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I am deriving the Cirac-Zoller CNOT gate as implemented by Schmidt-Kaler following the text of Nakahara & Ohmi's textbook Quantum Computing: From Liner Algebra to Physical Realizations. The gate sequence is as follows

$$ U = R_1^+(\pi,\pi) R_2(\pi/2,\pi)\Phi_2^c R_2(\pi/2,0)R_1^+(\pi,0), $$ where the subscript indicates which ion is being acted upon (either ion 1 or ion 2). This can be written as

$$ U = [R_1^+(\pi,\pi) \otimes I_2] [I_1\otimes R_2(\pi/2,\pi)\Phi_2^c R_2(\pi/2,0)][R_1^+(\pi,0)\otimes I_2] $$

$$ U = [R_1^+(\pi,\pi) R_1^+(\pi,0)] \otimes [R_2(\pi/2,\pi)\Phi_2^c R_2(\pi/2,0)], $$

where $I_1$ and $I_2$ are the $2 \times 2$ identity matrices acting on the first and second ions, respectively. $R_i^+(\theta,\phi)$ and $R_i(\theta,\phi)$ are blue-detuned and resonant transitions whose matrix representations are

$$ R_n^+(\theta,\phi) = \cos \left( \frac{\sqrt{n}\theta}{2} \right)I + \sin\left( \frac{\sqrt{n}\theta}{2} \right) \begin{pmatrix} 0 & -e^{-i\phi} \\ e^{i\phi} & 0\end{pmatrix}, $$

$$ R_n(\theta,\phi) = \cos \left( \frac{\sqrt{n}\theta}{2} \right)I - i \sin\left( \frac{\sqrt{n}\theta}{2} \right) \begin{pmatrix} 0 & e^{-i\phi} \\ e^{i\phi} & 0\end{pmatrix}, $$ in the subspaces spanned by $\{|0\rangle \otimes |n-1\rangle\, |1\rangle \otimes |n\rangle\}$ and $\{ |0\rangle \otimes |n\rangle, |1\rangle \otimes |n\rangle\}$, respectively. Here, $n$ is the index of the vibrational mode of the ion lattice. According to the book, $U$ should have the following matrix form

$$U = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & i \\ 0 & 0 & -i & 0 \end{pmatrix}. $$ I cannot get this result. The matrix representations for each gate, using the equations above, are $$ R_1^+(\pi,\pi) = \cos \left( \frac{\pi}{2} \right) I + \sin \left( \frac{\pi}{2} \right) \begin{pmatrix} 0 & -e^{-i\pi} \\ e^{i\pi} & 0\end{pmatrix}= \begin{pmatrix} 0 & -1 \\ -1 & 0\end{pmatrix},\\ $$

$$ R_1^+(\pi,0) = \cos \left( \frac{\pi}{2} \right) I + \sin \left( \frac{\pi}{2} \right) \begin{pmatrix} 0 & -e^{-i0} \\ e^{i0} & 0\end{pmatrix} = \begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix}, $$

$$ R_2(\pi/2,\pi) = \cos \left( \frac{\pi}{4} \right) I - i \sin \left( \frac{\pi}{4} \right) \begin{pmatrix} 0 & e^{-i\pi} \\ e^{i\pi} & 0\end{pmatrix}= \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & -i \\ i & 1\end{pmatrix}, $$

$$ R_2(\pi/2,0) = \cos \left( \frac{\pi}{4} \right) I - i \sin \left( \frac{\pi}{4} \right) \begin{pmatrix} 0 & e^{-i0} \\ e^{i0} & 0\end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & -i \\ -i & 1\end{pmatrix}, $$

$$ R_2^+(\pi/\sqrt{2},\pi/2) = \cos \left( \frac{\pi}{2\sqrt{2}} \right) I + \sin \left( \frac{\pi}{2\sqrt{2}} \right) \begin{pmatrix} 0 & -e^{-i\pi/2} \\ e^{i\pi/2} & 0\end{pmatrix} = \begin{pmatrix} a & ib \\ ib & a\end{pmatrix}, $$

$$ R_2^+(\pi,0) = \cos \left( \frac{\pi}{2} \right) I + \sin \left( \frac{\pi}{2} \right) \begin{pmatrix} 0 & -e^{-i0} \\ e^{i0} & 0\end{pmatrix} =\begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix}, $$ where $a$ ($b$) is $\cos(\pi/2\sqrt{2})$ ($\sin(\pi/2\sqrt{2})$). Putting these together, we obtain

$$ \Phi_2^c = \begin{pmatrix} a & ib \\ ib & a\end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix}\begin{pmatrix} a & ib \\ ib & a\end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix}=\begin{pmatrix} -a^2 - b^2 & 0 \\ 0 & -a^2-b^2\end{pmatrix} = - \begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix}, $$

$$ R_2(\pi/2,\pi)\Phi_2^c R_2(\pi/2,0) = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & -i \\ i & 1\end{pmatrix} \begin{pmatrix} -1 & 0 \\ 0 & -1\end{pmatrix}\frac{1}{\sqrt{2}}\begin{pmatrix} 1 & -i \\ -i & 1\end{pmatrix} = \begin{pmatrix} 0 & i \\ 0 & -1\end{pmatrix} $$

and

$$ R_1^+(\pi,\pi) R_1^+(\pi,0) = \begin{pmatrix} 0 & -1 \\ -1 & 0\end{pmatrix} \begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & 1\end{pmatrix}. $$

This finally leads to

$$ U = \begin{pmatrix} -1 & 0 \\ 0 & 1\end{pmatrix} \otimes \begin{pmatrix} 0 & i \\ 0 & -1\end{pmatrix} = \begin{pmatrix} 0 & -i & 0 & 0\\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & i \\ 0 & 0 & 0 & -1\end{pmatrix}, $$ which is obviously incorrect. I've repeated this a couple times and still get the wrong result every time. Where am I making a mistake?

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    $\begingroup$ Your U is not even unitary! You should trace down the origin of the non-unitarity. $\endgroup$ Commented May 24 at 11:53

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Your formula for $R_2(\pi/2,\pi)$ is incorrect: The resulting matrix is not unitary (but rather rank 1). (There should be +1 in the top right corner.)

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