Quantum Gate - Negating qubit whose state is known to lie in the equatorial x-y plane

Question

Given a qubit whose state is known to lie in the equatorial x–y plane in the Bloch sphere is it possible to find a quantum gate that will always negate this qubit? If so, exhibit such a gate. If not, explain why it is impossible.

Answer 1

Assuming that you are talking about a pure state, simply apply the $Z$ gate.

If you have a state $|\psi\rangle=|0\rangle+e^{i\phi}|1\rangle$ then let $|\phi\rangle=Z|\psi\rangle=|0\rangle-e^{i\phi}|1\rangle$. We have that $$ \langle\psi|\phi\rangle=(\langle 0|+e^{-i\phi}\langle 1|)(|0\rangle-e^{i\phi}|1\rangle)=1-1=0, $$ so they are orthogonal.

For a bit more insight about how to come with this: any state on the equator of the Bloch Sphere has a Bloch vector that comprises $X$ and $Y$ components but not $Z$ components. So, you're looking for something that anti-commutes with both $X$ and $Y$ (because that will introduce a minus sign, reversing the direction of the Bloch vector, giving the orthogonal state). That's the $Z$ operator.