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Say, I have a specific scheme, Fourier

where I need to specify inputs for controlled R logical gate, which here is $$ R(\theta)=\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & e^{\frac{2 \pi i}{2\theta}} \end{bmatrix} $$

But that is for the case when the first qbit is the control and the second is the operand.

I don't understand, how I can do it matrixwise, that is, making a matrix that'd use certain qbit of a register as control line and another as the resulting one, leaving everything else as is. I mean, it's okay, if I have a 2 lines where I just put the qregister in the gate and it works. But this is somewhat harder. How should I change the matrix represenation to specify the gate's inputs and outputs?

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The action of any controlled gate is to do nothing (i.e. apply the identity operation) if the control qubit is in $\vert 0\rangle$ and apply an operation $U$ on the target when the control is in $\vert 1\rangle$. All other qubits in the system are also left untouched (i.e. apply the identity operation).

Use the subscripts $c$ and $t$ for the control qubit and target qubit. The gate looks like this

$$ I_1\otimes I_2\otimes... \otimes\vert 0\rangle\langle 0\vert_c \otimes... \otimes\ I_t\ \otimes...\otimes I_n \ \ +\ \ I_1\otimes I_2\otimes... \otimes\vert 1\rangle\langle 1\vert_c \otimes... \otimes\ U_t\ \otimes...\otimes I_n$$

In matrix form, this is just

$$\begin{pmatrix}1&0\\ 0&1\end{pmatrix}_1\otimes \begin{pmatrix}1 & 0\\ 0&1 \end{pmatrix}_2 \otimes ... \otimes\begin{pmatrix}1 & 0\\ 0&0 \end{pmatrix}_c \otimes ... \otimes\begin{pmatrix}1 & 0\\ 0&1 \end{pmatrix}_t\otimes... \otimes \begin{pmatrix}1 & 0\\ 0&1 \end{pmatrix}_n \\ + \begin{pmatrix}1&0\\ 0&1\end{pmatrix}_1\otimes \begin{pmatrix}1 & 0\\ 0&1 \end{pmatrix}_2 \otimes ... \otimes\begin{pmatrix}0 & 0\\ 0 & 1 \end{pmatrix}_c \otimes ... \otimes\begin{pmatrix}u_{11} & u_{12}\\ u_{21}& u_{22} \end{pmatrix}_t\otimes... \otimes \begin{pmatrix}1 & 0\\ 0&1 \end{pmatrix}_n $$

A simple example to help is when you only have two qubits, $t$ corresponds to the first qubit and $c$ corresponds to the second qubit. Then you have

$$ I\otimes \vert 0\rangle\langle 0\vert + U\otimes \vert 1\rangle\langle 1\vert.$$

The matrix form is

$$\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & u_{11} & 0 & u_{12}\\ 0& 0& 1 & 0\\ 0 & u_{21} & 0 & u_{22} \end{pmatrix}$$

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  • $\begingroup$ Mathematics is beautiful and wonderful, yes, full of both. $\endgroup$ – Penter Pro Dec 9 '19 at 15:47
  • $\begingroup$ Please check if you have written the matrix correctly. Assume CNOT, i.e. $U=X$. In this case the matrix has following form: $\begin{pmatrix} I & 0 \\ 0 & X \end{pmatrix}$. This does not correspond to your answer. $\endgroup$ – Martin Vesely Dec 9 '19 at 23:11
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    $\begingroup$ @MartinVesely that form is when the first qubit is the control and the second is the target i.e. for $\vert 0\rangle\langle 0\vert\otimes I + \vert 1\rangle\langle 1\vert\otimes X$. In my example, the second qubit is the control and the first is the target and hence the matrix takes a different form. $\endgroup$ – rnva Dec 10 '19 at 11:55
  • $\begingroup$ Sorry, did not sebe that. Thanks for explanation. $\endgroup$ – Martin Vesely Dec 10 '19 at 13:23

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