# Compute the squared overlap between different given qubit states

I was checking this problem from the book. And here is an example, but I think it's wrong. If it is not wrong can you please explain how did they derive it?

As per my workout, it should be one. But It seems they are doing something fishy here.

To be honest I would be surprised if this book is giving wrong solutions.

By having in mind that $$\langle 0 | 0\rangle = \langle 1 | 1\rangle = 1$$ and $$\langle 0 | 1\rangle =\langle 1 | 0\rangle = 0$$, we have: \begin{align*} &\langle \psi_0 | \psi_1\rangle = \frac{-1}{2} \langle 0 | 0\rangle - \underbrace{\frac{\sqrt{3}}{2}\langle 0 | 1\rangle}_{=0} = -\frac{1}{2} \\ & \Longrightarrow |\langle \psi_0 | \psi_1\rangle|^2 = \left( \frac{-1}{2} \right)^2 = \frac{1}{4} \end{align*} You do the exact same thing for $$|\langle \psi_0 | \psi_2\rangle|^2$$, the only thing changing being the sign in front of $$\frac{\sqrt{3}}{2}$$, but since it "becomes" 0, nothing changes.
\begin{align*} \langle \psi_1 | \psi_2\rangle &= \left( \frac{-1}{2} \langle 0 | - \frac{\sqrt{3}}{2}\langle 1 |\right) \left( \frac{-1}{2} | 0 \rangle + \frac{\sqrt{3}}{2} |1 \rangle \right) \\ &= \frac{1}{4}\underbrace{\langle 0 | 0\rangle}_{=1} - \frac{3}{4}\underbrace{\langle 1 | 1\rangle}_{=1} -\underbrace{\frac{\sqrt{3}}{4}\langle 0 | 1\rangle + \frac{\sqrt{3}}{4}\langle 1 | 0\rangle}_{=0} \\ &= -\frac{1}{2} \\ & \Longrightarrow |\langle \psi_1 | \psi_2\rangle|^2 = \left( \frac{-1}{2} \right)^2 = \frac{1}{4} \end{align*}
• Embarassing that I missed the - sign. It would be (-3/4) not +3/4. Thanks for the help. Dec 23, 2020 at 9:07