How to construct a CU3 gate using only CX and U3 gates?

Question

Knowing that CX and U3 (taking 3 parameters $\theta, \phi$ and $\lambda$) form a set of universal gates how can I construct an arbitrary CU3 gate using a decomposition of only CX and arbitrary U3 gates?

Answer 1

I use the ideas from these slides, specifically slide 8,9,10.

We can decompose any $U_{3}(\theta,\phi,\lambda)$ into a rotation around the $Z,Y$ & again $Z$ axis, because for any $U \in SU(2)$ we can write: \begin{equation} U = \begin{bmatrix} e^{i(\alpha-\frac{\beta}{2}-\frac{\delta}{2})}\cos(\frac{\gamma}{2}) & e^{i(\alpha-\frac{\beta}{2}+\frac{\delta}{2})}\sin(\frac{\gamma}{2}) \\ e^{i(\alpha+\frac{\beta}{2}-\frac{\delta}{2})}\sin(\frac{\gamma}{2}) & e^{i(\alpha+\frac{\beta}{2}+\frac{\delta}{2})}\cos(\frac{\gamma}{2}) \end{bmatrix} = e^{i\alpha}R_{z}(\beta)R_{y}(\gamma)R_{z}(\delta), \end{equation}

where $\beta$, $\gamma$ & $\delta$ can be computed straightforwardly from $\theta$, $\phi$ and $\lambda$.

Then, let $A = R_{z}(\beta)R_{y}(\gamma/2)$, $B = R_{y}(-\gamma/2)R_{z}(-\delta/2-\beta/2)$ and $C = R_{z}(\delta/2 - \beta/2)$.

A straightforward calculation shows that: \begin{equation} \begin{split} ABC &= R_{z}(\beta)R_{y}(\gamma/2)R_{y}(-\gamma/2)R_{z}(-\delta/2-\beta/2)R_{z}(\delta/2 - \beta/2) = I\\ AXBXC &= R_{z}(\beta)R_{y}(\gamma/2)XR_{y}(-\gamma/2)R_{z}(-\delta/2-\beta/2)XR_{z}(\delta/2 - \beta/2) \\ &= R_{z}(\beta)R_{y}(\gamma/2)R_{y}(\gamma/2)XXR_{z}(\delta/2+\beta/2)R_{z}(\delta/2 - \beta/2) \\ &= R_{z}(\beta)R_{y}(\gamma)R_{z}(\delta) = e^{-i\alpha}U. \end{split} \end{equation}

We can use this fact to implement $CU$ by using two $CX$ gates that we apply between the $A$&$B$ and the $B$&$C$ gates: \begin{equation} \begin{split} &(I\otimes A)CX(I\otimes B)CX(I\otimes C) \\ = &\big(|0\rangle\langle0|\otimes ABC\big) + \big(|1\rangle\langle1|\otimes AXBXC\big) \\ = &\big(|0\rangle\langle0|\otimes I\big) + \big(|1\rangle\langle1|\otimes e^{-i\alpha}U\big) \\ = & CU \big(R_{z}(\alpha)\otimes I\big) \end{split} \end{equation} where the last phase gate on the control qubit is needed because we have the phase $\alpha$ in our equality $U = e^{i\alpha}AXBXC$.

This allows us to implement any controlled-$U$ gate.

Answer 2

Here is a construction of $CU3$ gate on IBM Q:

  u1((lambda+phi)/2) c;
  u1((lambda-phi)/2) t;
  cx c,t;
  u3(-theta/2,0,-(phi+lambda)/2) t;
  cx c,t;
  u3(theta/2,phi,0) t;

Where t is a target qubit and c is control qubit.

Note that $U1$ gate is a special case of $U3$, it holds that $U1(\lambda)=U3(0,0,\lambda)$.