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I am working with the set $\{\mathrm{CNOT}, \mathrm{H}, \mathrm{P}(\theta)\}$

where $\mathrm{H}$ is the Hadamard gate, and $\mathrm{P}(\theta)$ is the phase gate with angle $\theta$.

I want to build other gates with these gates, like $R_z(\theta)$, Control-$R_z(\theta)$, or Control-$P(\theta)$

How can I do this?

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    $\begingroup$ See N&C:Ch4 or Summary. You want to factor $U=e^{i\alpha}AXBXC$ with $ABC=Id$ and you will get an expression for Control-$U$. So do that for whichever $R_{\vec{n}} (\theta)$ you desire. $\endgroup$
    – AHusain
    Sep 30, 2018 at 18:16
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    $\begingroup$ This might be helpful: quantumcomputing.stackexchange.com/questions/4086/… $\endgroup$ Sep 30, 2018 at 20:32

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In your question, you don't define $P(\theta)$ or $R_z(\theta)$. I'm going to assume: $$ P(\theta)=\left(\begin{array}{cc} 1 & 0 \\ 0 & e^{i\theta} \end{array}\right)\qquad R_z(\theta)=\left(\begin{array}e^{-i\theta} & 0 \\ 0 & e^{i\theta} \end{array}\right). $$ In this case, you simply have that $$ R_z(\theta)=P(2\theta)e^{-i\theta}\equiv P(2\theta), $$ the point being that global phases are irrelevant. However, the difference is important when you look at the controlled-gates. Let's say we can create either controlled-$P$ or controlled-$R_z$. We can create the other via the identity

enter image description here

The extra $P(-\theta)$ is the gate that compensates for the extra phase.

To make controlled-$R_z$, the trick is to notice the identities $R_z(\theta_1)R_z(\theta_2)=R_z(\theta_1+\theta_2)$ and $XR_z(\theta)X=R_z(-\theta)$, where we will replace the Xs by controlled-not so that the change in rotation angle only happens if the control qubit is 1. Hence, we have

enter image description here

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