}|0_m ...


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