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 $[|i\rangle,|-i\rangle]$.

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

1 Answer 1


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

$$P_i = |\langle -| i \rangle|^2 = \big|\frac12 \big(\langle0| - \langle1|\big)\big(|0\rangle + i|1\rangle\big) \big|^2 = \big|\frac12 \big( \langle0|0\rangle - i \langle1|1\rangle\big)\big|^2 = \\ = \frac14|1-i|^2 = \frac14 \cdot (\sqrt2)^2 = \frac12$$


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