# Is it wrong to say that $a$ and $b$ are the square roots of the detection probabilities in a qubit state $|\psi \rangle = a|0 \rangle +b|1 \rangle$?

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?

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 $$|a|^2 + |b|^2 = 1$$.
• 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! Nov 1 '20 at 16:30