How can somebody use these gates to create other gates like pauli's or other gates? Also how can somebody go from pauli's to CNOT, Hadamard and Φ quantum gates?

  • 3
    $\begingroup$ By combining them in quantum circuits? I'm not really sure what you're asking, here. $\endgroup$ Feb 10, 2016 at 1:44
  • $\begingroup$ some gates are "functionally complete" & are enough to build full digital logic, suspect thats the point of the question... $\endgroup$
    – vzn
    Feb 10, 2016 at 16:24
  • $\begingroup$ see this related answer cs.stackexchange.com/a/345/157 $\endgroup$
    – Ran G.
    Feb 25, 2016 at 8:45

1 Answer 1


This is actually kind of complicated. Basically you can:

  • Decompose a huge multi-qubit operation into many single-qubit gates, but with lots of controls allowed on each single-qubit gate. Think of it as doing Gaussian elimination: you're factoring the matrix into a bunch of simple row operations.
  • Gradually reduce the number of controls to at-most-one-per-gate, by repeatedly using a construction based on the fact that every operation has a square root and an inverse.
  • Factor each single-qubit operation $U$ into $A$, $B$, $C$, $\theta$ such that $ABC = I$ but $AXBXC e^{i \theta} = U$. Use that to replace controlled-$U$ gates with uncontrolled single-qubit operations separated by $CNOT$ gates.
  • Approximate each uncontrolled single-qubit operation with a sequence of $H$ and $T$ gates (which is possible because they correspond to rotations that can approximate any other rotation).

What you're left with is a circuit containing only $CNOT$, $H$, and $T$ gates that approximates the original operation. You need more gates to get a better approximation, but you can get as close as desired. Also the overhead isn't too terrible.

For details of the constructions, and proofs that the errors don't compound in a way that destroys the efficiency of the approximation, you'll probably have to get a textbook or read papers. There's a lot of details.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.