Is measuring real qubit states in a complex basis such as $|0\rangle\pm i|1\rangle$ possible?

Say I want to measure a photon in state $$\dfrac{\sqrt 2}{2}|0\rangle-\dfrac{\sqrt 2}{2}|1\rangle$$ in the measurement basis $$[|i\rangle,|-i\rangle]$$.

• Can you define what you mean by $|i\rangle$? When you say a complex basis, do you mean something like $(|0\rangle+i|1\rangle)/\sqrt{2}$? Sep 4, 2019 at 8:17
• @DaftWullie Yes.
– apen
Sep 4, 2019 at 13:07

Yes, it is possible, same way as it is possible to measure a state with complex amplitudes in a basis with real amplitudes (say, a $$|i\rangle$$ state in the $$[|0\rangle, |1\rangle]$$ basis). Either way, the probability of measuring state $$|\psi\rangle$$ and getting measurement result corresponding to basis state $$|a_i\rangle$$ is defined as $$P_i = |\langle a_i| \psi \rangle|^2$$ - since you're taking absolute value of the amplitude before squaring it, it doesn't matter whether the value $$\langle a_i| \psi \rangle$$ is complex or real - the probability will still be a real number.
For your example, the probability of measuring $$| i \rangle$$ in the $$| - \rangle$$ state is
$$P_i = |\langle -| i \rangle|^2 = \big|\frac12 \big(\langle0| - \langle1|\big)\big(|0\rangle + i|1\rangle\big) \big|^2 = \big|\frac12 \big( \langle0|0\rangle - i \langle1|1\rangle\big)\big|^2 = \\ = \frac14|1-i|^2 = \frac14 \cdot (\sqrt2)^2 = \frac12$$