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I'm following the paper by Grover (L. K. Grover, Phys. Rev. Lett., 85:2000) and trying to synthesize the following superposition state $$|\psi\rangle = \frac{2|00\rangle - 3|01\rangle - 4i|10\rangle + 5i|11\rangle}{\sqrt{54}}$$

Following the paper, if I understand correctly, one should expect $$Q|00\rangle = |\psi\rangle$$ where $$Q = -I_sU^{-1}I_tU$$ with $$U = U_2U_1\;,\;\;U_1 = H^{\otimes2}$$ However, I don't understand the part where the author construct $U_2$.

Can anyone gives some tips?

Followups: Suppose the last qubit is the ancilla qubit. Then $$ccR_Y(00) = \begin{bmatrix} I_6 & \\ & \cos(\theta_{00}/2) & -\sin(\theta_{00}/2) \\ & \sin(\theta_{00}/2) & \cos(\theta_{00}/2) \end{bmatrix} $$ Then $$U_2 = ccR_Y(11)\cdot ccR_Y(10)\cdot ccR_Y(01)\cdot ccR_Y(00)$$ and so $$U = U_2U_1 = U_2\cdot(H^{\otimes2}\otimes I)$$

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I'd say that the paper is pretty clear about what you have to do. First, you introduce an extra qubit, so you start in $|0\rangle|00\rangle$. $U_2$ is specified (in your case) as a set of 4 controlled-$Y$ rotations, controlled off the two qubits on which you're creating the state and targeting the ancilla. (You'll be able to simplify this slightly).

Fr example, take the $|00\rangle$ term in your target state. Because the target amplitude is $2/\sqrt{54}$, you want to perform a controlled-$Y$ rotation, controlled off the two controls both being 0, implementing a $Y$ rotation such $R_Y$ such that $$ R_Y|0\rangle=\frac{2}{\sqrt{54}}|0\rangle+\sqrt{\frac{50}{54}}|1\rangle. $$ Now repeat for your other 3 basis states.

$I_t$ is a $Z$ gate applied to the ancilla qubit. $I_s$ is, effectively, a controlled-controlled-$Z$ gate, but it gives the -1 term only when all 3 qubits are in $|0\rangle$.

Simplification: Once you have calculated the four rotations $R_Y(\theta_i)$ which are all controlled off two qubits, you can simplify this a bit. Instead of c-c-$R_Y(\theta_{00})$, just apply $R_Y(\theta_{00})$ and set all other $\theta_i\rightarrow\theta_i-\theta_{00}$.

Then take one of the others, say c-c-$R_Y(\theta_{10})$ and instead apply c-$R_Y(\theta_{10})$ controlled off the first qubit being in the 1 state, and replace $\theta_{11}\rightarrow\theta_{11}-\theta_{10}$. Overall, that just reduces the number of multi-control gates you need.

To be more explicit, I am proposing the following circuit (the two different circuits are equivalent ways of writing the same thing), where the ancilla is at the bottom: enter image description here For this to work, you need $$ \cos\frac{\theta_{00}}2=\frac{2}{\sqrt{54}}, \cos\frac{\theta_{01}}2=-\frac{3}{\sqrt{54}}, \cos\frac{\theta_{10}}2=-\frac{4}{\sqrt{54}}, \cos\frac{\theta_{11}}2=\frac{5}{\sqrt{54}}. $$ Also note that I added an $S$ gate into the circuit. This was because I missed in the original problem specification that two of the amplitudes were imaginary. You could have alternatively used cc-$R_x$ gates.

Note that, in general, this is a really bad way to make a two-qubit state, where you never need more than one controlled-not gate (based on the Schmidt decomposition of the state).

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  • $\begingroup$ I still can't understand it well. I tried my best and wrote down $U$ based on your answer. Could you take a look? $\endgroup$
    – TurbPhys
    Commented Nov 16, 2021 at 22:37

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