# Prove a circuit with controlled $iR_x(\pi \alpha)$ is universal for quantum computation whenever $\alpha$ is irrational Show that the three qubit gate $$G$$ defined by the circuit is universal for quantum computation whenever $$\alpha$$ is irrational.

My Observations

The unitary gate on the third qubit is activated only when the first two qubits are both in the state $$1$$.

$$|\psi_{in}\rangle=|\psi_{1}\rangle\otimes|\psi_{2}\rangle\otimes|\psi\rangle=(a|0\rangle+b|1\rangle)\otimes(c|0\rangle+d|1\rangle)\otimes|\psi\rangle=(ac|00\rangle+ad|01\rangle+bc|10\rangle+bd|11\rangle)\otimes|\psi\rangle\\ =ac|00\rangle\otimes|\psi\rangle+ad|01\rangle\otimes|\psi\rangle+bc|10\rangle\otimes|\psi\rangle+bd|11\rangle\otimes|\psi\rangle$$ $$|\psi_{out}\rangle=ac|00\rangle\otimes|\psi\rangle+ad|01\rangle\otimes|\psi\rangle+bc|10\rangle\otimes|\psi\rangle+bd|11\rangle\otimes iR_x(\pi \alpha)|\psi\rangle$$

We can also prove that arbitrarily fine approximations of all angles can be achieved using a rotation by an irrational multiple of $$\pi$$.

I have no clue on how this circuit can be used for universal quantum gate construction ?

Firstly, imagine you start from a two-qubit state $$|00⟩$$, and apply an $$R_x$$ rotation with an angle equivalent to half that of a Pauli $$X$$ to the first qubit (I forget which convention N&C is using for their rotation gates). Then apply a controlled-not controlled off the first qubit and targeting the second qubit. Next, apply the inverse of the first rotation. Finally, measure the first qubit. If you get answer $$|1⟩$$, the second qubit is in the $$|−⟩$$ state. If it isn't, discard and repeat. So, we can produce the $$|−⟩$$ state. If you input this as the target qubit of the controlled-controlled-$$R_x$$ (of arbitrary rotation angle), and have one of the controls in the $$|1⟩$$ state, you get an arbitrary $$Z$$ rotation on the other control qubit.

So, we know we can do arbitrary $$X$$ and $$Z$$ rotations, meaning that you can make any single-qubit unitary. Combine that with controlled-not and you know you have universality.

This is not with any diagrams nor any equations and I do not get a clue of what it is suggesting to do ?

My Attempt

Imagine you start from a two-qubit state $$|00⟩$$, and apply an $$R_x$$ rotation with an angle equivalent to half that of a Pauli $$X$$ to the first qubit (I forget which convention N&C is using for their rotation gates). Then apply a controlled-not controlled off the first qubit and targeting the second qubit. Next, apply the inverse of the first rotation. Finally, measure the first qubit. If you get answer $$|1⟩$$, the second qubit is in the $$|−⟩$$ state $$M|00\rangle=\frac{i}{\sqrt{2}}\begin{bmatrix}1&-i\\-i&1\end{bmatrix}|00\rangle=\frac{1}{\sqrt{2}}[i|0\rangle+|1\rangle]\otimes|0\rangle=\frac{1}{\sqrt{2}}[i|00\rangle+|10\rangle]$$ $$M^{-1}[\frac{1}{\sqrt{2}}[i|00\rangle+|10\rangle]]=\frac{1}{2}[i(-i|0\rangle+|1\rangle)\otimes |0\rangle+(|0\rangle-i|1\rangle)\otimes|1\rangle]\\ =\frac{1}{2}[|00\rangle+i|10\rangle+|01\rangle-i|11\rangle]=\frac{1}{\sqrt{2}}[|0\rangle(\frac{|0\rangle+|1\rangle}{\sqrt{2}})+i|1\rangle(\frac{|0\rangle-|1\rangle}{\sqrt{2}})]\\ =\frac{1}{\sqrt{2}}[|0\rangle|+\rangle+i|1\rangle|-\rangle]$$

• Oct 24, 2021 at 14:45
• @Egretta.Thula Thanks for the hint. Can you assist me understand the solution there mathematically ? Oct 25, 2021 at 13:54