Measuring a state $\frac{1}{2}|0\rangle-\frac{\sqrt 3}{2}|1\rangle$ in the $X$ and $Z$-bases?

Question

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 $\{|+\rangle,|-\rangle\}$?

Answer 1

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

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

Answer 2

You can apply a Hadamard gate to your state, and then proceed to measure in the computational basis