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If a qubit is in the state $|\psi\rangle = \frac {1}{2}|0\rangle - \frac{\sqrt 3}{2} |1\rangle$, how do I measure it in the $Z$-basis, i.e. $\{|0\rangle,|1\rangle\}$, and the $X$ basis, i.e. $\{|+\rangle,|-\rangle\}$, and find the states and their probabilities?

What I thought of doing is for $\{|0\rangle,|1\rangle\}$ is: I take the squared absolute value of $\alpha$ and $\beta$ without any modification or conversion, and that would be my probabilities.

But I'm confused on what to do with $\{|+\rangle,|-\rangle\}$?

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    $\begingroup$ This sounds like homework, but you've copied the question in a way that doesn't make any sense, and you've not indicated what effort, if any, you have already taken. For example, it's clear that you intended to indicate the X basis in question (b). Also, your state is not normalized, it's not clear what the amplitudes for |0⟩ and |1⟩ are, because amplitudes of 21 and 23 don't make any sense. Please consider initially reviewing your question properly, and editing your question to include the details of what you've already done. – $\endgroup$ Feb 16, 2022 at 21:30
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    $\begingroup$ Good job on slowly turning your question around. I've retracted my vote to close. $\endgroup$ Feb 16, 2022 at 22:22

2 Answers 2

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Your approach to solving the case in the $Z$-basis $\{|0\rangle, |1\rangle\}$ is correct.

You can reduce the other case to the one you already know how to solve by expressing $|\psi\rangle$ in the $X$-basis $\{|+\rangle, |-\rangle\}$. To that end, you can use

$$ |0\rangle=\frac{|+\rangle+|-\rangle}{\sqrt2}\\ |1\rangle=\frac{|+\rangle-|-\rangle}{\sqrt2}. $$

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  • $\begingroup$ what about if I need it in z-basis? $\endgroup$
    – n22
    Feb 16, 2022 at 23:25
  • $\begingroup$ I am a beginner and I tried conversion this way : (1/2 + - 3^(1/2) / 2) / 2^(1/2) |+> + (1/2 + - 3^(1/2) / 2) / 2^(1/2) |-> and I keep getting 2+ 3^(1/2) / 4 $\endgroup$
    – n22
    Feb 16, 2022 at 23:28
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You can apply a Hadamard gate to your state, and then proceed to measure in the computational basis

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