Can you take infinitely many square roots of Pauli-X?

3

I am trying to find the cost for a n-bit Toffoli gate based on the recurrent circuit presented on Barenco's Work in Lemma 7.5 (Elementary gates for quantum computation)

The construction requires that we iteratively take the square root of Pauli X. I was wondering if there is some proof that we can always take the square root of Pauli X as many times as possible?

César Leonardo Clemente López

Posted 2020-08-16T19:42:16.393

Reputation: 541

Answers

3

Unitary matrices can be raised to any power, including fractional powers, so any root you want you can find. You find the root by eigendecomposing the matrix, modifying the eigenvalues (raising them to the desired power), then putting the matrix back together.

In the case of the Pauli X matrix, the eigenvectors are $|+\rangle\langle +|$ and $|-\rangle\langle -|$ so you can find roots like this:

$$X^s = \frac{1}{2}\begin{bmatrix} 1 & 1 \\ 1 &1\end{bmatrix} + \frac{e^{i \pi s}}{2}\begin{bmatrix} 1 & -1 \\ -1 & 1\end{bmatrix}$$

Once that's done the actual challenge is realizing $X^s$ gates using the gate set you have available on your computer. For example, if you're using the Clifford+T gate set then you could approximate the rotation using a series of H and T gates.

Note that, for performing a many-controlled NOT, there are more efficient ancilla-free constructions than the one you linked: https://algassert.com/circuits/2015/06/22/Using-Quantum-Gates-instead-of-Ancilla-Bits.html

Craig Gidney

Posted 2020-08-16T19:42:16.393

Reputation: 11 207

I found that website, It seems it yours by the icon :) Thank you so much for the info! I'll take a look. I found that using the recursive lemma on Barenco's we would need $O(3^{n-2})$ gates where n is the number of control qubits on the cnot. – César Leonardo Clemente López – 2020-08-17T06:23:18.947