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Is it wrong to say in $a$ and $b$ are the square roots of the probability of the qubit being in the state 0 and 1 when measured for a qubit in the state $|\psi \rangle = a|0 \rangle +b|1 \rangle $? And by that definition how can $a$ and $b$ be imaginary numbers?

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A qubit is a two-level quantum system, hence the state of a qubit can be written as $|\psi \rangle = a|0\rangle + b|1\rangle$ with $a, b \in \mathbb{C}$ and $|a|^2 + |b|^2 = 1 $. Note that $|a|^2$ is always a real number even if $a$ is complex because $|a|^2 = a\cdot \bar{a} $. And yes, the probability to observe the qubit in the state $|0\rangle$, $Pr(|0\rangle) = |a|^2$. Similarly the probability to observe the state $|1\rangle$, $Pr(|1\rangle ) = |b|^2$. This should shed some light into why we have the constraint $|a|^2 + |b|^2 = 1$.

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  • $\begingroup$ OK but what if a is i? $\endgroup$
    – Se1fie
    Nov 1 '20 at 13:36
  • $\begingroup$ Sure. You can have the state $|\psi \rangle = i|0\rangle $ then the probability you will observe the state $|0\rangle$ here is 1 since $|a|^2 = i \cdot (-i) = 1 $. Note that the probability of observing the state $|0\rangle$ is also had I have the state $|\psi \rangle = |0 \rangle$. This tells you something about the overall phase in a quantum state... Another thing, you can't have a state $|\psi \rangle = i |0 \rangle + | 1 \rangle$. This is because it is not normalized! $\endgroup$
    – KAJ226
    Nov 1 '20 at 16:30

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